Question

The percent error in hydronium ion concentration made by neglecting the ionisation of water in $1.0\times {10}^{-6}M$ $NaOH$ is

Answer 1

Approximate hydroxide concentration (by neglecting dissociation of water) $=1.0\times {10}^{-6}$

Actual hydroxide concentration (by considering dissociation of water) $=1.0\times {10}^{-6}+1.0\times {10}^{-7}$

Percentage error $=100\times \frac{1.0\times {10}^{-6}+1.0\times {10}^{-7}-1.0\times {10}^{-6}}{1.0\times {10}^{-6}+1.0\times {10}^{-7}}=1$%.

This is also equal to the percentage error in hydronium ion concentration.