JAMB Chemistry 2019 Paper

Concept:Neutralization of an acid by a base requires using the mole ratio from the balanced equation.Explanation:The balanced equation is:$$2\mathrm{NaOH} + \mathrm{H}_2\mathrm{SO}_4 \rightarrow \mathrm{Na}_2\mathrm{SO}_4 + 2\mathrm{H}_2\mathrm{O}$$From the equation, $1$ mole of tetraoxosulphate(VI) acid reacts with $2$ moles of sodium hydroxide.Therefore, the acid-to-base mole ratio is:$$\frac{n_A}{n_B} = \frac{1}{2}$$Use the neutralization formula:$$\frac{C_A V_A}{C_B V_B} = \frac{n_A}{n_B}$$Substitute $C_A = 1.25\,\mathrm{mol\,dm^{-3}}$, $V_A = 10\,\mathrm{cm^3}$, and $C_B = 0.5\,\mathrm{mol\,dm^{-3}}$:$$\frac{1.25 \times 10}{0.5 \times V_B} = \frac{1}{2}$$Simplify:$$\frac{12.5}{0.5 V_B} = \frac{1}{2}$$Cross-multiply:$$12.5 = 0.25 V_B$$$$V_B = 50\,\mathrm{cm^3}$$Answer:The required volume is $50\,\mathrm{cm^3}$, which is Option D.