Question

The degree of dissociation of water $1.8\text{ }\times \text{ 1}{{\text{0}}^{-9}}$ at 298K. Calculate the ionisation constant and ionic product of water at 298K.

Hint: Concentration of the water is defined as the ratio of moles of water per litre. Ionisation constant which is denoted by ${{\text{K}}_{a}}$ is the product of the degree of dissociation () and concentration of the water. Water cannot ionise completely because it is a weak electrolyte.

-Firstly, we have to calculate the concentration of water i.e no. of a mole per litre. Here, we know that the number of moles of water is 1000/18 as the molar mass of the water is 18 $(1\text{ }\times \text{ 2) + 16 = 18}$ and the mass of water is 1000g.
-So, water has the concentration = $\dfrac{1000}{18}=\text{ 55}\text{.66 mol }{{\text{L}}^{-1}}$.
-Now, we that a small amount is dissociated and give hydrogen ion and hydroxyl ion i.e.
${{\text{H}}_{2}}\text{O }\to \text{ }{{\text{H}}^{+}}\text{ + O}{{\text{H}}^{-}}$
-Here, hydrogen ion and hydroxyl ion are yielded in equal concentration. So, the ionic product of the water (${{\text{K}}_{\text{W}}}$) will be:
${{\text{K}}_{\text{W}}}\text{ = }\left( {{\text{H}}^{+}} \right)\left( \text{O}{{\text{H}}^{-}} \right)\text{ = }{{\text{C}}^{2}}{{\alpha }^{2}}$
-Here, C is the concentration of water and alpha is the degree of dissociation of water which is given in the question i.e. $1.8\text{ }\times \text{ 1}{{\text{0}}^{-9}}$
\begin{align} & {{\text{K}}_{\text{W}}}\text{ = (55}\text{.56}{{\text{)}}^{2}}\text{ }\times \text{ (1}\text{.8 }\times \text{ 1}{{\text{0}}^{-9}}{{)}^{2}} \\ & \text{ = 1 }\times \text{ 1}{{\text{0}}^{-14}} \\ \end{align}
-So, the ionic product of the water is $\text{1}{{\text{0}}^{-14}}$.
-Now, we have to calculate the ionic constant of the water, which is represented by ${{\text{K}}_{\text{a}}}$, it will be
${{\text{K}}_{\text{a}}}\text{ = }\dfrac{({{\text{H}}^{+}})(\text{O}{{\text{H}}^{-}})}{({{\text{H}}_{2}}\text{O)}}$
=$\dfrac{{{\text{C}}^{2}}{{\alpha }^{2}}}{\text{C}\alpha }$ here, the value of alpha of water is so much smaller than 1, so it can be neglected. The equation will now become:
$\text{C}{{\alpha }^{2}}\text{ = 55}\text{.56 }\times \text{ (1}\text{.8 }\times \text{ 1}{{\text{0}}^{-9}}{{)}^{2}}$ $\text{= 1}\text{.8 }\times \text{ 1}{{\text{0}}^{-16}}$.
Therefore, the value of the ionic constant and ionic product of water is $\text{1}{{\text{0}}^{-14}}$ and $\text{1}\text{.8 }\times \text{ 1}{{\text{0}}^{-16}}$ respectively.

Note: When the product of hydrogen ion and hydroxyl ion is constant at a constant temperature then it is known as the ionic product of water. Ionisation is the process in which the compound is broken down into its respective ion that is positive and negative.