f) Calculate the pH of a 0.01 mol dm⁻³ ethanoic acid solution given pKa = 4.7.

The relationship between $\text{pKa}$ and $\text{Ka}$ is:

$$\text{pKa} = -\log_{10}(\text{Ka})$$

Given $\text{pKa} = 4.7$, we can find $\text{Ka}$:

$$\text{Ka} = 10^{-\text{pKa}}$$ $$\text{Ka} = 10^{-4.7}$$ $$\text{Ka} \approx 1.995 \times 10^{-5}$$

Ethanoic acid ($\text{CH}_3\text{COOH}$) is a weak acid, so it partially dissociates in water:

$$\text{CH}_3\text{COOH} (\text{aq}) \rightleftharpoons \text{H}^+ (\text{aq}) + \text{CH}_3\text{COO}^- (\text{aq})$$

Initial concentrations:
$[\text{CH}_3\text{COOH}] = 0.01 \text{ mol dm}^{-3}$
$[\text{H}^+] = 0$
$[\text{CH}_3\text{COO}^-] = 0$

Change in concentrations: Let $x$ be the concentration of $\text{CH}_3\text{COOH}$ that dissociates.
$[\text{CH}_3\text{COOH}]$ changes by $-x$
$[\text{H}^+]$ changes by $+x$
$[\text{CH}_3\text{COO}^-]$ changes by $+x$

Equilibrium concentrations:
$[\text{CH}_3\text{COOH}] = 0.01 - x$
$[\text{H}^+] = x$
$[\text{CH}_3\text{COO}^-] = x$

The $\text{Ka}$ expression is:

$$\text{Ka} = \frac{[\text{H}^+][\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]}$$

Substitute the equilibrium concentrations:

$$1.995 \times 10^{-5} = \frac{x \cdot x}{0.01 - x}$$ $$1.995 \times 10^{-5} = \frac{x^2}{0.01 - x}$$

Since $\text{Ka}$ is small and the initial concentration is relatively large, we can assume that $x \ll 0.01$. This simplifies the denominator to 0.01.

$$1.995 \times 10^{-5} \approx \frac{x^2}{0.01}$$ $$x^2 = 1.995 \times 10^{-5} \times 0.01$$ $$x^2 = 1.995 \times 10^{-7}$$ $$x = \sqrt{1.995 \times 10^{-7}}$$ $$x \approx 4.466 \times 10^{-4} \text{ mol dm}^{-3}$$

So, $[\text{H}^+] \approx 4.466 \times 10^{-4} \text{ mol dm}^{-3}$.

The pH is given by the formula:

$$\text{pH} = -\log_{10}[\text{H}^+]$$ $$\text{pH} = -\log_{10}(4.466 \times 10^{-4})$$ $$\text{pH} \approx 3.35$$

The pH of the ethanoic acid solution is $\text{3.35}$.