10.0 ml Na2CO3 solution is titrated aginst 0.2 M HCL solution.

Question:

$10.0 \mathrm{ml}$ of $\mathrm{Na}_{2} \mathrm{CO}_{3}$ solution is titrated against $0.2$ $\mathrm{M} \mathrm{HCl}$ 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 $5.0 \mathrm{ml}$

Based on these readings, and convention of titrimetric estimation of concentration of $\mathrm{Na}_{2} \mathrm{CO}_{3}$ solution is_______ $\mathrm{mM}$.

(Round off to the Nearest integer)

Solution:

Most precise volume of $\mathrm{HCl}=5 \mathrm{ml}$ at equivalence point

Meq. of $\mathrm{Na}_{2} \mathrm{CO}_{3}=$ meq. of $\mathrm{HCl}$.

Let molarity of $\mathrm{Na} 2 \mathrm{CO}_{3}$

solution $=\mathrm{M}$, then

$\mathrm{M} \times 10 \times 2=0.2 \times 5 \times 1$

$\mathrm{M}=0.05 \mathrm{~mol} / \mathrm{L}$

$=0.05 \times 1000$

$=50 \mathrm{mM}$

 

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