Question

On neglecting the contribution of water, [H⁺] = 10^–6 M

When contribution of water is considered,

H₂O $\rightleftharpoons$ H⁺ + OH^–

Equilibrium (x + 10^–6) M xM

Now, (x + 10^–6)⋅x = 10^–14 ⇒ x = 9.9 × 10^–9

∴ [H⁺]₂ = (x + 10^–6) M = 1.0099 × 10^–6 M

Now,% error in[H⁺] =$\frac{\left\lbrack H^{+} \right\rbrack_{2} - \left\lbrack H^{+} \right\rbrack}{\left\lbrack H^{+} \right\rbrack_{2}}\left\lbrack H^{+} \right\rbrack \times 100\% = 0.98\%$