The pH of Weak Acids (OCR A-Level Chemistry A): Revision Notes

The pH of Weak Acids

Understanding weak acid dissociation

When a monobasic acid (HA) is dissolved in water, it can behave in two different ways depending on its strength:

• Strong acids completely dissociate in aqueous solution, meaning all acid molecules break apart. For a strong acid, the concentration of hydrogen ions equals the original acid concentration: $[\text{H}^+(\text{aq})] = [\text{HA}(\text{aq})]$

• Weak acids only partially dissociate in aqueous solution, meaning only some of the acid molecules break apart. For a weak acid, the hydrogen ion concentration is not equal to the original acid concentration: $[\text{H}^+(\text{aq})] \neq [\text{HA}(\text{aq})]$

Because weak acids only partially dissociate, they establish an equilibrium in solution. The dissociation can be represented by the following equilibrium equation:

$\text{HA}(\text{aq}) \rightleftharpoons \text{H}^+(\text{aq}) + \text{A}^-(\text{aq})$

When weak acid molecules dissociate, hydrogen ions and conjugate base ions are formed in equal quantities. This is a crucial point that will help simplify our calculations later.

The equilibrium expression for weak acids

To understand how to calculate the pH of weak acids, we need to consider what happens at equilibrium. An ICE (Initial, Change, Equilibrium) table helps us organise the concentrations:

At the start, we have only the undissociated weak acid present, with concentrations of $\text{H}^+(\text{aq})$ and $\text{A}^-(\text{aq})$ both equal to zero. At equilibrium, some of the acid has dissociated, producing equal concentrations of $[\text{H}^+(\text{aq})]_{\text{eqm}}$ and $[\text{A}^-(\text{aq})]_{\text{eqm}}$.

The acid dissociation constant, $K_\text{a}$, can be calculated using equilibrium concentrations:

$K_\text{a} = \frac{[\text{H}^+(\text{aq})]_{\text{eqm}} \times [\text{A}^-(\text{aq})]_{\text{eqm}}}{[\text{HA}(\text{aq})]_{\text{eqm}}}$

We can also express this as:

$K_\text{a} = \frac{[\text{H}^+(\text{aq})]_{\text{eqm}} \times [\text{A}^-(\text{aq})]_{\text{eqm}}}{[\text{HA}(\text{aq})]_{\text{start}} - [\text{H}^+(\text{aq})]_{\text{eqm}}}$

This expression looks complex, but we can simplify it significantly using two important approximations.

Simplifying calculations with approximations

The equilibrium expression shown above can be greatly simplified by making two reasonable approximations. These approximations are valid for most weak acids you'll encounter at A-Level, and they make pH calculations much more straightforward.

Approximation 1: Hydrogen ions from water are negligible

When a weak acid dissociates, it produces equal concentrations of $\text{H}^+(\text{aq})$ and $\text{A}^-(\text{aq})$ ions. However, water itself also dissociates slightly to produce a small concentration of hydrogen ions. In practice, this contribution from water is extremely small and can be neglected compared to the hydrogen ions produced by the weak acid.

Therefore, we can state:

$[\text{H}^+(\text{aq})]_{\text{eqm}} \approx [\text{A}^-(\text{aq})]_{\text{eqm}}$

Approximation 2: The acid concentration remains essentially constant

Because weak acids dissociate to such a small extent, the equilibrium concentration of the undissociated acid is very similar to the starting concentration. We can assume that:

$[\text{HA}(\text{aq})]_{\text{start}} \gg [\text{H}^+(\text{aq})]_{\text{eqm}}$

This means we can neglect any decrease in the concentration of undissociated acid:

$[\text{HA}(\text{aq})]_{\text{eqm}} = [\text{HA}(\text{aq})]_{\text{start}} - [\text{H}^+(\text{aq})]_{\text{eqm}} \approx [\text{HA}(\text{aq})]_{\text{start}}$

The simplified $K_\text{a}$ expression

By applying both approximations, we can dramatically simplify the $K_\text{a}$ expression:

Starting with the full expression and substituting in our approximations:

$K_\text{a} = \frac{[\text{H}^+(\text{aq})]_{\text{eqm}} \times [\text{A}^-(\text{aq})]_{\text{eqm}}}{[\text{HA}(\text{aq})]_{\text{start}} - [\text{H}^+(\text{aq})]_{\text{eqm}}}$

Using approximation 1: $[\text{H}^+(\text{aq})]_{\text{eqm}} = [\text{A}^-(\text{aq})]_{\text{eqm}}$

$K_\text{a} = \frac{[\text{H}^+(\text{aq})]_{\text{eqm}}^2}{[\text{HA}(\text{aq})]_{\text{start}} - [\text{H}^+(\text{aq})]_{\text{eqm}}}$

Using approximation 2: $[\text{HA}(\text{aq})]_{\text{start}} - [\text{H}^+(\text{aq})]_{\text{eqm}} \approx [\text{HA}(\text{aq})]_{\text{start}}$

Therefore:

$K_\text{a} = \frac{[\text{H}^+(\text{aq})]^2}{[\text{HA}(\text{aq})]}$

Calculating the pH of weak acids

To calculate the pH of a weak acid, we need to rearrange the simplified $K_\text{a}$ expression to make $[\text{H}^+(\text{aq})]$ the subject.

Starting with: $K_\text{a} = \frac{[\text{H}^+(\text{aq})]^2}{[\text{HA}(\text{aq})]}$

Rearranging: $[\text{H}^+(\text{aq})]^2 = K_\text{a} \times [\text{HA}(\text{aq})]$

$[\text{H}^+(\text{aq})] = \sqrt{K_\text{a} \times [\text{HA}(\text{aq})]}$

Once you have calculated $[\text{H}^+(\text{aq})]$, you can find the pH using: $\text{pH} = -\log_{10}[\text{H}^+(\text{aq})]$

Determining $K_\text{a}$ experimentally

The acid dissociation constant for a weak acid can be determined experimentally by following these steps:

• Prepare a standard solution of the weak acid with a known concentration
• Measure the pH of the standard solution using a pH meter

Once you have these two values, you can calculate $K_\text{a}$ by rearranging the simplified expression.

When do the approximations break down?

Both worked examples above use the two approximations we discussed earlier. But are these approximations always justified? Understanding when they fail is important for accurate calculations.

Limitations of approximation 1

The first approximation assumes that the dissociation of water is negligible: $[\text{H}^+(\text{aq})]_{\text{eqm}} \approx [\text{A}^-(\text{aq})]_{\text{eqm}}$

Limitations of approximation 2

The second approximation assumes that the concentration of acid is much greater than the hydrogen ion concentration at equilibrium: $[\text{HA}(\text{aq})]_{\text{start}} \gg [\text{H}^+(\text{aq})]_{\text{eqm}}$

This allows us to approximate: $[\text{HA}(\text{aq})]_{\text{eqm}} = [\text{HA}(\text{aq})]_{\text{start}} - [\text{H}^+(\text{aq})]_{\text{eqm}} \approx [\text{HA}]_{\text{start}}$

Checking your approximations

You can carry out weak acid pH calculations without making the $[\text{HA}(\text{aq})]_{\text{start}} - [\text{H}^+(\text{aq})]_{\text{eqm}}$ approximation, but you will then need to solve a quadratic equation. A far easier method is to use one of the many quadratic equation solver websites available online.