Question

H₂CO₃ $\rightleftharpoons$ H⁺ + HCO₃^-

$\frac{0.1}{100} \times 1000\text{ M}00$

Equilibrium (1 – x) M (x + y) M (x – y) M

$HCO_{3}^{-}$ $\rightleftharpoons$ $H^{+}$ + $CO_{3}^{2 -}$

Equilibrium (x – y) M (x + y) M y M

Now, $K_{a1} = \frac{\left\lbrack H^{+} \right\rbrack\left\lbrack HCO_{3}^{-} \right\rbrack}{\left\lbrack H_{2}CO_{3} \right\rbrack}$

$\Rightarrow 4 \times 10^{- 6} = \frac{(x + y).(x - y)}{(1 - x)} \approx \frac{x.x}{1}$

∴ x = 2 × 10^–3

and $K_{a2} = \frac{\left\lbrack H^{+} \right\rbrack\left\lbrack CO_{3}^{2 -} \right\rbrack}{\left\lbrack HCO_{3}^{-} \right\rbrack}$

∴ y = 5 × 10^–11

∴ [H⁺] = (x + y) ≈ x = 2 × 10^–3 M

[HCO₃⁻] = (x – y) ≈ x = 2 × 10^–3 M

[CO₃⁻] = y = 5 × 10^–11 M

$\left\lbrack OH^{-} \right\rbrack = \frac{K_{w}}{\left\lbrack H^{+} \right\rbrack} = \frac{10^{- 14}}{2 \times 10^{- 3}} = 5 \times 10^{- 12}\text{ M}$