# How to rationalise that when we dilute a weak acid, the pH increases but also the dissociation increases?

When water is added to a weak acid like ethanoic acid, the number of ethanoic acid molecules that dissociate increases, however, pH increases (less acidic), too. Why is this so?

It's simply due to dilution.

If it's true that adding water increases dissociation (Gerard has done a nice effort in Does the number of H+ ions in solution go up on dilution of a weak acid?), concentration of $$\ce{H3O+}$$ inevitably decreases. As a consequence, $$\mathrm{pH}$$, defined as $$-\log[\ce{H3O+}]$$, increases.

Start with $$\pu{1 L}$$ acetic acid concentration $$\pu{1 M}$$, $$\ce{[H3O+]} = \sqrt{K_c} = \pu{4.22E-3M},$$ that corresponds to $$\pu{4.22E-3mol}$$ of $$\ce{H3O+}$$. If you double the volume ($$\pu{2 L}$$), $$\ce{[H3O+]}=\sqrt{K_c/2}=\pu{2.98E-3M},$$ that corresponds to $$\pu{5.96E-3mol}$$ of $$\ce{H3O+}$$.

The amount of substance is higher, but concentration is less: dilution does play a role here.

Let's first examine the chemical reaction between our acids and bases. Ethanoic acid, commonly referred to as acetic acid reacts with our base, water. Water is amphiprotic, in that it can act as both a base and acid.

$$\ce{CH3COOH + H2O <=> CH3COO- + H3O+}$$

However, the acetate ion undergoes hydrolysis with water which now instead acts as an acid, reforming acetic acid as well has hydroxide ions:

$$\ce{CH3COO^- + H2O <=> CH3COOH + OH-}$$

Acetic acid is weak acid, however, the conjugate base, acetate, is a strong base/salt. Water acts as a strong base, however the hydronium ion is a weak acid, so we can except the pH to increase.

What many people often forget to realize is that once an acid/base reaction occurs, a salt is created along with water. However, a hydrolysis reaction between the salt and water can occur.

Taken from Chemical Principles: The Quest for Insight by Atkins et. al on page 494 in the tenth edition:

Calcium acetate, $$\ce{Ca(CH3CO2)2 (aq)}$$ is used in medicine to treat patients with a kidney disease that results in high levels of phosphate ions in the blood. The calcium binds with the phosphates so that they can be excreted. If you are using calcium acetate for this purpose, it is important to know the pH of the solution to avoid complications in treatment.

Say we have a $$\pu{0.15 M}$$ solution of $$\ce{Ca(CH3CO2)2 (aq)}$$.

Since the conjugate base of acetic acid is strong, we can expect the pH to increase.

Our expression can be written as this:

$$\ce{CH3COO^- + H2O <=> CH3COOH + OH-}$$

Calcium is a spectator ion and thus can be ignored.

$$K_b = \ce{\frac{[CH3COOH][OH-]}{[CH3COO-]}}$$

Constructing our RICE table:

$$\ce{CH3COO^- + H2O <=> CH3COOH + OH-}$$

$$\begin{array} {|c|c|c|c|c|} \hline &\ce{CH3COO^-} & \ce{H2O} & \ce{CH3COOH} & \ce{OH-}\\\hline \text{Initial conc.} & 0.30 & - & 0 & 0 \\ \hline \text{Change conc.} & -x & - & +x & +x\\ \hline \text{End conc.} & 0.30 - x & - & x & x \\ \hline \end{array}$$

\begin{align} K_b &= 5.6\times10^{-10} = \frac{[x][x]}{[0.30-x]}\\ x &= 1.3 \times 10^{-5}\\ \text{pOH} &= -\log([\ce{OH-}])\\ \text{pH} &= 14 - \text{pOH} = 14 - 4.89 = 9.11 \end{align}

On dilution, ionisation increases, although pH increases due to the fact that, volume increases dominantly over dissociation. Hence concentration of $$\ce{H+}$$ ions decreases instead of the extent of ionisation increases.

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