Question

For the diprotic weak acid H2A, Kal = 3.2 x 10- and K2 = 6.7 x 10-9. What is the pH of a 0.0450 M solution of H, A? pH = 3.42 What are the equilibrium concentrations of H, A and A2- in this solution? [H, A] = 0.0446 [A2-1 = 1.588 x100

Answer 1

For a diprotic acid , reactions are

H₂A $\rightleftharpoons$ HA^- + H⁺   Ka1 = 3.2 $\times$ 10^-6

HA^-  $\rightleftharpoons$ A^2- + H⁺   Ka2 = 6.7 $\times$ 10^-9

Now

for concentrations of H⁺ and HA^- , we will concider only first reaction

So if x moles of H₂A are dissolves then x moles of H⁺ and x moles of HA^- will produce and concentration of H₂A will reduce by x , which is

[H₂A] = 0.0450-x M

[H⁺] = x

[HA^-] = x

Now , for first order we have Ka1 ,and expression as

Ka1 = [H⁺][HA^-]/[H₂A]

So

3.2 $\times$ 10^-6 = (x)(x) / (0.0450- x )

3.2 $\times$ 10^-6 = x² / (0.0450-x)

x² + 3.2x $\times$ 10^-6 - 0.144 $\times$ 10^-6 = 0

by solving above quadratic equation , we get

x = 0.000364

So , we know [H⁺] = x = 0.000364

And pH = -log[H⁺]

pH = -log( 0.00036) = 3.43

And

[H₂A] = 0.0450-x = 0.0450- 0.00036 = 0.0446 M

Now for second reaction

Ka2 = [A^2-][H⁺] / [HA^-]

Ka2 = 6.7 $\times$ 10^-9 = [A^2-] $\times$ (x) / (x)

[A^2-] =  6.7 $\times$ 10^-9 M

So our answers are

pH =3.43

[H₂A] = 0.0446 M

[A^2-] =  6.7 $\times$ 10^-9 M