10.0 mL of $\mathrm{Na}_{2} \mathrm{CO}_{3}$ solution is titrated against 0.2 M HCl solution. The following titre values were obtained in 5 readings: $4.8 \mathrm{~mL}, 4.9 \mathrm{~mL} 5.0 \mathrm{~mL} 5.0 \mathrm{~mL}$ and 5.0 mL Based on these readings, and convention of titrimetric estimation the concentration of $\mathrm{Na}_{2} \mathrm{CO}_{3}$ solution is ______ mM. (Round off to the nearest integer).

JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Practical and Analytical Chemistry

10.0 mL of

Na_{2} CO_{3}

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

Na_{2} CO_{3}

solution is ______ mM.
(Round off to the nearest integer).

Answer: 50

Titrimetric practice uses only concordant readings, so the three agreeing values of

5.0

mL set the titre and the two earlier trial readings are discarded. Carbonate consumes two moles of acid per mole:

N a_{2} C O_{3} + 2 HCl ⟶ 2 NaCl + H_{2} O + C O_{2}

Moles of acid used

= 0.2 \times 5.0 \times 1 0^{- 3} = 1.0 \times 1 0^{- 3}

, so the

10.0

mL aliquot holds

0.5 \times 1 0^{- 3}

mol of carbonate. Its concentration is

\frac{0.5 \times 1 0 ^{- 3}}{0.0100} = 0.05 mol L^{- 1}

, that is

50

mM.

JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Integer answer