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At \[90^\circ \], pure water has [\[{H_3}{O^ + }\]] = \[{\text{1}}{{\text{0}}^{ - 6}}\]m on the value of \[{K_w}\]at this temperature is :
A. \[{10^{ - 6}}\]
B \[{10^{ - 12}}\]
C. \[{10^{ - 13}}\]
D. \[{10^{ - 14}}\]

Last updated date: 20th Jun 2024
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Hint: In case of pure water the amount of $[{H^ + }]$ and $[O{H^ - }]$ is the same. Water is a weak electrolyte and the amount of dissociation in pure water is very low. Due to this the amount of non-dissociated water molecules remains more or less constant. The product of $[{H^ + }]$ and $[O{H^ - }]$ is called an ionic product of water. This is denoted as \[{{\text{K}}_w}\].

Complete step by step solution:
Now Given,
Concentration of \[{{\text{H}}_{\text{3}}}{{\text{O}}^{\text{ + }}}{\text{ = 1}}{{\text{0}}^{{\text{ - 6}}}}{\text{mol/litre}}\]
The ionization reaction of water is,
\[{{\text{H}}_{\text{2}}}{\text{O}} \rightleftharpoons {{\text{H}}^{\text{ + }}}{\text{ + O}}{{\text{H}}^{\text{ - }}}\]
The expression for equilibrium constant is,
\[{\text{K = }}\dfrac{{\left[ {{{\text{H}}^{\text{ + }}}} \right]{\text{ }}\left[ {{\text{O}}{{\text{H}}^{\text{ - }}}} \right]}}{{\left[ {{{\text{H}}_{\text{2}}}{\text{O}}} \right]}}\]
After rearranging thisformula, we get,
\[{\text{K}} \times \left[ {{{\text{H}}_{\text{2}}}{\text{O}}} \right]{\text{ = }}\left[ {{{\text{H}}^{\text{ + }}}} \right]{\text{ }}\left[ {{\text{O}}{{\text{H}}^{\text{ - }}}} \right]\]
Due to the large amount of \[{{\text{H}}_{\text{2}}}{\text{O}}\] is large, that is why it can be taken as constant .
\[{{\text{K}}_w}{\text{ = }}\left[ {{{\text{H}}^{\text{ + }}}} \right]{\text{ }}\left[ {{\text{O}}{{\text{H}}^{\text{ - }}}} \right]\]
Where, \[{{\text{K}}_w}\] is the equilibrium constant.
Pure water means the concentration of hydronium ion and hydroxide ion are equal.
\[{\text{ }}\left[ {{{\text{H}}^{\text{ + }}}} \right]{\text{ = }}\left[ {{\text{O}}{{\text{H}}^{\text{ - }}}} \right]{\text{ = 1}}{{\text{0}}^{{\text{ - 6}}}}{\text{moles/litre}}\]
Now the values in the equation, \[{{\text{K}}_w}{\text{ = }}\left[ {{{\text{H}}^{\text{ + }}}} \right]{\text{ }}\left[ {{\text{O}}{{\text{H}}^{\text{ - }}}} \right]\]
  {K_{_W}} = {10^{ - 6}} \times {10^{ - 6}} \\
  {K_W} = {10^{ - 12}} \\

So, the correct option is B.

When a partly soluble salt dissolves into water it forms a dynamic equilibrium between the hydrated ions and its solid salt molecule. Due to this equilibrium the amount of solid salt is hydrolyzed remains unchanged after achieving an amount of hydration. Due to this a new term is introduced which is solubility product \[\left( {{{\text{K}}_{{\text{sp}}}}} \right)\].With increasing the temperature solubility of the salt increases as well as the concentrations of solvated ions. Therefore, with increasing temperature the value of solubility product increases and vice-versa.