Question

The % error in $\left[H^{+}\right]$ made by neglecting the ionisation of water in ${10}^{-6}M$ $NaOH$ is:

• A. $1$%
• B. $2$%
• C. $3$%
• D. $4$%

Answer 1

Answer: (A)

$1$%

$\left[H_3{O}^{+}\right]=\frac{Kw}{\left[O{H}^{-}\right]}=\frac{{10}^{-14}}{{10}^{-6}}={10}^{-8}M$ (neglecting ionisation of water)

Now considering ionisation of water:

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

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

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

$\Rightarrow y^2+{10}^{-6}y-{10}^{-14}=0$

On solving for $y$, $y=9.9\times {10}^{-9}$

$\therefore$ % of error= $\frac{{10}^{-8}-9.9\times {10}^{-9}}{9.9\times {10}^{-9}}\times 100\simeq 1$%