MCQ Solution: Ionic Equilibrium
The solubility product of silver chloride is 1.8x10-10 at 298 k. The soluability of AgCl in 0.01m HCl solution in more cm-3 is
1) 3.6 x 10-8
2) 2.4 x 10-9
3) 1.8 x 10-8
4) 0.9 x 10-9
Solution
Let solubility of AgCl in 0.01 M HCl =xmolL−1
∴[Ag+]=xmolL-1 ltbr [Cl−]=[HCl]=0.01M=10−2M
[Cl−] from AgCl can be neglected ]
Ksp=[Ag+][Cl−]
1.8×10−10=x×10−2⇒x=1.8×10−8
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