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Question: The unit of ionic product of water \({K_w}\) is: A. \(Mo{l^{ - 1}}{L^{ - 1}}\) B. \(Mo{l^2}{L^{ ...

The unit of ionic product of water Kw{K_w} is:
A. Mol−1L−1Mo{l^{ - 1}}{L^{ - 1}}
B. Mol2L−2Mo{l^2}{L^{ - 2}}
C. Mol−2L−1Mo{l^{ - 2}}{L^{ - 1}}
D. Mol2L2Mo{l^2}{L^2}

Explanation

Solution

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,
H2O(l)⇆H+(aq)+OH−(aq){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 Kw{K_w}.
Kw=[H+][OH−]{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∘C{25^ \circ }C we can experimentally obtain the ionic product in pure water as 1.0×10−141.0 \times {10^{ - 14}}.
The unit of concentration of hydrogen ion will be molL\dfrac{{mol}}{L}.
The unit of concentration of hydroxide ion will be molL\dfrac{{mol}}{L}.
Let us substitute the unit of concentrations of both hydrogen ion and hydroxide ion in the rate equation of Kw{K_w}.
We get the unit of Kw{K_w} as,
Kw=[H+][OH−]{K_w} = \left[ {{H^ + }} \right]\left[ {O{H^ - }} \right]
Kw=[molL][molL]{K_w} = \left[ {\dfrac{{mol}}{L}} \right]\left[ {\dfrac{{mol}}{L}} \right]
Kw=mol2L2{K_w} = \dfrac{{mo{l^2}}}{{{L^2}}}
Kw=mol2L−2{K_w} = mo{l^2}{L^{ - 2}}
The unit of ionic product of water is Mol2L−2Mo{l^2}{L^{ - 2}}.

∴\therefore The correct option is (B).

Note:
We have to remember that, the ion-product constant for water
(Kw)\left( {{{\text{K}}_{\text{w}}}} \right) is calculated using the formula, Kw=[H3O+][OH−]{K_w} = \left[ {{H_3}{O^ + }} \right]\left[ {O{H^ - }} \right]
The value of Kw{{\text{K}}_{\text{w}}} is 1.0×10−141.0 \times {10^{ - 14}} for all aqueous solution at 25oC{25^o}C
For a solution to be acidic, [H3O+]>[OH−]\left[ {{H_3}{O^ + }} \right] > \left[ {O{H^ - }} \right]
For a solution to be basic, [OH−]>[H3O+]\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,
[H3O+]=10−3M\left[ {{H_3}{O^ + }} \right] = {10^{ - 3}}M
Kw=1.0×10−14 M2{K_w} = 1.0 \times {10^{ - 14}}\,{M^2}
The concentration of OH−O{H^ - } is calculated as,
[OH−]=Kw[H3O+]\left[ {O{H^ - }} \right] = \dfrac{{{K_w}}}{{\left[ {{H_3}{O^ + }} \right]}}
[OH−]=1.0×10−1410−3\left[ {O{H^ - }} \right] = \dfrac{{1.0 \times {{10}^{ - 14}}}}{{{{10}^{ - 3}}}}
[OH−]=1.0×10−11M\left[ {O{H^ - }} \right] = 1.0 \times {10^{ - 11}}M
The value of [OH−]\left[ {O{H^ - }} \right] is 10−11M.{10^{ - 11}}M.
The solution is said to be acidic, since the concentration of H3O+{H_3}{O^ + } is greater than the concentration of hydroxide ion.