The of citric acid is 1.9 × at 25.0 °C.What is the pH of a 0.35 M aqueous solution of citric acid?

Question

The  of citric acid is 1.9 ×  at 25.0 °C.What is the pH of a 0.35 M aqueous solution of citric acid?

Quizplus · Accepted Answer

The equation for the dissociation of citric acid is:
H3C6H5O7 + H2O ⇌ H3O+ + H2C6H5O7-
The equilibrium constant expression for this reaction is:
Ka = [H3O+][H2C6H5O7-]/[H3C6H5O7]
The given value of Ka is 1.9 × 10^-4 at 25.0 °C. Since citric acid is a triprotic acid, we must use an ICE table and the expression for Ka and solve for the hydronium ion concentration to find the pH. 
We have:
[H3C6H5O7] = 0.35 M
[H2C6H5O7-] ≈ 0.35 M (since the dissociation is relatively small)
Let x be the hydronium ion concentration at equilibrium, then we have:
Ka = x^2/0.35
x^2 = Ka × 0.35
x = sqrt(Ka × 0.35) = 1.76 × 10^-3 M
The pH is given by:
pH = -log[H3O+] = -log(1.76 × 10^-3) = 2.59

The of Citric Acid Is 1