Question

What is the pH change of a 0.300 M solution of citric acid (pKa=4.77) if citrate is added to a concentration of 0.150 M with no change in volume?

Answer 1

Initial pH:

---citric acid <-----> citrate + H⁺(aq): Ka = 10^-4.77 = 1.70*10^-5

I: 0.300 M ------------ 0 M ----- 0 M

C: - X ----------------- +X ------- +X

E:(0.300 - X) --------- X --------- X

Ka = 1.70*10^-5 = [Citrate]*[H⁺(aq)] / [Citric acid​​​]

=> 1.70*10^-5 = X*X / (0.300 - X)

=> 1.70*10^-5 * (0.300 - X) = X²

=> 5.10*10^-6 - 1.70*10^-5X = X²

Solving the above quadratic equaiton gives:

X = [H⁺​​​​​​​(aq)] = 2.25*10^-3 M

=> pH = - log( 2.25*10^-3) = 2.64785

pH after addition of citrate:

Since citric acid is a weak acid and citrate is a conjugate base of citric acid, when we add citrate to citric acid, it acts as buffer solution.

Hence we can find pH by applying Henderson equation.

Since there is no change in volume, concentration also remain same. Hence

[citric acid] = 0.300 M

[Citrate] = 0.150 M

Applying Henderson equaiton:

pH = pKa + log([citrate] / [citric acid])

=> pH = 4.77 + log(0.150/0.300)

=> pH = 4.46897

pH change = 4.46897 - initial pH

=> pH change = 4.46897 - 2.64785 = 1.82 (Answer)