Question

Q. Given that the solubility product of PbBr₂ is 6.200×10^-6 determine the molar solubility of PbBr₂:
a) In pure water:

b)In a 0.201 M KBr solution:

c)In a 0.364 M Pb(NO₃)₂ solution:

Answer 1

a) In pure water:

PbBr2(s) --------------------> Pb^2+ (aq) + 2Br^- (aq)

                                             x                    2x

Ksp     =   [Pb^2+][Br^-]^2

6.2*10^-6    = x*(2x)^2

x    = 0.0116

The molar solubility of PbBr2 in water is 1.16*10^-2M

b)In a 0.201 M KBr solution:

    KBr(aq) ----------------> K^+ (aq) + Br^- (aq)

     0.201M                                        0.201M

PbBr2(s) --------------------> Pb^2+ (aq) + 2Br^- (aq)

                                             x                    2x+0.201

Ksp     =   [Pb^2+][Br^-]^2

6.2*10^-6    = x*(2x+0.201)^2

6.2*10^-6    = x*(0.201)^2                   [ 2x+0.201 , 2x<<<<0.201 ]

      x   = 0.000153

The solubility of PbBr2 in 0.201M KBr is 1.53*10^-4M

c)In a 0.364 M Pb(NO₃)₂ solution:

Pb(NO3)2(aq) ----------> Pb^2+ (aq) + 2NO3^- (aq)

0.364M                              0.364M

PbBr2(s) --------------------> Pb^2+ (aq) + 2Br^- (aq)

                                             x +0.364      2x

Ksp     =   [Pb^2+][Br^-]^2

6.2*10^-6    = (x+0.364)*(2x)^2

6.2*10^-6    = 0.364*(2x)^2                   [ x+0.364 , x<<<<0.364 ]

   x   =   0.00206

The solubility of PbBr2 in 0.364M Pb(NO3)2   = 2.06*10^-3M