One gram of a sample of CaCO₃ was strongly heated and the CO₂ liberated absorbed in 100 ml of 0.5 M NaOH. Assuming 90% purity for the sample, how much ml of 0.5 M HCl would be required to react with t

Question

One gram of a sample of CaCO₃ was strongly heated and the CO₂ liberated absorbed in 100 ml of 0.5 M NaOH. Assuming 90% purity for the sample, how much ml of 0.5 M HCl would be required to react with the solution of the alkali for the phenolphthalein end point?

TestHub · Accepted Answer

CaCO₃ = 0.9 g = 9 mmol, so CO₂ = 9 mmol. NaOH = 100 x 0.5 = 50 mmol. CO₂ + 2NaOH → Na₂CO₃ consumes 18 mmol NaOH and forms 9 mmol Na₂CO₃, leaving 32 mmol NaOH. At the phenolphthalein end point all OH⁻ is neutralized and CO₃²⁻ is converted only to HCO₃⁻: HCl = 32 + 9 = 41 mmol. Volume of 0.5 M HCl = 41/0.5 = 82 ml, option (c).

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