Question

Calculate [Cu2+] at equilibrium when 0.20 mol of CuSO4 is added to 1.000 L of 1.20 M NH3 solution. Kf for [Cu(NH3)4]2+ is 5.0 × 1013.

Answer 1

The molarity of Cuso, is calculated as follows: moles Molarity =- volume 0.20 mol IL = 0.20M The reaction is given below: [Cu(NH; )]** (aq)=Cu2+ (aq) + 4NH3(aq) 0.20 M 0 1.20M 0.20-X x 1.20 + 4x Therefore, K, for this reaction is given as follows: K. - (NH3)*(Cu?) [Cu(NH3).]" Since, K, is known, that means the K, is given as follows: 1 5x1013 = 2x10-14

Putting the values: x 2.0 × 10-14 - (1.20+4.x (0.20-x) Since, x will be small, therefore, 1.20+4x 1.20 0.20-x=0.20 2.0 x 10-14 - (1.20 )*.« (0.20) x=1.9290x10-15 Therefore, at equilibrium the concentration of Cu2+ is 11.9290 x 10-151.