Weak acids and bases

The ionisation constant of a weak acid

A weak acid HA sets up an equilibrium in water:

Its equilibrium constant is the acid ionisation constant:

A larger $K_a$ means more ionisation and a stronger acid. Because the values span many powers of ten, we also use $\text{p}K_a = -\log_{10} K_a$. A smaller pKa means a stronger acid.

The degree of ionisation

Start with $C$ mol/L of acid. Let a fraction $\alpha$ of it ionise. This $\alpha$ is the degree of ionisation.

Put these into $K_a$:

For a weak acid, $\alpha$ is small, so $1 - \alpha \approx 1$. Then

Worked example. Find the pH of $0.10\ \text{mol/L}$ acetic acid.

1. $[\text{H}_3\text{O}^+] \approx \sqrt{1.8 \times 10^{-5} \times 0.10} = 1.34 \times 10^{-3}\ \text{mol/L}$.
2. $\text{pH} = -\log_{10}(1.34 \times 10^{-3}) = 2.87$.
3. $\alpha \approx \dfrac{1.34 \times 10^{-3}}{0.10}$, about $1.3\%$.

A strong acid of the same concentration would have pH 1. The weak acid has fewer hydronium ions by a factor of about 75.

When the shortcut fails

The shortcut assumes $\alpha$ is small, say below about 5 per cent. If the acid is fairly strong or very dilute, $\alpha$ grows and the shortcut overshoots. Then solve the full equation. Write $x$ for $[\text{H}_3\text{O}^+]$:

The positive root is

Dilution and ionisation

Since $\alpha \approx \sqrt{K_a/C}$, a smaller $C$ gives a larger $\alpha$. Dilute a weak acid and a bigger share of it ionises. Yet the hydronium concentration $\sqrt{K_a C}$ still falls, so the pH rises. Both are true at once.

Practise with new numbers

Weak bases

A weak base B takes a proton from water:

The same algebra gives $[\text{OH}^-] \approx \sqrt{K_b C}$. Find the pOH from it, then the pH. Ammonia has $K_b = 1.8 \times 10^{-5}$ at $298\ \text{K}$.

Conjugate pairs: Ka times Kb

Write $K_a$ for $\text{NH}_4^+$ and $K_b$ for $\text{NH}_3$. Multiply them. The $[\text{NH}_4^+]$ and $[\text{NH}_3]$ cancel, leaving $[\text{H}_3\text{O}^+][\text{OH}^-]$:

So if you know one, you know the other. The conjugate base of acetic acid, the acetate ion, has $K_b = \dfrac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.6 \times 10^{-10}$. This explains the rule of lesson 1: the weaker the acid, the stronger its conjugate base.

Try it in Python

The program finds $[\text{H}_3\text{O}^+]$ of acetic acid at four concentrations. It uses both the shortcut and the exact quadratic.

Common mistakes

• Forgetting the square root. $[\text{H}_3\text{O}^+] \approx \sqrt{K_a C}$, not $K_a C$.
• Using the shortcut when $\alpha$ is large. Check $\alpha$ afterwards. Above about 5 per cent, solve the quadratic.
• Reading a big pKa as a strong acid. It is the other way round.
• Giving the pOH for a base. A weak base gives $[\text{OH}^-]$. Take the pOH away from 14.
• Thinking dilution lowers $\alpha$. Dilution raises the share ionised, though it lowers $[\text{H}_3\text{O}^+]$.

Remember these

• $K_a = \dfrac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA}]}$, and $\text{p}K_a = -\log_{10} K_a$.
• $K_a = \dfrac{C\alpha^2}{1 - \alpha} \approx C\alpha^2$, so $\alpha \approx \sqrt{K_a/C}$ and $[\text{H}_3\text{O}^+] \approx \sqrt{K_a C}$.
• If $\alpha$ is large, solve $x^2 + K_a x - K_a C = 0$.
• Weak base: $[\text{OH}^-] \approx \sqrt{K_b C}$.
• $K_a K_b = K_w$ and $\text{p}K_a + \text{p}K_b = 14$ for a conjugate pair.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Which is the stronger acid: one with pKa 3 or one with pKa 6?

2. A weak acid is diluted. What happens to its degree of ionisation and to its pH?

3. Why may we write $1 - \alpha \approx 1$ for a weak acid?

4. An acid has $\text{p}K_a = 9.2$. What is $\text{p}K_b$ of its conjugate base?

5. Which is the weaker acid in the table, and which conjugate base is the strongest?