Ionic Equilibrium
4.0 Ionization of Water
4.0 Ionization of Water
Pure water is weakly ionized to give ${H^ + }$ and $O{H^ - }$ ions. That is,
$${H_2}O{\text{ }} \rightleftharpoons {\text{ }}{{\text{H}}^{ + {\text{ }}}}{\text{ + O}}{{\text{H}}^ - }$$
The equilibrium constant is given as
$$K{\text{ = }}\frac{{\left[ {{H^ + }} \right]\left[ {O{H^ - }} \right]}}{{\left[ {{H_2}O} \right]}}{\text{ }}$$
$K$ has a value of $\frac{{{{10}^{ - 14}}}}{{55.5}}$ at ${25^ \circ }C$. The concentration of pure water is calculated as
$$1L{\text{ }} = {\text{ 1000 c}}{{\text{m}}^3}{\text{ = 1000 g (density assumed to be 1 g/cm)}}$$ which gives number of moles in $1L{\text{ }}$ of water equal to
$$\frac{{1000}}{{18}}{\text{ = }}55.5$$ Therefore,
$$\left[ {{H^ + }} \right]\left[ {O{H^ - }} \right]{\text{ }} = {\text{ }}K\left[ {{H_2}O} \right]{\text{ }}$$
$${K_{w{\text{ }}}} = {\text{ }}\frac{{{{10}^{ - 14}}}}{{55.5}}{\text{ }} \times {\text{ }}55.5{\text{ = 1}}{{\text{0}}^{ - 14}}$$
This value of ${\text{ 1}}{{\text{0}}^{ - 14}}$ is called ionic product of water and given by the symbol ${K_w}$. For pure water we have
$$\left[ {{H^ + }} \right]\left[ {O{H^ - }} \right]{\text{ = }}{10^{ - 14}}$$ or, $$\left[ {{H^ + }} \right]\left[ {{H^ + }} \right]{\text{ = }}{\left[ {{H^ + }} \right]^2}{\text{ = }}{10^{ - 14}}$$
Therefore, $$\left[ {{H^ + }} \right]{\text{ = }}{10^{ - 7}}$$
$pH$ of water is thus calculated as
$$pH{\text{ = - }}\log \left[ {{{10}^{ - 7}}} \right]{\text{ = 7}}$$ at ${25^ \circ }C$ and is defined as the neutral point of water at the same temperature.
Note: ${K_w}$ increases with increase in temperature. Accordingly, the neutral point of water also shifts to a value lower than $7$ with increase in temperature.
