What is the solubility (in M) of PbCl2 in a 0.15 M solution of HCl? The Ksp of PbCl2 is 1.6 x 10^-5?

7.11 × 10^-4M

Explanation:

PbCl2 will dissociate as follows

PbCl2 ⇌ Pb2+(x) + 2Cl- (2x)

HCl will dissociate as follows:

HCl ⇌H+ (0.15M) + Cl- (0.15M)

Ksp = {Pb2+} {Cl-}^2

Ksp = {x} {2x+ 0.15}^2

Due to common ion effect i.e. 2x<0.15, 2x will be neglected making the total concentration of chloride ion in the solution to be 0.15M

Ksp = {x} {0.15}^2

1.6 x 10^-5 = 0.0225x

x = 1.6 x 10^-5 ÷ 2.25 × 10^-2

x = 1.6/2.25 × 10^(-5+2)

x = 0.7111 × 10^-3

x = 7.11 × 10^-4

Therefore, the solubility is 7.11 × 10^-4M

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explanation:

the product of the pressure and volume is also constant