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Question: Given,\({K_w}\) for water at \(25^\circ {\rm{C}}\) is equal to \({10^{ - 14}}\). What is its value a...

Given,Kw{K_w} for water at 25∘C25^\circ {\rm{C}} is equal to 10−14{10^{ - 14}}. What is its value at 90∘C90^\circ {\rm{C}}?
(A) 10−15{10^{ - 15}}
(B)10−17{10^{ - 17}}
(C) 10−14{10^{ - 14}}
(D) 10−12{10^{ - 12}}

Explanation

Solution

We know that Kw{K_w} is the ionic product of water whose value is 10−14{10^{ - 14}} at 25∘C25^\circ {\rm{C}}. The Value of the ionic product of water is dependent on the temperature. If the temperature increases, the value of the ionic product of water also increases.

Complete step by step answer:
Let’s understand the ionization of water in detail.
We know that pure water is a very weak electrolyte. Due to this property it undergoes self ionization to a small extent as shown below:
H2O(l)+H2O(l)⇌H3O+(aq)+OH−(aq)H_{2}O\left ( l \right )+H_{2}O\left ( l \right )\rightleftharpoons H_{3}O^{+}\left ( aq \right )+OH^{-}\left ( aq \right )
Now, applying law of chemical equilibrium to the above equation,
K=[H3O+(aq)][OH−(aq)][H2O(l)]2K = \dfrac{{\left[ {{{\rm{H}}_{\rm{3}}}{{\rm{O}}^ + }\left( {aq} \right)} \right]\left[ {{\rm{O}}{{\rm{H}}^ - }\left( {aq} \right)} \right]}}{{{{\left[ {{{\rm{H}}_{\rm{2}}}{\rm{O}}\left( l \right)} \right]}^2}}}
As water is ionized to a very small extent, therefore, [H2O(l)]\left[ {{{\rm{H}}_{\rm{2}}}{\rm{O}}\left( l \right)} \right] may be supposed to remain constant. Therefore,
K[H2O(l)]2=[H3O+(aq)][OH−(aq)]K{\left[ {{{\rm{H}}_{\rm{2}}}{\rm{O}}\left( l \right)} \right]^2} = \left[ {{{\rm{H}}_{\rm{3}}}{{\rm{O}}^ + }\left( {aq} \right)} \right]\left[ {{\rm{O}}{{\rm{H}}^ - }\left( {aq} \right)} \right]
Or
Kw=[H3O+(aq)][OH−(aq)]{K_w} = \left[ {{{\rm{H}}_{\rm{3}}}{{\rm{O}}^ + }\left( {aq} \right)} \right]\left[ {{\rm{O}}{{\rm{H}}^ - }\left( {aq} \right)} \right]
So, the Kw{K_w} is the ionic product of water.
At room temperature (298 K or 25∘C25^\circ {\rm{C}}), the value of Kw{K_w} is 1.0×10−14 mol2L−21.0 \times {10^{ - 14}}\,{\rm{mo}}{{\rm{l}}^2}{{\rm{L}}^{ - 2}}
Due to the neutral nature of water, we can say that the concentration of hydronium ion is equal to the hydroxide ion.
[H3O+(aq)]=[OH−(aq)]\left[ {{{\rm{H}}_{\rm{3}}}{{\rm{O}}^ + }\left( {aq} \right)} \right] = \left[ {{\rm{O}}{{\rm{H}}^ - }\left( {aq} \right)} \right]
Therefore, we can say that concentration of both hydroxide and hydroxide ion is 1.0×10−7 M1.0 \times {10^{ - 7}}\,{\rm{M}}.
Let’s discuss the variation of the value of the ionic product of water with respect to temperature. The value of the ionic product of water increases with respect to temperature. The reason is due to the increasing temperature, ionization of water increases.
Now come to the question. Here, we have to identify the value of Kw{K_w} at 90∘ C90^\circ \,{\rm{C}}. The increase of temperature indicates the increase of temperature. So, we have to identify the value which is greater than 10−14{10^{ - 14}}.
So, Kw{K_w} at 90∘ C90^\circ \,{\rm{C}} is 10−12{10^{ - 12}}.
Hence, the correct choice is option D.

Note:
The increase of the value of Kw{K_w} on increasing temperature is explained by Le Chatelier’s principle. We know that the formation of hydroxide ion and hydrogen ion from water is an endothermic process (absorbs heat). According to Le Chatelier’s principle, increase in temperature causes a shift of equilibrium to lower temperature. It means forward reaction is favored and more production of hydrogen and hydroxide ions. So, an increase in the value of the ionic product of water takes place.