Equilibrium Question 907
Question: The solubility of $ AgCl $ in water at $ 25{}^\circ C $ is $ 1.79\times {10^{-3}}g/L $ . The $ K _{sp} $ of $ AgCl $ at $ 25{}^\circ C $ is
Options:
A) $ 1.68\times {10^{-12}} $
B) $ 1.55\times {10^{-10}} $
C) $ 12.4\times {10^{-8}} $
D) $ 1.73\times {10^{-14}} $
Show Answer
Answer:
Correct Answer: B
Solution:
Solubility of $ AgCl $ in water $ =1.79\times {10^{-3}}g/L $
$ =\frac{1.79\times {10^{-3}}}{143.5}mol/L $
$ =1.247\times {10^{-5}}mol/L $
For $ AgCl $
$ K _{sp}=S\times S=S^{2} $
$ ={{(1.247\times {10^{-5}})}^{2}} $
$ =1.55\times {10^{-10}}mol^{2}{L^{-2}} $
