Question

At ${80^o}C$, distilled water has \[\left[ {{H_3}{O^ + }} \right]\] concentration equal to $1 \times {10^{ - 6}}$ mol/litre. The value of ${K_w}$ at this temperature will be:
A) $1 \times {10^{ - 6}}$
B) $1 \times {10^{ - 12}}$
C) $1 \times {10^{ - 9}}$
D) $1 \times {10^{ - 15}}$

Answer 1

Hint: We are given distilled water and you must know that distilled water is a type of pure water and in pure water, concentration of ${H^ + }$ or ${H_3}{O^ + }$ ions equals the concentration of $O{H^ - }$ ions produced. Formula to calculate the value of ${K_w}$ at any temperature is:
${K_w} = \left[ {{H^ + }} \right]\left[ {O{H^ - }} \right]$

Complete step by step answer:
We are given that at ${80^o}C$, distilled water has \[\left[ {{H_3}{O^ + }} \right]\] concentration equal to $1 \times {10^{ - 6}}$ mol/litre. Distilled water contains same concentration of hydronium ions i.e.,$\left[ {{H^ + }} \right]$ and hydroxide ions i.e., $\left[ {O{H^ - }} \right]$.
Thus, we can write $\left[ {{H^ + }} \right] = \left[ {O{H^ - }} \right]$

Now, calculation for ${K_w}$:
To calculate ${K_w}$, let us first write the ionization reaction of water. The chemical reaction is as follows:
${H_2}O \rightleftharpoons {H^ + } + O{H^ - }$

The equilibrium constant for this reaction is called the water dissociation constant, ${K_w}$
${K_w} = \left[ {{H^ + }} \right]\left[ {O{H^ - }} \right]$
But, $\left[ {O{H^ - }} \right] = \left[ {{H^ + }} \right]$, because we are given distilled water. This means that we can replace concentration of hydroxide ions with hydronium ions.
Also, given that \[\left[ {{H_3}{O^ + }} \right] \equiv \left[ {{H^ + }} \right]\]= $1 \times {10^{ - 6}}$ mol/litre
Therefore, equation of ${K_w}$ would be:
${K_w} = \left[ {{H^ + }} \right]\left[ {{H^ + }} \right]$
Or, ${K_w} = {\left[ {{H^ + }} \right]^2} = {(1 \times {10^{ - 6}})^2}$
Hence, ${K_w} = 1 \times {10^{^{ - 12}}}mo{l^2}{L^{ - 2}}$
So, the correct answer is “Option B”.

Additional Information: The concentration of hydronium ions helps us to determine the pH value of a solution. The concentration of hydroxide ions determines the pOH value of a solution. Molecules of pure water auto dissociates (i.e., react with each other) into hydroxide and hydronium ions. The ionization reaction of water is an equilibrium reaction.

Note: One should know that distilled water is one type of purified water. Since pure water or distilled water contains equal numbers of hydronium and hydroxide ions, it is a neutral solution. Thus, pure water has a pH of 7 and also pOH of 7. Dissociation constant for water i.e., ${K_w}$ is also known as the ionic product for water. At room temperature, the value of ${K_w}$ is $1 \times {10^{ - 14}}mo{l^2}d{m^{ - 6}}$.