In a qualitative analysis procedure, a chemist adds HCI to an unknown group of i

Question

In a qualitative analysis procedure, a chemist adds HCI to an unknown group of ions and then saturates the solution with H2S so that the solution is 0.2 M in HCI and 0.10 M in H2S. (a) What is the [HS ] of the solution? (Note: The answer should be rounded to one significant figure. 6e careful, 3.5 rounds up to 4, but 4.5 rounds down to 4!) (b) If 0.01 M of each of the following ions is in the solution, which will form a precipitate? (Select all that apply.) Hint: First, calculate [HS] produced by the dissociation of H2S from the given concentrations of HCI and H2S. Second, calculate [OH] from the given concentration of HCI. Then calculate Qsp for each of the metal sulfides. Remember, sulfides have unique solubility equilibria.

Explanation / Answer

Ka1 for H2S = 1 x10^-7

a) H2S ----------------------> H+ +    HS-

       0.1                              0.2        x                             (note : 0.2 M [H+] from strong acid HCl)

    Ka1 = [H+][HS-]/[ H2S]

   1 x10^-7 = (0.2) x [HS--]/0.1   

[HS- ] = 5 x10^-8 M

b)   answer: Cu+2,Pb+2, Hg+2 these ions only form precipitate

Explanation :

if we want to solve the problem first we need to get S^-2 concentration

HS-   -----------------> H+ + S^-2

       5x 10^-8                    0.2     x

Ka2 for HS- = 9.1 x 10^-8

Ka2 = [H+][S^-2]/[HS]

9.1 x 10^-8 = (0.2 ) [S^-2]/5x 10^-8

[S^-2] = 2.27 x10^-14 M

from the above all metal ions Cu+2,Pb+2, Hg+2   these three are second group basic radical we need to check only these three because here second group reagent (HCl + H2S) given

for Cu+2

CuS ionic product = [Cu+2][S^-2]= (0.01)(2.27 x10^-14) = 2.27 x 10^-16

CuS Ksp = 6.3x10-36

here ioni product > Ksp

precipitate will form

for Pb+2

PbS ionic product = [Pb+2][S^-2]= (0.01)2.27 x10^-14 = 2.27 x10^-16

Ksp =8.0x10-28

here also ionic product > Ksp

precipitate will form

for Hg^+2

ionic product =2.27 x10^-16

Ksp for HgS = 1.6x10-52

here also ionic product > Ksp

precipitate will form

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