Question

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).

Hint:

Ignore outlier readings (like 4.8 or 4.9) and always perform calculations using the concordant volume.

Answer 1

Correct Answer: 50

To determine the concentration of the Na₂CO₃ solution, we first analyze the titration data. The volumes of HCl used in the titration are: 4.8 mL, 4.9 mL, 5.0 mL, 5.0 mL, and 5.0 mL. To ensure accuracy, the average volume is calculated, preferably excluding any anomalous values.
1. Calculate average titration volume excluding outliers:
The titre values 5.0 mL are consistent, while 4.8 mL appears as an outlier. Thus, average considering 4.9 mL, 5.0 mL, and 5.0 mL readings:
Average titre = (4.9 + 5.0 + 5.0)/3 = 4.97 mL
2. Analyze titration reaction:
Na₂CO₃ reacts with HCl:
Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂
This yields a 1:2 stoichiometric ratio. Therefore, moles of Na₂CO₃ = 1/2 × moles of HCl.
3. Calculate moles of HCl used:
Given concentration = 0.2 M HCl:
Moles of HCl = 0.2 M × 4.97 mL × 10^-3 L/mL = 0.000994 mol
4. Calculate moles of Na₂CO₃:
Moles of Na₂CO₃ = 0.000994 mol / 2 = 0.000497 mol
5. Calculate concentration of Na₂CO₃ solution:
Given volume of Na₂CO₃ solution is 10 mL (0.01 L):
Concentration (C) = 0.000497 mol / 0.01 L = 0.0497 M
Convert to mM (millimolar):
C = 0.0497 M × 1000 = 49.7 mM
Rounding to the nearest integer, the concentration of Na₂CO₃ solution is 50 mM.
This value falls within the specified range (50, 50), confirming its correctness.