Question

Mr. C claims he prepared 5 buffers with a weak acid HA (pKa = 4) and its potassium salt KA. The 5 “buffers” have pH = 2.9, pH = 3.5, pH = 4.1, pH = 4.2, pH = 5.1. Point out what he has done wrong. Then tell which buffer should have the highest capacity.

Answer 1

Ans. #1. Using Henderson- Hasselbalch equation -

            pH = pKa + log ([A^-] / [AH])                    - equation 1

where, [A^-] = conjugate base,

                        [AH] = Weak acid

A weak acid dissociates in water as follow-

            AH(aq) + H₂O(l) -----> A^-(aq) + H₃O⁺(aq)

Since the acid is weak, the dissociation is always less than 100.0 %.

Consider the example, there are 100 molecules of the AH initially taken in the solution. Some of AH would dissociate to produce A^- whereas most of it remains as AH.

Say, out of 100 AH molecules, 10 molecules (taking dissociation of extreme 10% for explanation. Normal dissociation would be less than 1%).

So, [A^-] = 10 , and [AH] = 90 (= Initial 100 – Change 10).

So,

            [A^-] / [AH] = 10 / 90 = 0.111

# Now, note the following points-

I. [AH] would always be greater than [A^-]

II. Since, [AH] > [A^-], the ratio [A^-] / [AH] would always be less than 1.00

III. log of a value less than 1.00 is always negative. For example, log 0.11 = -0.95.

IV. Therefore, log ([A^-] / [AH]) in equation 1 would always be a negative value. Here, log ([A^-] / [AH]) = -0.95

Now,

            pH = 4.2 + (-0.95) = = 4.2 – 0.95 = 3.25

Conclusion: The pH of a buffer consisting of a weak acid and its conjugate base can never be greater than the pKa value of weak acid.

Therefore, the pH 5.1 is NOT possible for the buffer because the pKa of acid is 4.2

So, the pH of 5.2 for the buffer is INCORRECT.

#2. The buffering capacity of a buffer is maximum when pH of the buffer is closest to the pKa value pf the weak acid.

Therefore, the buffer with pH = 4.2 would have maximum buffering capacity.