Question

10 ml of a solution containing 0.01 M $N{a}_{2}C{O}_3$ and 0.02 M $NaHC{O}_3$ requires what volume of 0.01 M $HCl$ for complete neutralization of the solution?

• A. 25 ml
• B. 35 ml
• C. 40 ml
• D. 50 ml

Answer 1

The correct option is C 40 ml

For neutralization reactions,
$N{a}_{2}C{O}_3+2HCl\to 2NaCl+H_{2}O+C{O}_2$
$NaHC{O}_3+HCl\to NaCl+H_{2}O+C{O}_2$

From stoichiometry,
2 moles of $HCl$ required to neutralize 1 mole of $N{a}_{2}C{O}_3$
Similarly,
1 mole of $HCl$ required to neutralize 1 mole of $NaHC{O}_3$

$\to$Let $M_1,M_2$ and $M_3$ are molarities of $HCl,N{a}_{2}C{O}_3$ and $NaHC{O}_3$ respectively.
Also, $V_1,V_2$ and $V_3$ are volumes of $HCl,N{a}_{2}C{O}_3$ and $NaHC{O}_3$ respectively.

$\therefore$ Total moles of $HCl$ consumed = 2$\times$moles of $N{a}_{2}C{O}_3$ neutralized + moles of $NaHC{O}_3$ neutralized
$\to$ $M_1{V}_1=2M_2{V}_2+M_3{V}_3$
Volume of $HCl$ consumed$\left(V_1\right)=\frac{2\times 0.01\times 10+0.02\times 10}{0.01}ml=40ml$