Question

For the diprotic weak acid H2A, Ka1 = 2.8 × 10-6 and Ka2 = 6.4 × 10-9.

What is the pH of a 0.0450 M solution of H2A?

What are the equilibrium concentrations of H2A and A2– in this solution?

Please show work ? I Have no idea how to do these types of problems.

Answer 1

H2A (aq) <---> HA- (aq) + H+ (aq) is 1st dissociation reaction , Ka1= 2.8 x 10^ -6

at equilibrium [H2A] = 0.045-X , [HA-] =[H+] = X

Ka1 = [HA-][H+]/[H2A]

2.8 x 10^ -6 = ( X) ( X) / ( 0.045-X)

2.8 x 10^ -6 = X^2 / ( 0.045-X)

solving quadratic equation we get X = 0.0003536 = 3.536 x 10^ -4 M = HA-

now HA- (aq) <---> A2- (aq) + H+ (aq) , ka2 = 6.4 x 10^ -9 M

at equilibrium [HA-] = 0.0003536-X , [A2-] =[H+]= X

Ka2 = [A2-][H+]/[HA-]

6.4 x 10^ -9 = ( X) (X) / ( 0.0003536-X)

solving qaudratic equation we get X = 1.5 x 10^ -6 M = [A2-]

now [H2A] = 0.045-0.00035356 = 0.04465 M

[A2-] = 1.5 x 10^ -6 M

[H+] = ( 0.00035356+ 1.5x10^-6) = 3.55 x 10^ -4

pH = -log [H+] = -log ( 0.000355) = 3.45