n a qualitative analysis procedure, a chemist adds HCl to an unknown group of io

Question

n a qualitative analysis procedure, a chemist adds HCl to an unknown group of ions and then saturates the solution with H2S so that the solution is 0.5 M in HCl and 0.10 M in H2S.

(a) What is the [HS -] of the solution? (Note: The answer should be rounded to one significant figure. Be 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.)Hg2+   Fe2+   Cu2+ Pb2+ Mn2+   Ni2+ K+ Ag+

C)n a qualitative analysis procedure, a chemist adds HCl to an unknown group of ions and then saturates the solution with H2S so that the solution is 0.1 M in HCl and 0.10 M in H2S.(a) What is the [HS -] of the solution? (Note: The answer should be rounded to one significant figure. Be careful, 3.5 rounds up to 4, but 4.5 rounds down to 4!)

(d) If 0.01 M of each of the following ions is in the solution, which will form a precipitate? (Select all that apply.)

Hg2+   Fe2+ Cu2+ Pb2+   Mn2+ Ni2+   K+   Ag+

Explanation / Answer

H2S ------> H+   +    HS-

Ka of H2S = 9.1 *10-8

9.1 *10-8 =   x2 / 0.1-x

x is small so 0.1-x ~ 0.1

x = 0.95 *10-4

[HS-] = 0.95 *10-4 = 0.000095 ~ 0.0001 M

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Group II metals will precipitate

Ag+ (with chloride and sulfide)
Pb2+ (with chloride and sulfide)
Cu2+ (with sulfide)
Hg2+ (with sulfide)
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