1 gram at a sample of CaCO 3 was strongly heated and the CO 2 liberated was 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 rea

Question

1 gram at a sample of CaCO 3 ​ was strongly heated and the CO 2 ​ liberated was 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 to reach the phenolphthalein end point?

TestHub · Accepted Answer

Moles of CO₂ liberated: Mass of pure CaCO₃ = $1 g \times 90% = 0.9 g$ , so moles of CO₂ = $\frac{0.9 g}{100 g / m o l} = 0.009 m o l = 9 mm o l$ .

Reaction with NaOH: $9 mm o l$ of CO₂ reacts with $2 \times 9 = 18 mm o l$ of NaOH to form $9 mm o l$ of Na₂CO₃.

Remaining components: Initial NaOH = $100 m L \times 0.5 M = 50 mm o l$ ; thus, unreacted NaOH = $50 - 18 = 32 mm o l$ .

Phenolphthalein endpoint titration: HCl neutralizes all remaining NaOH and half-neutralizes Na₂CO₃ (Na₂CO₃ $\to$ NaHCO₃).

Volume of HCl needed: Total mmol of HCl = $32 + 9 = 41 mm o l \to V o l u m e = \frac{41 mm o l}{0.5 M} = 82mL$ .

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