Question

The autoionization constant of water $\mathbf{\left(}{\mathbf{K}}_{\mathbf{w}}\mathbf{\right)}$isrole="math" localid="1663740263684" $\mathbf{1}\mathbf{.}\mathbf{139}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{15}}$at $\mathbf{0}\mathbf{°}\mathit{C}$and$\mathbf{9}\mathbf{.}\mathbf{614}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{14}}$at$\mathbf{60}\mathbf{°}\mathbf{C}$.

(a) Calculate the standard enthalpy of the autoionization of water.

$\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{\to }{\mathbf{H}}_{\mathbf{3}}{\mathbf{O}}^{\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{O}{\mathbf{H}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}$

(b) Calculate the standard entropy of the autoionization of water.
(c) At what temperature will the pH of pure water be 7.00, from these data?

Short answer

1. The standard enthalpy of autoionization of water is$\mathrm{\Delta H}°=55876.38\raisebox{1ex}{${\displaystyle \mathrm{J}}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.$
2. The standard entropy of autoionization of water is$\mathrm{\Delta S}°=-81.4\raisebox{1ex}{$\mathrm{J}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{K}$}\right.\mathrm{mol}$
3. The temperature of pure water is$\mathrm{T}=299.37\mathrm{K}$

Step-by-step solution

(a): The standard enthalpy.

The van' Hoff equation can be used to compute the standard enthalpy of autoionization of water

$\ln \left(\frac{{\mathrm{K}}_2}{{\mathrm{K}}_1}\right)=\frac{-\mathrm{\Delta H}°}{\mathrm{R}}\times \left(\frac{1}{{\mathrm{T}}_2}-\frac{1}{{\mathrm{T}}_1}\right)$

Therefore, equating the values:

$$\begin{gathered} \ln \left(\frac{9.614\times {10}^{-14}}{1.139\times {10}^{-15}}\right)=\frac{-\mathrm{\Delta H}°}{8.314\mathrm{K}/\mathrm{J}\times \mathrm{mol}}\times \left(\frac{1}{333}\mathrm{K}-\frac{1}{273}\mathrm{K}\right) \\ 4.4357=\frac{-\mathrm{\Delta H}°}{8.314\mathrm{K}/\mathrm{J}\times \mathrm{mol}}\times \left(-6.6\times {10}^{-4}\mathrm{K}\right) \\ \mathrm{\Delta H}°=\frac{4.4357\times 8.314\mathrm{K}/\mathrm{J}\times \mathrm{mol}}{6.6\times {10}^{-4}\mathrm{K}} \\ \mathrm{\Delta H}°=55876.38\mathrm{J}/\mathrm{mol} \end{gathered}$$

(b): Standard entropy of autoionization.

The formula for calculating the standard entropy of autoionization of water is

$\mathrm{\Delta S}°=\frac{\mathrm{\Delta H}°-\mathrm{\Delta G}°}{\mathrm{T}}$

First, we need to calculate $\mathrm{\Delta G}°$

$\begin{array}{l}\mathrm{\Delta G}°=-\mathrm{RTln}\left({\mathrm{K}}_{273}\right)\\ \mathrm{\Delta G}°=-8.314\mathrm{J}/\mathrm{K}\times \mathrm{mol}\times 273\mathrm{K}\times \ln \left(1.139\times {10}^{-15}\right)\\ \mathrm{\Delta G}°=78098.01\mathrm{J}/\mathrm{mol}\end{array}$

To get entropy

$\begin{array}{l}\mathrm{\Delta S}°=\frac{(55876.38-78098.01)\mathrm{J}/\mathrm{mol}}{273\mathrm{K}}\\ \mathrm{\Delta S}°=-81.4\mathrm{J}/\mathrm{K}\times \mathrm{mol}\end{array}$

(c): Calculation of the temperature

The of pure water${\mathrm{K}}_{\mathrm{w}}$ and is$1\times {10}^{-14}$

Use a formula to calculate the temperature

$\begin{array}{l}\mathrm{lnK}=\frac{-\mathrm{\Delta H}°}{\mathrm{R}\times \mathrm{T}}+\frac{\mathrm{\Delta S}°}{\mathrm{R}}\\ \ln \left(1\times {10}^{-14}\right)=\frac{-55876.38\mathrm{J}/\mathrm{mol}}{8.314\mathrm{J}/\mathrm{K}\times \mathrm{mol}\times \mathrm{T}}+\frac{-81.4\mathrm{J}/\mathrm{K}\times \mathrm{mol}}{8.314\mathrm{J}/\mathrm{K}\times \mathrm{mol}}\\ -32.24=-6720.76\mathrm{K}\times \frac{1}{\mathrm{T}}-9.79\\ -22.45=-6720.76\mathrm{K}\times \frac{1}{\mathrm{T}}\end{array}$

Solving further as:

$\begin{array}{l}22.45\mathrm{T}=6720.76\mathrm{K}\\ \mathrm{T}=299.37\mathrm{K}\end{array}$