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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?

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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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An easier approach would be that

concentration of hydromium and hydroxyl ions will increase with temperature

as tendency of

water to dissociate will increase

Hence, more quantity of products means greater Ionic Product.

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  • $\begingroup$ This is much oversimplified. The Wikipedia graphs referred to in Nicolau's comment shows a maximum at constant pressure, beyond which the ion product drops. Decreasing density, which kills the hydrogen bonding, eventually overrides the temperature effect we would normally expect. $\endgroup$ – Oscar Lanzi 2 days ago

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