Question

If 2.10 g of CuSO4 is dissolved in 8.31 × 102 mL of 0.350 M NH3, calculate the concentrations of the following species at equilibrium. Cu2+ × 10 MEnter your answer in scientific notation. NH3 M Cu(NH3)42+ M

Answer 1

As given the mass of CuSO4 used is = 2.1 grams

Wec an calcualte the moles as

Moles = mass / molar mass = 2.1g/ 159.6g/mol = 0.0132 moles

The molarity = Moles / Volume (L) = 0.0132 / 0.831 = 0.0159 M

The reaction for Formation of complex [Cu(NH₃)₄²⁺] is

               Cu²⁺(aq) + 4NH₃(aq) ------>Cu(NH₃)₄²⁺(aq)

Each mole of copper ion will react with four moles of NH3

[NH3] reacted = 4 X [Cu+2] = 4 X 0.0159 M = 0.0636 M

Given that the initial molarity of NH3 is 0.350 M so the ammoni is in excess than the required amount

[NH3] left = 0.350 - 0.0636 = 0.2864 M

the limiting reagent is copper ion so the moles of complex formed = moles of copper ions used

K_f = 5.0 X 10¹³ = [Cu(NH₃)₄²⁺] / [Cu²⁺][NH₃]⁴

Putting values

5.0 X 10¹³ = 0.0132 / [Cu²⁺][0.2864]⁴

[Cu²⁺] = 0.0132 / 5.0 X 10¹³ X [0.2864]⁴ '

[Cu²⁺] = 3.923 X 10^-14 M