Silver ion concentrations are needed for a solution. You have 0.050 M AgNO3, if you added 5.0mL of 12. M NH3 to 500.mL of this solution, how much silver ion in M will you have at equilibrium? [Kf for Ag(NH3)2+ is 1.5 x 10^7]

Answer 1

mols Ag^+ = 0.05M x 0.500L = 0.025 mols or 0.025/0.505 = 0.02475 M.
mols NH3 = 12M x 0.005L = 0.06 mols or 0.06/0.505 = 0.1188 M.
............Ag^+ + 2NH3 ==> Ag(NH3)2^+
Init.....0.025.0.06......0
change-0.025..-0.05..0...025
equil...0......0.010...0.025

equilibrium concns:
(Ag^+) = x
(NH3)2 = (0.010/0.505) = 0.0198 M
[Ag(NH3)2^+] = 0.025/0.505 = 0.0495 M

Set up Kf, substitute and solve for x = free (Ag^+).

Answer 2

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