Equilibrium Question 850
Question: A solution is saturated with respect to $ SrCO_2 $ and $ SrF_2 $ The $ [ CO_3^{2-} ] $ was found to be $ 1.2\times {10^{-3}}M. $ The concentration of $ {F^{-}} $ in the solution would be Given $ K _{sp},of,SrCO_3=7.0\times {10^{-10}}M^{2} $ , $ K _{sp} $ of $ SrF_2=7.9\times {10^{-10}}M^{3} $
Options:
A) $ 1.3\times {10^{-3}}M $
B) $ 2.6\times {10^{-2}}M $
C) $ 3.7\times {10^{-2}}M $
D) $ 5.8\times {10^{-7}}M $
Show Answer
Answer:
Correct Answer: C
Solution:
The two $ K _{sp} $ values do not differ very much. So it is a case of simultaneous equilibria, where the concentration of any species cannot be neglected. $ \frac{[ S{r^{2+}} ]{{[ {F^{-}} ]}^{2}}}{[ S{r^{2+}} ][ CO_3^{2-} ]}=\frac{{K _{s{p _{SrF_2}}}}}{{K _{s{p _{SrCO_2}}}}} $
$ =\frac{7.9\times {10^{-10}}}{7.0\times {10^{-10}}}=1.128 $
$ \therefore {{[{F^{-}}]}^{2}}=1.128\times 1.2\times {10^{-3}}=13.5\times {10^{-4}} $
$ \therefore [ {F^{-}} ]={{( 13.5\times {10^{-4}} )}^{1/2}} $
$ =3.674\times {10^{-2}}\approx 3.7\times {10^{-2}}M $
