Question

If 2.40g of CuSO4 is dissolved in 9.49x10^2 mL of 0.380M NH_​3, calculate the concentrations of the following species at equilibrium. Cu²⁺ (in scientific notation), NH_​3 , Cu(NH3)4^+​2

Answer 1

CuSO₄(aq) + 4NH₃(aq) -----> [Cu(NH₃)₄]SO₄

K_f of [Cu(NH₃)₄]²⁺ = 1.1*10¹³

Molar mass of CuSO₄ = 159.5 g/mole

Thus, moles of CuSO₄ in 2.4 g of it = mass/molar mass = 2.4/159.5 = 0.015

Molar concentration of CuSO₄ initially = moles/volume of solution in litres = 0.015/0.949 = 0.0159 M

Molar concentration of NH₃ initially = 0.38 M

Thus, let at eqb., [CuSO₄] = (0.0159-x) M; [NH₃] = (0.038-x) & [Cu(NH₃)₄]²⁺

Thus, K_f = {[Cu(NH₃)₄]SO₄}/{[CuSO₄]*[NH₃]⁴}

or, 1.1*10¹³ = x/{(0.0159-x)*(0.038-x)⁴}

solve for x