Fourth group of cations. Buffer Solutions.

 

“Biological control of the pH of cells and

body fluids is of central importance in all

aspects of intermediary metabolism and cellular function.”

Albert L. Lehninger

 

Buffers

A buffer is a solution characterised by the ability to resist changes in pH when limited amounts of acid or base are added to it. Buffer contains either a weak acid and its conjugate base or a weak base and its conjugate acid.

Suppose a buffer contains approximately equal molar amount of weak acid HA and its conjugate base A^–. When a strong acid is added to the buffer, it supplies hydrogen ions that react with the base A^–:

H⁺ + A^– ® HA

On the other hand, when a strong base is added to the buffer, it supplies hydroxide ions. Then ions react with the acid HA:

OH^– + HA ® H₂O + A^–

Thus a buffer solution resists changes in pH through its ability to combine with both H⁺ and OH^– ions.

There are three types of buffers which distinguish its components:

 

I.    Buffer contains weak acid and its salt (pH of buffer < 7):

HCOOH + HCOONa;            CH₃COOH + CH₃COONa.

II.               Buffer contains weak base and its salt (pH of buffer > 7):

H₃BO₃ + Na₂B₄O₇;                  NH₄OH + NH₄Cl

III. Buffer contains salts of polyprotic acids (pH of buffer » 7):

Na₂HPO₄ + NaH₂PO₄             Na₂CO₃ + NaHCO₃

Two important characteristics of a buffer are the pH and the buffer capacity.

 

The buffer capacity – is the amount of acid or base the buffer can react with before giving a significant pH change.

Buffer capacity (BC) depends on the amount of acid and conjugated base in the solution. The ratio of amounts of acid and conjugated base is also important. Unless this ratio is approximately 1 (between 1:10 and 10:1), the buffer capacity will be too low to be useful.

 

 

The other important characteristic of a buffer is its pH. Buffer always must be prepared from a conjugated acid-base pair in which the acid ionisation constant is approximately equal to the desired H⁺ ion concentration.

The Henderson-Hasselbalch equation relates the pH of a buffer for different concentrations of conjugate acid and base:

pH = pK_a + lg [base]/[acid]

By substituting the value of pK_a for the conjugate acid and the ratio [base]/[acid], we obtain the pH of the buffer.

 

Equations for Calculation [H⁺] and pH of Buffers

 

It must be remembered, however, that pH is not entirely established by ratio of conjugate base to conjugate acid bat can be affected by concentration. For typical buffers (i.e. concentration less than 0.1 M or with K values of 10^-3 or less) the Henderson-Hasselbalch equation can be used.

A buffer is a solution that can effectively resist pH change upon the addition of an acidic or basic component. It is able to neutralize small amounts of added acid or base, thus maintaining the pH of the solution relatively stable. This is important for processes and/or reactions which require specific pH ranges. Buffer solutions have a working pH range and capacity which dictate how much acid/base can be neutralized before pH changes, and the amount by which it will change.

To effectively maintain a pH range, a buffer must consist of a weak conjugate acid-base pair, meaning either a. a weak acid and its conjugate base, or b. a weak base and its conjugate acid. The use of one or the other will simply depend upon the desired pH when preparing the buffer. For example, the following could function as buffers when together in solution:

·                     Acetic acid (weak organic acid w/ formula CH₃COOH) and a salt containing its conjugate base, the acetate anion (CH₃COO^–), such as sodium acetate (CH₃COONa).

·                     Pyridine (weak base w/ formula C₅H₅N) and a salt containing its conjugate acid, the pyridinium cation (C₅H₅NH⁺), such as Pyridinium Chloride.

·                     Ammonia (weak base w/ formula NH₃) and a salt containing its conjugate acid, the ammonium cation, such as Ammonium Hydroxide (NH₄OH).

A buffer is able to resist pH change because the two components (conjugate acid and conjugate base) are both present in appreciable amounts at equilibrium and are able to neutralize small amounts of other acids and bases (in the form of H₃O⁺ and OH^–) when the are added to the solution.

In order to clarify this effect, we can consider the simple example of a Hydrofluoric Acid (HF) and Sodium Fluoride (NaF) buffer. Hydrofluoric acid is a weak acid due to the strong attraction between the relatively small F^– ion and solvated protons (H₃O⁺), which does not allow it to dissociate completely in water. Therefore, if we obtain HF in an aqueous solution, we establish the following equilibrium with only slight dissociation (K_a(HF) = 6.6×10^-4, strongly favors reactants):

We can then add and dissolve sodium fluoride into the solution and mix the two until we reach the desired volume and pH at which we want to buffer.

When Sodium Fluoride dissolves in water, the reaction goes to completion, thus we obtain:

Since Na⁺ is the conjugate of a strong base, it will have no effect on the pH or reactivity of the buffer. The addition of NaF to the solution will, however, increase the concentration of F^– in the buffer solution, and, consequently, by Le Chatlier’s principle, lead to slightly less dissociation of the HF in the previous equilibrium, as well.  The presence of significant amounts of both the conjugate acid, HF, and the conjugate base, F^–, allows the solution to function as a buffer.

This buffering action can be seen in the titration curve of a buffer solution.

As we can see, over the working range of the buffer. pH changes very little with the addition of acid or base. Once the buffering capacity is exceeded the rate of pH change quickly jumps. This occurs because the conjugate acid or base has been depleted through neutralization. This principle implies that a larger amount of conjugate acid or base will have a greater buffering capacity.

If acid is added:

In this reaction, the conjugate base, F^–, will neutralize the added acid, H₃O⁺, and this reaction goes to completion, because the reaction of F^– with H₃O⁺ has an equilibrium constant much greater than one.  (In fact, the equilibrium constant the reaction as written is just the inverse of the K_a for HF: 1/K_a(HF) = 1/(6.6×10^-4) = 1.5×10⁺³.)  So long as there is more F^– than H₃O⁺, almost all of the H₃O⁺ will be consumed and the equilibrium will shift to the right, slightly increasing the concentration of HF and slightly decreasing the concentration of F^–, but resulting in hardly any change in the amount of H₃O⁺ present once equilibrium is re-established. 

If base is added:

In this reaction, the conjugate acid, HF, will neutralize added amounts of base, OH^–, and the equilibrium will again shift to the right, slightly increasing the concentration of F^– in the solution and decreasing the amount of HF slightly. Again, since most of the OH^– is neutralized, little pH change will occur.

These two reactions can continue to alternate back and forth with little pH change.

Selecting proper components for desired pH

Buffers function best when the pK_a of the conjugate weak acid used is close to the desired working range of the buffer. This turns out to be the case when the concentrations of the conjugate acid and conjugate base are approximately equal (within about a factor of 10). For example, we know the K_a for hydroflouric acid is 6.6 x 10^-4 so its pK_a= -log(6.6 x 10^-4) = 3.18. So, a hydrofluoric acid buffer would work best in a buffer range of around pH = 3.18.

For the weak base ammonia (NH₃), the value of K_b is 1.8×10^-5, implying that the K_a for the dissociation of its conjugate acid, NH₄⁺, is K_w/K_b=10^-14/1.8×10^-5 = 5.6×10^-10.  Thus, the pK_a for NH₄⁺ = 9.25, so buffers using NH₄⁺/NH₃ will work best around a pH of 9.25. (It’s always the pK_a of the conjugate acid that determines the approximate pH for a buffer system, though this is dependent on the pK_b of the conjugate base, obviously.)

When the desired pH of a buffer solution is near the pK_a of the conjugate acid being used (i.e., when the amounts of conjugate acid and conjugate base in solution are within about a factor of 10 of each other), the Henderson-Hasselbalch equation can be applied as a simple approximation of the solution pH, as we will see in the next section.

Example

In this example we will continue to use the hydrofluoric acid buffer. We will discuss the process for preparing a buffer of HF at a pH of 3.0. We can use the Henderson-Hasselbalch equation to calculate the necessary ratio of F^– and HF.

This is simply the ratio of the concentrations of conjugate base and conjugate acid we will need in our solution. However, what if we have 100 ml of 1 M HF and we want to prepare a buffer using NaF? How much Sodium Fluoride would we need to add in order to create a buffer at said pH (3.0)?

We know from our Henderson-Hasselbalch calculation that the ratio of our base/acid should be equal to 0.66.  From a table of molar masses, such as a periodic table, we can calculate the molar mass of NaF to be equal to 41.99 g/mol.  HF is a weak acid with a K_a = 6.6 x 10^-4 and the concentration of HF is given above as 1 M.  Using this information, we can calculate the amount of F^– we need to add.

The dissociation reaction is:

We could use ICE tables to calculate the concentration of F^– from HF dissociation, but, since K_a is so small, we can approximate that virtually all of the HF will remain undissociated, so the amount of F^– in the solution from HF dissociation will be negligible.  Thus, the [HF] is about 1 M and the [F^–] is close to 0.  This will be especially true once we have added more F^–, the addition of which will even further suppress the dissociation of HF. 

We want the ratio of Base/Acid to be 0.66, so we will need [Base]/1M = 0.66.  Thus, [F^–] should be about 0.66 M.  For 100 mL of solution, then, we will want to add 0.066 moles (0.1 L x 0.66 M) of F^–.  Since we are adding NaF as our source of F^–, and since NaF completely dissociates in water, we need 0.066 moles of NaF.  Thus, 0.066 moles x 41.99 g/mol = 2.767 g.

Note that, since the conjugate acid and the conjugate base are both mixed into the same volume of solution in the buffer, the ratio of “Base/Acid” is the same whether we use a ratio of the “concentration of base over concentration of acid,” OR a ratio of “moles of base over moles of acid.”  The pH of the solution does not, it turns out, depend on the volume!  (This is only true so long as the solution does not get so dilute that the autoionization of water becomes an important source of H⁺ or OH^–.  Such dilute solutions are rarely used as buffers, however.)

 

Adding Strong Acids or Bases to buffer Solutions

Now that we have this nice F^–/HF buffer, let’s see what happens when we add strong acid or base to it. Recall that the amount of F^– in the solution is 0.66M x 0.1 L = 0.066 moles and the amount of HF is 1.0 M x 0.1L = 0.10 moles. Let’s double check the pH using the Henderson-Hasselbalch Approximation, but using moles instead of concentrations:

pH = pK_a + log(Base/Acid) = 3.18 + log(0.066 moles F^–/0.10 moles HF) = 3.00

Good.  Now let’s see what happens when we add a small amount of strong acid, such as HCl. When we put HCl into water, it completely dissociates into H₃O⁺ and Cl^–.  The Cl^– is the conjugate base of a strong acid so is inert and doesn’t affect pH, and we can just ignore it. However, the H₃O⁺ can affect pH and it can also react with our buffer components. In fact, we already discussed what happens. The equation is:

For every mole of H₃O⁺ added, an equivalent amount of the conjugate base (in this case, F^–) will also react, and the equilibrium constant for the reaction is large, so the reaction will continue until one or the other is essentially used up. If the F^– is used up before reacting away all of the H₃O⁺, then the remaining H₃O⁺ will affect the pH directly. In this case, the capacity of the buffer will have been exceeded – a situation one tries to avoid. However, for our example, let’s say that the amount of added H₃O⁺ is smaller than the amount of F^– present, so our buffer capacity is NOT exceeded.  For the purposes of this example, we’ll let the added H₃O⁺ be equal to 0.01 moles (from 0.01 moles of HCl). Now, if we add 0.01 moles of HCl to 100 mL of pure water, we would expect the pH of the resulting solution to be 1.00 (0.01 moles/0.10 L = 0.1 M; pH = -log(0.1) = 1.0). 

However, we are adding the H₃O⁺ to a solution that has F^– in it, so the H₃O⁺ will all be consumed by reaction with F^–. In the process, the 0.066 moles of F^– is reduced:

0.066 initial moles F^– – 0.010 moles reacted with H₃O⁺ = 0.056 moles F^– remaining

Also during this process, more HF is formed by the reaction:

0.10 initial moles HF + 0.010 moles from reaction of F^– with H₃O⁺ = 0.11 moles HF after reaction

Plugging these new values into Henderson-Hasselbalch gives:

pH = pK_a + log (base/acid) = 3.18 + log (0.056 moles F^–/0.11 moles HF) = 2.89

Thus, our buffer did what it should – it resisted the change in pH, dropping only from 3.00 to 2.89 with the addition of 0.01 moles of strong acid.

 

Buffer solutions are extremely important in biology and medicine because most biological reactions and enzymes need very specific Ph ranges in order to work properly.

Blood

Human blood contains a buffer of carbonic acid (H₂CO₃) and bicarbonate anion (HCO₃^–) in order to maintain blood pH between 7.35 and 7.45, as a value higher than 7.8 or lower than 6.8 can lead to death. In this buffer, hydronium and bicarbonate anion are in equilibrium with carbonic acid. Furthermore, the carbonic acid in the first equilibrium can decompose into CO₂ gas and water, resulting in a second equilibrium system between carbonic acid and water. Because CO₂ is an important component of the blood buffer, its regulation in the body, as well as that of O₂ , is extremely important. The effect of this can be  important when the human body is subjected to strenuous conditions.

In the body, there exists another equilibrium between hydronium and oxygen which involves the binding ability of hemoglobin. An increase in hydronium causes this equilibrium to shift towards the oxygen side, thus releasing oxygen from hemoglobin molecules into the surrounding tissues/cells. This system continues during exercise, providing continuous oxygen to working tissues.

In summation, the blood buffer is:

With the following simultaneous equilibrium:

Intravenous Infusion

As in the above section, all medical intravenous fluids must be buffered to maintain a pH similar to that of the human blood into which it is being infused.

Mediums for Biological Reaction

Buffers are used often in biological research to maintain ph of specific processes. This can be especially useful when culturing bacteria, as their metabolic waste can affect the pH of their medium, consequently killing the sample. For example, a buffer of cacodylic acid (C₂H₇AsO₂) and its conjugate base is used to make samples which will undergo electron microscopy. Another buffer, tricine (C₆H₁₃NO₅), is used to buffer.

Chemical Concepts Demonstrated

·                     Properties of buffers and buffer solutions

Demonstration

·                     All of the crystallizing dishes contain methyl orange and bromothymol blue indicator. ·                     The first 2 dishes contain water; the last 2 contain a buffersolution of 6M acetic acid and sodium acetate. ·                     HCl is added to the first and third dish. ·                     NaOH is added to the second and fourth dish. ·                     Compare the two solutions.

 

Observations

The water turns red with one drop of HCl and blue with one drop of NaOH.

Large amounts of HCl are added to the buffer solution and the color does not change. Large amounts of NaOH are added in another buffer solution and the color still does not change.  It should be noted that this will not go on forever.

Explanations

Buffers lessen or absorb the drastic changes in pH that occur when small amounts of acids and bases are added to solution. In this case, the buffer solution is made of water, acetic acid and sodium acetate. The acetate ions shift the equilibrium, depressing the ionization of the acetic acid.  The pH will remain essentially constant as long as the ratio of the concentrations of acids and bases are more or less constant.  When enough acid or base is added to exceed the buffer capacity of the solution, the pH will change significantly and its color will change.

OAc ^– (aq) + H₃O⁺(aq) —-> HOAc (aq) + H₂O (l)   (addition of acid)

HOAc (aq) + OH ^– (aq) —-> OAc ^– (aq) + H₂O (l)   (addition of base)

When it comes to buffer solution one of the most common equation is the Henderson-Hasselbalch approximation. An important point that must be made about this equation is it’s useful only if stoichiometric or initial concentration can be substituted into the equation for equilibrium concentrations.

Where did the Henderson-Hasselbalch Equation come from?

Where the Henderson-Hasselbalch approximation comes from:

Where,

= conjugate base

= weak acid 

We know that is equal to the products over the reactants and, by definition, H₂O is essentially a pure liquid that we consider to be equal to one.

Take the -log of both sides:

Using the following two relationships:

We can simplify the above equation:

If we add log[A^–] to both sides, we get the Henderson-Hasselbalch Equation:  

This equation is only valid when…

1.                 The conjugate base / acid falls between the values of 0.1 and 10

2.                 The molarity of the buffers exeeds the value of the K_a by a factor of at least 100

 

Preparing Buffer Solutions

There are two cases where we can use the Henderson-Hasselbalch Equation.

Suppose we needed to make a buffer solution with a pH of 2.11. In the first case, we would try and find a weak acid with a pK_a value of 2.11 but… at the same time the molarities of the acid and the its salt must be equal to one another. This will cause the two molarities to cancel; leaving the log [A^–]  equal to log (1) which is zero.

This is a very unlikely scenario, however, and you won’t often find yourself.

An Example:

What mass of NaC₇H₅0₂ must be dissolved in 0.200 L of 0.30 M HC₇H₅O₂ to produce a solution with pH = 4.78? (Assume solution volume is constant at 0.200L)

Solution:

 

Mass = 0.200 L x 1.14 mol C₇H₅O₂^– / 1L x 1mol NaC₇H₅O₂ / 1 mol C₇H₅O₂^– x 144 g NaC₇H₅O₂ / 1 mol NaC₇H₅O₂ = 32.832 g NaC₇H₅O₂.

 

         An important property of blood and other physiological components is that they resist change in pH. A buffer system occurs when a weak acid and its conjugate base are present in the same solution. For instance, blood has a pH of about 7.4, and complex chemical systems work to maintain that pH. The most important component of those systems is the carbonic acid bufeer system. Not by coincidence, this happens to be the same weak acid found in soft drinks.
         Buffer systems are an important application of acid–base equilibria. The study of acid–base equilibria is very useful because many other chemical systems can be understood through the same mathematical approach. The most common experimental method used to study acid–base systems is titration analysis, through which we can determine the pK_a of a weak acid and the pK_b of its conjugate base, the two essential components of a buffer.

The buffer Equation (Henderson-Hasselbach)

Let’s consider a weak acid equilibrium system and its corresponding equilibrium constant:
HA(aq) « H⁺(aq) + A^–(aq)

Ka = [H+]·[A–]/[HA]

         HA represents a weak monoprotic acid, and A^– is its conjugate base. Solving the equilibrium constant expression for hydrogen ion concentration,

[H+] = Ka·[HA]/[A^–]

         Taking the log of each side and multiplying by –1,

– log [H+] = – log (Ka·[HA]/[A^–])

Algebraically rearranging,

– log [H+] = – log Ka – log [HA]/[A^–]

– log [H+] = – log Ka + log [A–][HA]

Using the fact that pH = –log[H⁺], we arrive at the buffer equation:

pH = pKa + log[HA]/[A^–]

Note that since this equation is derived from the equilibrium constant expression, all concentrations must be equilibrium concentrations. However, we often find it useful and accurate to make the approximation that the weak acid is only slightly dissociated. Thus the equilibrium concentration of HA is approximately equal to the initial concentration, or

[HA]equilibrium = [HA]initial

         A similar assumption is also valid for weak bases:

[A^–]equilibrium = [A^–]initial.

 

The effectiveness of a buffer

         Consider a 100.0 mL solution containing 0.010 mol acetic acid, HC₂H₃O₂, and 0.010 mol sodium acetate, NaC₂H₃O₂. We have

[A^–]/[HA] = 1   and   log 1 = 0

therefore, pH = pKa. Looking up pKa for acetic acid, we find pH = pKa = 4.75. Now let’s consider what will happen if we add 0.005 mol of HCl to this solution. The strong acid will react with the acetate ion.

H⁺(aq) + C₂H₃O₂^–(aq) « HC₂H₃O₂(aq)

Initial Moles    0.005         0.010                0.010

Change         – 0.005       – 0.005            + 0.005

Final Moles           0           0.005               0.015

         The buffer equation caow be applied to determine the new solution pH:

pH = pKa + log [A–]/[HA] = 4.75 + log (0.005 mol/0.1000 L)(0.015 mol/0.1000 L) = 4.27

         The pH of the solution changes from 4.75 to 4.27 upon addition of the acid.

         Let’s compare this to what will happen if we add the same amount of HCl to a nonbuffered solution that begins at pH = 4.75.  A 1.8·10^–5 M HCl solution has a pH of 4.75. The number of moles of H⁺(aq) in this solution is

0.1000 L ´ 1.8·10^–5 mol/L = 1.810^–6 mol.

         The amount of HCl added was 0.005 mol, so after the acid is added, the number of moles is 0.005 + 0.0000018 = 0.005 mol. The new hydrogen ion concentration is

0.005 mol / 0.1000 L = 0.05 M

and the solution pH is

pH = – log [H+] = – log (0.05) = 1.3.

         In this unbuffered solution, the pH changes from 4.75 to 1.3, which is much a much larger change than in the buffered solution.

         A consideration that must be made when preparing a buffer is to have sufficient quantities of both the weak acid and its conjugate base to completely react with any base or acid that may be added to the system. The buffer capacity of a system is defined in terms of the concentrations of the acid–base conjugate pair. Greater concentrations will withstand greater additions of base or acid while still resisting a significant pH change. If we were to add so much acid so that it reacted with all of the base in a buffer system, the buffering capacity of the system would be exceeded, and further additions of acid would result in large changes in pH.
A guideline for preparing a buffer system is to choose an acid with a pK_a within one pH unit of the desired buffer. This ensures that the ratio of base to acid will range between 1 to 10 and 10 to 1, and thus sufficient quantities of both acid and base will be present in the buffering system.

 

pH titration

         A pH titration is performed by adding small amounts of a titrant to a solution and simultaneously monitoring the solution pH. A typical titration would be to add small amounts of sodium hydroxide solution to a weak acid solution. In this case, the pH during the titration is related to the pKa of the weak acid.

         During the titration, acid is being converted to its conjugate base, and a buffer solution is formed. Eventually, the quantity of base added is such that all of the acid has been converted to its conjugate base, and the equivalence point of the titration has been reached. The solution is no longer a buffer at the equivalence point.

A buffer is a special solution that stops massive changes in pH levels. Every buffer that is made has a certain buffer capacity, and buffer range. The buffer capacity is the amount of acid or base that can be added before the pH begins to change significantly. It can be also defined as the quantity of strong acid or base that must be added to change the pH of one liter of solution by one pH unit. The buffer range is the pH range where a buffer effectively neutralizes added acids and bases, while maintaining a relatively constant pH.

            The equation for pH also shows why pH does not change by much in buffers.

Where,

 is the concentration of the conjugate base

is the concentration of the acid

When the ratio between the conjugate base/acid is equal to 1, the pH = pK_a. If the ratio between the two is 0.10, the pH drops by 1 unit from pK_a since log (0.10) = -1. If a ratio increases to a value of 10, then the pH increases by 1 unit since log (10) = 1. The buffer capacity has a range of about 2. This means when a buffer is created, the pH can be changed by -1 by acid or +1 by base before the pH begins to change substantially. After the addition of base to raise the pH by 1 or more, most of the conjugate acid will have been depleted to try to maintain a certain pH, so the pH will be free to increase faster without the restraint of the conjugate acid. The same goes for the addition of acid, once the conjugate base has been used up, the pH will drop faster since most of the conjugate base has been used up.

 

Example 1

What is the effect on the pH of adding 0.006 mol HCL to 0.3L of a buffer solution that is 0.250M HC₂H₃O₂ and 0.560 M NaC₂H₃O₂?     pK_a= 4.74.

Calculate the starting amount of C₂H₃O₂^–

Calculate the starting amount of HC₂H₃O₂

 

Now calculate the new concentrations of C₂H₃O₂^– and HC₂H₃O₂:

Using the new concentrations, we can calculate the new pH:

Calculate the pH change:

Therefore, the pH dropped by 0.05 pH units.

Example 2

What is the effect on the pH of adding 0.006 mol NaOH to 0.3L of a buffer solution that is 0.250M HC₂H₃O₂ and 0.560 M NaC₂H₃O₂? pK_a= 4.74

Calculate the starting amount of HC₂H₃O₂

Calculate the starting amount of C₂H₃O₂^–

 

Now calculate the new concentrations of HC₂H₃O₂ and C₂H₃O₂-:

Using the new concentrations, we can calculate the new pH:

Calculate the pH change:

Therefore, the pH increased by 0.05 pH units.

Blood contains large amounts of carbonic acid, a weak acid, and bicarbonate, a base. Together they help maintain the bloods pH at 7.4. If blood pH falls below 6.8 or rises above 7.8, one can become sick or die. The bicarbonate neutralizes excess acids in the blood while the carbonic acid neutralizes excess bases.

Another example is when we consume antacids or milk of magnesia. After eating a meal with rich foods such as sea food, the stomach has to produce gastric acid to digest the food. Some of the acid can splash up the lower end of the esophagus causing a burning sensation. To relieve this burning, one would take an antacid, which when dissolved the bases buffer the excess acid by binding to them.

Buffering Blood

            Cell metabolism is based on the same general principle as the combustion of any fuel, whether it be in the automobile, power plant, or a home furnace. The general combustion reaction is:

CH₂O (fuel) + O₂ ===> CO₂ + HOH

The same reaction occurs in the cells. The “fuel” comes from food in the form of carbohydrates, fats, and proteins. The important principle to remember is that oxygen is needed by the cell and that carbon dioxide is produced as a waste product of the cell. Carbon dioxide must be expelled from the cells and the body. The lungs serve to exchange the two gases in the blood. Oxygen enters the blood from the lungs and carbon dioxide is expelled out of the blood into the lungs. The blood serves to transport both gases. Oxygen is carried to the cells. Carbon dioxide is carried away from the cells.

 

Gaseous Diffusion

         Partial pressures are used to designate the concentrations of gases. Dalton’s Law of Partial Pressures states that the total pressure of all gases is equal to the sum of the partial pressures of each gas. For example, the total atmospheric pressure of air is 760 mm Hg. In equation form:

P(total air) = P(O₂) + P(N₂) + P(CO₂) + P(H₂O)

760 = 160 + 594.7 + 0.3 + 5.0

         The partial pressures for oxygen and carbon dioxide in various locations are given in Figure 1. The movement or exchange of gases between the lungs, blood, and tissue cells is controlled by a diffusion process. The gas diffusion principle is: A gas diffuses from an area of higher partial pressure to an area of lower partial pressure.

Oxygen Transport

         In the lungs, oxygen diffuses from alveolar air into the blood because the venous blood has a lower partial pressure. The oxygen dissolves in the blood. Only a small amount is carried as a physical solution (0.31 ml per 100 ml). The remainder of the oxygen is carried in chemical combination with the hemoglobin in red blood cells (erthrocytes). Hemoglobin (molecular weight of 68,000) is made from 4 hemes, a porphyrin ring containing iron and globin, a 4 protein chains. Oxygen is bound to the iron for the transport process. Hemoglobin (HHgb) behaves as a weak acid (K = 1.4 x 10^-8; pKa = 7.85). Oxyhemoglobin (HHgbO₂) also behaves as a weak acid (K = 2.5 x 10^-7; pKa = 6.6)

         Because both forms of hemoglobin are weak acids, and a relationship of the numerical values of the equilibrium constants, the net reaction for the interaction of oxygen with hemoglobin results in the following equilibrium:

HHgb + O ₂ <===> HgbO ₂ + H⁺

         If HgbO ₂ is increased in the blood at the lungs, the equilibrium shifts to the right and H⁺ ions increase. Oxyhemoglobin can be caused to release oxygen by the addition of H⁺ ions at the cells. The difference in pH (7.44) of arterial blood and venous blood (pH = 7.35) is sufficient to cause release of oxygen from hemoglobin at the tissue cells.

 

Weak Acids & Bases

         Unlike strong acids/bases, weak acids and weak bases do not completely dissociate (separate into ions) at equilibrium in water, so calculating the pH of these solutions requires consideration of a unique ionization constant and equilibrium concentrations. Although this is more difficult than calculating the pH of a strong acid or base solution, most biochemically important acids and bases are considered weak, and so it is very useful to understand how to calculate the pH of these substances. The same basic method can be used to determine the pH of aqueous solutions of many different weak acids and bases.

         An aqueous solution of a weak acid or base contains both the protonated and unprotonated forms of the compound, so an ICE table can be made and used to plug in concentrations into an equilibrium constant expression. The ionization constant for the acid (K_a) or base (K_b) is a measure of how readily the acid donates protons or how readily a base accepts protons. Because you are calculating pH, you must solve for the unknown concentration of hydronium ions in solution at equilibrium.

         The first step in calculating the pH of an aqueous solution of any weak acid or base is to notice whether the initial concentration is high or low relative to 10^-7 M (the concentration of hydronium and hydroxide ions in water due to the autoionization of water). If the concentration of the acid or base is very close to or less than 10^-7 M, then the solution is considered dilute and additional steps must be taken to calculate pH.

         You must first be familiar with equilibrium constant expressions and how to write them for a chemical reaction. Then, by making an ICE table, you can find unknown concentration values that can be plugged into this equilibrium expression.

Example

Example 1 (Weak Acid):

What is the pH of 1.5 L of a vinegar that is 3% acetic acid by mass? (K_a = 1.8 x 10^-5)

         To start, you must find the initial concentration of acetic acid in the vinegar. Assume that the vinegar is really just a solution of acetic acid in water, and that density = 1 g/mL.

         So if the vinegar is 3% acetic acid by mass and the molar mass of HC₂H₃O₂ = 60.05 g/mol, then

= 0.75 mol HC₂H₃O₂

         Divide 0.75 mol by 1.5 L to get an initial concentration of 0.50 M.

         Now make an ICE table, considering the ionization of acetic acid in water into acetate ion and hydronium ion. Because only solutes and gases are incorporated into the equilibrium expression, you can ignore the concentration of water (a pure liquid) in our calculations.

HC₂H₃O₂(aq) + H₂O(l) C₂H₃O₂^–(aq) + H₃O⁺(aq)

        

            For every acetic acid molecule that dissociates, one acetate ion and one hydronium ion is produced. This can be represented by subtracting “x” from the original acetic acid concentration, and adding “x” to the original concentrations of the dissociated ions. You can create a modified equilibrium constant expression

and then plug in the concentration values you found in the ICE table

            so             

then use the quadratic formula to calculate

x = 0.0030 M = [H₃O⁺]

which can be plugged into the formula

pH = -log[H₃O⁺]

-log(.0030) = pH = 2.5

The same thing can be done for calculating the pH of a weak base.

Example

Example 2 (Weak Base):

What is the pH of a 7.0 x 10^-3 M NH₃ solution? (pK_b = 4.74)

NH₃(aq) + H₂O(l) NH₄⁺(aq) + OH^–(aq)

	NH3	H2O	NH4+	OH–
I	7.0 x 10-3	—	0	0
C	-x	—	+x	+x
E	7.0×10-3 – x	—	x	x

 

         Instead of K_b, you were given pK_b. So to get K_b

pK_b = -log(K_b) = 4.74

K_b = 10^-4.74 = 1.8 x 10^-5

         Plug these values into the equilibrium expression to get

and use the quadratic formula to find that

x = 3.46 x 10^-4 M = [OH^–]    so     pOH = -log(3.46 x 10^-4) = 3.46
and    pH + POH = 14
so   14 – 3.46 = pH = 10.54

 

Weak Polyprotic Acids and Bases

         Polyprotic acids have more than one proton to donate to water, and so they have more than one ionization constant (K_a1, K_a2, etc) that can be considered. Polyprotic bases take more than one proton from water, and also have more than one ionization constant (K_b1, K_b2, etc).

         Most often the first proton exchange is the only one that considerably affects pH. This is discussed more at the end of the first example.

Example

Example 1 (Polyprotic Acid):

What is the pH of a grapefruit that contains 0.007 M citric acid solution (C₆H₈O₇)?

(K_a1 = 7.5 x 10^-4, K_a2 = 1.7 x 10^-5, K_a3 = 4.0 x 10^-7)

         Make an ICE table for the first dissociation

C₆H₈O₇(aq) + H₂O(l) C₆H₇O₇^– + H₃O⁺(aq)

	C6H8O7	H2O	C6H7O7–	H3O+
I	0.007	—	0	0
C	-x	—	+x	+x
E	0.007 – x	—	x	x

and use the quadratic formula to find that

x = 0.00195 M = [H₃O⁺]

Then a second ICE table can be made for the second dissociation

C₆H₇O₇^–(aq) + H₂O(l) C₆H₆O₇^2- + H₃O⁺(aq)

	C6H7O7–	H2O	C6H6O72-	H3O+
I	0.00195	—	0	0.00195
C	-x	—	+x	+x
E	0.00195 – x	—	x	0.00195 + x

 

         Remember that, for the first dissociation, x = [H₃O⁺] = [C₆H₇O₇^–], so you can plug in the first value of x in for the initial concentrations of C₆H₇O₇^– and H₃O⁺.

 

and use the quadratic formula to find that

x = 1.67×10^-5

[H₃O⁺] = 0.00195 + 1.67×10^-5 = 0.00197 M

-log(0.00197) = pH = 2.71

         Note that if you ignored the addition of hydronium from the second dissociation, then [H₃O⁺] = 0.00195 M, and using this value to calculate pH still gives you the answer of 2.71.

So even though you made two ICE tables (you could even make a third table for K_a3), the protons donated in the second dissociation were negligible compared to the first dissociation. So you can see that it is really only the first dissociation that affects pH. Most often this is the case, and only one ICE table is necessary. It is up to you how certain you want to be and how many ICE tables you want to make when you calculate these problems.

Example

Example 2 (Polyprotic Base):
What is the pH of a saturated solution of sodium carbonate (Na₂CO₃)?
(solubility in water is 21.6g/100mL at room temperature)
(for carbonic acid, H₂CO₃, K_a1 = 4.5 x 10^-7, K_a2 = 4.7 x 10^-11).

First, you have to find the find the initial concentration of CO₃^2- which can be found from

= 0.204 mol Na₂CO₃ = 0.204 mol CO₃^2-
then divide 0.204 mol by 0.100 L to get 2.04 M CO₃^2-

         Plug into an ICE table

CO₃^2-(aq) + H₂O(l) HCO₃^–(aq) + OH^–(aq)

 

         But notice that the equilibrium constants are for carbonic acid.
         If you were considering the dissociation of carbonic acid, you would write the following expressions

for H₂CO₃ + H₂O HCO₃^– + H₃O⁺

for HCO₃^– + H₂O CO₃^2- + H₃O⁺

         The second acid ionization constant corresponds to the first base ionization constant (because the base reactions go backwards). To convert the second acid ionization constant to the first base ionization constant, you use the equation

K_a x K_b = K_w = 10^-14
so that
K_a2 x K_b1 = 10^-14
K_b1 = = 2.13 x 10^-4

         Use the same equation to convert the first acid ionization constant to the second base ionization constant

K_a1 x K_b2 = 10^-14

K_b2 = = 2.22 x 10^-8

         The expressions for the protonation of carbonate are now known to be

for CO₃^2- + H₂O HCO₃^– + OH^–

for HCO₃^– + H₂O H₂CO₃ + OH^–

Plug the ICE tables values into the first equilibrium expression

and use the quadratic formula to solve

x = 0.0207 M = [OH^–].

         You can ignore the second base ionization constant because it removes a negligible amount of protons from water. If you want to test this by making an ICE table, you should get that the total hydroxide concentration is 0.0207000222 M 0.0207 M

so pOH = -log(0.0207) = 1.68

pH + pOH = 14
14 – 1.68 = pH = 12.32

 

Dilute Weak Acids and Bases

         ‘Dilute’ refers to the concentration of the acid or base in water. If the concentration is close to or below 10^-7 M, then you must consider the donation of hydronium ions from water as well as from your acid or base. This is done by making an ICE table to find the protonation of the acid or base, while also incorporating the ion product of water.

Example

Example 1 (Dilute Weak Acid):

         If a bee sting excretes 5.00 micrograms of formic acid (HCO₂H), what would be the pH of a 500mL solution of this formic acid? (pK_a = 3.75)

HCO₂H(aq) + H₂O(l) CO₂H^–(aq) + H₃O⁺(aq)

	HCO2H	H2O	CO2H–	H3O+
I	2.17 x 10-7	—	0	0
C	-x	—	+x	+x
E	2.17×10-7 – x	—	x	x

 

         A second ICE table can be made for the autoionization of water

2H₂O(l) H₃O⁺(aq) + OH^–(aq)

 

	2H2O	H3O+	OH–
I	—	0	0
C	—	+y	+y
E	—	y	y

        

         Notice that total [H₃O⁺] = x + y

pK_a = 3.75
K_a = 10^-3.75 = 1.78 x 10^-4

        

So you can simultaneously solve both equations

 

and

K_w = [H₃O⁺][OH^–] = (x + y)(y) = 10^-14

 

         These calculations can be tricky, and it is very easy to make mistakes. It is usually easier to use variables to solve these problems instead of handling awkward numbers.

         For this problem use

a = K_a = 1.78 x 10^-4
c = initial [HCO₂H] = 2.17 x 10^-7
w = K_w = 10^-14

     and       

      and    

 

so when we plug back in the values

 

 

and use a graphing calculator to find that

y = 3.91 x 10^-8

and

 

and total

[H₃O⁺] = (x + y) = (2.17 x 10^-7 + 3.91 x 10^-8) = 2.56 x 10^-7 M

so       -log[H₃O⁺] = -log(2.56 x 10^-7) = pH = 6.59

         Compare this value to pH = 6.66, which is what would have been calculated if the autoprotonization of water was not considered.

Buffer Solutions

         Buffer solutions resist pH change when more acid or base is added. They are made from a weak acid and its conjugate base or a weak base and its conjugate acid. The Henderson-Hasselbalch equation can be used to find the pH of a buffer solution, and is derived from the acid equilibrium expression.

HA(aq) + H₂O(l) A^–(aq) + H₃O⁺(aq)

A similar equation can be used for bases

.

Example

Example 1:

         The pH of blood plasma is 7.40, and is maintained by a carbonic acid/hydrogen carbonate buffer system. What mass of sodium bicarbonate (NaHCO₃) should be added to a one liter solution of 0.250 M H₂CO₃ to maintain the solution at pH of 7.40? (pKa = 6.35)

1.05 = log[HCO₃^–] – log(0.250)

0.448 = log[HCO₃^–]

[HCO₃^–] = 0.356 M

0.356 M x 1 L solution = 0.356 mol HCO₃^–

= 29.9 g sodium bicarbonate.

         It is also possible to use the Henderson-Hasselbalch equation to find pK_a, pH, or [HA] if the other variables are given or calculated. Also notice that because will cancel out the unit of volume, moles of HA and A^– can be used instead of molarity.

 

Ion Exchange Chromatography

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         Imagine if we had a tube whose surfaces were coated with an immobilized cation. These would have electrostatic attraction for anions. If a solution containing a mixture of positively and negatively charged groups flows through this tube, the anions would preferentially bind, and the cations in the solution would flow through

·                     This is the basis of ion exchange chromatography. The example above is termed “anion exchange” because the inert surface is interacting with anions

·                     If the immobile surface was coated with anions, then the chromatography would be termed “cation exchange” chromatography (and cations would selectively bind and be removed from the solution flowing through

·                     Strength of binding can be affected by pH, and salt concentration of the buffer. The ionic species “stuck” to the column can be removed (i.e. “eluted”) and collected by changing one of these conditions. For example, we could lower the pH of the buffer and protonate anions. This would eliminate their electrostatic attraction to the immobilized cation surface. Or, we could increase the salt concentration of the buffer, the anions in the salt would “compete off” bound anions on the cation surface.

Capillary Electrophoresis

         Capillary electrophoresis is an analytical technique that separates ions based on their electrophoretic mobility with the use of an applied voltage. The electrophoretic mobility is dependent upon the charge of the molecule, the viscosity, and the atom’s radius. The rate at which the particle moves is directly proportional to the applied electric field–the greater the field strength, the faster the mobility. Neutral species are not affected, only ions move with the electric field. If two ions are the same size, the one with greater charge will move the fastest. For ions of the same charge, the smaller particle has less friction and overall faster migration rate. Capillary electrophoresis is used most predominately because it gives faster results and provides high resolution separation. It is a useful technique because there is a large range of detection methods available.

         Endeavors in capillary electrophoresis (CE) began as early as the late 1800’s. Experiments began with the use of glass U tubes and trials of both gel and free solutions.¹ In 1930, Arnes Tiselius first showed the capability of electrophoresis in an experiment that showed the separation of proteins in free solutions.² His work had gone unnoticed until Hjerten introduced the use of capillaries in the 1960’s. However, their establishments were not widely recognized until Jorgenson and Lukacs published papers showing the ability of capillary electrophoresis to perform separations that seemed unachievable. Employing a capillary in electrophoresis had solved some common problems in traditional electrophoresis. For example, the thin dimensions of the capillaries greatly increased the surface to volume ratio, which eliminated overheating by high voltages. The increased efficiency and the amazing separating capabilities of capillary electrophoresis spurred a growing interest among the scientific society to execute further developments in the technique.

            A typical capillary electrophoresis system consists of a high-voltage power supply, a sample introduction system, a capillary tube, a detector and an output device. Some instruments include a temperature control device to ensure reproducible results. This is because the separation of the sample depends on the electrophoretic mobility and the viscosity of the solutions decreases as the column temperature rises.³ Each side of the high voltage power supply is connected to an electrode. These electrodes help to induce an electric field to initiate the migration of the sample from the anode to the cathode through the capillary tube. The capillary is made of fused silica and is sometimes coated with polyimide.³ Each side of the capillary tube is dipped in a vial containing the electrode and an electrolytic solution, or aqueous buffer. Before the sample is introduced to the column, the capillary must be flushed with the desired buffer solution. There is usually a small window near the cathodic end of the capillary which allows UV-VIS light to pass through the analyte and measure the absorbance. A photomultiplier tube is also connected at the cathodic end of the capillary, which enables the construction of a mass spectrum, providing information about the mass to charge ratio of the ionic species.

Fig. Instrumental Setup

 

Biological importance of human erythrocyte Carbonic anhydrase B

         Carbonic anhydrases are abundant in mammalian tissues, plants and bacteria. The several distinct classes of this enzyme (α, β, γ etc.) share little similarity in terms of structure and sequence, but they all perform the same function of catalyzing the hydration of carbon dioxide and the dehydration of bicarbonate. Carbonic anhydrase B is one of the two major forms of carbonic anhydrase in human red blood cell (the other form is CA C). Like other members of the CA family, its main function is assisting the important chemical reaction that occurs in the blood

CO₂ + H₂O <—–> HCO₃^– + H⁺

         The above reaction is very slow in the absence of the enzyme. Hence, by assisting the inter-conversion between carbon dioxide and bicarbonate, Carbonic anhydrase B helps to maintain acid-base balance in blood, and transport carbon dioxide out of blood. Note that the carbonic acid (H₂CO₃) and hydrogen carbonate ions (HCO₃^–) is an important buffer system in the body. They equilibrate in the blood plasma to buffer pH. The inter-conversion between the two is shown in the second step of the reaction below

CO₂ + H₂O <——> H⁺(aq) + HCO₃^–(aq) <—–> H₂CO₃(aq)

            Another thing to note is that carbonic anhydrase B has very high catalytic power. For its catalytic activity of the hydration of dioxide, it increases the rate of the reaction by a factor of 10⁹ (with a turnover number for carbon dioxide is roughly 10⁶ s^-1 under saturated conditions). However, the apoenzyme was almost catalytically inactive, which means that the enzyme cannot function without the zinc ion. Experiments have shown that the activity of carbonic anhydrase B can be restored when zinc ions are added in 1:1 molar ratio. In fact, it is zinc-aquo complex (rather than zinc ion alone) that needs to be present in the active site for the catalytic functionality of the enzyme.

Water Autoionization and the Ionization Constant

Water, even pure water, has an amphiprotic nature. This means that a small amount of ions will form in pure water. Some molecules of H₂O will act as acids, each donating a proton to a corresponding H₂O molecule that acts as a base. Thus, the proton-donating molecule becomes a hydroxide ion, OH^–, while the proton-accepting molecule becomes a hydronium ion, H₃O⁺.

Water molecules can function as both acids and bases. One water molecule (acting as a base) can accept a hydrogen ion from a second one (acting as an acid). This will be happening anywhere there is even a trace of water – it doesn’t have to be pure.

A hydronium ion and a hydroxide ion are formed.

However, the hydroxonium ion is a very strong acid, and the hydroxide ion is a very strong base. As fast as they are formed, they react to poduce water again. The net effect is that an equilibrium is set up.

At any one time, there are incredibly small numbers of hydroxonium ions and hydroxide ions present. Further down this page, we shall calculate the concentration of hydroxonium ions present in pure water. It turns out to be 1.00 x 10^-7 mol dm^-3 at room temperature. This equilibrium written in a simplified form:

with H⁺(aq) actually refering to a hydronium ion.

It is important to remember that water contains VERY low concentration of these ions. In the reversible reaction:

H₂O + H₂O ↔ H₃O⁺ + OH^–

H₂O= Base(1) + H₂O= Acid(2) ↔ H₃O⁺= Acid(1) + OH^–= Base(2)

the reaction proceeds by far to the left. Pure water will dissociate to form equal concentrations (here, we are using molarities) of hydronium and hydroxide ions, thus:

[H₃0⁺] = [OH^–]

For this equation, we can find K, the equilibrium constant.

K= [H₃0⁺][OH^–]

At standard temperature and pressure, STP, the equilibrium constant of water, K_w, is equal to

K_w= [H₃0⁺][OH^–]

K_w=[1.0×10^-7][1.0×10^-7]

K_w=1.0×10^-14

In this equation [H₃O⁺] is the concentration of hydronium ions, which in a chemical equation is the acid concentration, K_a. The [OH^–] is the concentration of hydroxide ions, which in a chemical equation is the base concentration, K_b. If given a pH, then you can easily calculate the [H₃O⁺] by simply taking the negative reverse log of the pH:

[H₃O⁺] = 10^-pH.

The same formula applies to obtaining [OH^–] from the pOH:

[OH^–]=10^-pOH

Adding the pH’s gives you the pK_w

pK_w= pH + pOH =14.00

Since the reaction proceeds so heavily to the left, the concentration of these hydroxide and hydronium ions in pure water is extremely small. When making calculations determining involving acids and bases in solution, you do not need to take into account the effects of water’s autoionization unless the acid or base of interest is incredibly dilute. However, it is interesting to note that this water’s self-ionization is significant in that it makes the substance electrically conductive.

Problems

1. In the equation depicting the autoionization of water,

H₂O + H₂O ↔ H₃O⁺ + OH^–

The reaction proceeds far to the __________. Answer: left

The concentration of hydroxide and hydronium ions in pure water is very, very small. Although it is rarely something you need to worry about when looking at acids and bases in solution, it does help account for certain properties of water such as electrical conductivity.

If a solution has a pH of 2.1, determine the concentration of hydroxide ion, [OH^–]: to solve for this, you must first determine the concentration of the hydronium ion, [H₃O⁺]:

[H₃O⁺] =10^-pH =10^-2.1 = 7.94 x 10^-3

Then, you solve for [OH^–] using the K_w constant:

K_w = [H₃O⁺] [OH^–]

1.0 x 10^-14 = [OH-][7.94 x 10^-3]

[OH-] = (1 x 10^-14)/ (7.94 x 10^-3)

= 1.26 x 10^-12

If a solution has a pOH of 11.2, determine the concentration of hydronium ion, [H₃O⁺]: to solve for this, you must first determine the concentration of the hydroxide ion, [OH^–]:

[OH^–]=10^-pOH =10^-11.2 = 6.31 x 10^-12

Then, you solve for [H₃O⁺] using the K_w constant:

K_w = [H₃O⁺] [OH^–]
1.0 x 10^-14 = [H₃O⁺][6.31 x 10^-12]
[H₃O⁺]= (1 x 10^-14)/ (6.31 x 10^-12)= .00158

 

 

In IV analytical group cations on the acid-base classification are ions Al³⁺, Zn²⁺, Cr³⁺, As^III, As^V, Sn^II, Sn^IV. Compounds with Al, Zn, As are applied as drugs; compounds of Al, Cr, As, Sn are used in the pharmaceutical analysis. Therefore the future pharmacist should own knowledge of chemical-analytical properties of the given group of ions.

The fourth analytical group includes cations Al³⁺, Zn²⁺, Cr³⁺, As^III, As^V, Sn^II and Sn^IV.

Ions As^III and As^V in water solutions corresponding are present as аrsenyte AsО₃^3-or AsО₂^– and arsenat – AsО₄^3-. In the concentrated solutions of chloridic acid they are present as  complex [AsCl₄]^– and [AsCl₆]^–.

2 mol/L solution of Sodium hydroxide (or Potassium hydroxide) at presence the Hydrogene peroxide are a group reagent on cations of IV analytical group; sometimes – without the Hydrogene peroxide.

The common reactions of IV analytical group cations

 Sodium hydroxide (or Potassium hydroxide) precipitate Al³⁺, Cr³⁺, Zn²⁺, Sn²⁺ ions as amphoteric hydroxides which are dissolved in excess of alkali with formation complexes, for example:

Al³⁺ + 3OH^–®Al(OH)₃¯, Al(OH)₃ + 3OH^– ® [Al(OH)₆]^3-;

Cr^{3 +} + 3OH^– ®Cr(OH)₃¯, Cr(OH)₃ + 3OH^– ® [Cr(OH)₆]^3- or Cr(OH)₃ + OH^– ®[Cr(OH)₄]^–;

Zn^{2 +} + 2OH^– ®Zn(OH)₂¯, Zn(OH)₂ + 2OH^–® [Zn(OH)₄]^2-;

Sn²⁺ + 2OH^– ®Sn(OH)₂¯, Sn(OH)₂ + 4OH^– ®[Sn(OH)₆]^4-.

At the presence of Hydrogene peroxide cations Cr³⁺, As³⁺, Sn²⁺ are oxidised accordingly to chromat – arsenat – and hexahydroxostanat (ІV) – ions:

2 [Cr(OH)₄]^– + 3Н₂О₂ + 2OH^– ®2CrO₄^2- + 8Н₂О;

[Sn(OH)₆]^4- + Н₂О₂ ®[Sn(OH)₆]^2- + 2OH^–;

AsО₃^3-+ Н₂О₂ ®AsО₄^3-+ Н₂О.

Reaction performance. Into four test tubes select on 3-4 drops of salts solutions of Aluminium, Chrome (ІІІ), Zinc and Tin (ІІ) and add in each some drops 2 mol/L a solution Sodium hydroxide. At first observe formation of white precipitate of Al(OH)₃, Zn(OH)₂, Sn(OH)₂ and dirty green Cr(OH)₃, and then their dissolution is an exess of alkali solution. Into a test tube with hexahydroxochromate (ІІІ) add 5-6 drops of Hydrogene peroxide solution and heat a test tube on a water-bath, observing change colouring from green to yellow (form chromate ions).

Alkaline metals and ammonium hydrogenphosphate form white crystal precipitates of AlPO₄, Zn₃(PO₄)₂, CrPO₄, Sn₃(PO₄)₂:

Al³⁺ + HPO₄^2- + NH₃ ® AlPO₄ + NH₄⁺

Aluminium, Zinc, Chrome and Tin phosphates are dissolved in solutions of alkalis. Zinc phosphate, also is well dissolved in ammonia, therefore it forms complex:

3Zn²⁺ + 2HPO₄^2- ® Zn₃(PO₄)₂ + 2H⁺;

Zn₃(PO₄)₂ + 12NH₃ ® 3[Zn(NH₃)₄]^{2 +} + 2PO₄^3-.

Reaction performance. Into four test tubes select on 3-4 drops of salts solutions of Aluminium, Zinc, Chrome (ІІІ) and Tin (ІІ), add in each some drops of 0,5 mol/L Sodium hydrogenphosphate solution and observe occurrence of precipitates. Each of them investigate on solubility in alkalis, and Zinc phosphate – also in 2 mol/L ammonia solution.

Characteristic reactions of ions Al³⁺

The solution of ammonia NH₃ (pharmacopeia’s reaction) with salts of Aluminium in the neutral medium forms white amorphous precipitate Аl(OH)₃:

Al³⁺ + 3NH₃ + 3H₂O АAl(OH)₃↓ + 3NH₄⁺,

Al³⁺ + 3OH^– ®Аl(OH)₃↓.

Aluminium hydroxide has amphoteric properties. It is dissolved in acids:

Аl(OH)₃ + 3НNO₃ ®Аl(NO₃)₃ + 3Н₂O,

And also in alkalis:

Аl (OH)₃ + 3NaOH® Na₃[Аl(OH)₆].

After heating of complex it forms metaaluminate:

Na₃[Аl(OH)₆] NNaАlО₂ + 2NaOH + 2H₂O.

In the presence of NH₄Cl at heating aluminate forms precipitate Аl(OH)₃:

NaАlО₂ + NH₄Cl + H₂O АAl(OH)₃¯ + NH₃ + NаCl.

Reaction performance. To 3-4 drops of an investigated solution add 1-2 drops of 1 mol/L chloridic acid and 2-3 drops of tioacetammide solution. After that add some drops of 1 mol/L a Sodium hydroxide solution. If there are ions Аl³⁺, white precipitate forms. It is dissolved in excess of Sodium hydroxide solution. To the received solution add 1 mol/L ammonium chloride solution and observe formation white precipitate.

Alizarin (1, 2-dioksiantrahinon) with ions Al³⁺ in weak acidic medium forms red complex of Aluminium alizarinate Al(OH)₂[C₁₄H₆O₃(OH)] which is not dissolved in acitic acid. It is called as an aluminium varnish.

The ions of Fe^{3 +}, Bi^{3 +}, Cu^{2 +} interfere with the exposure of ions of Al³⁺.

Reaction performance. Reaction are executed in the drop way. On a strip of a filtering paper put a drop of solution K₄[Fe(CN)₆], and in its centre a drop of an investigated solution. After that a strip of a filtering paper are keeped above ammonia, and put on it some drops of alizarin solution. After that a strip of a filtering paper are keeped above ammonia. After that paper are dried, and put on it some drops of 2 mol/L CH₃COOH solution. If there are ions Al^{3 +}, there is a pink ring.

Cobalt nitrate with Aluminium salts (by length heating) forms mixed оxide Aluminium and Cobalt (Cobalt aluminate) Co(AlО₂)₂ dark blue colours:

2Al₂(SO₄)₃ + 2Co(NO₃)₂ ®2Со(AlО₂)₂ + 4NO₂­ + 6SO₃­ + O₂­.

Reaction performance. On a strip of a filtering paper put some drops of 0,1 mol/L investigated solution, and add 2-3 drops of Cobalt nitrate solution. A paper is dried, place into a porcelain crucible and length heating. If there are ions Al^{3 +}, ash will be dark blue colour.

Аluminon with ions Al^{3 +} forms red complex.

The ions of Ca^{2 +}, Cr^{3 +}, Fe^{3 +} interfere with the exposure of ions of Al³⁺.

Reaction performance. To some drops of an investigated solution add 2-3 drops of 2 mol/L CH₃COOH solution, 3-5 drops of aluminon solution and a mix are heated. Then add solution of NH₃ to basic medium (occurrence of a smell of ammonia) and 2-3 drops of 1 mol/L Na₂CO₃ solution. If there are ions Al^{3 +}, the red precipitate forms.

Characteristic reactions of ions Cr^{3 +}

Chrome (ІІІ) in the basic medium is oxidised to CrО₄^2-, and in acidic – to Cr₂O₇^2-. Ions Cr³⁺ to Cr₂O₇^2- are oxidised only strong oxidizers (H₂O₂, Na₂O₂, KMnО₄, Cl₂, Br₂, (NH₄)₂S₂O₈).

Chrome (ІІІ) in the basic medium is oxidised by H₂O₂, chloric or bromic water. In the basic medium chrome (ІІІ) with oxidizing reagents gives complex of tetrahydroxichromit (ІІІ):

2 [Cr(OH)₄]^– + 3Br₂ + 8OH ^– = 2CrO₄^2- + 6Br^– + 8H₂O

Reaction performance. To 2-3 drops of an investigated solution add 3-4 drops of 2 mol/L NaOH solution, 2-3 drops of bromic (chloric) water or 3 % Hydrogene peroxide solution and heat it on a water-bath to change colouring of a solution from green to yellow. Presence of CrО₄^2- ions check by reaction formation peroxichromatic acid H₂CrО₆.

Chrome (ІІІ) in the acidic medium is oxidised by KMnО₄, (NH₄)₂S₂O₈ and many other strong oxidizers:

2Cr^{3 +} + 2MnO₄^– + 5H₂O = Cr₂O₇^2- + 2MnО(OH)₂¯ + 6H⁺.

Reaction passes at heating and is accompanied by formation of brown precipitate MnО(OH)₂.

2Cr³⁺ + 3S₂O₈^2- + 7H₂O = Cr₂O₇^2- + 6SO₄^2- + 14H⁺.

Reaction is accelerated in the presence of Silver salts (catalyst).

Reaction performance. To 2-3 drops of an investigated solution add 2-3 drops of 1 mol/L HNO₃ or H₂SO₄ solution, 10-15 drops KMnО₄ and heat this solution. If there are ions Cr^{3 +}, there is an orange solution. A part of the solution is investigated on formation peroxichromatic acid.

Formation of peroxichromatic acid. If to asidic solution of chromate or bichromate add H₂O₂, dark blue solution of peroxichromatic acid forms:

Cr₂O₇^2- + 4H₂O₂ + 2H⁺ = 2H₂CrО₆ + 3H₂O.

In the aqueous solution peroxichromatic acid is very unstable (it is displayed with formation Cr³⁺), therefore to a solution add organic solvent (amyl alcohol or diethyl ether).

Reaction performance. To 2-3 drops of received solution (Cr₂O₇^2-) add 1 mol/L H₂SO₄ to acidic medium, 0,5 mL amyl alcohol and 4-5 drops H₂O₂. If there are Chrome ions the top bed of a solution are painted in dark blue colour. Combining this reaction with any reaction of oxidation Cr³⁺ to Cr₂O₇^2- it is possible to use it for fractional determination of Chrome ions in the presence of all others cations.

Characteristic reactions of ions Zn^{2 +}

Ammonium tetrarhodanomercurate (NH₄)₂[Hg(SCN)₄] with Zn²⁺ ions forms a white crystal precipitate:

Zn²⁺ + [Hg(SCN)₄]^2- = Zn[Hg(SCN)₄]¯.

Reaction passe in the acidic medium. Concentration of acid (it is better H₂SO₄) should not exceed 1 mol/L.

The ions of Fe²⁺, Fe^{3 +} interfere with the exposure of ions of Zn²⁺.

Reaction performance. To 2-3 drops of an investigated solution, add some drops of 1 mol/L sulphatic acid and 2-3 drops of ammonium tetrarhodanomercurate. If there are ions Zn²⁺, the white crystal precipitate forms.

Sodium sulphide Na₂S (pharmacopeia’s reaction) with solutions of Zinc salts forms white precipitate ZnS which is not dissolved in acetic acid, but is dissolved in diluted chloric acid:

Zn²⁺ + S^2- ® ZnS¯

ZnS + 2HCl ZZnCl₂ + H₂S­

The ions of Ag ⁺, Pb^{2 +}, Hg^{2 +} interfere with the exposure of ions of Zn²⁺.

Reaction performance. To 2-3 drops of an investigated solution (рН=5-6) add 1-2 drops of Sodium sulphide solution. If there are ions Zn²⁺, the white precipitate forms.

Potassium hexacyanoferrate (II) K₄[Fe(CN)₆] (pharmacopeia’s reaction) with Zn²⁺ ions forms white precipitate Potassium-Zinc hexacyanoferrate (ІІ) which is not dissolved in diluted chloridic acid:

3Zn²⁺ + 2К₄[Fe (CN)₆] = К₂Zn₃[Fe(CN)₆]₂¯ + 6К⁺.

This reaction may to distinguish Aluminium and Zinc ions.

Reaction performance. To 3-4 drops of an investigated solution add to 1-2 drop Potassium hexacyanoferrate (II). If there are ions Zn²⁺, the white precipitate forms. It is not dissolved in diluted chloric acid.

Dithizon (diphenylthiocarbasol) in solution CCl₄ (or chloroform CHCl₃) with ions Zn²⁺ forms complex bright red colour – Zinc dithizoate, which is extracted by ССl₄ or CHCl₃.

The ions of Ag⁺, Bi³⁺, Pb²⁺, Cu²⁺ interfere with the exposure of ions of Zn²⁺. So before deremination of Zn²⁺ ions their linkage in complexes with 0,5 mol/L Sodium thiosulphate.

Reaction performance. To 2-3 drops of an investigated solution add 1 mL acetic buffer (pН=5) and 1-2 mL of 10 % of dithizon solution in CCl₄ (or CHCl₃). If there are ions Zn²⁺ the organic layer is painted in red colour.

Cobalt nitrate with Zinc salts (by length heating) forms mixed оxide Zinc and Cobalt СоZnО₂ green colour – so-called „Renmarn’s greens”:

Zn(NO₃)₂ + Co(NO₃)₂ ® СоZnО₂ + 4NO₂­ + O₂­.

Reaction performance. On a strip of a filtering paper put some drops of 0,1 mol/L investigated solution, and add 2-3 drops of Cobalt nitrate solution. A paper are dried, place into a porcelain crucible and length heating. If there are ions Zn²⁺, ash will be green colour.

Characteristic reactions of ions Sn^II

Potassium (Sodium) hydroxide with ions Sn^{2 +} forms white precipitate Sn(OH)₂ which is dissolved in excess of alkali with formation tetrahydroxostannate (ІІ):

Sn²⁺ + 2OH^– ®Sn(OH)₂¯;

Sn(OH)₂ + 2OH^–® [Sn(OH)₄]^2-.

Precipitate hydroxide well dissolves in strong mineral acids.

If to solution of Sn (II) salts add some drops of ammonium salt, precipitate Sn(OH)₂ forms.

Reaction performance. To 3-4 drops of an investigated solution add drops of 2 mol/L Sodium hydroxide solution. If there are ions Sn (II), the white precipitate forms, it is dissolved in excess of alkali.

Sodium (Potassium) sulphide and hydrogen sulphide with ions Sn²⁺ forms dark brown precipitate Tin (ІІ) sulphide:

Sn²⁺ + S^2- ®SnS¯.

The precipitate is not dissolved in alkalis, excess of Sodium sulphide.

The ions of Cu^{2 +}, Hg^{2 +} interfere with the exposure of ions of Sn²⁺.

Reaction performance. To 3-4 drops of an investigated solution add 2-3 drops of 0,5 mol/L Sodium sulphide solution. If there are Sn (II) ions, brown precipitate forms.

Salts of Bismuth (ІІІ) with Sn^II ions forms metal Bismuth:

2Bi^{3 +} + 3[Sn(OH)₄]^2- +6ОН^–= 2Bi¯ + 3[Sn(OH)₆]^2-.

Reaction pass in basic medium. Metal Bismuth forms a precipitate of black colour.

The ions which in basic medium form precipitates of hydroxides interfere with the exposure of ions of Sn²⁺.

Reaction performance. To 5-6 drops of an investigated solution add some drops of 2 mol/L a Sodium hydroxide solution to dissolution of a precipitate which can be formed from the first drops. To the received solution add 1-2 drops of 0,5 mol/L Bi(NO₃)₃ solution. If there are Sn (II) ions, black precipitate of metal Bismuth forms.

Mercury (ІІ) chloride HgCl₂. In an acidic solution Sn²⁺ ions reducte HgCl₂ to formation of white precipitate Hg₂Cl₂. If to precipitate add excess of HgCl₂ solution, black precipitate of metal mercury forms:

[SnCl₄]^2- + 2HgCl₂ ®[SnCl₆]^2- + Hg₂Cl₂¯;

[SnCl₄]^2- + Hg₂Cl₂ ®[SnCl₆]^2- + 2Hg¯.

The chloride ions interfere with the exposure of ions of Sn²⁺.

Reaction performance. To 2-3 drops of an investigated solution add 2-3 drops of concentrated HCl and 2-3 drops of solution HgCl₂. If there are Sn (II) ions, white precipitate forms which gradually blackens.

Ammonium tetramolybdatophosphat (NH₄)₃[P(Mo₃O₁₀)₄]×H₂O. Ions Sn²⁺ reducte molybdenum (VI) to dark blue compound (molybdenic blue). In this compound Molybdenum has the lowest oxidation state. The reducer interfere with the exposure of ions of Sn²⁺.

Reaction performance. To 2-3 drops of Na₂HPO₄ solution add 2-3 drops of molybdenic liquids (a mix of (NH₄)₂MoО₄ and NH₄NO₃) and heat it. The yellow precipitate ammonium tetramolybdatophosphat (NH₄)₃[P(Mo₃O₁₀)₄] forms. Into other test tube add 2-3 drops of an investigated solution, 3-4 drops of concentrated HCl and a strip of iron (some milligrammes of a iron powder). A mix heats for 2-3 minutes. Some drops of receved solution add to the first test tube with a yellow precipitate. If there are Sn (II) ions, blue precipitate.

Characteristic reactions of ions Sn^IV

Sn (IV) ions usually determinate after preliminary are reducted by metal Fe, Mg, Al etc. to Sn (II). Then spend reactions which are characteristic for ions Sn (II).

For determination ions Sn (IV) it is possible to use reaction with hydrogen sulphide.

Hydrogen sulphide with Sn (IV) ions forms yellow precipitate Tin (IV) sulphide:

H₂[SnCl₆] + 2H₂S S SnS₂¯ + 6HCl.

Precipitate Tin (IV) sulphide, unlike Tin (II) sulphide, is dissolved in excess of ammonium or Sodium sulphide with formation tiosalt:

SnS₂ + (NH₄)₂S ( (NH₄)₂SnS₃.

Therefore, if to acidic solutions of Sn (IV) salts add a solution of ammonium or Sodium sulphide precipitate SnS₂ will not be formed.

Reaction performance. To 3-4 drops of the investigated solution acidified by 1-2 drops concentrated hydrochloric acid, add some drops hydrosulphuric water. If there are Sn (IV) ions the yellow precipitate is formed. If to this precipitate add excess of Sodium or ammonium sulphide solution, the precipitate is dissolved.

Reduction of ions Sn^IV. The most characteristic reactions of Tin are reactions Sn (ІІ), therefore at first it is necessary to reducte Sn^IV to Sn^II by metal iron:

SnCl₆^2- + Fe S SnCl₄^2- + Fe²⁺ + 2Cl^–.

More active reducers (metal zinc, magnesium) reducte Sn^II and Sn^IV to metal tin.

Reaction performance. To 7-8 drops of an investigated solution add a drop concentrated HCl. 2-3 strips of the iron are dipped into solution; Sn^IV passes at Sn^II.

Characteristic reactions of ions AsО₃^3- or AsО₂^–

Arsene compounds is very toxic! At work with them it is necessary to show extra care!

Silver nitrate with ions AsО₃^3- forms yellow precipitate Ag₃AsО₃ which is dissolved in HNO₃ and NH₄OH.

AsО₂^–  + 3Ag⁺ + H₂O A Ag₃AsО₃¯ + 2H⁺;

Ag₃AsО₃ + 6NH₄OH 3  [Ag(NH₃)₂]⁺ + AsО₃^3–  + 6H₂O.

The РО₄^3-, J ^– Br ^– ions interfere with the exposure of ions of As^III.

Reaction performance. To 2-3 drops of an investigated solution add 1-2 drop of 0,1 mol/L AgNO₃ solution. If there are Аs (III) yellow precipitate Ag₃AsО₃ forms.

Iodine in a neutral or basic medium becomes colourless by arsenit ions (forms arsenat ions):

AsО₂^–  + I₂ + 2H₂O H H₂AsО₄^– + 2I^– + 2H ⁺.

Reaction is carry out in the presence of NaHCO₃.

The other reducers interfere with the exposure of ions of As^III.

Reaction performance. To 2-3 drops of a acidic investigated solution add a some crystals of NaHCO₃, and after its dissolution – one drop of a solution of iodine. If at a solution there are arsenit-ions iodine becomes colourless.

Sodium sulphide Na₂S (pharmacopeia’s reaction) in the acidic medium reacts with arsenits with formation of a yellow precipitate, insoluble in concentrated hydrochloric acid, but soluble in ammonia solution:

AsО₃^3- + 6H⁺ ®As³⁺ + 3H₂O;

2As³⁺ + 3S^2- ®As₂S₃¯.

Reaction performance. To 4-5 drops of an investigated solution add 3-4 drops of 2 mol/L chloric acid solution and a solution of Sodium sulphide. If there are As (ІІІ) ions, the yellow precipitate forms.

Sodium hypophosphite NaH₂PO₂ (pharmacopeia’s reaction) (reaction Bugo and Tille) in the acidic medium reductes compounds of As (III) and As (V) to elementary Arsene which is formed black-brown precipitate:

4H₃AsО₃ + 3H₂PO₂^– ® 4As¯ + 3H₂PO₄^– + 6H₂O.

Reaction performance. To 5-7 drops of an investigated solution add 5-7 drops of Sodium hypophosphite solution. If there are As (III) or As (V) ions, a black-brown precipitate forms.

Characteristic reactions of ions AsО₄^3–

Silver nitrate with ions AsО₄^3– forms a chocolate precipitate:

AsО₄^3–  + 3Ag⁺ ®Ag₃AsО₄¯.

All ions which form with Ag⁺ ions precipitate interfere with the exposure of ions of As^V.

Reaction performance. To 2-3 drops of an investigated solution add 4-5 drops of Silver nitrate solution. If there are arsenat-ions, the precipitate of chocolate colour is formed.

Ammonium molybdat (NH₄)₂MoО₄. Arsenitic acid and its salts at presence nitric acid and ammonium nitrate by heating with ammonium molybdat are formed a yellow crystal precipitate (NH₄)₃[As(Mo₃O₁₀)₄]HH₂O

H₃AsО₄ + 12(NH₄)₂MoО₄ + 21HNO₃ ® (NH₄)₃[As(Mo₃O₁₀)₄]×H₂O ¯+ 21NH₄NO₃ + 11H₂O.

The precipitate is not dissolved iitric acid, considerably dissolved in excess of molybdat and it is easy dissolved – in alkalis and ammonia.

Reaction performance. To 2-3 drops of an investigated solution add 4-5 drops ammonium molybdat, 3-4 drops concentrated nitric acid, some crystals of NH₄NO₃ and heat it to a boiling on the water-bath. At presence in a solution of ions AsО₄^3– the yellow precipitate is formed.

Magnesian mix (MgCl₂+NH₄Cl+NH₄OH) (pharmacopeia’s reaction). Arsenats form with magnesian mix a white crystal precipitate MgNH₄AsO₄:

AsО₄^3– + Mg²⁺ + NH₄⁺ ® MgNH₄AsО₄¯.

This precipitate is similar to Magnesium and ammonium phosphate MgNH₄PO₄. MgNH₄AsO₄ is dissolved in acids and practically insoluble in the diluted ammonia solution.

Reaction performance. To 2-3 drops of an investigated solution add some drops of magnesian mix and wait 5-10 minutes. If the precipitate was not formed, it is necessary rub the wall-side of test tube a glass stick. The white crystal precipitate is formed in the presence of ions AsО₄^3–.

Potassium iodide. Compounds of As (V) in an acidic solution oxidise iodides to free iodine:

AsО₄^3– + 2I^– + 2H⁺ Û AsО₃^3– + I₂ + H₂O.

The other oxidizers interfere with the exposure of As^V ions.

Sensitivity of reaction can be raised, adding to a solution starch or benzene.

Reaction performance. To 2-3 drops of an investigated solution add 2-3 drops acetic acid, same quantity of Potassium iodide and some drops of starch. At presence in solution As (V) it is formed I₂ which paints starch in dark blue colour.

The common reactions of determination As^III and As (V)

High sensitivity reaction of determination As^III and As (V) is reduction them to AsН₃ and element Arsene.

Reaction of reduction to AsН₃ (it is used for determination of small quantities of As^III and As^V). Analytical signal of this reaction is formation black paper moistened by AgNO₃ solution. The paper blackens because AsH₃ reduces ions Ag⁺ to metal silver (in basic or acidic medium).

Reduction in the acidic medium.

Reaction performance. First of all check cleanliness of used reagents on Arsene. In a microcrucible place some drops of solution HCl. A filtering paper moistens 0,1 mol/L AgNO₃ solution. This paper is put on Petri cup, on it put crucible. After that in a crucible place a strip of metal magnesium or zinc and quickly cover a crucible with a small funnel with the closed end. If paper has not black colour, reagents pure and can be found for determination of Arsene. After that to crucible (with zinc (or Mg) and HCl) add 2-3 drops of an investigated solution. In the presence of Arsene compounds the paper blackens. Thus there are reactions:

In crucible:

AsО₃^3- + 9H⁺ + 3Mg ­­­­® AsН₃­ + 3Mg²⁺ + 3H₂O;

AsО₄^3- + 11H⁺ + 4Mg ® AsН₃­ + 4Mg²⁺ + 4H₂O;

On a paper:

AsH₃ + 6Ag⁺ + 3H₂O ® 6Ag¯ + H₃AsО₃ + 6H⁺.

But Sb^III,V ions, H₂S interfere with the exposure of ions of As^III.

If Sb^III,V ions are in an investigated solution, determination of Arsene compounds carry out in very basic medium by heating.

Reduction in the basic medium.

Use 8 mol/L solution NaOH. In an investigated solution first of all reduce As (V) to As^III by Potassium iodide in the presence of 2 mol/L H₂SO₄ solution. Iodine which is formed thus, delete evaporation of a solution to a dry condition, after that to the rest add 8 mol/L NaOH solution and add metal zinc. Further experience continue how at reduction in the acidic medium, heating up Petri cup on a warm water-bath and periodically moisten a paper which acts from under a funnel by distilled water.

The equation of reaction:

AsО₃^3– + 3Zn + 3OH^– ®AsН₃­ + 3ZnO₂^2-.

 

Systematic analysis of a cations mix of the IV analytical group

1. Preliminary tests:

Determination of Al³⁺ ions by the drop way with alizarine;

Determination of Cr³⁺ ions by oxidation to H₂CrO₆;

Determination of Sn (II) ions with Bismuth salt or Mercury (II) salt;

Determination of Zn²⁺ ions with ammonium sulphide in case of absence Sn (II);

Determination of Arsene compounds with Silver nitrate, hydrosulphuric acid (or ammonium sulphide) in the acidic medium.

2. Systematic analysis of a cations.

2.1. To mix of cations IV groups add a group reagent (2 mol/L NaOH and H₂O₂).

As a result of reaction receive a solution 1.

          

Solution 1

SnО₃^2-, CrО₄^2-, AsО₄^3-, [Al(OH)₆]^3-, [Zn(OH)₄]^2-

2.2. Sedimentation of Aluminium and Tin. To a solution 1 add ammonium chloride crystal and a solution heat to dry rest. The precipitate hydroxides Al(OH)₃, Sn(OН)₄ and a solution arsenat, tetrahydroxozincat, chromate is formed.

          Precipitate 1                                                      Solution 2

Al (OH)₃, Sn(OН)₄                                                                               CrО₄^2-, AsО₄^3-, [Zn(OH)₄]^2-

2.3. Determination of Aluminium and Tin. A precipitate 1 dissolve in 6 mol/L a solution chlorid acid. In the received solution are determinated Al^{3 +} and Sn (IV)ions.

2.4. In a solution 2 determinate of ions of Arsene, Zinc and Chrome (in the form of chromat-ions).

The scheme of the analysis of IV analytical group cations mix

 

 

Systematic analysis of of I-IV analytical groups cations mix

Preliminary tests.

1.  Determination NH₄ ⁺ (with alkalis by heating).

2.  Determination Hg₂^{2 +} (with metal copper).

3.  Determination Сr^{3 +} (oxidation in the basic medium and formation peroxichromatic acids).

4.  Determination Al^{3 +} (drop way with alizarine).

5.  Determination Nа ⁺ (with Potassium hexahydroxoantimonate (V), you do it after determination of NH₄ ⁺ ions, if there aren’t this ions, than you do preliminary test).

6.  Determination As^{III, V} (restoration to arsine).

7.  Determination Sn^{2 +} (reaction with Bi(NO₃) ₃ in basic medium).

 

The scheme of the analysis of a cation mix І – ІV groups

 

Determination of Arsenic

         The spectrophotometric methods of determination of metals have been very popular but recently have become obsolete, in particular for analyses of environmental samples, as they are time-consuming and do not ensure a sufficiently low detection limit. At present the methods are used for direct determination of inorganic metal compounds. They are based on more or less specific reactions of colour metal complexes which can be assessed on the basis of the absorption of a certain wavelength of radiation passing through a studied solution. The too high detection limits of this method can be reduced by special sample preparation, e.g. extraction, sorption, chelation, in order to concentrate the metal to be determined. Speciation analyses are

possible with the use of different reaction conditions (e.g. different pH values) or a combination of different methods.

Spectrophotometric Methods of Determination of Arsenic

The molybdenum blue method is based on the formation of molybdenum blue by ammonium molybdate (VI) with As(V) ions (in the environment of hydrazine sulphate). The sample to be determined is mineralised by inorganic acids, extracted by arsenic and then arsenic iodide is extracted by a mixture of chloroform with 10 M hydrochloric acid. The As (III) ions are oxidized by cerium sulphate to As (V), which takes part in the reaction of molybdenium blue formation and is detected at the wavelength 825 nm. In this method there is a possibility of interference from the presence of germanium, but this can be eliminated by co-precipitation with hydrated iron oxide or extraction with bromides applied by Rao.

The method based on the use of silver dithiocarbaminian (DDTC-Ag). In this method arsenic liberated from a solution as arsenic hydride forms complexes with DDTC in the presence of a strong reducing agent (sodium borohydride) and can be detected at a wavelength of 525 nm. Different kinetics of the reaction of hydride generation, depending on the pH of the environment, allows speciation determination of As (III) and As (V). At pH above 6 the hydrides are formed with As (III). At pH 1-4 As (V) is reduced to As (III)

As (V) + 2BH₄^– + 6H₂O → As (III) + 2B(OH)₃ + 7H₂ (1)

and then hydrides are generated

As (III) + 3BH₄^– + 9H₂O → AsH₃ + 3B(OH)₃ + 9H₂ (2)

The method is subjected to different interference, first of all from the elements forming volatile hydrides (Sb, Se, Sn, Te, Bi) but also from transition metals (Ag, Au, Cr, Cu, Ni, Pt), but only when present at concentrations much higher thaatural. The interference can be eliminated by

the addition of 1,10-phenantroline (Pt) or extraction of all other elements mentioned above by dithiozine prior to determination of arsenic. The limit of linearity of calibration curve 2-40 μg for 20 mL of sample, which corresponds to a detection limit of 100 ng/mL.

         The method based on the formation of DDTC-Ag complex has been recommended by International Standards ISO 6595 for determination of total arsenic in water samples. After oxidation of organic compounds or sulphates by heating with potassium manganate and potassium sulphate peroxide, arsenic is reduced to As(III) by hydroxylamine hydrochloride and then As(III) is reduced to hydride by free hydrogen appearing in an acidic medium as a product of the reaction of hydrochloric acid with tin chloride. Arsenic hydride is absorbed in a solution of silver diethyldithiocarbaminate in chloroform or pyridine. The red-violet complex formed as a product is determined at the wavelength of 510 or 525 nm. In this reaction interference from antimony can be expected as it forms a red complex with silver diethyldithiocarbaminate. The method is unreliable at its high concentrations of above 500 ng/mL. The method based on the reaction of complex formation with silver dicarbaminate was recommended for determination of antimony in samples of natural water. The methods used for separation of arsenic from the solution or for the concentration of arsenic are: distillation after reduction to arsenic hydride, extraction of halogenides by CCl₄ from an acidic environment and extraction of the diethyldithiocarbaminate complex with As (III).

        

         A number of novel arsenic compounds have been discovered in marine plants and animals, and identification of the compounds has been made after their complete isolation. Water-soluble arsenic compounds have been separated and purified by gel permeation chromatography(GPC) (Sephadex LH-20, G-15, G-10 and other resins), ion exchange chromatography (diethylaminoethyl-, carboxymethyl-, and other ion exchange resins), thin layer chromatography (TLC) (silica, cellulose), and high performance liquid chromatography (HPLC) (GPC, silica, reverse phase, ion exchange).

         Separation of lipid-soluble arsenic compounds has been achieved by GPC (LH-20) and HPLC (GPC, ordinary phase and reverse phase).

         X-ray diffraction analysis gives a complete identification. It requires, however, a large amount of sample (usually >10 mg) and a good crystal. NMR spectroscopy (lH, and 3C) usually provides adequate structural information for identification. Several tens of micrograms are usually necessary for a good quality 13C NMR spectrum.

         Mass spectrometry (MS) may provide useful data when the amount of the sample available is less than several micrograms. Both field desorption (FD) and fast atom bombardment (FAB) MS have played a role in the identification of highly polar arsenicals.          Often these method are most useful in identifying small quantities of known arsenicals, but are also valuable when used in conjunction with NMR spectroscopy for providing structural information oew compounds. The identification of arsenobetaine in shark, plaice, shrimp and ivory shell was based on the detection of the protonated molecular ion of arsenobetaine at m/z 179 in their FD spectra. Several groups have experienced difficulties in obtaining good-quality FD and FAB mass spectra from biological extracts. It has been reported that there are differences between the FD mass spectral behaviour of pure organoarsenicals and those contained in biological extracts. The base peak at m/z 134 of the arsenobetaine standard was reduced to a very low level in extracts of shark, plaice and sole and the m/z 135 ion became the dominant peak. The molecular ion was also very weak or even absent in these samples, It is common in FAB-MS that peaks are masked by matrix ions and ionization of the compounds of interest will also be suppressed by impurities in the extract. It is therefore necessary to purify samples before MS analysis. For unequivocal identification, it may be necessary to use high resolution MS or a tandem mass spectrometric technique which requires more than microgram quantities. In performing quantitative determinations, isotope dilution methods using stable isotope labelled compounds as internal standards will be the only acceptable method.

        

         Arsenic is a naturally occurring element widely distributed in the earth’s crust. In the environment, arsenic is combined with oxygen, chlorine, and sulfur to form inorganic arsenic compounds. Arsenic in animals and plants combines with carbon and hydrogen to form organic arsenic compounds.

         Inorganic arsenic compounds are mainly used to preserve wood. Copper chromated arsenate (CCA) is used to make “pressure-treated” lumber. CCA is no longer used in the U.S. for residential uses; it is still used in industrial applications. Organic arsenic compounds are used as pesticides, primarily on cotton fields and orchards.

What happens to arsenic when it enters the environment?

·                     Arsenic occurs naturally in soil and minerals and may enter the air, water, called arsenobetaine that is much less harmful. and land from wind-blown dust and may get into water from runoff and leaching.

·                     Arsenic cannot be destroyed in the environment. It can only change its form.

·                     Rain and snow remove arsenic dust particles from the air.

·                     Many common arsenic compounds can dissolve in water. Most of the arsenic in water will ultimately end up in soil or sediment.

·                     Fish and shellfish can accumulate arsenic; most of this arsenic is in an organic form

How might I be exposed to arsenic?

·                     Ingesting small amounts present in your food and water or breathing air containing arsenic.

·                     Breathing sawdust or burning smoke from wood treated with arsenic.

·                     Living in areas with unusually high natural levels of arsenic in rock.

·                     Working in a job that involves arsenic production or use, such as copper or lead smelting, wood treating, or pesticide application.

How can arsenic affect my health?

         Breathing high levels of inorganic arsenic can give you a sore throat or irritated lungs.

Ingesting very high levels of arsenic can result in death. Exposure to lower levels can cause nausea and vomiting, decreased production of red and white blood cells, abnormal heart rhythm, damage to blood vessels, and a sensation of “pins and needles” in hands and feet.

Ingesting or breathing low levels of inorganic arsenic for a long time can cause a darkening of the skin and the appearance of small “corns” or “warts” on the palms, soles, and torso.

Skin contact with inorganic arsenic may cause redness and swelling.

Almost nothing is known regarding health effects of organic arsenic compounds in humans. Studies in animals show that some simple organic arsenic compounds are less toxic than inorganic forms. Ingestion of methyl and dimethyl compounds can cause diarrhea and damage to the kidneys.

How likely is arsenic to cause cancer?

         Several studies have shown that ingestion of inorganic arsenic can increase the risk of skin cancer and cancer in the liver, bladder, and lungs. Inhalation of inorganic arsenic can cause increased risk of lung cancer. The Department of Health and Human Services (DHHS) and the EPA have determined that inorganic arsenic is a known human carcinogen. The International Agency for Research on Cancer (IARC) has determined that inorganic arsenic is carcinogenic to humans.

How does arsenic affect children?

         There is some evidence that long-term exposure to arsenic in children may result in lower IQ scores. There is also some evidence that exposure to arsenic in the womb and early childhood may increase mortality in young adults.

         There is some evidence that inhaled or ingested arsenic can injure pregnant women or their unborn babies, although the studies are not definitive. Studies in animals show that large doses of arsenic that cause illness in pregnant females, can also cause low birth weight, fetal malformations, and even fetal death. Arsenic can cross the placenta and has been found in fetal tissues. Arsenic is found at low levels in breast milk.