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

Step 1: In titrimetric analysis, the concordant value (the most frequent consistent reading) is used. Here, it is $5.0$ mL.
Step 2: Reaction: $Na_2CO_3 + 2HCl \rightarrow 2NaCl + H_2O + CO_2$. The n-factor for $Na_2CO_3$ is 2 and for $HCl$ is 1.
Step 3: Using $n_1 M_1 V_1 = n_2 M_2 V_2$: \[ 2 \times M_{Na_2CO_3} \times 10.0 = 1 \times 0.2 \times 5.0 \implies 20 M_{Na_2CO_3} = 1.0 \] \[ M_{Na_2CO_3} = 0.05 \text{ M} \]
Step 4: $0.05 \text{ M} = 0.05 \times 1000 = 50 \text{ mM}$.