Question

A diprotic acid, H2A, has acid dissociation constants of ?a1=3.69×10−4 and ?a2=4.08×10−12. Calculate the pH and molar concentrations of H2A, HA−, and A2− at equilibrium for each of the solutions.

A 0.102 M solution of H2A.

A 0.102 M solution of NaHA.

A 0.102 M solution of Na2A

Answer 1

Answer - We are given, weak diprotic acid H₂A, Ka1 = 3.69*10^-4 , Ka2 = 4.08*10^-12

A) 0.102 M solution of H2A.

We need to put ICE chart for the first dissociation -

       H₂A + H₂O -------> HA^-   + H₃O⁺

I    0.102                         0             0

C     -x +x            +x

E    0.102-x                 +x             +x

Ka1 = [HA^- ] [H₃O⁺] / [H₂A]

3.69*10^-4 = (x) * (x) / (0.102-x)

3.69*10^-4 * (0.102-x)= x²

x² + 3.69*10^-4 – 3.76 *10^-5

Using quadratic equation

x = 0.00595 M

[H₃O⁺] = x = 0.00595 M

[HA^-] = x = 0.00595 M

[H₂A] = 0.102-x

           = 0.102-0.00595

           = 0.0960 M

Now second dissociation-

       HA^- + H₂O -------> A^2-   + H₃O⁺

I    0.00595                       0            0

C      -x                           +x          +x

E 0.00595–x                +x           +x

Ka2 = [A^2-] [H₃O⁺] / [HA^-]

4.08*10^-12 = (x) * (x) / (0.00595-x)

We can neglect x in 0.00595-x , since Ka2 value is too low

4.08*10^-12 * 0.00595 = x²

x = 1.56*10^-7 M

[H₃O⁺] = x = 1.56*10^-7 M

x = [A^2-] = 1.56*10^-7 M

[HA^-] = 0.00595-x

          = 0.00595 – 1.56*10^-7

           = 0.005949 M

So total [H₃O⁺] = 0.00595M + 1.56*10^-7 M

                         = 0.00595 M

We know pH formula

pH = -log [H₃O⁺]

      = -log 0.00595 M

      = 2.23

So, the pH of 0.102 M weak acid H₂A is 2.23

At equilibrium molar concentration of each as follow –

[H₂A] = 0.0960 M

[HA^-] = 0.005949 M

[A^2-] = 1.56*10^-7 M