Answer:
a) <u>Molarity NH3 = 0.307 M</u>
b)<u>Molarity Cu(NH3)4^2+= 0.0100 M</u>
<u>
</u>
c) [Cu2+] = 2.3 * 10^-13 M
Explanation:
Step 1: Data given
(Kf = 4.8 *10^12)
volume of a 1.10 M CuSO4 = 11.0 mL = 0.011 L
Volume of a 0.350 M NH3 = 1.20 L
Step 2: The balanced equation
Cu2+ + 4NH3 ⇆ Cu(NH3)42+
Step 3: Calculate moles
Moles = molarity * volume
Moles CuSO4 = 1.10 M * 0.011 L
Moles CuSO4 = 0.0121 moles
⇒ For 1 mol CuSO4 we'll have 1 mol Cu^2+
⇒ For 0.0121 moles CuSO4 we have 0.0121 moles Cu^2+
Moles NH3 = 0.350 M * 1.20 L
Moles NH3 = 0.42 moles
Step 3: Initial moles
n(Cu^2+) = 0.0121 moles
n(NH3) = 0.42 moles
n(Cu(NH3)4^2+ = 0 moles
Step 4: Calculate the limiting reactant
For 1 mol Cu^2+ we need 4 moles NH3 to produce 1 mol Cu(NH3)4^2+
Cu^2+ is the limiting reactant. It will completely be consumed (0.0121 moles). NH3 is in excess. There will react 4 * 0.0121 = 0.0484 moles
There will remain 0.42 - 0.0484 = 0.3716 moles
Step 5: Calculate the final molarity of NH3
Molarity = moles / volume
Molarity = 0.3716 moles / 1.211 L
<u>Molarity NH3 = 0.307 M</u>
Step 6: Calculate the final amount of moles Cu(NH3)4^2+
For 1 mol Cu^2+ we need 4 moles NH3 to produce 1 mol Cu(NH3)4^2+
For 0.0121 moles Cu^2+ well have 0.0121 moles Cu(NH3)4^2+
Step 7: Calculate molarity of Cu(NH3)4^2+
Molarity = moles / volume
Molarity = 0.0121 moles / 1.211 L
<u>Molarity Cu(NH3)4^2+= 0.0100 M</u>
<u>
</u>
Step 8: What is the concentration of Cu2+ in the resulting solution
Kf = [products] / [reactants]
Kf = [Cu(NH3)4^2+] / ([Cu2+] * [NH3]
^4)
[Cu2+] = [Cu(NH3)4^2+] / ( Kf * [NH3]^4)
[Cu2+] = (0.01000 / ( 4.8 *10^12 * 0.307^4)
[Cu2+] = 2.3 * 10^-13 M
