Question

Calculate the percent error in the $\left[H_3{O}^{\ominus }\right]$ concentration made by neglecting the ionization of water in ${10}^{-6}MNaOH$ solution.

Answer 1

$\left[H3O^{+}\right]=K_w/\left[O{H}^{-}\right]={10}^{-14}/{10}^{-6}={10}^{-8}$M

(Neglecting ionization of water)

Consider ionization of water

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

$[H_{3}O+]\left[O{H}^{-}\right]=K_w={10}^{-14}$

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

On solving for y

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

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