Chemistry Solubility Math Question 2

Question

How much potassium oxalate can be added to 90 mL of 0.4 M [Cu(NH₃)₄]Cl₂ before a precipitate of CuC₂O₄ begins to form? K_sp(CuC₂O₄)=2.9x10^-8 and K_f(Cu(NH₃)₄²⁺)=1.1x10¹²

Answer 1

Formation of [Cu(NH₃)₄]²⁺ complex:

Cu²⁺(aq) + 4NH₃(aq) <==> [Cu(NH₃)₄]²⁺(aq)

Kf = 1.1x10¹² = [Cu(NH₃)₄]²⁺ / [Cu²⁺][NH₃]⁴

1.1x10¹² = 0.4 M/(x)(4x)⁴

x = 1.7x10^-2 M = [Cu²⁺]

CuC₂O₄(s) <==> Cu²⁺(aq) + C₂O₄^2-(aq)

Ksp = 2.9x10^-8 = [Cu²⁺][C₂O₄^2-]

2.9x10^-8 = (1.7x10^-2)(C₂O₄^2-)

[C₂O₄^2-] = 1.7x10^-6 M

To obtain 1.7x10^-6 M C₂O₄^2- in 90 ml using K₂C₂O₄ with a molar mass of 184 g/mole, we have...

x mol/0.090 L = 1.7x10^-6 moles/L

x = 1.5x10^-7 moles

1.5x10^-7 moles K₂C₂O₄ x 184 g/mol = 2.8x10-5 g of K₂C₂O₄