The materials to prepare students for practical lessons of inorganic chemistry

LESSON № 10

Theme. Photolytic processes.

Plan

Types of chemical reactions. Proctolitic processes. Types of protolytic reactions: neutralization, hydrolysis and ionization reactions. Hydrolysis of compounds. Hydrolysis of salts. Degree of hydrolysis, dependence ofits on the concentration and temperature. Constant about hydrolysis reaction. Role of hydrolysis in biochemical processes.

 

Acid–base reactions

Proton-transfers

As already mentioned (The Brønsted–Lowry definition), the reaction expressed by the Brønsted–Lowry definition, A ⇄ B + H⁺, does not actually occur in any solution processes. This is because H⁺, the bare proton, has an enormous tendency to add to almost all chemical species and cannot exist in any detectable concentrations except in a high vacuum. Apart from any specific chemical interaction, the very small size of the proton (about 10⁻¹⁵ metre) means that it exerts an extremely powerful electric field, which will polarize and therefore attract any molecule or ion it comes into contact with. It has been estimated that the dissociation of 19 grams of the hydronium ion H₃O⁺ to give 1 gram of protons and 18 grams of water would require the expenditure of about 1,200,000 joules (290,000 calories) of energy, and thus it is an extremely unlikely process indeed.

Typical acid–base reactions may be thought of as the combination of two reaction schemes, A₁ ⇄ B₁ + H⁺ and H⁺ + B₂ ⇄ A₂, leading to the combined form A₁ + B₂ ⇄ B₁ + A₂. This represents a proton-transfer reaction from A₁ to B₂, producing B₁ and A₂. A large number of reactions in solution, often referred to under a variety of names, can be represented in this way. This is illustrated by the following examples, in each of which the species are written in the order A₁, B₂, B₁, A₂.

DISSOCIATION OF MOLECULAR ACIDS IN WATER

In this instance, water acts as a base. The equation for the dissociation of acetic acid, for example, is CH₃CO₂H + H₂O ⇄ CH₃CO₂⁻ + H₃O⁺.

DISSOCIATION OF BASES IN WATER

In this case, the water molecule acts as an acid and adds a proton to the base. An example, using ammonia as the base, is H₂O + NH₃ ⇄ OH⁻ + NH₄⁺. Older formulations would have written the left-hand side of the equation as ammonium hydroxide, NH₄OH, but it is not now believed that this species exists, except as a weak, hydrogen-bonded complex.

DISSOCIATION OF ACIDS AND BASES IN NONAQUEOUS SOLVENTS

These situations are entirely analogous to the comparable reactions in water. For example, the dissociation of acetic acid in methanol may be written as CH₃CO₂H + CH₃OH ⇄ CH₃CO₂⁻ + CH₃OH and the dissociation of ammonia in the same solvent as CH₃OH + NH₃ ⇄ CH₃O⁻ + NH₄⁺.

SELF-DISSOCIATION OF AMPHOTERIC SOLVENTS

In this case, one solvent molecule acts as an acid and another as a base. Self-dissociation of water and liquid ammonia may be given as examples:

NEUTRALIZATION

For a strong acid and a strong base in water, the neutralization reaction is between hydrogen and hydroxide ions—i.e., H₃O⁺ + OH⁻ ⇄ 2H₂O. For a weak acid and a weak base, neutralization is more appropriately considered to involve direct proton transfer from the acid to the base. For example, the neutralization of acetic acid by ammonia may be written as CH₃CO₂H + NH₃ → CH₃CO₂⁻ + NH₄⁺. This equation does not involve the solvent; it therefore also represents the process of neutralization in an inert solvent, such as benzene, or in the complete absence of a solvent. (If one of the reactants is present in large excess, the reaction is more appropriately described as the dissociation of acetic acid in liquid ammonia or of ammonia in glacial acetic acid.)

HYDROLYSIS OF SALTS

Many salts give aqueous solutions with acidic or basic properties. This is termed hydrolysis, and the explanation of hydrolysis reactions in classical acid–base terms was somewhat involved. In terms of the Brønsted–Lowry concept, however, hydrolysis appears to be a natural consequence of the acidic properties of cations derived from weak bases and the basic properties of anions derived from weak acids. For example, hydrolysis of aqueous solutions of ammonium chloride and of sodium acetate is represented by the following equations:

The sodium and chloride ions take no part in the reaction and could equally well be omitted from the equations.

The acidity of the solution represented by the first equation is due to the presence of the hydronium ion (H₃O⁺), and the basicity of the second comes from the hydroxide ion (OH⁻). The reverse reactions simply represent, respectively, the neutralization of aqueous ammonia by a strong acid and of aqueous acetic acid by a strong base.

A superficially different type of hydrolysis occurs in aqueous solutions of salts of some metals, especially those giving multiply charged cations. For example, aluminum, ferric, and chromic salts all give aqueous solutions that are acidic. This behaviour also can be interpreted in terms of proton-transfer reactions if it is remembered that the ions involved are strongly hydrated in solution. In a solution of an aluminum salt, for instance, a proton is transferred from one of the water molecules in the hydration shell to a molecule of solvent water. The resulting hydronium ion (H₃O⁺) accounts for the acidity of the solution:

In the Brønsted–Lowry theory, an acid donates a proton and the base accepts it. The ion or molecule remaining after the acid has lost a proton is known as that acid’s conjugate base, and the species created when the base accepts the proton is known as the conjugate acid. This is expressed in the following reaction:

acid + base is in equilibrium with conjugate base + conjugate acid.

Notice how this reaction can proceed in either forward or backward direction; in each case, the acid donates a proton to the base.

With letters, the above equation can be written as:

HA + B is in equilibrium with A− + HB+

The acid, HA, donates a H+ ion to become A−, its conjugate base. The base, B, accepts the proton to become HB+, its conjugate acid. In the reverse reaction, A− it accepts a H+ from HB+ to recreate HA in order to remain in equilibrium. In the reverse reaction, as HB+ has donated a H+ to A−, it therefore recreates B and remains in equilibrium.

In 1923 the Danish chemist J.N. Brønsted, building on Franklin’s theory, proposed that

An acid is a proton donor; a base is a proton acceptor.

In the same year the English chemist T.M. Lowry published a paper setting forth some similar ideas without producing a definition; in a later paper Lowry himself points out that Brønsted deserves the major credit, but the concept is still widely known as the Brønsted-Lowry theory.

These definitions carry a very important implication: a substance cannot act as an acid without the presence of a base to accept the proton, and vice versa. As a very simple example, consider the equation that Arrhenius wrote to describe the behavior of hydrochloric acid:

HCl → H⁺ + A^–

This is fine as far as it goes, and chemists still write such an equation as a shortcut. But in order to represent this more realistically as a proton donor-acceptor reaction, we now depict the behavior of HCl in water by in which the acid HCl donates its proton to the acceptor (base) H₂O.

 

“Nothing new here”, you might say, noting that we are simply replacing a shorter equation by a longer one. But consider how we might explain the alkaline solution that is created when ammonia gas NH₃ dissolves in water. An alkaline solution contains an excess of hydroxide ions, so ammonia is clearly a base, but because there are no OH^– ions in NH₃, it is clearly not an Arrhenius base. It is, however, a Brønsted base:

In this case, the water molecule acts as the acid, donating a proton to the base NH₃ to create the ammonium ion NH₄⁺.

The foregoing examples illustrate several important aspects of the Brønsted-Lowry concept of acids and bases:

·                        A substance cannot act as an acid unless a proton acceptor (base) is present to receive the proton;

·                        A substance cannot act as a base unless a proton donor (acid) is present to supply the proton;

Water plays a dual role in many acid-base reactions; H₂O can act as a proton acceptor (base) for an acid, or it can serve as a proton donor (acid) for a base (as we saw for ammonia.

The hydronium ion H₃O⁺ plays a central role in the acid-base chemistry of aqueous solutions.

Hydrogen ions cannot exist in water

There is another serious problem with the Arrhenius view of an acid as a substance that dissociates in water to produce a hydrogen ion. The hydrogen ion is no more than a proton, a bare nucleus. Although it carries only a single unit of positive charge, this charge is concentrated into a volume of space that is only about a hundred-millionth as large as the volume occupied by the smallest atom. (Think of a pebble sitting in the middle of a sports stadium!) The resulting extraordinarily high charge density of the proton strongly attracts it to any part of a nearby atom or molecule in which there is an exess of negative charge. In the case of water, this will be the lone pair (unshared) electrons of the oxygen atom; the tiny proton will be buried within the lone pair and will form a shared-electron (coordinate) bond with it, creating a hydronium ion, H₃O+. In a sense, H₂O is acting as a base here, and the product H₃O+ is the conjugate acid of water:

Atoms can gain or lose electrons in order to form ions in a process called ionization (compounds formed in this way are called ionic compounds).  When ionic compounds dissolve in water, their ions separate from one another in a process called dissociation.  One interesting feature of water and many other covalent compounds is that they too can dissociate into ions.  Unlike ionic compounds, such as sodium chloride, they are not ionized before they dissociate; they accomplish ionization and dissociation at the same time.

Acetic acid is a monoprotic acid, with one acidic hydrogen atom. It contains the acid group of carbon-containing molecules, -COOH and is called a carboxylic acid. The Lewis structure of acetic acid and its conjugate base acetate ion are shown below.

For convenience, we will write CH₃COOH to denote the acid formula. In water, acetic acid establishes an equilibrium between the weak acid, acetic acid, and the conjugate base, acetate ion.

CH₃COOH + H₂O H₃O⁺ + CH₃COO^– (1)

The equilibrium constant, K_a, for this reaction is written

(2)

For acetic acid, the value of K_a equals 1.76 x 10^Р5 and pK_a equals 4.75. The magnitudes of the K_a and pK_a values of different weak acids give us a comparison of their relative strength. A weaker acid has less dissociation to the conjugate base and the equilibrium favors the undissociated weak acid form (as in Eqn. 1). This results in a smaller K_a value (in Eqn 2). A smaller K_a value corresponds to a larger pKa, since pK_a = -log K_a. In other words, the weaker the acid, the larger the pK_a value. Table 1, at the end of this week’s experiment compares the K_a and pK_a values for several weak monoprotic and polyprotic acids. From this table you can see that acetic acid is a stronger acid than those listed with a larger pKa and a weaker acid than those with a smaller pK_a value.

When a weak acid is titrated, carefully measured volumes of strong base solution are added to it. Some titrations are done for the purpose of determining the concentration of a weak acid solution. In this case, base solution is added until the equivalence point is reached. At this point, enough base has been added to react with all the weak acid present in the sample. The endpoint can be determined with a suitable indicator, as you did in the titration of KHP and your seawater ion exchange effluent solutions in Week 4’s experiment. In this week’s experiment, you will determine the concentration of acetic acid in a commercial vinegar solution. You will use a pH meter, rather than an indicator, to determine the equivalence point of the titration. When the equivalence point is reached, the pH of the solution will change rapidly, because all the acid has reacted with the added base. The following example illustrates the stoichiometry of a titration of vinegar.

Example: Titration of Vinegar

10.0 mL of vinegar solution is titrated with 0.1800 M NaOH solution. After 45.87 mL of base are added, the pH increases sharply, indicating that the endpoint has been reached. What is the concentration of acetic acid in the vinegar solution?

Solution: First we calculate the moles of base required to reach the equivalence point: 45.87 mL x 0.1800 mmoles/mL = 8.257 mmol OH^– Since acetic acid and sodium hydroxide react in a 1 to 1 ratio, this is also the number of moles of acetic acid that reacted with the base.

8.257 mmol OH^– x 1 H₃O⁺ / 1 OH^– = 8.257 mmol H₃O⁺

From the moles of acid and the volume of vinegar in the original sample, we can calculate the concentration of acetic acid in vinegar:

8.257 mmol H₃O⁺ / 10.0 mL = 0.826 M

In addition to finding the concentration of acetic acid in vinegar, your titration will provide information about the behavior of this weak acid throughout the pH scale. You will measure the pH vs. volume base (mL) added to the vinegar solution, to determine the titration curve. A theoretical titration curve for acetic acid is shown below.

Figure. Titration curve of acetic acid

For this titration, a 10 mL volume of 0.5 molar acetic acid is titratated with 0.5 M NaOH. Because the concentration of acid and base are the same, the equivalence point is reached when 10 mL of base are added to the solution. The equivalence point, when enough base is added to react with all the acid present, exhibits the sharp increase in pH, as discussed above. This happens when the moles of base added equals the moles of acid in the sample. It is this point that you will use to determine the concentration of acetic acid in vinegar.

The flat portion of the titration curve before the equivalence point is called the buffer region. In this part of the pH scale, the acid and conjugate base are both present in significant concentrations and the solution resists changes in pH. As base is added to a solution in this buffer region, acetic acid reacts with it to form acetate ion, without a large change in pH. If additional acid were added to a solution in the buffer region, it would react with the conjugate base, acetate ion, and, again, the pH would not change appreciably.

In the middle of the buffer region lies the half-equivalence point. Here the volume of base added is half that required to reach the equivalence point and half the acetic acid has been converted to the conjugate base, acetate ion. This means that the concentrations of acetic acid and acetate ion are equal. If we examine the equilibrium expression at the half-equivalence point, we find something interesting:

At the half-equivalence point, [CH₃COOH] = [CH₃COO^–], so

K_a = [H₃O⁺]

And taking the log and multplying both sides by -1 yields

pK_a = pH

So at the half-equivalence point, half the acetic acid has been converted to the conjugate base, acetate ion, and the pH will be equal to the pK_a of the acid. This gives us an experimental way to determine the pK_a of a weak acid. In this week’s experiment, you will determine the pK_a of acetic acid and compare it to the literature value. Next week, you may find this method useful for identifying unknown solutions of weak acids, bases, or buffers.

More About Buffers

In the buffer region of a titration, little pH change occurs upon addition of OH^– or H₃O⁺. In this region, the acid and its conjugate base are present in similar (within a factor of 10) concentrations. Such a solution has the important property of being able to resist changes in pH and is called a buffer solution.

Buffers and buffering ability play a central role in much of biology and medicine. Living systems require that their physiological fluids be maintained at certain pH values for the proper biochemical reactions to occur. Blood, because of its role iutrient transport, must be maintained at pH 7.4, in spite of moving large quantities of CO₂ from the cells to the lungs. This pH is maintained primarily by the buffering action of carbon dioxide, hydrogen carbonate, and carbonate ions. Two major components involved in the buffering of intracellular fluids are the dihydrogen phosphate and hydrogen phosphate ions involved in one of the acid-base equilibria that you will study in this week’s experiment.

For experimental work in aqueous solutions, it is fundamentally important to be able to prepare a buffer solution at a desired pH. Based on what you know about the buffer region of a titration curve, a buffer solution capable of moderating H₃O⁺ and OH^– additions equally well would have equal concentrations of an acid and its conjugate base. The dissociation equilibrium relationship for the general weak acid HA can be written:

(3)

Rearranging (3) gives an expression for the hydronium ion concentration:

(4)

We then take the logarithm of both sides to obtain:

(5)

Multiplying both sides by -1 and rearranging gives an expression for the pH in terms of the log of the ratio [A-]/[HA] and the pK_a value of the acid: (remember that log(a/b) = -log(b/a) and pK_a = -log K_a)

(6)

Equation (6) is called the Henderson-Hasselbalch relationship. Using it, we can predict the pH of a solution from the concentration of an acid and its conjugate base. It can also be used as a guide to making a buffer at a given pH. There are a large number of acids with pK_a‘s that span the common pH range, and generally one can be chosen with a pK_a near the pH value which one wishes to buffer.

It is useful to consider the half-equivalence point again, using the Henderson-Hasselbalch relation. As we saw above, at the half equivalence point for a weak acid, [A-] = [HA] and pH = pK_a (because log [A-] / [HA] = log 1 = 0). This solution will be equally effective as a buffer towards either H₃O⁺ or OH^– because equal amounts of acid and conjugate base are present. If you consider the titration curve, in Figure 1, you can see that the pH equals the pK_a at a point midway between equivalence points, at the center of the buffer region.

The Henderson-Hasselbalch relationship is very useful in preparing buffer solutions. It certainly is possible to prepare a buffer solution starting from the pure acid (HA) and adding a strong base (i.e. NaOH) until the desired pH (in the buffer region of the acid) is achieved; conversely one could start with the pure base (A^–) and add the appropriate amount of strong acid (i.e. HCl) until the same point is reached. The Henderson-Hasselbalch equation, however, allows us to calculate the correct ratio of basic form to acidic form which can be mixed to achieve the desired buffered pH. An example is acetic acid and its conjugate base, acetate ion, from sodium acetate.

Example: Acetic acid/acetate buffer

A buffer solution with pH 5 is to be prepared by adding sodium acetate and acetic acid to enough water to make 1.00 L of solution. The pK_a of acetic acid is 4.75. Given 0.600 mol of sodium acetate, what amount of acetic acid should be added to produce 1.00 L of a buffer solution at pH = 5.00?

Solution: We first rewrite the Henderson-Hasselbalch equation as

In the buffer solution, virtually all 0.60 mol of sodium acetate, CH₃COONa, will be completely dissociated to acetate ion, CH₃COO^–, and the acetic acid we add will be present as undissociated CH₃COOH (to a very good approximation). Thus we solve the equation above for [CH₃COOH], finding

[CH₃COOH] = 10^-0.25[CH₃COO^–] = (0.56) (0.60 mol/L of buffer solution) = 0.34 mol acetic acid per L of buffer solution

We can check the calculation with the original Henderson-Hasselbalch equation:

Note that the Henderson-Hasselbalch relationship indicates that the pH of a buffer solution does not depend on the total concentration of the buffering acid and conjugate base but only on the pK_a and the ratio of the concentration of these two species. On the other hand, the buffering capacity of a solution quantifies the amount of H₃O⁺ or OH^– the solution is capable of neutralizing before the acid or conjugate base form is saturated and the pH begins to fall or rise precipitously. This will depend on the total concentration of the acid and conjugate base buffer ions. For example, a solution 0.2 M in total acetate, distributed between acetic acid, CH₃COOH, and acetate ion, CH₃COO^–, will be able to neutralize more H₃O⁺ or OH^– than a solution at the same pH that is 0.1 M in total acetate. Also, the buffering capacity may be different towards addition of acid than towards base. This will be true unless the pH of the buffer solution is identical to the pK_a of the buffering acid-base equilibrium.

We will use the following definition, applicable to any solution, for the buffering capacity:

• the acidic buffering capacity of a solution is the number of moles of H₃O⁺ per liter of buffer which are required to lower the pH by one unit

• the basic buffering capacity of a solution is the number of moles of OH^– per liter of buffer which will raise the pH by one unit.

The figure below indicates how the acidic and basic buffering capacity of a buffer solution can be graphically determined from a titration curve, using the pH = 5.00 acetate buffer solution from the Example above.

Figure. Buffer Capacity

We take the measured volume of base added to raise the pH by one unit, multiply by the base concentration, and divide by the initial volume of buffer. This gives the base buffer capacity as moles of base per liter of buffer.

Polyprotic Acids

In contrast to a simple monoprotic acid like acetic acid, with only one equilibrium between the acid and conjugate base, a polyprotic acid contains more than one acidic hydrogen. For a polyprotic acid,acidic hydrogens will exist in solution in equilibrium with conjugate base forms (for a total of n+1 species). For example, when phosphoric acid (n = 3, a triprotic acid) is dissolved in solution, the following equilibria are established among the four species H₃PO₄ (phosphoric acid itself), H₂PO₄^-1 (dihydrogen phosphate anion), HPO₄^-2 (hydrogen phosphate anion), and PO₄^-3 (phosphate anion):

(7-9)

Note that each successive pair of species is linked in an independent equilibrium with H₃O⁺ and that each link has a unique equilibrium constant. If we want to understand how polyprotic acid solutions respond to chemical changes (such as addition of OH^– or H₃O⁺), we need to learn how to predict the concentrations of the four phosphate species, as well as H₃O⁺ and OH^–, under a variety of conditions. To account for all six species, we will use relationships such as chemical equilibrium reactions and associated equilibrium constant equations (Eq. (7)-(9)), the mass balance, electroneutrality and water auto-ionization equilibrium relationships (Eq.’s (10) – (12)):

mass balance (total phosphate):

[ H₃PO₄]_initial = [ H₃PO₄] + [ H₂PO₄^-1] + [ HPO₄^-2] + [ PO₄^-3] (10)

electroneutrality (charge balance):

[Na⁺] + [H₃O⁺] = [ H₂PO₄^-1] + 2[ HPO₄^-2] + 3[ PO₄^-3] + [OH^–] (11)

The concentrations are multiplied by species’ charges here. For example, one phosphate ion carries a -3 charge; thus, its concentration is multiplied by 3 in the charge balance equation.

water dissociation (auto-ionization) equilibrium:

2 H₂O OH^– + H₃O⁺

K_w = [OH^–][H₃O⁺] = 1.0 x 10^-14 (12)

Therefore, we have six (not very simple) equations with six unknowns that describe our triprotic acid solution. Because these equations are non-linear (i.e. they are algebraic equations in which the variables–our species’ concentrations–may be raised to powers other than 1), solving these equations for the concentrations is most easily done by a computer program. Simple elementary algebra of the “six equations in six unknowns” variety wonХt work.

Despite this complicated set of equations, we can still understand the relationship between the concentration of acid and conjugate base pairs. Note that the ratio of concentrations of any two species in a polyprotic acid solution can be determined by using only the dissociation equilibrium constant (such as Eq.’s (7)-(9)) and [H₃O⁺]. For example, from Eq. (7), we can write and find the phosphoric acid to dihydrogen phosphate concentration ratio at any pH.

For polyprotic acids with equilibrium constants well separated in magnitude (i.e. different by factors of 10³ to 10⁴ or so), approximations can be made which allow us to consider only those species involved in one of the deprotonation equilibria. For example, K_a1 for phosphoric acid is so much larger than either K_a2 or K_a3 that the initial part of a titration of H₃PO₄ against standardized NaOH involves almost entirely equilibrium (7) between H₃PO₄ and its conjugate base H₂PO₄^-1. At pH values just past the first equivalence point, H₂PO₄^-1 will be the dominant species in solution, and the next region of the base titration will involve the second deprotonation equilibrium, (8). Thus, a titration curve for a polyprotic acid which meets these criteria will appear to be a sequence of individual titration curves, with the post-equivalence region of one equilibrium overlapped with the buffer region of the next. In polyprotic acid systems such as these, at any pH value, one acidic form and its conjugate base will be present in much higher concentration than all other polyprotic acid forms. The concentrations of these minor forms can usually be ignored.

A theoretical titration curve for the titration of 10 mL of 0.1 M phosphoric acid with 0.1 M NaOH is shown below. Because the initial acid and the titrating base concentrations are equal and the initial acid volume was 10 mL, the first equivalence point should appear at 10 mL added base. The graph shows this point clearly, along with the clear second equivalence point at 20 mL added base. But the third equivalence point (at 30 mL added base) is not distinct. This point is washed out because K_a3 is close to K_w, the water dissociation constant. Phosphate ion is a fairly strong base and competes with OH^– for hydrogen ions, at high pH. Note that the half-equivalence points are found in each buffer region, midway between adjacent equivalence points.

Figure 3. Titration Curve for Phosphoric Acid

This type of plot can be produced by acid/base software available in the General Chemistry folder on the Public File Server. This program allows you to select mono-, di-, or tri-protic acids by name and to explore the theoretical titration curves for such acids with any strong base. You may want to run this program with a series of “artificial” K_a values to see how the shape of the titration curve changes with the absolute magnitude and the relative closeness of Ka’s. You can do this by creating your own acid or base. Instructions for using the software are at the end of this section of the manual.

The equilibrium expressions and the total phosphate mass balance expression can be manipulated to give equations for the fraction of total phosphate in any one form as a function of pH. The program also uses these equations to produce what are referred to as alpha plots, which look like the Figure 4.

Figure 4. Alpha Plot for Phosphoric Acid

This figure shows that at any pH, only one acidic form and its conjugate base dominate the concentration of all phosphate species. At, for instance, pH = 2, the concentrations of H₃PO₄ and H₂PO₄^-1 are roughly equal while the concentrations of HPO₄^-2 and PO₄^-3 are negligibly small. We can use this graph to make the following table, showing the dominant acid-conjugate base pair at various pH ranges:

pH range Dominant species

0 – 4.7      H₃PO₄, H₂PO₄^–1

4.7 – 9.7       H₂PO₄^–1, HPO₄^-2

9.7 – 14        HPO₄^-2, PO₄^-3

This situation is fairly common, not only among polyprotic inorganic acids such as H₃PO₄, but also among biologically important acids, such as amino acids.

 

Table 1: Weak Acids, K_a, and pK_a values

 

POLYPROTIC ACIDS

Many acids contain two or more ionizable hydrogens. There are two in carbonic acid, H₂CO₃, and three in phosphoric acid, H₃PO₄. For any such multiple hydrogen acid, the first hydrogen is most easily removed, and the last hydrogen is removed with the greatest difficulty. These acids are called polyprotic (many protons) acids. The multiple acid ionization constants for each acid measure the degree of dissociation of the successive hydrogens.

Table gives ionization data for four series of polyprotic acids. The integer in parentheses after the name denotes which hydrogen is being ionized, where (1) is the first and most easily ionized hydrogen.

Remember: The strongest acids dissociate most readily. Of the nine acids listed in Table , the strongest is sulfuric (1), with the highest acid ionization constant, and the weakest is phosphoric (3).

Here are the chemical equations for the three successive ionizations of phosphoric acid:

Consequently, an aqueous solution of phosphoric acid contains all the following molecules and ions in various concentrations:

Consulting the table of the dissociation constants K_a‘s for phosphoric acid shows that the first dissociation is much greater than the second, about 100,000 times greater. This means nearly all the H₃O⁺ ( aq) in the solution comes from the first step of dissociation. The second and third steps add very little H₃O⁺ ( aq) to the solution. So a solution of phosphoric acid will contain H₃PO₄ molecules in highest concentration with smaller, and nearly equal, concentrations of H₃O⁺ and . The and ions are present in very small concentrations.

Example: H₃PO₄

H₃PO_{4 (aq)} H₂PO₄^–_(aq) + H⁺_(aq)        K_a1 = 7.11×10^-3     pK_a1 = 2.15

H₂PO₄^–_(aq) HPO₄^2-_(aq) + H⁺_(aq)        K_a2 = 6.34×10^-8     pK_a2 = 7.20

HPO₄^2-_(aq) PO₄^3-_(aq) + H⁺_(aq)        K_a3 = 4.20×10^-13     pK_a3 = 12.38

Alanine Hydrochloride (ref .1)

 

As shown in Table 15.3, the values of stepwise dissociation constants of polyprotic acids decrease in the order . Because of electrostatic forces, it’s more difficult to remove a positively charged proton from a negative ion.

Example. Calculate the pH and the concentrations of all species present (H₂CO₃, HCO₃^–,CO₃^2-, H₃O⁺and OH^– ) in a 0.020 M carbonic acid solution.

Use the eight-step procedure summarized in Example 3.

Step 5. Substituting the equilibrium concentrations into the equilibrium equation for the principal reaction gives

Assuming that 0.020 –x = 0.02, we have:

X² = (4.3 * 10^-7) 0.020

X = 9.3 * 10^-5

Step 6. The big concentrations are 

Step 7. The small concentrations are obtained from the subsidiary equilibria—(1) dissociation of HCO₃^– and (2) dissociation of water—and from the big concentrations already determined:

In general, for a solution of a weak diprotic acid  

NOTE The second dissociation of H₂CO₃ produces a negligible amount of H₃O⁺ compared with the obtained H₃O⁺ from the first dissociation.

Step 8.  

 

Calculating Equilibrium Concentrations of Polyprotic Acids

Polyprotic acids have complex equilibriums due to the dissociation of hydrogen species at different pHs and the presence of multiple species in solution.

Fractional Ion Calculations for Polyprotic Acids

The above complex equations can be used to determine the fractional concentration of various ions from polyprotic acids.

· Polyprotic acids contain multiple acidic protons that can dissociate from the compound sequentially with unique acid dissociation constants for each proton.

· Because of the increased variety of possible ionic species in solution for each acid, the calculations to precisely determine the concentrations of different species at equilibrium can be very complicated.

· Certain simplifications can be made to make the calculations easier. What the simplifications are vary with the specific acid and the solution conditions.

· Equilibrium. The state of a reaction in which the rates of the forward and reverse reactions are the same

· polyprotic acids. Referring to an acid with multiple protons capable of dissociating from the compound.

· ionic species. Referring to chemical species with a residual charge; in acid-base equilibria, this concept refers to the charge arising from the loss or addition of electrons from chemical compounds.

Polyprotic acids are acids that can lose more than one proton. The dissociation constant of the first proton may be denoted as K_a1 and the constants for dissociation of successive protons as K_a2, etc. Common polyprotic acids include sulfuric acid, H₂SO₄ and phosphoric acid, H₃PO₄.

To determine equilibrium concentrations of different ions produced by polyprotic acids, the equations can get very complicated to account for all of the different components. For instance, one can calculate the fractional concentration (alpha) of different ions using complex equation that account for hydrogen concentration (pH) and equilibrium constants, as seen in Figure 0.

However, one can simplify the problem, depending on the polyprotic acid. The following examples indicate the mathematics and simplifications for a few polyprotic acids under specific conditions.

Sulfuric Acid

If water is the solvent, sulfuric acid, H₂SO₄, loses one proton as a strong acid with an immeasurably large dissociation constant.

H₂SO₄ → H⁺ + HSO₄^–

It also can lose a second proton as a weak acid with a measurable dissociation constant.

K_a2 = 1.20×10^-2   pKa2 = 1.92

Therefore, one can assume that there is no measurable H₂SO₄ in the solution. Instead, it has completely dissociated to H⁺ and HSO₄^–, which in turn has dissociated to more protons and SO₄^2-.

Phosphoric Acid

Phosphoric acid, H₃PO₄, has three dissociations, as can be viewed in Figure 1:

pK_a1 = 2.12

pK_a2 = 7.21

pK_a3 = 12.67

Thus, in an aqueous solution of phosphoric acid, there will theoretically be seven ionic and molecular species present: H₃PO₄, H₂PO^–, HPO^2-, PO^3-, H₂O, H⁺, and OH^–. Life might appear impossibly complicated, were we not able to make some approximations.

At a pH equal to the pKa for a particular dissociation, the two forms of the dissociating species are present in equal concentrations, due to the following mathematical observation. For the second dissociation of phosphoric acid, for which pK_a2 = 7.21:

· pK_a2 = -log(K_a2) = -log([H⁺]*[HPO₄^2-]/[H₂PO₄^–])

· pH = -log[H⁺]

Therefore, pH – pK_a2 = log([HPO₄^2-]/[H₂PO₄^–])

When pH = pKa2, we have the ratio [HPO₄^2-]/[H₂PO₄^–] = 1.00. Hence, in a neutral solution, H₂PO₄^– and HPO₄^2- are present in about the same concentrations. Very little undissociated H₃PO₄ and fully dissociated PO₄^3- will be found as can be determined through simliar equation with their given K_a‘s.

The only phosphate species that we have to consider near pH = 7 are H₂PO₄^– and HPO₄^2-. Similarly, in strong acid solutions near pH = 3, only H₃PO₄ and H₂PO₄^– are important. As long as the pKa’s of successive dissociations are separated by three or four units (as they almost always are), matters are simplified.

Carbonic Acid

There is still another simplification. When a weak polyprotic acid, such as carbonic acid, H₂CO₃, dissociates, most of the protons present come from the first dissociation:

pK_a1 = 6.37

Since the second dissociation constant is smaller by four orders of magnitude (pK_a2 = 10.25 is larger by four units), the contribution of hydrogen ions from the second dissociation will be only one ten-thousandth as large. Correspondingly, the second dissociation has a negligible effect on the concentration of the product of the first dissociation, HCO₃^–.

Consider the following chemical equation as the molecule acetic acid equilibrates in the solution:

Although acetic acid carries a four hydrogen atoms, only a single becomes ionized. Not to get into too much detail between monoprotic and polyprotic acids, but if you desire to find the pH given a concentration of a weak acid (in this case, acetic acid), you would create and complete an ICE Table adjusting for how much acetic acid disassociates.

However, if you wish to find the pH of a solution after a polyprotic acid disassociates, there are extra steps that would need to be done. Let’s first take a look at a unique example: 

Find the pH of a polyprotic acid

(Note:  refers to the ionization constant for the first hydrogen atom and  refers to the ionization constant for the second hydrogen atom)

As an example, let’s take a look at the polyprotic acid sulfuric Acid, or H₂SO₄, briefly. In sulfuric Acid, there are two ionizeable hydrogen atoms. What makes this molecule interesting is that its ionization constant for the first hydrogen ionized is significantly larger than is the second ionization constant.

The  constant for sulfuric acid is conveniently dubbed “very large” while the constant is 1.1 x 10^-2. As such, the sulfuric acid will completely disassociate into HSO₄^– and H₃O⁺ ions as if it were a strong acid.

Since the sulfuric acid completely disassociates in the solution, we can skip the ICE table process for sulfuric acid, and assert that the concentration HSO₄^– and H₃O⁺ are the same as that of HSO₄^–, that is 0.75M.

Equation:

ICE Table:

Assume x in the denominator is negligible.  Therefore,

Since we know the value of x, we can use the equation from the ICE table to find the value of [HSO₄^–]. 

We can also find [H₃O⁺] using the equation from the ICE table.

We can then find the pH from the calculated [H₃O⁺] value.

Finding the pH of a polyprotic base

Let’s say our task is to find the pH given a polyprotic base which gains protons in water. Thankfully, the process is essentially the same as finding the pH of a polyprotic acid except in this case we deal with the concentration of OH^– instead of H₃O⁺. Let’s take a look at how to find the pH of C₂₀H₂₄O₂N₂, a diprotic base with a concentration of 0.00162 M, and a  of 10^-6 and a   of 10^-9.8.

Equation:

Again, assume x in the denominator is negligible.  Therefore,

Then,

We can then find the pH.

As we determine the pH of the solution, we realize that the OH^–gained using the second ionization constant is so insignificant that it does not impact the final pH value. For good measure, the following is the process to determine the pH in case the second use of the ICE table would indeed make a difference.

Equation:

ICE Table:

 

A buffer is an aqueous solution consisting of a mixture of a weak acid and its conjugate base or a weak base and its conjugate acid. Its pH changes very little when a small amount of strong acid or base is added to it and thus it is used to prevent changes in the pH of a solution. Buffer solutions are used as a means of keeping pH at a nearly constant value in a wide variety of chemical applications. Many life forms thrive only in a relatively small pH range so they utilize a buffer solution to maintain a constant pH. One example of a buffer solution found iature is blood.

Principles of buffering

Simulated titration of an acidified solution of a weak acid (pK_a = 4.7) with alkali.

Addition of hydroxide to a mixture of a weak acid and its conjugate base

Buffer solutions achieve their resistance to pH change because of the presence of an equilibrium between the acid HA and its conjugate base A^–.

HA H⁺ + A^–

When some strong acid is added to an equilibrium mixture of the weak acid and its conjugate base, the equilibrium is shifted to the left, in accordance with Le Chatelier’s principle. Because of this, the hydrogen ion concentration increases by less than the amount expected for the quantity of strong acid added.

Similarly, if strong alkali is added to the mixture the hydrogen ion concentration decreases by less than the amount expected for the quantity of alkali added. The effect is illustrated by the simulated titration of a weak acid with pK_a = 4.7. The relative concentration of undissociated acid is shown in blue and of its conjugate base in red. The pH changes relatively slowly in the buffer region, pH = pK_a ± 1, centered at pH = 4.7 where [HA] = [A^–]. The hydrogen ion concentration decreases by less than the amount expected because most of the added hydroxide ion is consumed in the reaction

OH^– + HA → H₂O + A^–

and only a little is consumed in the neutralization reaction which results in an increase in pH.

OH^– + H⁺ → H₂O

Once the acid is more than 95% deprotonated the pH rises rapidly because most of the added alkali is consumed in the neutralization reaction.

Applications

Buffer solutions are necessary to keep the correct pH for enzymes in many organisms to work. Many enzymes work only under very precise conditions; if the pH moves outside of a narrow range, the enzymes slow or stop working and can denature. In many cases denaturation can permanently disable their catalytic activity. A buffer of carbonic acid (H₂CO₃) and bicarbonate (HCO₃⁻) is present in blood plasma, to maintain a pH between 7.35 and 7.45.

Industrially, buffer solutions are used in fermentation processes and in setting the correct conditions for dyes used in colouring fabrics. They are also used in chemical analysis and calibration of pH meters.

The majority of biological samples that are used in research are made in buffers, especially phosphate buffered saline (PBS) at pH 7.4.

Simple buffering agents

For buffers in acid regions, the pH may be adjusted to a desired value by adding a strong acid such as hydrochloric acid to the buffering agent. For alkaline buffers, a strong base such as sodium hydroxide may be added. Alternatively, a buffer mixture can be made from a mixure of an acid and its conjugate base. For example, an acetate buffer can be made from a mixture of acetic acid and sodium acetate. Similarly an alkaline buffer can be made from a mixture of the base and its conjugate acid.

“Universal” buffer mixtures

By combining substances with pK_a values differing by only two or less and adjusting the pH,a wide-range of buffers can be obtained. Citric acid is a useful component of a buffer mixture because it has three pK_a values, separated by less than two. The buffer range can be extended by adding other buffering agents. The following two-component mixtures (McIlvaine’s buffer solutions) have a buffer range of pH 3 to 8

A mixture containing citric acid, potassium dihydrogen phosphate, boric acid, and diethyl barbituric acid can be made to cover the pH range 2.6 to 12.

Other universal buffers are Carmody buffer and Britton-Robinson buffer, developed in 1931.

Common buffer compounds used in biology

** Values are approximate. Biological buffers cover 1.9 to 11 pH range.

Buffer capacity

Buffer capacity for a 0.1 M solution of an acid with pK_a of 7

Buffer capacity, β, is a quantitative measure of the resistance of a buffer solution to pH change on addition of hydroxide ions. It can be defined as follows.

where dn is an infinitesimal amount of added base and d(p[H⁺]) is the resulting infinitesimal change in the cologarithm of the hydrogen ion concentration. With this definition the buffer capacity of a weak acid, with a dissociation constant K_a, can be expressed as

where C_A is the analytical concentration of the acid. pH is approximately equal to -log₁₀[H⁺].

There are three regions of high buffer capacity.

· At very low p[H⁺] the first term predominates and β increases in proportion to the hydrogen ion concentration. This is independent of the presence or absence of buffering agents and applies to all solvents.

· In the region p[H⁺] = pK_a ± 2 the second term becomes important. Buffer capacity is proportional to the concentration of the buffering agent, C_A, so dilute solutions have little buffer capacity.

· At very high p[H⁺] the third term predominates and β increases in proportion to the hydroxide ion concentration. This is due to the self-ionization of water and is independent of the presence or absence of buffering agents.

The buffer capacity of a buffering agent is at a maximum p[H⁺] = pK_a. It falls to 33% of the maximum value at p[H⁺] = pK_a ± 1 and to 10% at p[H⁺] = pK_a ± 1.5. For this reason the useful range is approximately pK_a ± 1.

Calculating buffer pH

Monoprotic acids

First write down the equilibrium expression.

HA A^– + H⁺

This shows that when the acid dissociates equal amounts of hydrogen ion and anion are produced. The equilibrium concentrations of these three components can be calculated in an ICE table.

The first row, labelled I, lists the initial conditions: the concentration of acid is C₀, initially undissociated, so the concentrations of A^– and H⁺ would be zero; y is the initial concentration of added strong acid, such as hydrochloric acid. If strong alkali, such as sodium hydroxide, is added y will have a negative sign because alkali removes hydrogen ions from the solution. The second row, labelled C for change, specifies the changes that occur when the acid dissociates. The acid concentration decreases by an amount -x and the concentrations of A^– and H⁺ both increase by an amount +x. This follows from the equilibrium expression. The third row, labelled E for equilibrium concentrations, adds together the first two rows and shows the concentrations at equilibrium.

To find x, use the formula for the equilibrium constant in terms of concentrations:

Substitute the concentrations with the values found in the last row of the ICE table:

Simplify to:

With specific values for C₀, K_a and y this equation can be solved for x. Assuming that pH = -log₁₀[H⁺] the pH can be calculated as pH = -log₁₀x.

Polyprotic acids

% species formation calculated for a 10 millimolar solution of citric acid.

Polyprotic acids are acids that can lose more than one proton. The constant for dissociation of the first proton may be denoted as K_a1 and the constants for dissociation of successive protons as K_a2, etc. Citric acid, H₃A, is an example of a polyprotic acid as it can lose three protons.

When the difference between successive pK values is less than about three there is overlap between the pH range of existence of the species in equilibrium. The smaller the difference, the more the overlap. In the case of citric acid, the overlap is extensive and solutions of citric acid are buffered over the whole range of pH 2.5 to 7.5.

Calculation of the pH with a polyprotic acid requires a speciation calculation to be performed. In the case of citric acid, this entails the solution of the two equations of mass balance

C_A is the analytical concentration of the acid, C_H is the analytical concentration of added hydrogen ions, β_q are the cumulative association constants

K_w is the constant for Self-ionization of water. There are two non-linear simultaneous equations in two unknown quantities [A^3-] and [H⁺]. Many computer programs are available to do this calculation. The speciation diagram for citric acid was produced with the program HySS.

In general the two mass-balance equations can be written as

In this general expression [A] stands for the concentration of the fully deprotonated acid and the electrical charge on this species is not specified.

A buffer is a solution that can maintain a nearly constant pH when diluted, or when strong acids or bases are added. A buffer solution consists of a mixture of a weak acid and its conjugate base (or a weak base and its conjugate acid).

Example: Consider a solution containing both acetic acid, CH₃COOH, and acetate ions, CH₃COO^–.

Any strong base that is added to the solution is neutralized by acetic acid:

CH₃COOH _(aq) + OH^–_(aq) CH₃COO^–_(aq) + H₂O _(aq)

Any strong acid that is added to the solution is neutralized by acetate:

CH₃COO^–_(aq) + H⁺_(aq) CH₃COOH _(aq)

The amount of strong acid or base that a buffer caeutralize is called the buffer capacity. After the strong base or acid is neutralized, equilibrium will be reestablished.

Since the concentrations of the conjugate acid-base pair in a buffer are usually high, there is very little change in the acid-base pair concentrations as the system establishes equilibrium. Therefore, the [H⁺] concentration can be calculated from the equilibrium expression using the pre-equilibrium concentrations of the conjugate acid-base pair. For our problem-solving method, step 4 can be omitted.

In the example above how could we produce the solution containing CH₃COOH and CH₃COO^–? In general a buffer can be prepared by:

1.                    Mixing a weak acid and a salt of its conjugate base in solution.

2.                    Mixing a weak acid and enough strong base to neutralize a portion of the weak acid.

3.                    Mixing a weak base and enough strong acid to neutralize a portion of the weak base.

 

Sample Problem

What is the pH of the resulting solution when 50.0 mL of 0.700 M NaOH is mixed with 50.0 mL of 0.500 M H₃PO₄?

1. It is easiest to consider the initial reactions stepwise:

moles of H₃PO₄ = (0.0500 L)(0.500 M) = 0.0250 moles

moles of OH^– = (0.0500 L)(0.700 M) = 0.0350 moles

0.0250 moles of H₃PO₄ neutralizes 0.0250 moles of OH^–. We now have 0.0100 moles of OH^– and 0.0250 moles of H₂PO₄^–. The original amount of H₃PO₄ is completely consumed.

0.0100 moles of H₂PO₄^– neutralizes the remaining OH^–, leaving 0.0150 moles of H₂PO₄^– and 0.0100 moles of HPO₄^2-.

2. So the equilibrium is:     H₂PO₄^–_(aq) HPO₄^2-_(aq) + H⁺_(aq)        K_a2 = 6.34×10^-8

and the equilibrium expression is:

        [H⁺][HPO₄^2-]

K_a2  =  ————

          [H₂PO₄^–]

3. and 4. We have formed a buffer so steps 3. and 4. are not needed explicitly.

5. We caow calculate the equilibrium concentrations using the equilibrium constant expression:

        [H⁺][HPO₄^2-]

K_a2  =  ————

          [H₂PO₄^–]

 

             [H⁺](0.0100 moles/0.100 L)

6.34×10^-8  =  ————————

               (0.0150 moles/0.100 L)

Note that the volume appears in the numerator and denominator and cancels out.

[H⁺] = (6.34×10^-8) * (0.0150) / (0.0100)

[H⁺] = 9.51×10^-8

This problem asked for the pH of the solution.

pH = -log[H⁺]

pH = -log(9.51×10^-8)

What is the pH of a buffer solution containing 0.100 moles of both CH₃COOH and CH₃COO^– in 0.100 L of water?

1. The only species present in solution that take part in the equilibrium are acetic acid and acetate ion. The pre-equilibrium concentrations of acetic acid and acetate ion are given as 1.00 M (0.100 moles/0.100 L). In buffers, the concentrations of the weak acid and conjugate base are high, so the [H⁺] of water (1.0×10^-7 M) can always be neglected.

2. The balanced equilibrium reaction is: CH₃COOH_(aq) H⁺_(aq) + CH₃COO^–_(aq)

and the equilibrium constant expression is:

     [H⁺][CH₃COO^–]

K_a = ————-

      [CH₃COOH]

3. It is seldom necessary to calculate Q in buffer problems. See explanation in step 4.

4. As stated in the introduction above, this step is not necessary in buffer problems. The concentrations of CH₃COOH and CH₃COO^– do not change appreciably as equilibrium is established.

5. We caow calculate the equilibrium concentrations using the equilibrium constant expression:

     [H⁺][CH₃COO^–]

K_a = ————-

      [CH₃COOH]

 

     (x)(1.00 M)

K_a = ———–

      (1.00 M)

x = K_a

x = 1.8×10^-5

This problem asked for the pH of the solution.

[H⁺] = x = 1.8×10^-5 M

pH = -log[H⁺] = -log(1.8×10^-5)

pH = 4.74

 

Equilibrium in Solutions of Weak Bases

Weak bases, such as ammonia, accept a proton from water to give the conjugate acid of the base and OH^– ions:

The equilibrium constant K_b is called the base-dissociation constant:

 

Relation Between K_a and K_b

For any conjugate acid–base pair, the product of the acid-dissociation constant for the acid and the base-dissociation constant for the base always equals the ionproduct constant for water:

For example,

 

Acid–Base Properties of Salts

When an acid neutralizes a base, an ionic compound called a salt is formed. Salt solutions can be neutral, acidic, or basic, depending on the acid–base properties of the constituent cations and anions. As a general rule, salts formed by reaction of a strong acid with a strong base are neutral, salts formed by reaction of a strong acid with a weak base are acidic, and salts formed by reaction of a weak acid with a strong base are basic. It’s as if the influence of the stronger partner dominates:

Salts That Yield Neutral Solutions

Salts such as NaCl that are derived from a strong base (NaOH) and a strong acid (HCl) yield neutral solutions because neither the catioor the anion reacts appreciably with water to produce H₃O⁺ or OH^– ions.

The following ions do not react appreciably with water to produce either H₃O⁺ or OH^– ions:

Salts that contain only these ions give neutral solutions in pure water pH=7

 

Salts That Yield Acidic Solutions

Salts such as NH₄Cl that are derived from a weak base NH₃ and a strong acid (HCl) produce acidic solutions. In such a case, the anion is neither an acid nor a base, but the cation is a weak acid:

Example.

In aqueous solution, the Al³⁺ ion bonds to six water molecules to give the hydrated cation Al(H₂O)₆³⁺

In general, the acidity of hydrated main-group cations increases from left to right in the periodic table as the metal ion charge increases and the metal ion size decreases  Transition metal cations, such as Zn²⁺ Cr³⁺and Fe³⁺ also give acidic solutions hydrated cation are much stronger proton donors than are free solvent water molecules.

 

 

Step 5. The value of x is obtained from the equilibrium equation:

Salts That Yield Basic Solutions

Salts such as NaCN that are derived from a strong base (NaOH) and a weak acid (HCN) yield basic solutions. In this case, the cation is neither an acid nor a base, but the anion is a weak base:

Example 5.

Salts That Contain Acidic Cations and Basic Anions

         Example. Finally, let’s look at a salt such as (NH₄)₂CO₃ in which both the cation and the anion can undergo proton-transfer reactions. Because NH₄⁺ is a weak acid and CO₃^2- is a weak base, the pH of an (NH₄)₂CO₃ solution depends on the relative acid strength of the cation and base strength of the anion:

We can distinguish three possible cases:

Because K_a < K_b the solution is basic (pH > 7).

Buffer Solutions

Solutions , which contain a weak acid and its conjugate base, are called buffer solutions because they resist drastic changes in pH. If a small amount of OH^– is added to a buffer solution, the pH increases, but not by much because the acid component of the buffer solutioeutralizes the added OH^– If a small amount of H₃O⁺ is added to a buffer solution, the pH decreases, but agaiot by much because the base component of the buffer solutioeutralizes the added H₃O⁺

Buffer solutions are very important in biological systems. Blood, for example, is a buffer solution that can soak up the acids and bases produced in biological reactions. The pH of human blood is carefully controlled at a value very close to 7.4 by conjugate acid–base pairs, primarily H₂CO₃ and its conjugate base HCO₃^–. The oxygen-carrying ability of blood depends on control of the pH to within 0.1 pH unit.

Example. To see how a buffer solution works, let’s return to the 0.10 M acetic acid–0.10 M sodium acetate solution.

The principal reaction and the equilibrium concentrations for the solution are

Note that in calculating this result we have set the equilibrium concentrations, (10-x) and (10+x) equal to the initial concentrations, 0.10, because x is negligible compared with the initial concentrations, 0.10.

Addition of OH^– to a Buffer

Suppose that we add 0.01 mol of solid NaOH to 1.00 L of the 0.10 M acetic acid–0.10 M sodium acetate solution. Initially, we have (1.00 L) (0.1 mol/L) = 0.1 mol of acetic acid and an equal amount of acetate ion. When we add 0.01 mol of NaOH, the neutralization reaction will alter the numbers of moles:

If we assume that the solution volume remains constant at 1.00 L, the concentrations of the buffer components after neutralization are

Substituting these concentrations into the expression for [H₃O⁺] we can then calculate

the pH:

The corresponding change in pH, from 4.74 to 4.82, is only 0.08 pH unit.

 

Addition of H₃O⁺ to a Buffer

Now suppose that we add 0.01 mol of HCl to 1.00 L of the 0.10 M acetic acid–0.10 M sodium acetate buffer solution. The added strong acid will convert 0.01 mol of acetate ions to 0.01 mol of acetic acid because of the neutralization reaction.

Again, the change in pH, from 4.74 to 4.66, is small because the concentration ratio [weak acid]/[conjugate base] remains close to its original value.

Buffer Capacity

To appreciate the ability of a buffer solution to maintain a nearly constant pH. We sometimes talk about the buffering ability of a solution using the term buffer capacity as a measure of the amount of acid or base that the solution can absorb without a significant change in pH. Buffer capacity is also a measure of how little the pH changes with the addition of a given amount of acid or base. Buffer capacity depends on how many moles of weak acid and conjugate base are present. For equal volumes of solution, the more concentrated the solution, the greater the buffer capacity. For solutions having the same concentration, the greater the volume, the greater the buffer capacity.

 

Buffer calculation example 1

A buffer solution was prepared which had a concentration of 0.20 mol dm^-3 in ethanoic acid and 0.10 mol dm^-3 in sodium ethanoate. If the K_a for ethanoic acid is 1.74 x 10^-5 mol dm^-3, calculate the theoretical hydrogen ion concentration and pH of the buffer solution.

• K_a = [H⁺_(aq)] [salt_(aq)]_/[acid_(aq)]

• 1.74 x 10^-5 = [H⁺_(aq)] x 0.10 / 0.20

• [H⁺_(aq)] = 1.74 x 10^-5 x 0.20_/0.10 = 3.48 x 10^-5 mol dm^-3

• pH = -log(3.48 x 10^-5) = 4.46

Buffer calculation example 2

In what ratio should a 0.30 mol dm^-3 of ethanoic acid be mixed with a 0.30 mol dm^-3 solution of sodium ethanoate to give a buffer solution of pH 5.6?

• K_a for ethanoic acid is 1.74 x 10^-5 mol dm^-3

• [H⁺_(aq)] = 10^-pH = 10^-5.6 = 2.51 x 10^-6 mol dm^-3

• K_a = [H⁺_(aq)] [salt_(aq)]_/[acid_(aq)]

• [salt]_/[acid] = K_a/[H⁺_(aq)] = 1.74 x 10^-5_/2.51 x 10^-6  = 6.93

• Therefore volume ratio is 6.93 : 1 for salt : acid, e.g. 6.93 cm³ of 0.30M sodium ethanoate is mixed with 1.0 cm³ of 0.30 M ethanoic acid to give a buffer solution of pH 5.6.

• Note that the pH is determined by the ratio of concentrations, but the buffering capacity of the solution can be increased by increasing the concentrations of both components in the same molar concentration ratio.

Buffer calculation example 3

What is the pH of a buffer solution made from dissolving 2.0g of benzoic acid and 5.0g of sodium benzoate in 250 cm³ of water?

• K_{a benzoic acid} = 6.3 x 10^-5 mol dm^-3, A_r‘s: H = 1, C = 12, O = 16, Na = 23

• Molecular masses: M_r(C₆H₅COOH) = 122, M_r(C₆H₅COO^–Na⁺) = 144, 250 cm³ = 0.25dm³

• moles acid C₆H₅COOH = 2.0/122 = 0.0164 mol, molarity = 0.0656 mol dm^-3

• moles salt C₆H₅COO^–Na⁺ = 5.0/144 = 0.0347 mol, molarity = 0.139 mol dm^-3

• [H⁺_(aq)] = K_a [acid_(aq)]_/[salt_(aq)]

• [H⁺_(aq)] = 6.3 x 10^-5 x 0.0656 / 0.139 = 2.97 x 10^-5 mol dm^-3

• pH = -log[H⁺_(aq)] = -log(2.97 x 10^-5) = 4.53

Buffer calculation example 4

Calculate the pH of a buffer made by mixing 100 cm³ of a 0.40 M sodium propanoate and 50 cm³ of 0.2 M propanoic acid solution.

• K_{a propanoic acid} = 1.3 x 10^-5 mol dm^-3, total volume of buffer = 150 cm³

• molarities in the mixture:

• [salt] = 0.40 x 100_/150 = 0.267 mol dm^-3

• [acid] = 0.20 x 50_/150 = 0.0667 mol dm^-3

• [H⁺_(aq)] = K_a [acid_(aq)]_/[salt_(aq)]

• [H⁺_(aq)] = 1.3 x 10^-5 x 0.0667 / 0.267 = 3.24 x 10^-6 mol dm^-3

• pH = -log[H⁺_(aq)] = -log(3.24 x 10^-6) = 5.48

 

Buffer calculation example 5

Using the Henderson equation pH_buffer = pK_a + log₁₀([salt_(aq)] / [acid_(aq)]) Calculate the pH of buffer solution made by mixing together 100 cm³ of 0.100M ethanoic acid and 50 cm³ of 0.400M sodium ethanoate, given that K_a for ethanoic acid is 1.74 x 10^-5 mol dm^-3 Now because the volumes are not equal, the real concentrations in the mixture must be worked out.

• The total volume is 150 cm³, therefore the dilutions are given by

• [acid] = 0.1 x 100/150 = 0.06667

• [salt] = 0.4 x 50/150 = 0.1333

• Substituting in the Henderson Equation gives

• pH_buffer = -log₁₀(1.74 x 10^-5) + log₁₀(0.1333/0.0667)

• pH_buffer = -log₁₀(1.74 x 10^-5) + log₁₀(2)

• pH_buffer = 4.76 + 0.3010

• pH_buffer = 5.06

 

 

 TYPES OF CHEMICAL REACTIONS

There are six types of chemical reactions:

1.                    ADDITION REACTION (SYNTHESIS).

2.                    DECOMPOSITION.

3.                    DISPLACEMENT.

4.                    DOUBLE DISPLACEMENT.

5.                    NEUTRALIZATION.

HYDROLYSIS

ADDITION REACTIONS OR SYNTHESIS

Chemical reactions in which two or more substances react to produce one compound are known as addition reaction.

They can be represented as:

A + B = AB

EXAMPLE:

H₂ + Cl₂ = 2HCl

SO₂ + Cl₂  = SO₂Cl₂

DECOMPOSITION

A chemical reaction in which a compound splits into two or more fragment is called decomposition.

They can be represented as:

AB =A+B

EXAMPLE:

2KClO₃ + HEAT = 2KCl + 3O₂

CH₄ + HEAT = C + 2H₂

Usually for decomposition heat is required.

DISPLACEMENT REACTION

A chemical reaction in which a radical or group replaces another radical or group form a compound and    takes its position is called a displacement reaction.

They can be represented as:

A + BC = AC + B

EXAMPLE:

Zn + H₂SO₄ = ZnSO₄ + H₂

Zn + CuSO₄ = ZnSO₄ + Cu

DOUBLE DISPLACEMENT

A chemical reaction in which two substances are decomposed to form new substances by exchanging their radicals is called double displacement reaction.

They can be represented as:

AB+CD = AD+CB

EXAMPLE:

BaCl₂ + Na₂SO₄ = BaSO₄ + 2NaCl

AgNO₃+ NaCl = AgCl + NaNO₃

NEUTRALIZATION

A chemical reaction in which an acid and a base react to form salt and water is called Neutralization.

Acid + Base = Salt + Water

EXAMPLE:

HCl + NaOH = NaCl + H₂O

HNO₃ + NaOH = NaNO₃ + H₂O

 

HYDROLYSIS

The type of double displacement reaction in which one reactant is water is called “Hydrolysis”.

Reaction of a substance with water in which P^H of water is changed is referred to as “Hydrolysis”.

EXAMPLE:

FeCl₃ + 3H₂O = Fe (OH)₃ + 3HCl

Na₂CO₃+2H₂O = 2NaOH+ H₂CO₃

 

Hydrolysis Reactions

 

 A reaction with water that results in water being split into a hydrogen and a hydroxide ion.

 

 

SPECTATORS– ions which do NOT hydrolyze (need periodic table and acid table to find these)

 

 

Hydrolysis When BOTH Cation and Anion hydrolyze 

 

Eg. Is the salt ammonium nitrite  NH4NO₂ acidic, basic or neutral?

 

 

 

ROLE OF HYDROLYSIS IN BIOCHEMICAL PROCESSES.

Usually hydrolysis is a chemical process in which a molecule of water is added to a substance. Sometimes this addition causes both substance and water molecule to split into two parts. In such reactions, one fragment of the target molecule (or parent molecule) gains ahydrogen ion.

Salts

A common kind of hydrolysis occurs when a salt of a weak acid or weak base (or both) is dissolved in water. Water spontaneously ionizes into hydroxyl anions and hydrogen cations. The salt, too, dissociates into its constituent anions and cations. For example, sodium acetate dissociates in water into sodium and acetate ions. Sodium ions react very little with the hydroxyl ions whereas the acetate ions combine with hydrogen ions to produce acetic acid. In this case the net result is a relative excess of hydroxyl ions, giving a basic solution.

Strong acids also undergo hydrolysis. For example, dissolving sulfuric acid (H2SO4) in water is accompanied by hydrolysis to give hydronium and bisulfate, the sulfuric acid’s conjugate base. For a more technical discussion of what occurs during such a hydrolysis, see Brønsted–Lowry acid–base theory.

Esters and amides

Acid–base-catalysed hydrolyses are very common; one example is the hydrolysis of amides or esters. Their hydrolysis occurs when the nucleophile (a nucleus-seeking agent, e.g., water or hydroxyl ion) attacks the carbon of the carbonyl group of the ester or amide. In an aqueous base, hydroxyl ions are better nucleophiles than polar molecules such as water. In acids, the carbonyl group becomes protonated, and this leads to a much easier nucleophilic attack. The products for both hydrolyses are compounds with carboxylic acid groups.

Perhaps the oldest commercially practiced example of ester hydrolysis is saponification (formation of soap). It is the hydrolysis of a triglyceride (fat) with an aqueous base such as sodium hydroxide (NaOH). During the process, glycerol is formed, and the fatty acids react with the base, converting them to salts. These salts are called soaps, commonly used in households.

In addition, in living systems, most biochemical reactions (including ATP hydrolysis) take place during the catalysis of enzymes. The catalytic action of enzymes allows the hydrolysis of proteins, fats, oils, and carbohydrates. As an example, one may consider proteases (enzymes that aid digestion by causing hydrolysis of peptide bonds in proteins). They catalyse the hydrolysis of interior peptide bonds in peptide chains, as opposed to exopeptidases (another class of enzymes, that catalyse the hydrolysis of terminal peptide bonds, liberating one free amino acid at a time).

However, proteases do not catalyse the hydrolysis of all kinds of proteins. Their action is stereo-selective: Only proteins with a certain tertiary structure are targeted as some kind of orienting force is needed to place the amide group in the proper position for catalysis. The necessary contacts between an enzyme and its substrates (proteins) are created because the enzyme folds in such a way as to form a crevice into which the substrate fits; the crevice also contains the catalytic groups. Therefore, proteins that do not fit into the crevice will not undergo hydrolysis. This specificity preserves the integrity of other proteins such as hormones, and therefore the biological system continues to functioormally.

Upon hydrolysis, an amide converts into a carboxylic acid and an amine or ammonia. The carboxylic acid has a hydroxyl group derived from a water molecule and the amine (or ammonia) gains the hydrogen ion. The hydrolysis of peptides gives amino acids.

Many polyamide polymers such as nylon 6,6 hydrolyse in the presence of strong acids. The process leads to depolymerization. For this reasoylon products fail by fracturing when exposed to small amounts of acidic water. Polyesters are also susceptible to similar polymer degradation reactions. The problem is known as stress corrosion cracking.

ATP

Hydrolysis is related to energy metabolism and storage. All living cells require a continual supply of energy for two main purposes: for the biosynthesis of micro and macromolecules, and for the active transport of ions and molecules across cell membranes. The energy derived from the oxidation of nutrients is not used directly but, by means of a complex and long sequence of reactions, it is channelled into a special energy-storage molecule, adenosine triphosphate (ATP). The ATP molecule contains pyrophosphate linkages (bonds formed when two phosphate units are combined together) that release energy wheeeded. ATP can undergo hydrolysis in two ways: the removal of terminal phosphate to form adenosine diphosphate (ADP) and inorganic phosphate, or the removal of a terminal diphosphate to yield adenosine monophosphate (AMP) and pyrophosphate. The latter usually undergoes further cleavage into its two constituent phosphates. This results in biosynthesis reactions, which usually occur in chains, that can be driven in the direction of synthesis when the phosphate bonds have undergone hydrolysis.

Polysaccharides

Figure. Sucrose. The glycoside bond is represented by the central oxygen atom, which holds the two monosaccharide units together.

Monosaccharides can be linked together by glycosidic bonds, which can be cleaved by hydrolysis. Two, three, several or many monosaccharides thus linked form disaccharides, trisaccharides, oligosaccharides or polysaccharides, respectively. Enzymes that hydrolyse glycosidic bonds are called “glycoside hydrolases” or “glycosidases”.

The best-known disaccharide is sucrose (table sugar). Hydrolysis of sucrose yields glucose and fructose. Invertase is a sucrase used industrially for the hydrolysis of sucrose to so-called invert sugar. Lactase is essential for digestive hydrolysis of lactose in milk; many adult humans do not produce lactase and cannot digest the lactose in milk (not a disorder).

The hydrolysis of polysaccharides to soluble sugars is called “saccharification”. Malt made from barley is used as a source of β-amylase to break downstarch into the disaccharide maltose, which can be used by yeast to produce beer. Other amylase enzymes may convert starch to glucose or to oligosaccharides. Cellulose is first hydrolyzed to cellobiose by cellulase and then cellobiose is further hydrolyzed to glucose by beta-glucosidase. Animals such as cows (ruminants) are able to hydrolyze cellulose into cellobiose and then glucose because of symbiotic bacteria that produce cellulases.

Metal aqua ions

Metal ions are Lewis acids, and in aqueous solution they form aqua ions of the general formula M(H₂O)_n^m+. The aqua ions undergo hydrolysis, to a greater or lesser extent. The first hydrolysis step is given generically as

M(H₂O)_n^m+ + H₂O   M(H₂O)_n−1(OH)^(m−1)+ + H₃O⁺

 

References:

1. The abstract of the lecture.

2. intranet.tdmu.edu.ua/auth.php

3. Atkins P. W. Physical chemistry / P.W. Atkins. – New York, 1994. – P.299‑307.

4. Cotton F. A. Chemical Applications of Group Theory / F. A. Cotton. ‑ John Wiley & Sons : New York, 1990.

5. Girolami G. S. Synthesis and Technique in Inorganic Chemistry / G. S. Girolami, T. B. Rauchfuss, R. J. Angelici. ‑ University Science Books : Mill Valley, CA, 1999.

6. Russell J. B. General chemistry / J B. Russell. New York.1992. – P. 550‑599.

7. Lawrence D. D. Analytical chemistry / D. D. Lawrence. –New York, 1992. – P. 218–224.

8. http://www.lsbu.ac.uk/water/ionish.html

 

Prepared by PhD Falfushynska H.