Mixture of $ 1 \mathrm{~g} $ each of $ \mathrm{Na}_{2} \mathrm{CO}_{3} $ and $ \mathrm{NaHCO}_{3} $ is reacted with $ 0.1 \mathrm{NHCl} $. The quantity of $ 0.1 \mathrm{M} \mathrm{HCl} $ required to react completely with the above mixture is

Question

Mixture of $ 1 \mathrm{~g} $ each of $ \mathrm{Na}_{2} \mathrm{CO}_{3} $ and $ \mathrm{NaHCO}_{3} $ is reacted with $ 0.1 \mathrm{NHCl} $. The quantity of $ 0.1 \mathrm{M} \mathrm{HCl} $ required to react completely with the above mixture is, $ 15.78 \mathrm{~mL} $, $ 157.8 \mathrm{~mL} $, $ 198.4 \mathrm{~mL} $, $ 308 \mathrm{~mL} $

MARKS App · Accepted Answer

Na 2 CO 3 + 2 HCl → 2 NaCl + H 2 O + CO 2 1 mol 2 mol Na 2 CO 3 = 1 g = 1 106 m o l . N a 2 C o 3 ≡ 2 106 m o l . H C l NaHCO 3 + HCl → NaCl + H 2 O + CO 2 NaHCO 3 = 1 g = 1 8 4 mol ≡ 1 8 4 mol HCl Total HCl required = 1 5 3 + 1 8 4 mol HCl Let volume of 0.1 N HCl = V mL then, Moles = 0.1 × V 1 0 0 0 = V 1 0 0 0 0 ∴ V 1 0 0 00 = 1 5 3 + 1 8 4 = 0.0308 mol V = 0.0308 × 1 0 0 0 0 = 308 mL

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