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

Question

In 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 0.2 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!) WebAssign will check your answer for the correct number of significant figures. M (b) If 0.01 M of each of the following ions is in the solution, which will form a precipitate? (Select all that apply.) Mn2+ Pb2+ Cu2+ Ni2+ Fe2+ Ag+ K+ Hg2+

Hint: First, calculate [HS-] produced by the dissociation of H2S from the given concentrations of HCl and H2S. Second, calculate [OH-] from the given concentration of HCl. Then calculate Qsp for each of the metal sulfides. Remember, sulfides have unique solubility equilibria.

Explanation / Answer

Since HCL is a strong acid we can assume that it will fulyy ionise

HCl -----> H+ + Cl-

0.2 0 0

0 0.2 0.2

H2S ------- > H+ + HS-

Initial 0.10 0.2 0

Final 0.10-x 0.2+x x

I am taking pKa = 6.9

=> Ka = 1.259 *10-7 = x*(0.2+x)/(0.10-x)

=> x2 + 0.2x - 1.259*10-8 = 0 => x = 6.295* 10-8

=> Concentration of HS- = [HS-] = 6 * 10-8 M

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