Some Basic Concepts of Chemistry 2 Question 6
6. $50 \mathrm{~mL}$ of $0.5 \mathrm{M}$ oxalic acid is needed to neutralise $25 \mathrm{~mL}$ of sodium hydroxide solution. The amount of $\mathrm{NaOH}$ in $50 \mathrm{~mL}$ of the given sodium hydroxide solution is
(2019 Main, 12 Jan I)
(a) $40 \mathrm{~g}$
(b) $80 \mathrm{~g}$
(c) $20 \mathrm{~g}$
(d) $10 \mathrm{~g}$
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Solution:
- The reaction takes place as follows,
$$ \mathrm{H}{2} \mathrm{C}{2} \mathrm{O}{4}+2 \mathrm{NaOH} \longrightarrow \mathrm{Na}{2} \mathrm{C}{2} \mathrm{O}{4}+2 \mathrm{H}_{2} \mathrm{O} $$
Now, $50 \mathrm{~mL}$ of $0.5 \mathrm{M} \mathrm{H}{2} \mathrm{C}{2} \mathrm{O}_{4}$ is needed to neutralize $25 \mathrm{~mL}$ of $\mathrm{NaOH}$.
$\therefore$ Meq of $\mathrm{H}{2} \mathrm{C}{2} \mathrm{O}_{4}=$ Meq of $\mathrm{NaOH}$
$$ \begin{aligned} 50 \times 0.5 \times 2 & =25 \times M_{\mathrm{NaOH}} \times 1 \ M_{\mathrm{NaOH}} & =2 \mathrm{M} \end{aligned} $$
Now, $\quad$ molarity $=\frac{\text { Number of moles }}{\text { Volume of solution (in L) }}$
$=\frac{\text { Weight } / \text { molecular mass }}{\text { Volume of solution (in L) }}$
$2=\frac{w_{\mathrm{NaOH}}}{40} \times \frac{1000}{50}$
$w_{\mathrm{NaOH}}=\frac{2 \times 40 \times 50}{1000}=4 \mathrm{~g}$
Thus, $(*)$ none option is correct.
