What volume of 0.2 mol dm$^{-3}$ NaOH is required to neutralise 25 cm$^3$ of 0.1 mol dm$^{-3}$ H$_2$SO$_4$?
A100 cm$^3$
B50 cm$^3$
C12.5 cm$^3$
D25 cm$^3$CORRECT
JUPEBChemistryStoichiometry and Volumetric Analysis
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H$_2$SO$_4$ + 2NaOH → Na$_2$SO$_4$ + 2H$_2$O. Moles of acid $= 0.1 \times 0.025 = 0.0025$; NaOH needed $= 2 \times 0.0025 = 0.005$ mol; volume $= \dfrac{0.005}{0.2} = 0.025$ dm$^3 = 25$ cm$^3$. Forgetting the 1:2 ratio gives 12.5 cm$^3$.
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