Question

$125$ mL of $63$ % ($w/v$) $H_2{C}_2{O}_4.2H_{2}O$ is made to react with $125$ mL of a $40$ % ($w/v$) $NaOH$ solution. The resulting solution is:

• A. neutral
• B. acidic
• C. strongly acidic
• D. alkaline

Answer 1

The correct option is A neutral

Molecular weight of $H_2{C}_2{O}_4.2H_{2}O$ $=126$ g/mol.
Moles of $H_2{C}_2{O}_4.2H_{2}O$ = $\frac{0.63\times 125}{126}=0.625$
Molecular weight of $NaOH$ $=40$ g/mol
Moles of $NaOH$ = $\frac{0.4\times 125}{40}=1.25$
$2$ moles of $NaOH$ neutralizes $1$ mole of $H_2{C}_2{O}_4.2H_{2}O$. So, the solution contains neither base nor acid. Hence, solution is neutral.