A volume of 40 ml of 0.05 M solution of sodium sesquicarbonate (Na₂CO₃·NaHCO₃·2H₂O) is titrated against 0.05 M HCl solution. x ml of HCl solution is used when phenolphthalein is the indicator and y ml

Question

A volume of 40 ml of 0.05 M solution of sodium sesquicarbonate (Na₂CO₃·NaHCO₃·2H₂O) is titrated against 0.05 M HCl solution. x ml of HCl solution is used when phenolphthalein is the indicator and y ml of HCl is used when methyl orange is the indicator in two separate titrations. Hence (y-x) is

TestHub · Accepted Answer

Salt = 40 x 0.05 = 2 mmol, containing 2 mmol CO₃²⁻ and 2 mmol HCO₃⁻. Phenolphthalein: half-neutralization of CO₃²⁻ only, H⁺ = 2 mmol, x = 2/0.05 = 40 ml. Methyl orange: full neutralization, H⁺ = 2(2) + 2 = 6 mmol, y = 6/0.05 = 120 ml. y - x = 120 - 40 = 80 ml, option (a).

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