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Question: At \({80^o}C\), distilled water has \[\left[ {{H_3}{O^ + }} \right]\] concentration equal to \(1 \ti...

At 80oC{80^o}C, distilled water has [H3O+]\left[ {{H_3}{O^ + }} \right] concentration equal to 1×10−61 \times {10^{ - 6}} mol/litre. The value of Kw{K_w} at this temperature will be:
A) 1×10−61 \times {10^{ - 6}}
B) 1×10−121 \times {10^{ - 12}}
C) 1×10−91 \times {10^{ - 9}}
D) 1×10−151 \times {10^{ - 15}}

Explanation

Solution

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+{H^ + } or H3O+{H_3}{O^ + } ions equals the concentration of OH−O{H^ - } ions produced. Formula to calculate the value of Kw{K_w} at any temperature is:
Kw=[H+][OH−]{K_w} = \left[ {{H^ + }} \right]\left[ {O{H^ - }} \right]

Complete step by step answer:
We are given that at 80oC{80^o}C, distilled water has [H3O+]\left[ {{H_3}{O^ + }} \right] concentration equal to 1×10−61 \times {10^{ - 6}} mol/litre. Distilled water contains same concentration of hydronium ions i.e.,[H+]\left[ {{H^ + }} \right] and hydroxide ions i.e., [OH−]\left[ {O{H^ - }} \right].
Thus, we can write [H+]=[OH−]\left[ {{H^ + }} \right] = \left[ {O{H^ - }} \right]

Now, calculation for Kw{K_w}:
To calculate Kw{K_w}, let us first write the ionization reaction of water. The chemical reaction is as follows:
H2O⇌H++OH−{H_2}O \rightleftharpoons {H^ + } + O{H^ - }

The equilibrium constant for this reaction is called the water dissociation constant, Kw{K_w}
Kw=[H+][OH−]{K_w} = \left[ {{H^ + }} \right]\left[ {O{H^ - }} \right]
But, [OH−]=[H+]\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 [H3O+]≡[H+]\left[ {{H_3}{O^ + }} \right] \equiv \left[ {{H^ + }} \right]= 1×10−61 \times {10^{ - 6}} mol/litre
Therefore, equation of Kw{K_w} would be:
Kw=[H+][H+]{K_w} = \left[ {{H^ + }} \right]\left[ {{H^ + }} \right]
Or, Kw=[H+]2=(1×10−6)2{K_w} = {\left[ {{H^ + }} \right]^2} = {(1 \times {10^{ - 6}})^2}
Hence, Kw=1×10−12mol2L−2{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., Kw{K_w} is also known as the ionic product for water. At room temperature, the value of Kw{K_w} is 1×10−14mol2dm−61 \times {10^{ - 14}}mo{l^2}d{m^{ - 6}}.