Calculate the enthalpy change when $ 50 \mathrm{~mL} $ of $ 0.01 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2} $ reacts with $ 25 \mathrm{~mL} $ of $ 0.01 \mathrm{M} \mathrm{HCl} $. Given that, $ \Delta \mathrm{H}^{\circ} $ of neutralization of a strong acid and a strong base is $ 140 \mathrm{kcal} \mathrm{mol}^{-1} $.

Question

Calculate the enthalpy change when $ 50 \mathrm{~mL} $ of $ 0.01 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2} $ reacts with $ 25 \mathrm{~mL} $ of $ 0.01 \mathrm{M} \mathrm{HCl} $. Given that, $ \Delta \mathrm{H}^{\circ} $ of neutralization of a strong acid and a strong base is $ 140 \mathrm{kcal} \mathrm{mol}^{-1} $., $ 14 \mathrm{kcal} $, $ 35 \mathrm{cal} $, $ 10 \mathrm{cal} $, $ 7.5 \mathrm{cal} $

MARKS App · Accepted Answer

Number of moles of HCl = M V 1000 = 0.01 × 25 1000 = 25 × 10 - 5 H C l → H + + C l - Moles of H + = 25 × 10 - 5 Number of moles of C a O H 2 = M V 1000 = 0.01 × 50 1000 = 50 × 10 - 5 Moles of O H - = 2 × 50 × 10 - 5 = 10 - 3 In the process of neutralization, 25 × 10 - 5 moles of H + will be completely neutralized. ∴ ∆ H = 140 × 25 × 10 - 5 k c a l = 0.035 k c a l = 35 c a l

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