Equilibrium Question 699
Question: Solubility product of silver bromide is $ 5.0\times {10^{-13}} $ . The quantity of potassium bromid (molar mass taken as 120 g $ mo{l^{-1}} $ ) to be added to 1 L of 0.05 M solution of silver nitrate to start the precipitatin of AgBr is
Options:
A) $ 1.2\times {10^{-10}}g $
B) $ 1.2\times {10^{-9}}g $
C) $ 6.2\times {10^{-5}}g $
D) $ 5.0\times {10^{-8}}g $
Show Answer
Answer:
Correct Answer: B
Solution:
$ [AgBr]=[A{g^{+}}]=0.05M $
$ K _{sp}[AgBr]=[A{g^{+}}][B{r^{-}}] $
$ \Rightarrow [B{r^{-}}]=\frac{K _{sp}[AgBr]}{[A{g^{+}}]} $
$ =\frac{5.0\times {10^{-13}}}{0.05}={10^{-11}}M[mol{L^{-1}}] $ Moles of KBr needed to precipitate AgBr $ =[B{r^{-}}]\times V={10^{-11}}mol,{L^{-1}}\times 1,L={10^{-11}}mol $ Therefore, amount of KBr needed to precipitate AgBr $ ={10^{-11}}mol\times 120gmo{l^{-1}}=1.2\times {10^{-9}}g $
BETA
