What is the solubility for PbS including hydrolysis of ions (assume pH of the solution tobe equal to 7).

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Given that Ksp(PbS)=7x10^-29
Kb (Pb(OH))+=1.5x10^-8
Ka1 H2S=1.1x10^-1
Ka2 HS- = 10^-14

I got the concept but I just cant solve the equations.
They are

y^2/(x-y) = 6.7 x 10^-7
(z-a)(z+a)/(x-z)=1
a(a+z)/(z-a)=9.1 x 10^-8
(x-y)(x-z)= 7 x 10^-29

Please help!!

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2 Answers

Euclid ·2010-09-22 17:44:22

i m unable to grasp wat u hav done....

multiply the two dissociation constants of the H₂S and u can get the concentration of S^2- using other given data...

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swordfish ·2010-09-22 22:23:33

We cannot multiply them as the amount of reactants dissociated in 2 steps may be different.
I will tell you what I have done.

Let the solubility of PbS be x moles/litre.
so Initially, Pb2+ and S2- formed will be x (they will change on further reaction)
x mole of Pb2+ gives y mole of Pb(OH)+ and H3O+
x mole of S2- gives z mole of HS- and OH-
z mole of HS- gives a mole of H2S

Since HS- and OH- is present in both equlibriums, HS- present is z-a and OH- is z+a

At equilibrium,
following are the concentrations LEFT.
PB2+ = x-y Pb(OH)+ and H3O+ = y
S2- = x-z HS-=z-a and OH- =z+a
HS-= z-a H2S=a and OH-=z+a

I hope you got it. Please tell me how to solve the equations.

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What is the solubility for PbS including hydrolysis of ions (assume pH of the solution tobe equal to 7).

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