Question

What is the pH p H change of a 0.200 M M solution of citric acid (pKa=4.77 p K a = 4.77 ) if citrate is added to a concentration of 0.140 M M with no change in volume?

Answer 1

Let citric acid be denoted by HA.

Dissociation of HA:

HA(aq) + H₂O (l) <=> A^- (aq) + H₃O⁺ (aq) ; where Ka is the dissociation constant.

0.2 0 0 [Initial concentration]

0.2-x x x [Concentration at equilibrium]

Ka = [A^-] [H₃O⁺] / [HA] ; [H₃O⁺ ] = x

Also, Ka of citric acid = 10^-pka = 10^-4.77 = 1.7 * 10^-5[ as given in data]

Ka = 1.7 * 10^-5 = x * x / (0.200 - x) = x² / 0.200 - x

To avoid solving a quadratic equation we can neglect x in the denominator which is negligible with respect to 0.200. This can be understood as 0.2 is itself a small value and subtracting a value smaller than that will not make a huge difference in the denominator.

Hence. 1.7 * 10^-5 = x² / 0.200 => 1.7 * 10^-5 * 0.200 = x²

x² = 3.4 * 10^-6 => x = 1.84 * 10^-3 = [H₃O⁺]

pH = - log [H₃O⁺] = - log [1.84 * 10^-3 ] = 2.73 _(1) [This is the initial pH before citrate is added]

Using Henry Hasselbalch equation we can find out the final pH after adding 0.14 M citrate (conjugate base) in the solution.

pH = pKa + log [conjugate base] / [acid]

pH = 4.77 + log [0.140] / [0.200] = 4.77 - 0.15 = 4.62_(2) [Final pH after addition of citrate]

pH change = 4.62 - 2.73 = 1.89 [ subtracting 1 from 2]