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).
Answer (integer)
50
Solution
From the given value of HCl, it is clear that most appropriate volume of HCl used is 5 ml because it occurs most number of times.<br><br>Na<sub>2</sub>CO<sub>3</sub> + 2 HCl $\to$ 2 NaCl + CO<sub>2</sub> + H<sub>2</sub>O<br><br>n factor of Na<sub>2</sub>CO<sub>3</sub> = 2<br><br>n factor of HCl = 1<br><br>equivalent of Na<sub>2</sub>CO<sub>3</sub> = equivalent of HCl<br><br>$\Rightarrow$ ${{10} \over {1000}} \times 2 \times M = {5 \over {1000}} \times 1 \times 0.2$<br><br>$\Rightarrow M = {1 \over {20}}M$<br><br>= 0.05 M<br><br>= 50 $\times$ 10<sup>$-$3</sup> M<br><br>= 50 mM
About this question
Subject: Chemistry · Chapter: Some Basic Concepts of Chemistry · Topic: Mole Concept
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