Chemical And Ionic Equilibrium Result Question 8
8. An aqueous solution contains an unknown concentration of $\mathrm{Ba}^{2+}$. When 50 mL of a 1 M solution of $\mathrm{Na}_2 \mathrm{SO}_4$ is added, $\mathrm{BaSO}_4$ just begins to precipitate. The final volume is 500 mL . The solubility product of $\mathrm{BaSO}_4$ is $1 \times 10^{-10}$. What is the original concentration of $\mathrm{Ba}^{2+}$ ?
(2018 Main)
(a) $5 \times 10^{-9} \mathrm{M}$
(b) $2 \times 10^{-9} \mathrm{M}$
(c) $1.1 \times 10^{-9} \mathrm{M}$
(d) $1.0 \times 10^{-10} \mathrm{M}$
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Answer:
Correct Answer: 8. ( c )
Solution:
- Its given that the final volume is $500$ mL and this final volume was arrived when $50$ mL of $1$ $ \mathrm{M} $ $\mathrm{Na}_2 \mathrm{SO}_4$ was added to unknown $\mathrm{Ba}^{2+}$ solution.
So, we can interpret the volume of unknown $\mathrm{Ba}^{2+}$ solution as $450$ mL i.e.
$\underset{\substack{\mathrm{Ba}^{2+} \ \text { solution }}}{450 \mathrm{~mL}}+\underset{\substack{\mathrm{Na}_2 \mathrm{SO}_4 \ \text { solution }}}{50 \mathrm{~mL}} \longrightarrow \underset{\substack{\mathrm{BaSO}_4 \ \text { solution }}}{5000 \mathrm{~mL}}$
From this we can calculate the concentration of $\mathrm{SO}_4^{2-}$ ion in the solution via
$\begin{gathered} M_1 V_1=M_2 V_2 \\ 1 \times 50=M_2 \times 500 \end{gathered}$
(as $1 $ $\mathrm{M}$ $\mathrm{Na}_2 \mathrm{SO}_4$ is taken into consideration)
$M_2=\frac{1}{10}=0.1 \mathrm{M}$
Now for just precipitation,
Ionic product $=$ Solubility product $\left(K_{\text {sp }}\right)$
i.e. $\quad\left[\mathrm{Ba}^{2+}\right]\left[\mathrm{SO}4^{2-}\right]=K{\mathrm{sp}}$ of $\mathrm{BaSO}_4$
Given $K_{\text {sp }}$ of $\mathrm{BaSO}_4=1 \times 10^{-10}$
So, $\left[\mathrm{Ba}^{2+}\right][0.1]=1 \times 10^{-10}$
or $\left[\mathrm{Ba}^{2+}\right]=1 \times 10^{-9} \mathrm{M}$
Remember This is the concentration of $\mathrm{Ba}^{2+}$ ions in final solution. Hence, for calculating the $\left[\mathrm{Ba}^{2+}\right]$ in original solution we have to use
$\begin{aligned} M_1 V_1 & =M_2 V_2 \\ as \hspace {10mm} M_1 \times 450 & =10^{-9} \times 500 \\ so \hspace {10mm}M_1 & =1.1 \times 10^{-9} \mathrm{M} \end{aligned}$
BETA
