Question

In: Chemistry

For the diprotic weak acid H2A, Ka1 = 3.1 × 10-6 and Ka2 = 6.6 ×...

For the diprotic weak acid H2A, Ka1 = 3.1 × 10-6 and Ka2 = 6.6 × 10-9. What is the pH of a 0.0700 M solution of H2A? What are the equilibrium concentrations of H2A and A2– in this solution?

pH=

[H2A]=

[A^2-]=

Expert Solution

H2A --------------> H^+ (aq) + HA^-

I 0.07M 0 0

C -x +x +x

E 0.07-x +x +x

Ka1 = [H^+][A^-]/[HA]

3.1*10^-6 = x*x/0.07-x

3.1*10^-6*(0.07-x) = x^2

x = 0.000464

[H^+] = 0.000464M

PH = -log[H+]

= -log0.000464

= 3.3334

[H2A] = 0.07-x =0.07-0.000464 = 0.0695M

[HA^-] = x = 0.000464M

HA^- ----------> A^2- + H^+

Ka2 = [A^2-][H^+]/[HA^-]

6.6*10^-9 = [A^2-]*0.000464/0.000464

[A^2-] = 6.6*10^-9M

queen_honey_blossom answered 2 years ago

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