Question

What is the pH of a solution which begins to precipitate PbS (Ksp = 8.4 x I 0-28 from a solution which is 0.01 M with respect to Pb2+ and has an H2S concentration of 0.02M. Im totally stuck so if you could walk me through to the answer, that would be awesome.

Answer 1

Answer –

Given, [Pb²⁺] = 0.01 M, [H₂S]= 0.02M , Ksp of PbS = 8.4*10^-28

Now in this problem first we need to calculate the [S^2-]

We know the Ksp expression for the PbS

Ksp = [Pb²⁺][S^2-]

So, [S^2-] = Ksp / [Pb²⁺]

              = 8.4*10^-28 / 0.01

              = 8.4*10^-26 M

Now we know, the H₂S is the weak acid and there are two Ka values

H₂S ------> H⁺ + HS^- Ka1 = 1.1*10^-7

HS^- ------> H⁺ + S^2- , Ka2 = 1.3*10^-13

When we added the both equation then we will get total [H⁺]

H₂S ------> H⁺ + HS^- Ka1 = 1.1*10^-7

HS^- ------> H⁺ + S^2- , Ka2 = 1.3*10^-13

H₂S ------> 2H⁺ + S^2- , Ka = Ka1*Ka2 = 1.40*10^-20

So, Ka = [H⁺]²[S^2-] /[H₂S]

1.40*10^-20 =[H⁺]² * 8.4*10^-26 M / 0.020 M

So, [H⁺]² = 1.40*10^-20 *0.020 M /8.4*10^-26 M

             =3140.0

[H⁺]       = 58.35

We know , pH = -log [H⁺]

                     = -log 58.35

                    =-1.77