Question

Ionic product of water at 310 K is $2.7\times {10}^{-14}$. What is the $pH$ of neutral water at this temperature?

Answer 1

$K_w=\left[H^{+}\right]\left[O{H}^{-}\right]=2.7\times {10}^{-14}=[H^{+}{]}^2$

As, in neutral $\left[H^{+}\right]=\left[O{H}^{-}\right]$

$\left[H^{+}\right]=\sqrt{2.7\times {10}^{-14}}=1.643\times {10}^{-7}$

$pH=-log\left[H^{+}\right]=-log(1.643\times {10}^{-7})=6.78$