Question

Percentage ionisation of water at certain temperature is $3.6\times {10}^{-7}$%. Calculate $K_w$ and $pH$ of water.

• A. ${10}^{-14},pH=6/7$
• B. $4\times {10}^{-14},pH=6.7$
• C. $2\times {10}^{-14},pH=7$
• D. ${10}^{-14},pH=7$

Answer 1

The correct option is B $4\times {10}^{-14},pH=6.7$

Given that

$\alpha =\frac{3.6\times {10}^{-7}}{100}$

No. of ionized species $=c\times \alpha$

Molarity of water, C $=\frac{1000}{18}$ $M$

$C\times \alpha =\frac{1000}{18}\times \frac{3.6\times {10}^{-7}}{100}=2\times {10}^{-7}M$

$\left[H^{+}\right]\left[O{H}^{-}\right]=2\times {10}^{-7}\times 2\times {10}^{-7}=4\times {10}^{-14}$

$\therefore K_w=4\times {10}^{-14}$

$pkW=14-\log 4$

$pH=\frac{pkW}{2}=7-\log 2=6.7$