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]
give me the answer please!
To find the equilibrium concentration of silver ions (Ag+) in the solution, we need to consider the formation of the complex ion Ag(NH3)2+.
The balanced equation for the reaction of Ag+ with NH3 to form Ag(NH3)2+ is:
Ag+(aq) + 2NH3(aq) ⇌ Ag(NH3)2+(aq)
According to the information provided, we have 0.050 M AgNO3 and we are adding 5.0 mL of 12.0 M NH3 to 500.0 mL of this solution.
To solve this problem, we need to follow these steps:
Step 1: Determine the initial moles of Ag+ present in the AgNO3 solution.
Moles = Volume (L) x Concentration (M)
Initial moles of Ag+ = 0.050 M x 0.500 L = 0.025 moles
Step 2: Determine the moles of Ag+ that react with NH3.
From the balanced equation, we can see that for every 1 mole of Ag+ that reacts, 2 moles of NH3 react. Therefore, the moles of Ag+ that react will be half the moles of NH3 added.
Moles of Ag+ that react = (5.0 mL / 1000 mL/L) x 12.0 M x 2 = 0.120 moles
Step 3: Determine the moles of Ag(NH3)2+ formed.
Since the reaction occurs in a 1:1 stoichiometric ratio (1 mole Ag+ to 1 mole Ag(NH3)2+), the moles of Ag(NH3)2+ formed will be equal to the moles of Ag+ that react.
Moles of Ag(NH3)2+ = 0.120 moles
Step 4: Determine the total volume of the final solution.
The total volume of the final solution will be the sum of the volumes of the original solution and the volume of NH3 added.
Total volume = 0.500 L + (5.0 mL / 1000 mL/L) = 0.505 L
Step 5: Calculate the equilibrium concentration of Ag+.
Equilibrium concentration (M) = Moles / Total volume (L)
Equilibrium concentration of Ag+ = 0.120 moles / 0.505 L ≈ 0.238 M
Therefore, the equilibrium concentration of silver ions (Ag+) in the solution will be approximately 0.238 M.
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^+).