Why is the ionic product for water ($K_\text{w}$) temperature dependent? The concentration of $\ce{H+}$ and $\ce{OH-}$ ions in pure water is $[\ce{H+}] = [\ce{OH-}] = 1.0\times 10^{-7}\ \mathrm{mol/l}$ at $25\ \mathrm{^\circ C}$, and therefore the ionic product of water at $25\ \mathrm{^\circ C}$ is $1.0\times 10^{-14}$. Why it is so? What is the effect of temperature on this value?


I think this link should really help you out in understanding how the equilibrium constant depends on temperature.

Now, considering the case of water dissociation: $$\ce{H2O <=> H+ + HO-}$$

We can calculate the free energy of the reaction ($\Delta_\text{r}G$), using the Gibbs energy for the reactant and the products (in aqueous solution):

$$\Delta G(\ce{H2O}) = -237.13\ \mathrm{kJ/mol}$$ $$\Delta G(\ce{H+}) = 0\ \mathrm{kJ/mol}$$ $$\Delta G(\ce{HO-}) = -157.24\ \mathrm{kJ/mol}$$

$$\Delta_\text{r}G = \Delta G(\ce{H+}) + \Delta G(\ce{HO-}) - \Delta G(\ce{H2O}) = -157.24\ \mathrm{kJ/mol} + 237.13\ \mathrm{kJ/mol} = 79.89\ \mathrm{kJ/mol}$$ The values of the Gibbs energies were taken from the 8th edition of Atkins' Physical Chemistry.
Now, applying the formula below (which can be obtained from $\Delta_\text{r}G = -RT\ln K_\text{eq}$) for $T = 298\ \mathrm{K}$ (or $25\ \mathrm{^\circ C}$): $$K_\text{eq} = \operatorname{e}^{-\frac{\Delta_\text{r}G}{RT}}$$ we get something like this : $$K_\text{eq} = 2.71^{-\frac{79890\ \mathrm{J\ mol^{-1}}}{8.3145\ \mathrm{J\ mol^{-1}\ K^{-1}} \times 298\ \mathrm{K}}} = 9.93 \times 10^{-15}$$ This value can be approximated to ${10^{-14}}$.


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