Question
Chemistry Question on coordination compounds
The solubility product of silver chloride is 1.8 × 10-10 at 298 k. The solubility of AgCl in 0.01 M KCl aqueous solution in mol dm-3 is
A
(A) 9.0 × 10-9
B
(B) 1.8 × 10-8
C
(C) 3.6 × 10-8
D
(D) 2.4 × 10-9
Answer
(B) 1.8 × 10-8
Explanation
Solution
Explanation:
Let solubility of AgCl in water = sAgCI(s) inwater ⟶Ag(aq)++C(aq)−kspAgCl=s2s2=1.8×10−10Now in 0.01MKCl aqueous solution Complete hydrolysis of KCl will take place and there will be a concentration of [Cl−]=0.01M0.01MKCl⟶0.01MK++0.01MCl−Now if solubility of AgCl in KCI solution is s' ThenAgCI(s)inKC1⟶Ag(aq)++CI(aq)−s′0.01+s′kspAgCl=s′(0.01+s′)s′ is small compare to 0.01 so neglecting it with respect to 0.01 kspAgCl=s′(0.01)1.8×10−10=s′×10−2s′=1.8×10−8molL−1=1.8×10−8moldm−3(1dm3=1Liter)