2. Consider the equilibria shown below: Pb 3 (PO 4 ) 2(s) 3 Pb 2+ (aq) + 2 PO 4

Question

2. Consider the equilibria shown below:

Pb3(PO4)2(s)     3 Pb2+(aq) + 2 PO43-(aq)                                 Keq = 1.0 x 10-54

Pb2+(aq) + 3 OH-(aq)     Pb(OH)3-(aq)                                               Keq = 8.0 x 1013

Pb3(PO4)2(s) + 9 OH-(aq)     3 Pb(OH)3-(aq) + 2 PO43-(aq)                                    

a) Calculate the equilibrium constant for the third equilibrium shown.

b) Calculate the equilibrium concentrations of Pb(OH)3- and of PO43- in a pH 10 solution containing the third equilibrium.

Explanation / Answer

we need to:

multiply rxn 2 by 3

Pb3(PO4)2(s)      3 Pb2+(aq) + 2 PO43-(aq)                                 Keq = 1.0 x 10-54 = 10^-54

3Pb2+(aq) + 9OH-(aq) 3Pb(OH)3-(aq)                                               Keq = (8*10^13)^3 = 5.12*10^41

add all

Pb3(PO4)2(s) + 9OH-(aq)    3Pb(OH)3-(aq) + 2 PO43-(aq) Ktotal Keq1Keq2 = (10^-54)(5.12*10^41) =5.12*10^-13

then

K = [Pb(OH)3-]^3 * [PO4-3]^2 / [OH-]^9

pH = 10 ; pOH = 4, [OH-] = 10^-4

5.12*10^-13 = (3S)^3*(2S)^2 / (10^-4)^9

108*S^5 = (5.12*10^-13)((10^-4)^9)

S = ((5.12*10^-49)/(108))^(1/5)

S = 8.613*10^-11

[Pb(OH)3-] = 3S = 3* 8.613*10^-11 =2.583*10^-10 M

[PO4-3] = 2S = 2* 8.613*10^-11 = 1.722*10^-10

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