Question

Given:

Ka1= 7.5*10^-3

Ka2=6.2*10^-8

Ka3=4.2*10^-13

What is the pH of a solution of 0.550 M K2HPO4, potassium hydrogen phosphate?

What is the pH of a solution of 0.400 M KH2PO4, potassium dihydrogen phosphate?

Answer 1

K_a1 = 7.5 x 10^-3

pK_a1 = -log(K_a1)

pK_a1 = -log(7.5 x 10^-3)

pK_a1 = 2.125

Similarly, pK_a2 = 7.208

pK_a3 = 12.377

(a) 0.550 M K₂HPO₄

K₂HPO₄ dissociates in the solution as : K₂HPO₄ (aq) 2 K⁺ (aq) + HPO₄^2- (aq)

HPO₄^2- undergoes two reactions

HPO₄^2- (aq) H⁺ (aq) + PO₄^3- (aq)

HPO₄^2- (aq) + H₂O (l) H₂PO₄^- (aq) + OH^- (aq)

pH = (pK_a2 + pK_a3) / 2

pH = (7.208 + 12.377) / 2

pH = 9.79

This pH does not depend upon concentration of K₂HPO₄

(b) 0.400 M KH₂PO₄

KH₂PO₄ dissociates in the solution as : KH₂PO₄ (aq) K⁺ (aq) + H₂PO₄^- (aq)

H₂PO₄^- undergoes two reactions

H₂PO₄^- (aq) H⁺ (aq) + HPO₄^2- (aq)

H₂PO₄^- (aq) + H₂O (l) H₃PO₄ (aq) + OH^- (aq)

pH = (pK_a1 + pK_a2) / 2

pH = (2.125 + 7.208) / 2

pH = 4.67