Equilibrium Question 809
Question: On addition of increasing amount of $ AgNO_3 $ to 0.1 M each of $ NaCl $ and $ NaBr $ in a solution, what % of $ B{r^{-}} $ ion get precipitated when $ C{l^{-}} $ ion starts precipitating. $ K _{sp}( AgCl )=1.0\times {10^{-10}} $ , $ K _{sp}(AgBr)=1\times {10^{-13}} $
Options:
A) 0.11
B) 99.9
C) 0.01
D) 9.99
Show Answer
Answer:
Correct Answer: B
Solution:
To precipitate the $ AgCl $
$ [A{g^{+}}] $ required $ =\frac{K _{sp}(AgCl)}{[C{l^{-}}]}=\frac{1.0\times {10^{-10}}}{0.1}=1.0\times {10^{-9}}M $
$ [ B{r^{-}} ] $ left at this stage $ =\frac{K _{sp}(AgBr)}{[A{g^{+}}]} $
$ =\frac{1.0\times {10^{-13}}}{1.0\times {10^{-9}}}=1.0\times {10^{-4}}M $ % of remaining $ [B{r^{-}}]=\frac{1.0\times {10^{-4}}}{0.1}\times 100=0.1 $ % of $ B{r^{-}} $ to be precipitated $ =100-0.1=99.9 $
BETA
