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Hint: The ionic product of water is the product of concentrations of ${{\text{H}}^{\text{ + }}}$ and ${\text{O}}{{\text{H}}^ - }$ ions in water at a particular temperature. It is denoted by ${{\text{K}}_{\text{w}}}$
Step-by-Step Explanation: Water is a weak electrolyte and ionizes only to a small extent. ‘The ionic product of the water is the product of concentrations of ions in which it is dissociated at a particular temperature’. The water is dissociated in the following manner-
$ \Rightarrow {{\text{H}}_2}{\text{O}} \rightleftharpoons {{\text{H}}^{\text{ + }}} + {\text{O}}{{\text{H}}^ - }$
Then ${{\text{K}}_{\text{w}}}$=$\dfrac{{\left[ {{{\text{H}}^{\text{ + }}}} \right]\left[ {{\text{O}}{{\text{H}}^ - }} \right]}}{{\left[ {{{\text{H}}_2}{\text{O}}} \right]}}$
$ \Rightarrow {{\text{K}}_{\text{w}}}\left[ {{{\text{H}}_2}{\text{O}}} \right] = \left[ {{{\text{H}}^{\text{ + }}}} \right]\left[ {{\text{O}}{{\text{H}}^ - }} \right]$
Because $\left[ {{{\text{H}}_2}{\text{O}}} \right] = 1$ then,
$ \Rightarrow {{\text{K}}_{\text{w}}} = \left[ {{{\text{H}}^{\text{ + }}}} \right]\left[ {{\text{O}}{{\text{H}}^ - }} \right]$
Ionic products are temperature dependent because the Ionic product is directly proportional to the temperature.
By Le-Chatelier’s principle, as the temperature increases, the position of equilibrium shifts to the right of the equation to minimize the effect of increased temperature.
This means with the increase in temperature the concentration of$\left[ {{{\text{H}}^{\text{ + }}}} \right]$ and$\left[ {{\text{O}}{{\text{H}}^ - }} \right]$ ions will increase which will result in increased value of${{\text{K}}_{\text{w}}}$ as forward reaction is favoured.
This means that the ionic product depends only on temperature and is unaffected by any other changes.
Hence correct answer is ‘D’.
Note: Since concentration of hydrogen ion increases when temperature is increased and we know that pH is inversely proportional to concentration of hydrogen ion then the pH of water will decrease with increase in temperature. Also,
When,$\left[ {{{\text{H}}^{\text{ + }}}} \right] = \left[ {{\text{O}}{{\text{H}}^ - }} \right]$ then the solution is neutral [pure water is neutral].
When $\left[ {{{\text{H}}^{\text{ + }}}} \right] > \left[ {{\text{O}}{{\text{H}}^ - }} \right]$ then the solution becomes acidic.
When $\left[ {{{\text{H}}^{\text{ + }}}} \right] < \left[ {{\text{O}}{{\text{H}}^ - }} \right]$ then the solution becomes basic.
Step-by-Step Explanation: Water is a weak electrolyte and ionizes only to a small extent. ‘The ionic product of the water is the product of concentrations of ions in which it is dissociated at a particular temperature’. The water is dissociated in the following manner-
$ \Rightarrow {{\text{H}}_2}{\text{O}} \rightleftharpoons {{\text{H}}^{\text{ + }}} + {\text{O}}{{\text{H}}^ - }$
Then ${{\text{K}}_{\text{w}}}$=$\dfrac{{\left[ {{{\text{H}}^{\text{ + }}}} \right]\left[ {{\text{O}}{{\text{H}}^ - }} \right]}}{{\left[ {{{\text{H}}_2}{\text{O}}} \right]}}$
$ \Rightarrow {{\text{K}}_{\text{w}}}\left[ {{{\text{H}}_2}{\text{O}}} \right] = \left[ {{{\text{H}}^{\text{ + }}}} \right]\left[ {{\text{O}}{{\text{H}}^ - }} \right]$
Because $\left[ {{{\text{H}}_2}{\text{O}}} \right] = 1$ then,
$ \Rightarrow {{\text{K}}_{\text{w}}} = \left[ {{{\text{H}}^{\text{ + }}}} \right]\left[ {{\text{O}}{{\text{H}}^ - }} \right]$
Ionic products are temperature dependent because the Ionic product is directly proportional to the temperature.
By Le-Chatelier’s principle, as the temperature increases, the position of equilibrium shifts to the right of the equation to minimize the effect of increased temperature.
This means with the increase in temperature the concentration of$\left[ {{{\text{H}}^{\text{ + }}}} \right]$ and$\left[ {{\text{O}}{{\text{H}}^ - }} \right]$ ions will increase which will result in increased value of${{\text{K}}_{\text{w}}}$ as forward reaction is favoured.
This means that the ionic product depends only on temperature and is unaffected by any other changes.
Hence correct answer is ‘D’.
Note: Since concentration of hydrogen ion increases when temperature is increased and we know that pH is inversely proportional to concentration of hydrogen ion then the pH of water will decrease with increase in temperature. Also,
When,$\left[ {{{\text{H}}^{\text{ + }}}} \right] = \left[ {{\text{O}}{{\text{H}}^ - }} \right]$ then the solution is neutral [pure water is neutral].
When $\left[ {{{\text{H}}^{\text{ + }}}} \right] > \left[ {{\text{O}}{{\text{H}}^ - }} \right]$ then the solution becomes acidic.
When $\left[ {{{\text{H}}^{\text{ + }}}} \right] < \left[ {{\text{O}}{{\text{H}}^ - }} \right]$ then the solution becomes basic.
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