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JAMB Chemistry 1980 Question 8

Question 8

Volumetric Analysis

1.0 g of CaCO3 was treated with excess 0.1 M HCl. At the end of the reaction, the acid still left unreacted required 25 cm3 of 0.1 M Na2CO3 to neutralize it. What was the original volume of the acid? (Na = 23, Ca = 40, C = 12, O = 16, Cl = 35.5)

  1. A.220 cm3
  2. B.230cm3
  3. C.240 cm3
  4. D.250cm3
  5. E.260cm3.
Show the answer and explanation

Answer: D (250cm3)

Explanation

1.0 g CaCO3 = 0.01 mol, which uses 0.02 mol HCl (200 cm3). The leftover acid reacted with 0.0025 mol Na2CO3, i.e. 0.005 mol HCl (50 cm3). Total = 250 cm3.