A lake of volume 1.0x10^6 L has pH 4.0 owing to acid rain. To neutralise the H+ present by adding limestone (CaCO3, molar mass 100, CaCO3 + 2H+ -> Ca^2+ + H2O + CO2), and ignoring buffering and later acid input, the mass of CaCO3 needed is

JEE Advanced levelSingle correct optionDifficulty 3 of 3ev048

A lake of volume $1.0 \times 1 0^{6}$ L has pH 4.0 owing to acid rain. To neutralise the $H X^{+}$ present by adding limestone ( $CaCO X_{3}$ , molar mass 100, $CaCO X_{3} + 2 H X^{+} Ca X^{2 +} + H X_{2} O + CO X_{2}$ ), and ignoring buffering and later acid input, the mass of $CaCO X_{3}$ needed is

[H X^{+}] = 1 0^{- 4}

M, so moles of

H X^{+}

1 0^{- 4} \times 1.0 \times 1 0^{6} = 100

mol. Two

H X^{+}

react with each

CaCO X_{3}

, so 50 mol

CaCO X_{3}

50 \times 100 = 5000 g = 5.0 kg

. Forgetting the factor of 2 gives 10 kg.