Question

Use the Ksp value in the table to calculate the molar solubility of AgBr, Fe(OH)2, and PbBr2 in pure water

Answer 1

Molar solubility

Ksp

AgBr = 5.0 E-13

Fe(OH)2 = 2.5E-15

Pb(Br)2= 4.6E-6

AgBr (s) -- > Ag+ (aq) + Br- (aq)

I                                     0                    0

C     -x (molar sol.)      +x                  +x

E                                    x                  x

Ksp = [Ag+][Br-]

5.0E-13 = x²

x = 7.07 E-7 M

Molar solubility of AgBr = 7.07E-7 M

Fe(OH)2

            Fe(OH)2(s) --- > Fe²⁺ (aq) + 2 OH- (aq)

I                                               0                      0

C         -x                                 +x                    +2x

E                                              x                      2x

2.5 E-15 = x (2x)²

2.5E-15 = 4x³

x = 8.55 E-6 M

PbBr2

            PbBr2 (s)     ----- > Pb²⁺(aq) + 2 Br- (aq)

I                                               0                      0

C                                             +x                    +2x

E                                              x                      2x

4.6 E-6 = 4x³

x = 1.05 E-2 M