Question

If 2.35 g of Cuso, is dissolved in 0.51 x 102mL of 0.380 M NH3, calculate the concentrations of the following species at equilibrium. x 10M Enter your answer in scientific notation. NH3 Cu(NH3)22+

Please show all work so that it's easy to follow and understand.

Answer 1

i)Cu²⁺ ion form complex with NH3

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

K_f = [Cu(NH3)4²⁺]/([Cu²⁺][NH3]⁴) = 1.1×10¹³

ii) Let us assume the reaction is complete

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

stoichiometrically, 1mole of Cu²⁺ react with 4moles of NH3 to give 1mole of Cu(NH3)4²⁺

given moles CuSO4 = 2.35g/ 159.61g/mol = 0.01472mol

given moles of Cu²⁺ = 0.01472mol

[Cu²⁺] = (0.01472mol/951mol)× 1000ml = 0.01548M

[NH3] = 0.380M

0.01548M of Cu²⁺ react with 0.06192M of NH3 to give 0.01548M of Cu(NH3)4²⁺

After completion of reaction remaining concentration of NH3 = 0.380M - 0.06192M = 0.3181M

iii) Now, consider the dissociation equillibrium of Cu(NH3)4²⁺

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

Kd = [NH3]⁴[Cu²⁺]/[Cu(NH3)4²⁺]

Kd = 1/K_f = 1/ (1.1 ×10¹³) = 9.09×10^-14

iv) Initial concentration

[Cu(NH3)4²⁺] = 0.01548

[Cu²⁺] = 0

[NH3] = 0.3181

change in concentration

[Cu(NH3)4²⁺] = -x

[Cu²⁺] = + x

[NH3] = + 4x

Equilibrium concentration

[Cu(NH3)4²⁺] = 0.01548 - x

[Cu²⁺] = x

[NH3] = 0.3181 + 4x

so,

x(0.3181 + 4x)⁴ / (0.01548 - x ) = 9.09×10^-14

we can assume 0.3181 + 4x = 0.3181 and 0.01548 - x = 0.01548 because x is small value

x ( 0.3181)⁴ / 0.01548 = 9.09 ×10^-14

x 0.6614 = 9.09×10^-14

x = 1.37 × 10^-13

Therefore,

at equillibrium

[Cu²⁺] = 1.37 ×10^-13M

[NH3] = 0.3181M

[Cu(NH3)4²⁺] = 0.01548M