Question

Ionic product of water is $1\times {10}^{-12}$. pH of water will be:

Answer 1

$H_{2}O\rightleftharpoons H^{+}+O{H}^{-}$

$C$

$C-x$ $x$ $x$

$K_w=\left[H^{+}\right]\left[O{H}^{-}\right]={10}^{-12}=x^2$

$x={10}^{-6}=\left[H^{+}\right]$

As $pH=-\log \left(\right[H^{+}\left]\right)=6$