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I read in a textbook that
the water ionization constant ($K_\mathrm{w}$) increases as the ionic strength of the solution increases
and it confused me. If the ionic strength increased, then the activity of each of the ions $\ce{H3O+}$ and $\ce{OH-}$ would decrease, leading to a decrease in the value of $K_\mathrm{w}$.
Can someone please explain the mistake in my way of thinking?
The equilibrium constant is given though $$K_x = \prod_{\ce{B}} x_{\ce{B}}^{\nu_{\ce{B}}},$$ where $x$ is a specific quantity. We can choose that to be the activity $a$.
For the auto-ionisation $$\ce{H2O (aq) <=>[H2O] H+ (aq) + OH- (aq)}\tag{1}\label{water}$$ we find therefore $$K_a = \frac{a(\ce{H+})\cdot a(\ce{OH-})}{a(\ce{H2O})}.$$
Assuming the equilibrium $\eqref{water}$ is largely on the left side and therefore there is an excess of water, a reasonable simplification is to treat $a(\ce{H2O})$ as constant. $$K_a' = K_a\cdot a(\ce{H2O}) = a(\ce{H+})\cdot a(\ce{OH-})$$
In ideally diluted solutions we also find $$\lim_{c\to0}\gamma_i = \frac{a_i}{c_i}=1,$$ and therefore $$K_c = c(\ce{H+})\cdot c(\ce{OH-}) \overset{\mathrm{def}}{=} K_\mathrm{w}.$$
The ionic strength is defined as $$I_c = \frac12\sum_{\ce{B}}c_{\ce{B}}z_{\ce{B}}^2,$$ or for our case $$I_{c,\mathrm{w}} = \frac12 \bigg[c(\ce{H+}) + c(\ce{OH-})\bigg].$$
Because of $\eqref{water}$ we can write $$I_{c,\mathrm{w}} = c(\ce{H+})$$ and $$K_\mathrm{w} = c(\ce{H+})^2, $$ therefore $$I_{c,\mathrm{w}} = \sqrt{K_\mathrm{w}}.$$
From this relation you can see, that when $K_\mathrm{w}$ increases, the ionic strength also increases, vice versa.
If the ionic strength increased, then the activity of each of the ions $\ce{H3O+}$ and $\ce{OH−}$ would decrease, leading to a decrease in the value of $K_\mathrm{w}$.
The first part is correct, the activity of each ion would decrease. However, there are more ions, so the activity of the ensemble increases more (given ideal dilution), therefore also leading to an increase in $K_\mathrm{w}$.
