when 20 ml of a .10 M silver nitrate solution is mixed with 25 ml of a 0.080 M solution of sodium sulfide, producing a precipitate of silver sulfide, what is the final concentration of sulfide ion?

Answer 1

This is a limiting reagent problem as well as a solubility produce problem.
2AgNO3 + Na2S ==> Ag2S + 2NaNO3
AgNO3 initially = 20*0.1M = 2 millimols.
It will produce 2 * 1/2 = 1 mmol Ag2S

Na2S initially = 25 x 0.08M = 2 mmols.
It will produce 2 mmols Ag2S.

You will produce the smaller amount of Ag2S (1 mmol) with some Na2S remaining. How much Na2S is left behind?
Na2S initially = 2 mmols.
Na2S used = 2 mmols AgNO3 x (1 mol Na2S/2 mol AgNO3) = 2 * 1/2 = 1 mmol Na2S used.
Remaining is 2 mmol - 1 mmol = 1 mmols.
(Na2S) = (S^2-) = (1 mmol/45 mL) = ?M