(a) The balanced chemical equation for the reaction between sodium hydroxide (NaOH) and sulfuric acid (H₂SO₄) is: $$ 2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O} $$ (b) The number of moles of sulfuric acid used is calculated as: $$ \begin{aligned} n_{\text{H}_2\text{SO}_4} &= V \times S \\ &= 22.3 \times 10^{-3} \text{ dm}^3 \times 0.10 \text{ mol/dm}^3 \\ &= 2.23 \times 10^{-3} \text{ mol} \end{aligned} $$ (c) From the balanced equation, 1 mole of H₂SO₄ reacts with 2 moles of NaOH. Therefore, the amount of NaOH present is: $$ \begin{aligned} n_{\text{NaOH}} &= 2 \times n_{\text{H}_2\text{SO}_4} \\ &= 2 \times 2.23 \times 10^{-3} \text{ mol} \\ &= 4.46 \times 10^{-3} \text{ mol} \end{aligned} $$ (d) The molarity (concentration) of the sodium hydroxide solution is: $$ \begin{aligned} S_{\text{NaOH}} &= \frac{n_{\text{NaOH}}}{V_{\text{NaOH}}} \\ &= \frac{4.46 \times 10^{-3} \text{ mol}}{50 \times 10^{-3} \text{ dm}^3} \\ &= 0.0892 \text{ M} \end{aligned} $$ The concentration of the sodium hydroxide solution is 0.0892 M.