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?
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
To find the final concentration of sulfide ion, we need to use the concept of stoichiometry and the balanced chemical equation for the reaction between silver nitrate (AgNO3) and sodium sulfide (Na2S) to form silver sulfide (Ag2S).
The balanced chemical equation for the reaction is:
2 AgNO3 + Na2S → Ag2S + 2 NaNO3
From the balanced equation, we can see that 2 moles of silver nitrate react with 1 mole of sodium sulfide to form 1 mole of silver sulfide. Therefore, the ratio of AgNO3 to Na2S is 2:1.
Given that the initial volume of the silver nitrate solution is 20 mL and the initial volume of the sodium sulfide solution is 25 mL, we can calculate the number of moles of AgNO3 and Na2S.
Number of moles of AgNO3 = Volume (in L) × Molarity = 20 mL × 0.10 M = 0.0020 moles
Number of moles of Na2S = Volume (in L) × Molarity = 25 mL × 0.080 M = 0.0020 moles
Since the stoichiometric ratio is 2:1, the number of moles of AgNO3 and Na2S are equal, and hence, all the sulfide ion from Na2S will react with AgNO3.
Therefore, the final concentration of sulfide ion (S2-) is the same as its initial concentration, which is 0.080 M.