Some Basic Concepts of Chemistry 2 Question 47
47. $5 \mathrm{~mL}$ of $8 \mathrm{~N}$ nitric acid, $4.8 \mathrm{~mL}$ of $5 \mathrm{~N}$ hydrochloric acid and a certain volume of $17 \mathrm{M}$ sulphuric acid are mixed together and made up to $2 \mathrm{~L} .30 \mathrm{~mL}$ of this acid mixture exactly neutralise $42.9 \mathrm{~mL}$ of sodium carbonate solution containing one gram of $\mathrm{Na}{2} \mathrm{CO}{3} \cdot 10 \mathrm{H}_{2} \mathrm{O}$ in $100 \mathrm{~mL}$ of water. Calculate the amount in gram of the sulphate ions in solution.
(1985, 4M)
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Solution:
- Molecular weight of $\mathrm{Na}{2} \mathrm{CO}{3} \cdot 10 \mathrm{H}_{2} \mathrm{O}=286$
$\Rightarrow$ Molarity of carbonate solution $=\frac{1}{286} \times \frac{1000}{100}=0.035$
$\Rightarrow$ Normality of carbonate solution $=2 \times 0.035=0.07 \mathrm{~N}$
In acid solution : Normality of $\mathrm{HNO}_{3}=\frac{8 \times 5}{2000}=0.02$
Normality of $\mathrm{HCl}=\frac{5 \times 4.8}{2000}=0.012$
Let normality of $\mathrm{H}{2} \mathrm{SO}{4}$ in final solution be $N$.
$$ \begin{aligned} \Rightarrow & & (N+0.02+0.012) \times 30 & =0.07 \times 42.9 \ \Rightarrow & & N & =0.0681 \end{aligned} $$
$\Rightarrow$ Gram equivalent of $\mathrm{SO}_{4}^{2-}$ in $2 \mathrm{~L}$ solution $=2 \times 0.0681$
$\begin{aligned} & =0.1362 \ \Rightarrow \text { Mass of } \mathrm{SO}_{4}^{2-} \text { in solution }=0.1362 \times \frac{96}{2} & =6.5376 \mathrm{~g}\end{aligned}$
