Question

 

a) The molar solubility of iron(III) sulfide in a 0.249 M iron(III) nitrate solution is __________ M.

b) The maximum amount of iron(II) sulfide that will dissolve in a 0.164 M sodium sulfide solution is ___________ M.

c) The molar solubility of lead iodide in a 0.170 M sodium iodide solution is ______________ M.

Answer 1

a. Fe(NO3)3 ---------------> Fe^3+ (aq) + 3NO3^-

0.249M                        0.249M

       Fe2S3(s) ---------------> 2Fe^3+ (aq) + 3S^2-

                                           2s+0.249           3s

    [Fe^3+]    = 2s+0.249    = 0.249                 [2s<<<<<0.249]

    Ksp    = [Fe^3+]^2[S^2-]^3

1.4*10^-88            = (0.249)^2*(3s)^3

              s     = 4.37*10^-30

    solubility of Fe2S3 is 4.37*10^-30M

b.     Na2S -------------------> 2Na^+ (aq) + S^2- (aq)

           0.164M                                          0.164M

FeS(s) ------------------> Fe^2+ (aq) + S^2- (aq)

                                      s                 s+0.174

Ksp = [Fe^2+][S^2-]

4.9*10^-18          = s*(0.174)

    s   = 2.82*10^-17

solubility of FeS = 2.82*10^-17 M

c.     NaI -------------------> Na^+ (aq) + I^- (aq)

        0.17M                                          0.17M

        PbI2 -------------------> Pb^2+ (aq) + 2I^- (aq)

                                          s                    2s+0.17

       Ksp = [Pb^2+][I^-]^2

8.7*10^-9        = s*(0.17)^2

     s   = 3*10^-7

solubility of PbI2 = 3*10^-7 M