CHM 1046

CHM 1046
General Chemistry II
Dr. Michael Blaber

Additional Aspects of Aqueous Equilibria

Buffered Solutions

Solutions of a weak acid, and its conjugate base, establish an equilibrium:

HA ó H⁺ + A^-

From Le Chatelier's principle, the equilibrium opposes changes in the [H⁺] of the solution

Addition of H⁺ will shift the equilibrium to the left (reducing [H⁺])

Removal of H⁺ will shift the equilibrium to the right (increasing [H⁺])

Solutions that resist a change in pH upon addition of small amounts of acid or base are called "Buffered" solutions (or just "Buffers")

Buffered solutions of weak acid/base conjugate pairs, and salts of that acid or base

Consider the weak acid HA:

HA(aq) ó H⁺(aq) + A^-(aq)

The corresponding acid dissociation constant, K_a, is:

We are interested in the possible buffering effects upon [H⁺], so solving for [H⁺] gives:

This simple analysis provides a clue as to the various entities that can influence the [H⁺], and therefore, the pH:

The value of the acid dissociation constant K_a
The ratio of the concentration of acid to conjugate base

How does the pH change with the addition of either H⁺ or OH^- ions?

OH^- ions will react with H⁺ ions to produce H₂O(l). This is the neutralization reaction, or the reverse of the water ionization reaction. In any case, it results in removal of H⁺ ions. The removal of H⁺ ions will increase the pH (remember, pH = -log[H⁺]; the smaller the value of [H⁺] the greater the value of -log)

However, from Le Chatelier's principle the equilibrium will respond by shifting to the right, increasing the [H⁺]
If the equilibrium shifts to the right, however, [HA] will decrease and [A^-] will increases. Thus, the ratio [HA]/[A^-] will get smaller. This is another way of saying that even though the equilibrium shifts to the right to oppose the loss of [H⁺], it can't make up for the loss entirely and the [H⁺] has to decrease somewhat

If the concentrations of [HA] and [A^-] are large to begin with, and if the added concentration of OH^- is small, the change in pH upon addition of the OH^- will be small (the pH of the solution will be buffered)

What about the addition of H⁺ ions?

From Le Chatelier's principle, the reaction will respond by shifting to the left, to reduce the [H⁺]
This is another way of saying the H⁺ reacts with the conjugate base, A^-, to produce the acid HA
Thus, in shifting to the left, A^- is consumed and HA is produced. Therefore, the ratio [HA]/[A^-] will get larger. This is another way of saying that even though the system responds by reducing [H⁺], it can't reduce the added H⁺ entirely and the [H⁺] has to rise somewhat

Again, if the concentrations of [HA] and [A-] are large to begin with, and if the added concentration of H⁺ is small, the change in pH upon addition of the H⁺ will be small (the pH of the solution will be buffered)

Weak acid/conjugate base pairs can therefore effectively buffer pH changes in either direction (i.e. buffer against either an increase or decrease in [H⁺])

However, since the important aspect of buffering is to have the change in [HA] and [A^-] be small in response to the added [H⁺] or [OH^-], buffers work best when the ratio of [HA]/[A^-] is 1.0. (In other words, if the concentration of either HA or A^- is small, the solution can't buffer very well)
From the above equation, if [HA]/[A^-] = 1.0, then [H⁺] = K_a

Thus, for any give weak acid/conjugate base pair, there will be a pH at which it has the greatest buffering capacity.
This pH will be where the [H⁺] = K_a.

In other words, best buffer pH = -log(K_a) = pKa

For example, if you want to buffer a solution at pH = 8.5 choose a weak acid/base conjugate pair whose pK_a = 8.5. (log K_a = -8.5, K_a = 3.16 x 10^-9; hypobromous acid is sort of close, pK_a = 2.5 x 10^-9)

Buffer Capacity and pH

The two important characteristics of a buffer are:

The pH at which the buffer is most effective (this will be equal to the pK_a for that buffer)
How much acid or base can be added before the pH starts to change

The greater the concentrations of both [HA] and [A^-] (i.e. the acid/conjugate base-pair) the greater the buffering capacity

How do the other components (i.e. conjugate base and undissociated acid when considering weak acids) affect the pH?

From the equation relating [H⁺] to the concentration of acid/conjugate base:

It can be seen that [H⁺], and therefore, the pH, is dependent upon the relative concentrations of the acid (HA) and conjugate base (A^-)

Thus, pH is determined not only by the pK_a of the acid/conjugate base-pair, but also the concentration of the acid and conjugate base

This relationship is known as the Henderson-Hasselbalch equation.

It allows one to determine the pH of a solution from knowing the pK_aof the buffer and the concentration of the acid and conjugate base components

A buffer works best when the concentration of the acid component (HA) is equal to the concentration of the conjugate base (A^-)
However, we know that for weak acids, only a small percentage of the acid will dissociate at equilibrium. Therefore, the concentration of conjugate base will be quite small

How can we get the concentration of conjugate base to be equivalent to the concentration of the acid component?

We can use the common ion effect

The acid and conjugate base components share a common ion

HC₃H₅O₃(aq) ó H⁺(aq) + C₃H₅O₃^-(aq)
(lactic acid) ó (protons) + (lactate ion)

The conjugate base ion can be added as a salt, where the cation component is from a strong base

NaC₃H₅O₃Sodium Lactate salt

The sodium lactate salt will dissociate completely, releasing Na⁺ and C₃H₅O₃^-

NaC₃H₅O₃(aq) ® Na⁺(aq) + C₃H₅O₃^-(aq)

The Na⁺(aq) is the conjugate acid of a strong base (NaOH), and will therefore, not change the pH (it has no desire to react with an OH^- group)
The C₃H₅O₃^-(aq) will raise the concentration of this component in the solution. (Normally, increasing this conjugate base will shift the equilibrium to the left. However, this will only happen if an equivalent amount of H⁺ is around. Since we added the C₃H₅O₃^- as the sodium salt, no H⁺ was added (or is available)).

Finally, in the Henderson-Hasselbalch equation, the concentrations of [conjugate base] and [acid] refer to the concentrations at equilibrium. However, since such a small percentage of the acid will dissociate in solution, and such a small percentage of added conjugate base will be protonated, we can use the starting concentrations of acid and conjugate base as relatively accurate values in the calculation

Thus, most buffer solutions are made of a weak acid with the conjugate base (in approximately equal concentration) supplied as a salt (where the cation is the conjugate acid of a strong base). [similarly with buffers of a weak base]

Addition of Strong Acids or Bases to Buffers

As long as we do not exceed the buffering capacity of the buffer, we can assume that a strong acid, or strong base, reacts completely with the buffer

To calculate the effect upon the pH, assume that all the H⁺ of an added strong acid reacts with the conjugate base of the buffer, to produce an equivalent concentration of the acid form of the buffer.
Likewise, to calculate the effect upon the pH, assume that all the OH^- of a strong base reacts with the acid form of the buffer (i.e. it neutralizes an equivalent concentration of H⁺, this in turn is obtained by a stoichiometrically equivalent amount of buffer acid that must dissociate) to produce an equivalent concentration of conjugate base form of the buffer.

What is the pH of a buffer that is 0.15M in acetic acid (CH₃COOH) and 0.05M in sodium acetate (NaCH₃COO)? The K_a for acetic acid is 1.8 x 10^-5.

We can use the Henderson-Hasselbalch equation to solve this problem:

pH = pK_a + log ([base]/[acid])

We assume that a very small percentage of the acid will ionize as the process goes to equilibrium. Therefore, a very small amount of the conjugate base (acetate ion) can combine with protons (i.e. the reverse reaction) as the process goes to equilibrium (in other words, a small percentage of acid ionizing yields a low concentration of protons. Thus, there are not many protons available for the reverse reaction). In any case, all these assumptions lead to the conclusion that we can use the starting concentrations of acid and conjugate base as a good estimate of the equilibrium concentrations (and can therefore plug them into the Henderson-Hasselbalch equation:

pH = -log(1.8 x 10^-5) + log(0.05/0.15)

pH = 4.74 - 0.477

pH = 4.26

We can also work this out another way. The balanced equation for the ionization of acetic acid is:

CH₃COOH(aq) ó CH₃COO^-(aq) + H⁺(aq)

The sodium acetate is a salt derived from a strong base (NaOH) and a weak acid (acetic acid). The Na⁺ ion therefore will not affect the pH, and the added sodium acetate will contribute to the starting concentration of the conjugate base of acetic acid (i.e. acetate ion)

Initial concentration of CH₃COOH = 0.15M
Initial concentration of CH₃COO^- = 0.05M
Initial concentration of H⁺ = 1 x 10^-7M i.e. the situation in pure H₂O. We can consider it to be 0M

Amount of acetic acid that ionizes as the solution goes to equilibrium = XM

Equilibrium concentration of CH₃COOH = 0.15M - XM
Equilibrium concentration of CH₃COO^- = 0.05M + XM
Equilibrium concentration of H⁺ = XM

K_a = 1.8 x 10^-5 = (0.05 + X)(X) / (0.15 - X)

(1.8 x 10^-5)(0.15 - X) = X² + 0.05X

2.7 x 10^-6 - 1.8 x 10^-5X = X² + 0.05X

X² + 0.05X - 2.7 x 10^-6 = 0

This is a quadratic, a = 1, b = 0.05, c = -2.7 x 10^-6

X = (-.05 + (.0025 - (4*1*-2.7 x 10^-6))^0.5) / 2

5.39 x 10^-5

X = (-.05 - (.0025 - (4*1*-2.7 x 10^-6))^0.5) / 2

-0.05

A negative value would imply a negative concentration of H⁺, therefore, the value of X = 5.39 x 10^-5. Thus,

[H⁺] = 5.39 x 10^-5

pH = -log(5.39 x 10^-5)

pH = 4.27

Compare above and you will see that the value agrees well with the Henderson-Hasselblach equation. The Henderson-Hasselblach equation is a lot simpler, don't you agree?

In the above problem, what would the pH be if 0.01M of NaOH is added to the buffer?

The NaOH is a strong base and will dissociate completely. The Na⁺ ion will not affect the pH, but the OH^- certainly will. The OH^- ion will make the solution more basic. In particular, it will react with protons to produce H₂O in a neutralization reaction. From the balanced equation, we see that removal of H⁺ will shift the equilibrium to the right:

CH₃COOH(aq) ó CH₃COO^-(aq) + H⁺(aq)

The 0.01M of OH^- ions can react stoichiometrically with 0.01M of H⁺, and 0.01M of H⁺ is produced from the ionization of 0.01M of acetic acid. Therefore, addition of 0.01M of NaOH will reduce the concentration of acetic acid by 0.01M, and increase the concentration of conjugate base (acetate ion) by 0.01M.

We can apply the Henderson-Hasselbalch equation to this problem as well:

pH = -log(1.8 x 10^-5) + log((0.05+0.01)/(0.15-0.01))

pH = 4.74 + log(0.06/0.14)

pH = 4.74 - 0.368

pH = 4.37

We can also work this out the long way:

Initial concentration of CH₃COOH = 0.14M
Initial concentration of CH₃COO^- = 0.06M
Initial concentration of H⁺ = 1 x 10^-7M i.e. the situation in pure H₂O. We can consider it to be 0M

Amount of acetic acid that ionizes as the solution goes to equilibrium = XM

Equilibrium concentration of CH₃COOH = 0.14M - XM
Equilibrium concentration of CH₃COO^- = 0.06M + XM
Equilibrium concentration of H⁺ = XM

K_a = 1.8 x 10^-5 = (0.06 + X)(X) / (0.14 - X)

(1.8 x 10^-5)(0.14 - X) = X² + 0.06X

2.52 x 10^-6 - 1.8 x 10^-5X = X² + 0.06X

X² + 0.06X - 2.52 x 10^-6 = 0

Solving the quadratic:

X = (-0.06 + (0.0036 + 1.01 x 10^-5)^0.5) / 2

X = 4.20 x 10^-5

Therefore:

[H⁺] = 4.20 x 10^-5M

pH = -log(4.20 x 10^-5)

pH = 4.38

This compares well with the results using the Henderson-Hasselbalch equation. (I don't know about you, but I prefer to solve these types of problems using the H-H equation)