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Question: \({{\text{K}}_{\text{w}}}\) for water at \({25^ \circ }{\text{C}}\) is equal to \({10^{ - 14}}\). Wh...

Kw{{\text{K}}_{\text{w}}} for water at 25∘C{25^ \circ }{\text{C}} is equal to 10−14{10^{ - 14}}. What is its value at 90∘C{90^ \circ }{\text{C}}-
(A) 10−15{10^{ - 15}}
(B) 10−17{10^{ - 17}}
(C) 10−14{10^{ - 14}}
(D) 10−12{10^{ - 12}}

Explanation

Solution

We are given Kw{{\text{K}}_{\text{w}}} which is the ionic product of water. We know that the ionic product of water is the dissociation of a molecule of water into hydrogen ions and hydroxyl ions. As the temperature increases the number of hydrogen ions and hydroxyl ions increases.

Complete Answer:
We know that water is a weak electrolyte. The dissociation reaction of water is as follows:
H2O⇌H++OH−{{\text{H}}_{\text{2}}}{\text{O}} \rightleftharpoons {{\text{H}}^ + } + {\text{O}}{{\text{H}}^ - }
Apply the law of chemical equilibrium to the above reaction. Thus,
K=[H+][OH−][H2O]{\text{K}} = \dfrac{{[{{\text{H}}^ + }][{\text{O}}{{\text{H}}^ - }]}}{{[{{\text{H}}_2}{\text{O}}]}}
Where K{\text{K}} is the dissociation constant of water.
K[H2O]=[H+][OH−]{\text{K}}[{{\text{H}}_2}{\text{O}}] = [{{\text{H}}^ + }][{\text{O}}{{\text{H}}^ - }]
Kw=[H+][OH−]{{\text{K}}_{\text{w}}} = [{{\text{H}}^ + }][{\text{O}}{{\text{H}}^ - }] …… K[H2O]=Kw{\text{K}}[{{\text{H}}_2}{\text{O}}] = {{\text{K}}_{\text{w}}}
Where Kw{{\text{K}}_{\text{w}}} is the ionic product of water.
Thus, the ionic product of water is the product of molar concentration of hydrogen ions and the molar concentration of hydroxyl ions.
We are given that Kw{{\text{K}}_{\text{w}}} for water at 25∘C{25^ \circ }{\text{C}} is equal to 10−14{10^{ - 14}}.
The degree of dissociation of water increases with an increase in temperature. As the temperature increases, the molar concentration of hydrogen ions and hydroxyl ions increases. Thus, the ionic product of water increases with increase in temperature.
Thus, as the temperature increases, the value of the ionic product of water increases.
We are given four values 10−15{10^{ - 15}}, 10−17{10^{ - 17}}, 10−14{10^{ - 14}} and 10−12{10^{ - 12}}.
The value greater than 10−14{10^{ - 14}} is 10−12{10^{ - 12}}. Thus, at 90∘C{90^ \circ }{\text{C}}, the Kw{{\text{K}}_{\text{w}}} for water is 10−12{10^{ - 12}}.

Thus, the correct option is (D) 10−12{10^{ - 12}}.

Note: While writing the expression for the ionic product of water, the molar concentration of water molecules is not included. Water undergoes self-ionization. Thus, water can act as acid as well as base.
When an acid is added to water the ionic product of water remains constant but the concentration of hydrogen ion increases. Thus, the concentration of hydrogen ions is higher than the concentration of the hydroxyl ions.
When a base is added to water the ionic product of water remains constant but the concentration of hydroxyl ion increases. Thus, the concentration of hydroxyl ions is higher than the concentration of the hydrogen ions.