Question

For the diprotic weak acid H2A, Ka1 = 3.0 × 10^-6 and Ka2 = 7.6 × 10^-9. What is the pH of a 0.0600 M solution of H2A? What are the equilibrium concentrations of H2A and A2– in this solution?

pH=?

[H2A]=?

[A2-]=?

Answer 1

We have to use two equations:

            H₂A ------------- HA^- +   H⁺

I        0.0600             0            0

C           -x                +x          +x

E         0.0600-x       x             x

Ka1= [HA^-][H⁺]/[H₂A]

3.0 10^-6= x x / (0.0600-x)

3.0 10^-6 (0.0600-x)= x²

1.8 10^-7 - 3.0 10^-6 x = x²

x² +3.0 10^-6 x -1.8 10^-7 = 0, using our quadratic formula x= 4.22 10^-4 M,

H₂A= 0.0600-x = 0.0600-4.22 10^-4 = 0.059578 M

HA^-= 4.22 10^-4 M

H⁺ = 4.22 10^-4 M

Our second reaction:

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

I    4.22 10^-4                0        4.22 10^-4

C        -x                      +x          +x

E 4.22 10^-4-x               x         4.22 10^-4+x

ka2=[H⁺][A^-2]/[HA^-]

7.6 10^-9= (4.22 10^-4+x) x / (4.22 10^-4-x)

7.6 10^-9 (4.22 10^-4-x) = (4.22 10^-4+x) x

3.20 10^-12 - 7.6 10^-9x = 4.22 10^-4x + x²

x² + 4.22 10^-4x + 7.6 10^-9x - 3.20 10^-12 = 0

x² + 4.220076 10^-4x -3.20 10^-12= 0 Using our quadratic formula again

x= 7.58 10^-9 M, this is the concentration of A^-2, for H+= 4.22 10^-4+x = 4.2200758 10^-4

Now, the answers=

pH= -log ([H⁺])= -log (4.2200758 10^-4) = 3.37

[H₂A]= 0.059578 M

[A^2-]=7.58 10^-9 M