Question

Calculate the molar solubility and the solubility in g/L of each salt at 25^oC:

a) PbF2 Ksp = 4.0 x 10^-8

b) Ag2CO3 Ksp = 8.1 x 10^-12

c) Bi2S3 Ksp = 1.6 x 10^-72

Answer 1

a) PbF2 ------ Pb(2+) + 2F-

     x               0          0

     0               x         2x

Molar mass of PbF2 = 245.2 gm/mol

Hence the solubility in g/L will be equal to 2.1544 * 245.2 gm/L * 10^(-3) = 0.528 g/L

b) Ag2CO3 ------ 2Ag(+) + CO3(2-)

     x               0          0

     0               2x         x

Molar mass of Ag2CO3= 275.74 gm/mol

Hence the solubility in g/L will be equal to 1.265 * 275.74 gm/L * 10^(-4) = 0.0348 g/L

c)

Bi2S3 ------ 2Bi(3+) + 3S(2-)

x                  0            0

0                2x           3x

Ksp = (2x)^2(3x)^3 = 108x^5

108x^5 = 1.6 * 10^(-72)

x^5 = 14.8148 * 10^(-75)

x = 1.714 * 10^(-15) M

Molar mass of Bi2S3 = 514.16 gm/mol

Hence solubility in g/L = 8.8153 * 10^(-13) g/L