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Chemistry

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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?

  • Chemistry -

    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

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