10.0 mL of Na₂CO₃ solution is titrated against 0.2 M HCl solution. The following titre values were obtained in 5 readings :
4.8 mL, 4.9 mL, 5.0 mL, 5.0 mL and 5.0 mL
Based on these readings, and convention of titrimetric estimation the concentration of Na₂CO₃ solution is ___________ mM.
(Round off to the Nearest Integer).

Solution

Chemistry ma’am · 4 steps

Throw out the stray readings and use the concordant 5.0 mL, then remember carbonate needs two HCl.

1. Pick the titre. By convention you use the concordant readings, the ones that agree with each other. Here 5.0 mL appears three times. The 4.8 mL and 4.9 mL readings are rough early trials, so the titre is 5.0 mL, not the average of all five.
2. Mole ratio. The reaction is Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂. One mole of carbonate uses two moles of acid, so the millimoles of HCl are twice the millimoles of Na₂CO₃.

   2 × M_(Na₂CO₃) × 10.0 = 0.2 × 5.0

3. Put in numbers. HCl used is 0.2 × 5.0 = 1.0 mmol, so Na₂CO₃ is 0.5 mmol in 10.0 mL. That gives M = 0.5/10.0 = 0.05 M.
4. So. 0.05 M = 0.05 × 1000 mM, so the concentration is 50 mM.

The common slip: 49. Averaging all five readings gives 4.94 mL and a concentration of 49.4 mM, which rounds to 49, not 50.

The idea to keep. In titration data, take the concordant (repeated, agreeing) reading as the titre, never the average of everything. Then balance the equation before you equate millimoles.

Answer from the NTA final answer key. Solution by PrepWiser.