Question

Consider the following equilibrium: Cu2+ + 4NH3 Cu(NH3)42+ (Kf = 4.8e+12) A solution is made by mixing 11.0 mL of 1.10 M CuSO4 and 1.20 L of 0.350 M NH3. Assume additive volumes to answer the following questions. Give all answers to three sig figs. a) What is the concentration of NH3 in the resulting solution? [NH3] = M b) What is the concentration of Cu(NH3)42+ in the resulting solution? [Cu(NH3)42+] = M c) What is the concentration of Cu2+ in the resulting solution? [Cu2+] = M

Answer 1

Answer:

a) [NH₃ ] = 0.0396 M

b) [Cu(NH₃)₄²⁺] = 0.0099 M

c) [Cu²⁺] = 8.39x10⁻¹⁰ M

Explanation:

Given data:

VCuSO₄ = 11 mL = 0.011 L

[CuSO₄] = 1.1 M

[NH₃] = 0.35 M

VNH₃ = 1.2 L

$[CuSO_{4} ]=\frac{1.1*0.011}{0.011+1.2} =0.0099M$

$[NH_{3} ]=\frac{0.35*1.2}{0.011+1.2} =0.347M$

The chemical reaction is:

             Cu²⁺        +          4NH₃          =          Cu(NH₃)₄²⁺

I           0.0099                  0.347                      0

C         -0.0099                -4*0.0099              +0.0099

E          x                            0.0396

The [Cu²⁺] is:

$[Cu^{2+} ]= \frac{[Cu(NH_{3})_{4} ^{2+} ] }{Kf*[NH_{3}]^{4} } =\frac{0.0099}{4.8x10^{12}*0.0396^{4} } =8.39x10^{-10} M$

[NH₃ ] = 0.0396 M

[Cu(NH₃)₄²⁺] = 0.0099 M

Answer 2

Answer:

a) Molarity NH3 = 0.307 M

b)Molarity Cu(NH3)4^2+= 0.0100 M

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

Molarity NH3 = 0.307 M

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

Molarity Cu(NH3)4^2+= 0.0100 M

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