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A solution is saturated with \(\mathrm{SrCO}_{3}\) and \(\mathrm{SrF}_{2} .\) 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) \(5.1 \times 10^{-7} \mathrm{M}\)
(D) \(3.7 \times 10^{-2} \mathrm{M}\)
Given: \(\mathrm{K}_{\mathrm{sp}}\left(\mathrm{SrCO}_{3}\right)=7.0 \times 10^{-10}, \mathrm{~K}_{\mathrm{sp}}\left(\mathrm{SrF}_{2}\right)=7.9 \times 10^{-10}\)

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Ans: (D)
 \(\mathrm{SrCO}_{3}(\mathrm{~s}) \rightleftarrows \mathrm{Sr}^{2+}(\mathrm{aq})+\mathrm{CO}_{3}{ }^{2-}(\mathrm{aq})\)
$$
\begin{array}{r}
\mathrm{S}_{1}+\mathrm{S}_{2} \quad \mathrm{~S}_{1} \\
\mathrm{SrF}_{2}(\mathrm{~s}) \rightleftarrows \mathrm{Sr}^{2+}(\mathrm{aq})+2 \mathrm{~F}-(\mathrm{aq}) \\
\mathrm{S}_{1}+\mathrm{S}_{2} \quad 2 \mathrm{~S}_{2}
\end{array}
$$
Where \(\mathrm{S}_{1}\) and \(\mathrm{S}_{2}\) are solubilities of \(\mathrm{Sr} \mathrm{CO}_{3}(\mathrm{~s})\) and \(\mathrm{SrF}_{2}(\mathrm{~s})\) respectively.
\begin{aligned}
&\mathrm{S}_{1}=1.2 \times 10^{-3} \mathrm{M}(\mathrm{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} \\
&\mathrm{~K}_{\mathrm{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} \ldots \text { (ii) } \\
&\text { Dividing equation (ii) by equation (i) } \\
&\frac{7.9 \times 10^{-10}}{7 \times 10^{-10}}=\frac{4 \mathrm{~S}_{2}^{2}}{\mathrm{~S}_{1}} \\
&\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}^{-}\right]=2 \mathrm{~S}_{2}=2 \times 1.84 \times 10^{-2}=3.68 \times 10^{-2} \mathrm{M}
\end{aligned}
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