Calculate the concentration of the H3O+ ions in the final solution and the initial concentration of the NaOH(aq).

The relationship between pH and H₃O⁺ concentration is:

$$\text{pH} = -\log[\text{H}_3\text{O}^+]$$

Given $\text{pH} = 2$.

$$2 = -\log[\text{H}_3\text{O}^+]$$ $$[\text{H}_3\text{O}^+] = 10^{-2} \text{ mol} \cdot \text{dm}^{-3}$$ $$\boxed{1.0 \times 10^{-2} \text{ mol} \cdot \text{dm}^{-3}}$$

7.2.2 Initial concentration of the NaOH(aq)

$$V_{\text{NaOH}} = 150 \text{ cm}^3 = 0.150 \text{ dm}^3$$ $$V_{\text{HCl}} = 200 \text{ cm}^3 = 0.200 \text{ dm}^3$$

Using the concentration from 7.2.1:

$$n_{\text{H}_3\text{O}^+ \text{ (final)}} = [\text{H}_3\text{O}^+] \times V_{\text{total}}$$ $$n_{\text{H}_3\text{O}^+ \text{ (final)}} = (1.0 \times 10^{-2} \text{ mol} \cdot \text{dm}^{-3}) \times (0.350 \text{ dm}^3)$$ $$n_{\text{H}_3\text{O}^+ \text{ (final)}} = 0.0035 \text{ mol}$$

$$n_{\text{HCl (initial)}} = C_{\text{HCl}} \times V_{\text{HCl}}$$ $$n_{\text{HCl (initial)}} = (0.03 \text{ mol} \cdot \text{dm}^{-3}) \times (0.200 \text{ dm}^3)$$ $$n_{\text{HCl (initial)}} = 0.006 \text{ mol}$$

$$n_{\text{HCl (reacted)}} = n_{\text{HCl (initial)}} - n_{\text{H}_3\text{O}^+ \text{ (final)}}$$ $$n_{\text{HCl (reacted)}} = 0.006 \text{ mol} - 0.0035 \text{ mol}$$ $$n_{\text{HCl (reacted)}} = 0.0025 \text{ mol}$$

From the balanced equation $\text{HCl(aq)} + \text{NaOH(aq)} \rightarrow \text{NaCl(aq)} + \text{H}_2\text{O(l)}$, the mole ratio of HCl to NaOH is $1:1$.
Therefore, the moles of NaOH that reacted are equal to the moles of HCl that reacted. Since NaOH was the limiting reactant, this is its initial number of moles.

$$n_{\text{NaOH (initial)}} = n_{\text{HCl (reacted)}} = 0.0025 \text{ mol}$$

$$C_{\text{NaOH (initial)}} = \frac{n_{\text{NaOH (initial)}}}{V_{\text{NaOH (initial)}}}$$ $$C_{\text{NaOH (initial)}} = \frac{0.0025 \text{ mol}}{0.150 \text{ dm}^3}$$ $$C_{\text{NaOH (initial)}} = 0.01666... \text{ mol} \cdot \text{dm}^{-3}$$ $$\boxed{0.0167 \text{ mol} \cdot \text{dm}^{-3}}$$