Question

Solubility of AgCl in 0.1 M CaCl2 . KSP= 1.8×10^-10

Answer 1

Answer:

Let the solubility of AgCl be S. 

Then,

AgCl⇌Ag++Cl−

S             S           S

Potassium chloride is a strong electrolyte and is completely ionised. It shall provide Cl− ion concentration =0.1M

[Ag+]=S

[Cl−]=(S+0.1)M

Ksp=[Ag+][Cl−]=S×(S+0.01)=S2+0.1S

Neglecting S3 and S2

1.20×10−10=0.1S

or S=0.11.20×10−10=1.20×10−9molL−1

Answer 2

Answer:

1.2×10

−9

M

Let the solubility of AgCl be S.

Then,

AgCl

⇌Ag

+

+Cl

−

S S S

Potassium chloride is a strong electrolyte and is completely ionised. It shall provide Cl

−

ion concentration =0.1M

[Ag

+

]=S

[Cl

−

]=(S+0.1)M

K

sp

=[Ag

+

][Cl

−

]=S×(S+0.01)=S

2

+0.1S

Neglecting S

3

and S

2

1.20×10

−10

=0.1S

or S=

0.1

1.20×10

−10

=1.20×10

−9

molL

−1