# Relationship between the water ionization constant and ionic strength

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?

• Mother Nature is not always easy to understand. I am afraid that your textbook is telling the truth. I have not yet understood all the details myself, but here is a link and a reference that perhaps can enlighten us both: www1.lsbu.ac.uk/water/water_dissociation.html#prod T. Vilarifio and M. E. Sastre de Vicente, J. Solution Chem. 26 (1997) 833-846. Theoretical calculations of the ionic strength dependence of the ionic product of water based on a mean spherical approximation.
– Bive
Mar 28, 2017 at 23:11

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 $K_\mathrm{w}$.

• "The first part is correct" I don't see how. Jul 27, 2017 at 8:07