Question

The unit of ionic product of water ${K_w}$ is:
A. $Mo{l^{ - 1}}{L^{ - 1}}$
B. $Mo{l^2}{L^{ - 2}}$
C. $Mo{l^{ - 2}}{L^{ - 1}}$
D. $Mo{l^2}{L^2}$

Answer 1

Hint: We know that molecules of water can act as both acids and bases. Water molecule acting as a base accepts a hydrogen ion from the other molecule that acts as an acid. We know that pure water is a weak electrolyte and it undergoes autoprotolysis (or) self-ionization.

Complete step by step answer:
We know that water undergoes self-ionization to form hydronium ions and hydroxide ions. When collision happens between two molecules of water, and transfer of hydrogen ions from one molecule to another takes place. The products obtained are positively charged hydronium ion and negatively charged hydroxide ion. We can write the equation as,
${H_2}O\left( l \right)\overset {} \leftrightarrows {H^ + }\left( {aq} \right) + O{H^ - }\left( {aq} \right)$
The equilibrium constant for the self-ionization of water is known as the ionic product of water and it is represented by the symbol ${K_w}$.
${K_w} = \left[ {{H^ + }} \right]\left[ {O{H^ - }} \right]$
The mathematical product of the concentration of hydrogen ions and hydroxide ions is the ionic product of water. We do not consider water in the ionic product expression because it is a pure liquid.
At ${25^ \circ }C$ we can experimentally obtain the ionic product in pure water as $1.0 \times {10^{ - 14}}$.
The unit of concentration of hydrogen ion will be $\dfrac{{mol}}{L}$.
The unit of concentration of hydroxide ion will be $\dfrac{{mol}}{L}$.
Let us substitute the unit of concentrations of both hydrogen ion and hydroxide ion in the rate equation of ${K_w}$.
We get the unit of ${K_w}$ as,
${K_w} = \left[ {{H^ + }} \right]\left[ {O{H^ - }} \right]$
${K_w} = \left[ {\dfrac{{mol}}{L}} \right]\left[ {\dfrac{{mol}}{L}} \right]$
${K_w} = \dfrac{{mo{l^2}}}{{{L^2}}}$
${K_w} = mo{l^2}{L^{ - 2}}$
The unit of ionic product of water is $Mo{l^2}{L^{ - 2}}$.

$\therefore $ The correct option is (B).

Note:
We have to remember that, the ion-product constant for water
$\left( {{{\text{K}}_{\text{w}}}} \right)$ is calculated using the formula, ${K_w} = \left[ {{H_3}{O^ + }} \right]\left[ {O{H^ - }} \right]$
The value of ${{\text{K}}_{\text{w}}}$ is $1.0 \times {10^{ - 14}}$ for all aqueous solution at ${25^o}C$
For a solution to be acidic, $\left[ {{H_3}{O^ + }} \right] > \left[ {O{H^ - }} \right]$
For a solution to be basic, $\left[ {O{H^ - }} \right] > \left[ {{H_3}{O^ + }} \right]$
Example: Let us calculate the concentration of hydroxide ion from ionic product of water and concentration of the hydronium ion.
Given,
$\left[ {{H_3}{O^ + }} \right] = {10^{ - 3}}M$
${K_w} = 1.0 \times {10^{ - 14}}\,{M^2}$
The concentration of $O{H^ - }$ is calculated as,
$\left[ {O{H^ - }} \right] = \dfrac{{{K_w}}}{{\left[ {{H_3}{O^ + }} \right]}}$
$\left[ {O{H^ - }} \right] = \dfrac{{1.0 \times {{10}^{ - 14}}}}{{{{10}^{ - 3}}}}$
$\left[ {O{H^ - }} \right] = 1.0 \times {10^{ - 11}}M$
The value of $\left[ {O{H^ - }} \right]$ is ${10^{ - 11}}M.$
The solution is said to be acidic, since the concentration of ${H_3}{O^ + }$ is greater than the concentration of hydroxide ion.