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rakesh61

18.The solubility product of AgCl is 4 x 10^-10 at 298 K.The solubility of AgCl in 0.04M CaCl₂ will be :

(1)	2 x 10^-5 M	(2)	5 x 10^-9 M
(3)	10^-4 M	(4)	2.2 x 10^-4 M

magiclko

here normal solubility of AgCl , S = sqrrt (Ksp) = 2 X 10 ^ -5 [normal refers to the solubility of AgCl, when no CaCl2 is present ]

now in the presence of Cl(-) ions from CaCl2, this solubilty will be depressed, while the concentration of Cl- will be due to both CaCl2 and AgCl...so the equilibrium will be like this

AgCl <====> Ag(+) + Cl (-)

S' S'+ 0.08

now S' (S'+0.08) = Ksp

here as u mst have noticed S <<< 0.1

and S' will even be more smaller than S, so S'<<<<<0.1, so we can neglect S' in (S'+0.08)

so the last equation reduces to

S' (0.08) = 4X 10^-10

=> S' = 5 X 10^-9

arun-rashi

this is based on common ion effect funda where chloride ion of CaCl2 and AgCl are there .solubility product of AgCl is constant at fixed temperature so total Cl ion are (S+.o8) Ksp of AgCl =S (S+.08) so calculate S by neglecting S2.Solubility of silver chloride decreases by common ion effect.

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