Question

Calculate changes in the concentration of ion in one litre of water, when temperature changes from $\text{ 298 K }$ to $\text{ 310 K }$. Given $\text{ }{{\text{K}}_{\text{w}}}\text{(298 K) = 1}{{\text{0}}^{-14}}\text{ }$and$\text{ }{{\text{K}}_{\text{w}}}\text{(300 K) = 2}\text{.56}\times \text{1}{{\text{0}}^{-14}}\text{ }$.

Answer 1

Water is dissociated to a very small extent into hydrogen and hydroxyl ions, as represented by the equation,
$\text{ }{{\text{H}}_{\text{2}}}\text{O (}l)\text{ }\rightleftharpoons \text{ }{{\text{H}}^{\text{+}}}\text{(}aq)\text{ + O}{{\text{H}}^{-}}(aq)\text{ }$
The dissociation constant for the water $\text{ }{{\text{K}}_{\text{w}}}\text{ }$ is written as the product of the concentration of hydroxyl ion and hydrogen ion.It is as written as below,
$\text{ }{{\text{K}}_{\text{w}}}\text{ }=\text{ }\left[ {{\text{H}}^{+}} \right]\left[ \text{O}{{\text{H}}^{-}} \right]\text{ }$
In pure water, the concentration of hydrogen and hydroxyl ion must be equal to one .Thus,
$\text{ }\left[ {{\text{H}}^{+}} \right]=\left[ \text{O}{{\text{H}}^{-}} \right]\text{ }$

Complete step by step answer:
We are provided with the following data:
Volume of solution is 1 litre
Temperature of the solution changes from $\text{ }{{\text{T}}_{\text{1}}}\text{ = 298 K }$ to$\text{ }{{\text{T}}_{2}}\text{ = 300 K }$.
Dissociation constant for the water $\text{ }{{\text{K}}_{\text{w}}}\text{(298 K) = 1}{{\text{0}}^{-14}}\text{ }$ and $\text{ }{{\text{K}}_{\text{w}}}\text{(300 K) = 2}\text{.56}\times \text{1}{{\text{0}}^{-14}}\text{ }$
We have to calculate the change in the hydrogen ion $\text{ }{{\text{H}}^{\text{+}}}\text{ }$ concentration in the given volume of solution.
The dissociation constant for the water $\text{ }{{\text{K}}_{\text{w}}}\text{ }$ is written as the product of the concentration of hydroxyl ion and hydrogen ion.It is as written as below,
$\text{ }{{\text{K}}_{\text{w}}}\text{ }=\text{ }\left[ {{\text{H}}^{+}} \right]\left[ \text{O}{{\text{H}}^{-}} \right]\text{ }$
In pure water, the concentration of hydrogen and hydroxyl ion must be equal to one .Thus,
$\text{ }\left[ {{\text{H}}^{+}} \right]=\left[ \text{O}{{\text{H}}^{-}} \right]\text{ }$
a) At $\text{ 300 K }$ temperature:
The hydrogen ion concentration is calculated from the ionic product of the water. The ionic product of the water at $\text{ 298 K }$is equal to $\text{1}{{\text{0}}^{-14}}\text{ }$.Thus, from ionic product of water the concentration of hydrogen ion is as follows,
$\begin{aligned} & \text{ }{{\text{K}}_{\text{w}}}\text{ }=\text{ }\left[ {{\text{H}}^{+}} \right]_{(298\text{K)}}^{2}\text{ } \\\ & \Rightarrow {{\left[ {{\text{H}}^{+}} \right]}_{(298\text{K)}}}\text{ = }\sqrt{{{\text{K}}_{\text{w}}}}\text{ = }\sqrt{{{10}^{-14}}}\text{ } \\\ & \therefore {{\left[ {{\text{H}}^{+}} \right]}_{(298\text{K)}}}\text{ = }{{10}^{-7}}\text{ } \\\ \end{aligned}$
Thus, the hydrogen ion concentration at the $\text{ 298 K }$is equal to $\text{ 1}{{\text{0}}^{-7\text{ }}}$ .
b) At $\text{ 310 K }$ temperature:
The hydrogen ion concentration is calculated from the ionic product of the water. The ionic product of the water at $\text{ 310 K }$is equal to $\text{ 2}\text{.56 }\times \text{ 1}{{\text{0}}^{-14}}\text{ }$.Thus, from ionic product of water the concentration of hydrogen ion is as follows,
$\begin{aligned} & \text{ }{{\text{K}}_{\text{w}}}\text{ }=\text{ }\left[ {{\text{H}}^{+}} \right]_{(310\text{K)}}^{2}\text{ } \\\ & \Rightarrow {{\left[ {{\text{H}}^{+}} \right]}_{(310\text{K)}}}\text{ = }\sqrt{{{\text{K}}_{\text{w}}}}\text{ = }\sqrt{2.56\times {{10}^{-14}}}\text{ } \\\ & \therefore {{\left[ {{\text{H}}^{+}} \right]}_{(310\text{K)}}}\text{ = 1}\text{.6}\times {{10}^{-7}}\text{ } \\\ \end{aligned}$
Thus, the hydrogen ion concentration at the $\text{ 310 K }$is equal to $\text{ 1}\text{.6}\times {{10}^{-7}}\text{ }$ .
So, change in the hydrogen ion concentration at the $\text{ 310 K }$and $\text{ 298 K }$would be,
$\text{ }{{\left[ {{\text{H}}^{+}} \right]}_{(310\text{K)}}}-{{\left[ {{\text{H}}^{+}} \right]}_{(298\text{K)}}}\text{ = }\left( \text{1}\text{.6}\times {{10}^{-7}} \right)-\left( 1.0\times {{10}^{-7}} \right)\text{ = 0}\text{.6}\times {{10}^{-7}}\text{ }$

Therefore change in concentration of hydrogen ion is equal to $\text{ 0}\text{.6}\times {{10}^{-7}}\text{ }$.

Note: Note that, pure water means a water which is free from any kind of impurities. For pure water that ionic product or the dissociation constant of water is equal to $\text{ 1}{{\text{0}}^{-14}}\text{ }$. However, with increase in temperature more number of water molecules undergo the dissociation into its corresponding hydroxyl and hydrogen ion thus the dissociation constant increases. We can say dissociation of water is directly related to the temperature.