Question

The solubility of pbcl2 in water is 0.01 m at 25°c . Its maximum concentration in 0.1 m hcl will be

Answer 1

solubility of pbcl2 is 0.01 M  

$K_s_p = 1.6 * 10^-^5$

The dissociation of PbCl_2 is,    

$PbCl_2 -> Pb^+^2 + 2 Cl^-</p> <p>\\ K_s_p = [ Pb^+^2 ] [ 2 s]^2$

Chloride ion concentration is coming partially from lead chloride and partially from HCl = (0.01 + 0.1) M

$1.6 * 10^-^5 = [ Pb^+^2 ] [ 2 (0.01 + 0.1) ]^2</p> <p>\\ [ Pb^+^2 ] = (1.6 * 10^-^5) /[ 0.22]^2</p> <p>\\ [ Pb^+^2 ] = 3.3 * 10^-^4 mol /L$

Thus, concentration of Pb^+^2 in 0.1 m hcl is $3.3 * 10^-^4 mol /L$