Q33 / 155 · WBJEE 2020 · Chemistry
A solution is saturated with $\mathrm{SrCO}_{3}$ and $\mathrm{Sr} \mathrm{F}_{2}$. The $\left[\mathrm{CO}_{3}^{2-}\right]$ is found to be $1.2 \times 10^{-3} \mathrm{M}$. The concentration of $\mathrm{F}$ - in the solution would be
- A$3.7 \times 10^{-6} \mathrm{M}$
- B$3.2 \times 10^{-3} \mathrm{M}$
- C$5.1 \times 10^{-7} \mathrm{M}$
- Correct answerD$3.7 \times 10^{-2} \mathrm{M}$ Given : $\mathrm{K}_{\mathrm{sp}}\left(\mathrm{SrCO}_{3}\right)=7.0 \times 10^{-10}, \mathrm{~K}_{\mathrm{sp}}\left(\mathrm{SrF}_{2}\right)=7.9 \times 10^{-10}$
Solution
Hint:
$\begin{aligned}
\mathrm{Sr} \mathrm{CO}_{3}(\mathrm{~s}) \stackrel{\mathrm{S}}{\rightleftarrows} \mathrm{Sr}^{2+}(\mathrm{aq})+\mathrm{CO}_{3}^{2-}(\mathrm{aq}) \\
\mathrm{S}_{1}+\mathrm{S}_{2} \quad \mathrm{~S}_{1} \\
\mathrm{SrF}_{2}(\mathrm{~s}) \rightleftarrows \mathrm{Sr}^{2+}(\mathrm{aq})+2 \mathrm{~F}^{-}(\mathrm{aq}) \\
\mathrm{S}_{1}+\mathrm{S}_{2} \quad 2 \mathrm{~S}_{2}
\end{aligned}$
Where $\mathrm{S}_{1}$ and $\mathrm{S}_{2}$ are solubilities of $\mathrm{SrCO}_{3}(\mathrm{~s})$ and $\mathrm{Sr} \mathrm{F}_{2}(\mathrm{~s})$ respectively.
$\begin{array}{l}
\mathrm{S}_{1}=1.2 \times 10^{-3} \mathrm{M}(\text { Given }) \\
\mathrm{K}_{\mathrm{sp}}\left(\mathrm{SrCO}_{3}\right)=\left[\mathrm{Sr}^{2+}\right]\left[\mathrm{CO}_{3}^{-2}\right] \\
7 \times 10^{-10}=\left(\mathrm{S}_{1}+\mathrm{S}_{2}\right) \mathrm{S}_{1}-\ldots-\text { (i) } \\
\mathrm{K}_{\text {sp }}\left(\mathrm{SrF}_{2}\right)=\left[\mathrm{Sr}^{2+}\right]\left[\mathrm{F}^{-}\right]^{2} \\
7.9 \times 10^{-10}=\left(\mathrm{S}_{1}+\mathrm{S}_{2}\right)\left(2 \mathrm{~S}_{2}\right)^{2}
\end{array}$
Dividing equation (ii) by equation (i)
$\begin{array}{l}
\frac{7.9 \times 10^{-10}}{7 \times 10^{-10}}=\frac{4 \mathrm{~S}_{2}^{2}}{\mathrm{~S}_{1}} \\
\begin{array}{l}
\therefore \mathrm{S}_{2}=\sqrt{\frac{7.9}{7} \times \frac{1.2 \times 10^{-3}}{4}} \\
\quad=1.84 \times 10^{-2} \\
\therefore\left[\mathrm{F}-\mathrm{]}=2 \mathrm{~S}_{2}=2 \times 1.84 \times 10^{-2}=3.68 \times 10^{-2} \mathrm{M}\right.
\end{array}
\end{array}$
Hence, the correct answer is option: 'D'.