Question

In: Chemistry

2. Consider the equilibria shown below: Pb3(PO4)2(s)→← 3 Pb2+(aq) + 2 PO43-(aq) &nbsp

2. Consider the equilibria shown below:

Pb₃(PO₄)_2(s)^→_← 3 Pb²⁺_(aq) + 2 PO₄^3-_(aq) K_eq = 1.0 x 10^-54

Pb²⁺_(aq) + 3 OH^-_(aq)^→_← Pb(OH)₃^-_(aq) K_eq = 8.0 x 10¹³

Pb₃(PO₄)_2(s) + 9 OH^-_(aq)^→_← 3 Pb(OH)₃^-_(aq) + 2 PO₄^3-_(aq)

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

b) Calculate the equilibrium concentrations of Pb(OH)₃^- and of PO₄^3- in a pH 10 solution containing the third equilibrium.

Expert Solution

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

queen_honey_blossom answered 3 years ago

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