Question

Percentage ionisation of water as follows at certain temperature is $3.6\times {10}^{-7}$. Calculate $K_w$ and $pH$ of water this temperature. $2H_{2}O\rightleftharpoons H_3{O}^{+}+O{H}^{-}$.

Answer 1

Actual ionization of water at certain temperature = Percentage ionisation/100 $=3.6\times {10}^{-7}/100=3.6\times {10}^{-9}$

Ionization of water occurs as follows

$2H_{2}O\to O{H}^{-}+H_3{O}^{+}$

Initial moles 1 0 0

Now

$K_w=\left[O{H}^{-}\right]\left[H_3{O}^{+}\right]$

$K_w=3.6\times {10}^{-9}\times 3.6\times {10}^{-9}$

$K_w=12.96\times {10}^{-18}$

And

$pH=log\left[H_3{O}^{+}\right]$

$pH=log3.6\times {10}^{-9}$

$pH=8.4$