Asked by Chris
50 mL of 1M silver Nitrate is combined with 25 mL of 2 M potassium Chloride. What is the concentration of a resulting solution
Answers
Answered by
DrBob222
The question makes no sense. What's the concn of what? Ag^+, Cl^-, K^+, or NO3^-. And AgCl is insoluble.
Answered by
Chris
Idk my teacher gave me this question for homework and i had no idea how to do this. But this has to do with molarity and stoichiometrty if that helps
Answered by
DrBob222
OK. So we give a complete answer.
AgNO3 + KCl ==> AgCl + KNO3
moles AgNO3 initially = L x M = 0.050 x 1 = 0.050 moles.
moles KCl initially = 0.025 x 2 = 0.050 moles.
Since the ratio is 1:1, there are exactly enough moles AgNO3 to react with KNO3. Therefore, moles AgCl formed equals 0.050 moles and the same for KNO3. AgCl is insoluble and KNO3 ionizes completely.
(KNO3) = 0.050 moles/0.100 L = 0.50 M and since it ionizes completely we know (K^+) = 0.5M and (NO3^-) = 0.5M.
AgCl is insoluble, the solution will be saturated with respect to AgCl. Have you gone through how to solve for Ag^+ and Cl^-.
AgCl(s) ==> Ag^+ + Cl^-
If we let solubility AgCl = X, then Ag^+ = X and Cl^- = X
The solubility product = Ksp = (Ag^+)(Cl^-). You can look it up in your text or notes but the number in my text (look up yours for it may very well be different) is 1.1 x 10^-10 so
(X)(X) = 1.1 x 10^-10.
Solve for X and it is 1.05 x 10^-5 M; therefore, the (Ag^+) = (Cl^-) = 1.05 x 10^-5M. You can look up the value in your text and adjust the answer accordingly. This is about as complete an answer as you an get.
AgNO3 + KCl ==> AgCl + KNO3
moles AgNO3 initially = L x M = 0.050 x 1 = 0.050 moles.
moles KCl initially = 0.025 x 2 = 0.050 moles.
Since the ratio is 1:1, there are exactly enough moles AgNO3 to react with KNO3. Therefore, moles AgCl formed equals 0.050 moles and the same for KNO3. AgCl is insoluble and KNO3 ionizes completely.
(KNO3) = 0.050 moles/0.100 L = 0.50 M and since it ionizes completely we know (K^+) = 0.5M and (NO3^-) = 0.5M.
AgCl is insoluble, the solution will be saturated with respect to AgCl. Have you gone through how to solve for Ag^+ and Cl^-.
AgCl(s) ==> Ag^+ + Cl^-
If we let solubility AgCl = X, then Ag^+ = X and Cl^- = X
The solubility product = Ksp = (Ag^+)(Cl^-). You can look it up in your text or notes but the number in my text (look up yours for it may very well be different) is 1.1 x 10^-10 so
(X)(X) = 1.1 x 10^-10.
Solve for X and it is 1.05 x 10^-5 M; therefore, the (Ag^+) = (Cl^-) = 1.05 x 10^-5M. You can look up the value in your text and adjust the answer accordingly. This is about as complete an answer as you an get.
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