Question

The pH of a neutral solution at ${50}^{o}C$ is:
($K_w={10}^{-13.26}$ at ${50}^{o}C$)

• A. $7$
• B. $6.0$
• C. $7.23$
• D. $6.63$

Answer 1

The correct option is D $6.63$

Ionic product of water is $K_w={10}^{-13.26}$

$K_w=\left[H^{+}\right]\left[O{H}^{-}\right]$

Since $\left[H^{+}\right]=\left[O{H}^{-}\right]$

$[H^{+}{]}^2=K_w$

${\left[H^{+}\right]}^2=K_w$

$\left[H^{+}\right]=\sqrt{K_w}=\sqrt{{10}^{-13.26}}=2.344\times {10}^{-7}pH=-log\left(\right[H^{+}\left]\right)=-log(2.344\times {10}^{-7})=6.628$

$pH=6.63$