Question

Calculate the pOH of 10^-8 M of HCl.

Answer 1

10^-8 M indicates a very dilute solution. Thus, H⁺ ions of water cannot be ignored. But dissociation of water is suppressed due to common ion effect.

∴ [H⁺] ≠ 10^-7 M but less than 10^-7 M.

\[\ce{H2O ⇌ \underset{\ce{(10^{-8} + α)}}{H^+ } + \underset{\text{α}}{OH-}}\]

K_w = [H⁺] [OH^–] = (10^–8 + α) α

∴ α² + α × 10^-8 - 10^-14 = 0

Solving above quadratic equation, we get

α = 0.95 × 10^–7

∴ [H⁺] = 10^-8 + 0.95 × 10^-7 = 1.05 × 10^-7 M

∴ pH = – log₁₀[H⁺]

= - log₁₀[1.05 × 10^-7]

= 6.9788

∴ pOH = 14 – 6.9788 = 7.0212 ≈ 7