Explain the concept of hydrolysis with examples and discuss the suitability of indicators for titrations involving strong acids and weak bases.

$$
V_{\text{NH}_3} = 25 \text{ cm}^3 = 25 \times 10^{-3} \text{ dm}^3 = 0.025 \text{ dm}^3
$$

$$
n_{\text{NH}_3} = C_{\text{NH}_3} \times V_{\text{NH}_3}
$$
$$
n_{\text{NH}_3} = 0.1 \text{ mol} \cdot \text{dm}^{-3} \times 0.025 \text{ dm}^3
$$
$$
n_{\text{NH}_3} = \boxed{\text{0.0025 mol}}
$$

$$
n_{\text{NH}_3} = C_{\text{NH}_3} \times V_{\text{NH}_3} = 0.1 \text{ mol} \cdot \text{dm}^{-3} \times 0.025 \text{ dm}^3 = 0.0025 \text{ mol}
$$
$$
n_{\text{HCl}} = C_{\text{HCl}} \times V_{\text{HCl}} = 0.1 \text{ mol} \cdot \text{dm}^{-3} \times 0.040 \text{ dm}^3 = 0.0040 \text{ mol}
$$

The reaction is $\text{HCl(aq)} + \text{NH}_3\text{(aq)} \to \text{NH}_4\text{Cl(aq)}$. The molar ratio is 1:1.  
Since $\text{NH}_3$ is the limiting reactant, 0.0025 mol of HCl will react.
$$
n_{\text{HCl, remaining}} = n_{\text{HCl, initial}} - n_{\text{NH}_3, \text{reacted}}
$$
$$
n_{\text{HCl, remaining}} = 0.0040 \text{ mol} - 0.0025 \text{ mol} = 0.0015 \text{ mol}
$$

$$
V_{\text{total}} = V_{\text{NH}_3} + V_{\text{HCl}} = 0.025 \text{ dm}^3 + 0.040 \text{ dm}^3 = 0.065 \text{ dm}^3
$$

$$
[\text{H}^+] = \frac{n_{\text{HCl, remaining}}}{V_{\text{total}}} = \frac{0.0015 \text{ mol}}{0.065 \text{ dm}^3}
$$
$$
[\text{H}^+] \approx 0.023077 \text{ mol} \cdot \text{dm}^{-3}
$$

$$
\text{pH} = -\log_{10}[\text{H}^+]
$$
$$
\text{pH} = -\log_{10}(0.023077)
$$
$$
\text{pH} \approx \boxed{\text{1.64}}
$$