Equilibrium Question 187
Question: The solubility of $ BaSO_4 $ in water is $ 2.33\times {10^{-3}},gm/litre. $ Its solubility product will be (molecular weight of $ BaSO_4=233) $ [AIIMS 1998]
Options:
A) $ 1\times {10^{-5}} $
B) $ 1\times {10^{-10}} $
C) $ 1\times {10^{-15}} $
D) $ 1\times {10^{-20}} $
Show Answer
Answer:
Correct Answer: B
Solution:
The solubility of $ BaSO_4 $ in g/litre is given $ 2.33\times {10^{-3}} $
$ \because $ in mole/litre. $ n=\frac{W}{m.,wt} $
$ =1\times {10^{-5}} $
$ =,\frac{2.33\times {10^{-3}}}{233} $ Because $ BaSO_4 $ is a compound $ K _{sp}=S^{2}={{[1\times {10^{-5}}]}^{2}} $
$ =1\times {10^{-10}} $
