Question

In: Chemistry

2. For the diprotic weak acid H2A, Ka1 = 2.7 × 10-6 and Ka2 = 8.8...

2. For the diprotic weak acid H2A, Ka1 = 2.7 × 10-6 and Ka2 = 8.8 × 10-9. What is the pH of a 0.0500 M solution of H2A? What are the equilibrium concentrations of H2A and A2– in this solution?

Expert Solution

H2A ---------> HA^- + H⁺

I 0.05 0 0

C -x +x +x

E 0.05-x +x +x

ka1 = [H+][HA-]/[H2A]

2.7*10^-6 = x*x/0.05-x

2.7*10^-6 *(0.05-x) = x²

x = 0.000366

[H+] = x = 0.000366M

[HA-] = x = 0.000366M

HA^- -------> A^2- + H⁺

Ka2 = [A^2-][H⁺]/[HA^-]

8.8*10^-9 = [A^2-]*0.000366/0.000366

[A^2-] = 8.8*10^-9 M

PH = -log[H+]

= -log0.000366 = 3.4365

queen_honey_blossom answered 9 months ago

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