Question

A. The maximum amount of chromium(III) hydroxide that will dissolve in a 0.293 M chromium(III) acetate solution is (in M).

B. The molar solubility of lead bromide in a 0.116 M lead nitrate solution is (in M)

Answer 1

A )

[ Cr+3 ] = 0.293 M

Cr(OH)3 ------------------> Cr+3 + 3OH-

                                       S              3S

Ksp = [Cr+3][3S]^3

1.6 x 10^-30 =0.293 x 27 x S^3

S = 5.87 x 10^-11 M

The maximum amount of chromium(III) hydroxide = 5.87 x 10^-11 M

B)

[Pb+2] = 0.116 M

PbBr2 -----------------> Pb+2 + 2Br-

                                  0.116       2S

Ksp = [Pb+2] [Br-]^2

6.60 x 10^-6 = 0.116 (2S)^2

S = 3.77 x 10^-3 M

The molar solubility = 3.77 x 10^-3 M