Question

Calculate the percentage error in hydronium ion concentration made by neglecting ionisation of water in ${10}^{-6}M$ $NaOH$.

Answer 1

$\left[H_3{O}^{+}\right]=\frac{K_w}{\left[{OH}^{-}\right]}=\frac{{10}^{-14}}{{10}^{-6}}={10}^{-8}M$

(Neglecting ionization of water)

Consider ionization of water

$\left[H_3{O}^{+}\right]=y\left[{OH}^{-}\right]=(y+{10}^{-6})$

$\left[H_3{O}^{+}\right]\left[{OH}^{-}\right]=K_w={10}^{-14}$

$y(y+{10}^{-6})={10}^{-14}$

On solving for $y$

$y=9.9\times {10}^{-9}$

% error $=\frac{{10}^{-8}-9.9\times {10}^{-9}}{9.9\times {10}^{-9}}\times 100=1.01$