Question

At ${100}^{o}C$ the $K_w$ of water is 55 times its value at ${25}^{o}C$. What will be the pH of neutral solution?
$(log55=1.74)$

• A. 6.13
• B. 7.00
• C. 7.87
• D. 5.13

Answer 1

The correct option is A 6.13

$K_w$ at ${25}^{o}C$ = $1\times {10}^{–14}$
At ${25}^{o}C$
$K_w=\left[H^{+}\right]\left[O{H}^{–}\right]={10}^{–14}$
At ${100}^{o}C$ (given)
$K_w=\left[H^{+}\right]\left[O{H}^{–}\right]=55\times {10}^{–14}$
For a neutral solution
$\left[H^{+}\right]=\left[O{H}^{–}\right]$
$[H^{+}{]}^2=55\times {10}^{–14}$
or $\left[H^{+}\right]=(55\times {10}^{–14}{)}^{1/2}$
On taking negative log on both side
$–log\left[H^{+}\right]=–log(55\times {10}^{–14}{)}^{1/2}$
$pH=–log\left[H^{+}\right]$
$pH=-\frac{1}{2}[log55-14log10]$
$pH=\frac{1}{2}\times [-1.74+14]$
$pH=6.13$

Hence, option (a) is correct.