Unlike strong acids, weak acids dissociate only partially in aqueous solution. This partial dissociation sets up a dynamic equilibrium. To quantify the strength of a weak acid and perform pH calculations, we use the equilibrium constant for this reaction: the acid dissociation constant, Ka.
The equilibrium constant for the dissociation of a weak acid HA:
\( \text{K}_a = \frac{[\text{H}^+\text{(aq)}][\text{A}^-\text{(aq)}]}{[\text{HA(aq)}]} \)
Units are typically mol dm^-3. The larger the Ka, the greater the extent of dissociation and the stronger the weak acid.
The negative logarithm to the base 10 of Ka, used to compare acid strengths on a simpler scale:
\( \text{pK}_a = -\log_{10}(\text{K}_a) \quad \text{and} \quad \text{K}_a = 10^{-\text{pK}_a} \)
A smaller pKa indicates a stronger weak acid.
Visualising Partial Dissociation
In a weak acid solution, the vast majority of acid molecules remain undissociated. Only a tiny fraction dissociate into ions:
🔑 Key Principle
When calculating the pH of a weak acid, we make two critical simplifying approximations:
1. \( [\text{H}^+] \approx [\text{A}^-] \): We assume all H+ ions come only from the acid HA (neglecting the dissociation of water).
2. \( [\text{HA}]_{\text{equilibrium}} \approx [\text{HA}]_{\text{initial}} \): We assume the dissociation is so small that the equilibrium concentration of HA is virtually unchanged from its initial concentration.
Using these, the Ka expression simplifies to: \( \text{K}_a = \frac{[\text{H}^+]^2}{[\text{HA}]_{\text{initial}}} \).
Rearranging the Simplified Ka Expression
From the simplified equation, we can find any of the variables:
- To calculate \( [\text{H}^+] \) and pH from Ka and acid concentration:
\( [\text{H}^+] = \sqrt{\text{K}_a \times [\text{HA}]_{\text{initial}}} \)
- To calculate Ka from pH and acid concentration:
\( \text{K}_a = \frac{[\text{H}^+]^2}{[\text{HA}]_{\text{initial}}} \quad \text{where} \quad [\text{H}^+] = 10^{-\text{pH}} \)
You may be asked to state the approximations made when calculating the pH of a weak acid. Make sure you can write down both assumptions clearly: first, that the concentration of hydrogen ions is equal to the conjugate base concentration; second, that the dissociation of the acid is so small that the initial concentration of the acid is equal to its concentration at equilibrium. If you are asked to state why these approximations might fail, write that for stronger weak acids (with a relatively high Ka) or for very dilute solutions, the percentage dissociation increases, making the second approximation invalid.
(\( \text{K}_a = 1.74 \times 10^{-5}\text{ mol dm}^{-3} \) at 298 K)
Step 1: Write down the simplified Ka expression and solve for [H+]:
\[ \text{K}_a = \frac{[\text{H}^+]^2}{[\text{CH}_3\text{COOH}]} \]
Rearrange to solve for \( [\text{H}^+] \):
\[ [\text{H}^+] = \sqrt{\text{K}_a \times [\text{CH}_3\text{COOH}]} \]
\[ [\text{H}^+] = \sqrt{(1.74 \times 10^{-5}) \times 0.150} = \sqrt{2.61 \times 10^{-6}} = 1.616 \times 10^{-3}\text{ mol dm}^{-3} \]
Step 2: Calculate the pH:
\[ \text{pH} = -\log_{10}(1.616 \times 10^{-3}) = 2.79 \]
The pH of the ethanoic acid solution is 2.79 (to 2 decimal places).
Step 1: Find the hydrogen ion concentration:
\[ [\text{H}^+] = 10^{-\text{pH}} = 10^{-3.82} = 1.514 \times 10^{-4}\text{ mol dm}^{-3} \]
Step 2: Calculate Ka using the simplified expression:
\[ \text{K}_a = \frac{[\text{H}^+]^2}{[\text{HX}]_{\text{initial}}} = \frac{(1.514 \times 10^{-4})^2}{0.050} = \frac{2.291 \times 10^{-8}}{0.050} = 4.58 \times 10^{-7}\text{ mol dm}^{-3} \]
Step 3: Calculate pKa:
\[ \text{pK}_a = -\log_{10}(\text{K}_a) = -\log_{10}(4.58 \times 10^{-7}) = 6.34 \]
Thus, \( \text{K}_a = 4.58 \times 10^{-7}\text{ mol dm}^{-3} \) and \( \text{pK}_a = 6.34 \).
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