Question

10 mL of a 1 M solution of sodium hydroxide is pipetted into a 100 mL standard flask and made upto the mark. 10 mL of this solution requires 20 mL of HCl for complete neutralization. Calculate the molarity of HCl solution.

• A. 0.05 M
• B. 0.01 M
• C. 0.001 M
• D. 0.005 M

Answer 1

The correct option is A 0.05 M

Molarity of the original NaOH solution $V_1$= 1 M
Volume of the original NaOH solution $M_1$ = 10 mL
Volume of the diluted NaOH solution $V_2$= 100 mL
Molarity of the diluted NaOH solution $M_2$ = $\frac{V_1{M}_1}{V_2}=\frac{10\times 1}{100}=0.1$ M

The chemical reaction involved in the titration is,
$NaOH+HCl\to NaCl+H_{2}O$
Given,
Volume of the HCl solution required for the titration $V_3$ = 20 mL
Let, the molarity of $HCl$ solution be = $M_3$
Thus, $M_2{V}_2=M_3{V}_3$
Molarity of the HCl solution $M_3$ = $\frac{V_2{M}_2}{V_3}$
Molarity of HCl $M_3$ = $\frac{10\times 0.1}{20}=0.05$ M