# Effect of dilution on weak acids pH [closed]

My textbook has a question which says what would happen to the pH of a weak acid if it is diluted with water. The answer is its pH would decrease.

My teacher’s argument is that it would indeed decrease because water ionizes to hydronium ions which would increase the concentration of H+ ions and therefore, decrease the pH.

My argument is that it would increase because when the acid is diluted, the concentration of H+ ions will decrease and hence, the pH will increase. Further details in the attached photo.

My father, who is a chemist, argues that it would decrease then increase. He says that it varies upon the amount of water added and that it is not a direct relation. Instead, it would form a graph similar to the letter V.

I am honestly lost. What would exactly happen?

• Simplified formula is pH = 0.5 ( pKa - log c ) Dec 12, 2020 at 19:06
• Things get weird when there is very little water, as in glacial acetic acid. See e.g. sciencedirect.com/science/article/abs/pii/S0167732216329154
– Karsten
Dec 13, 2020 at 1:06
• This would be crystal clear with a derivation of the appropriate pH equation from scratch, with no approximations. This question is closely related. The first graph in the top answer directly shows the shape of the curve you seek - the pH of a weak acid solution rises smoothly and monotonically until it becomes so dilute that the pH is pegged by the autoionisation of water at pH ~ 7. You are correct. Dec 13, 2020 at 4:43
• The pH will increase because if you dilute enough you will have a solution with acid concentration and its $\ce{[H]^+} \lt 10^{-7}$ and then the dissociation of water will keep the pH close to 7. Dec 14, 2020 at 16:48

Whatever the nature of the acids, weak or strong, dilution of an acidic solution produces an increase of pH. The dissociation coefficient may increase, but the pH will always increase. The expression "like a Y letter" is pure fantasy.

Your calculations about acetic acid are correct. Just a small remark : in the second part, you say :"if you double the volume". It should be stated that this operation is made by adding pure water. In this case, the total concentration of acid is divided by 2. Surprisingly, the concentration of $$\ce{H+}$$ is not divided by $$2$$ : $$\ce{[H^+]}$$ = $$2.99 ·10^{-3} mol L^{-1}$$. So it has been divided by $$√2$$ (and not by $$2$$).

Now, let's speak about the number of moles. During the dilution, the total number of moles of acetic acid has not changed, but the total number of moles of $$\ce{H^+}$$ has slightly increased during the dilution, which means that the acid is more dissociated in the diluted solution.

The degree of dissociation will increase in this case, so more ionization will take place,but it wont b able to counter the decreased concentration.

Actually, if you write the equilibrium equation, taking c as the initial concentration of acetic acid, and x as the degree of dissociation, then the acid dissociation constant, Kc will approximately be equal to $$cx^2$$, and the concentration of H+ will be cx.

Now, Kc is constant for a given acid at a given temperature, therefore, $$cx^2= cx*x$$ is constant. So if x increases on dilution, the cx will decrease, hence the concentration of H+ will decrease.

So, the Ph has to increase in this case.