Question

You mix a 115.0 −mL sample of a solution that is 0.0104 M in NiCl2 with a 195.0 −mL sample of a solution that is 0.300 M in NH3.

After the solution reaches equilibrium, what concentration of Ni2+(aq) remains? The value of Kf for Ni(NH3)62+ is 2.0×108.

Express the concentration to two significant figures and include the appropriate units.

Answer 1

Ans.

Molarity of NiCl₂ = 0.0104M

Volume of NiCl₂ = 115.0mL = 115.0 x 10^-3L

No. of mol of NiCl₂ is

The chemical equilibrium for the formation of the complex is

1 mol of NiCl₂reacts with 6 mol of NH₃

Since the value of K_f is very high, assume that all NiCl₂ will react.

Therefore, 1.196 x 10^-3 mol of NiCl₂ will react with no. of mol of NH₃ given by

Molarity of NH₃ = 0.300M

Volume of NH₃ = 195.0mL

No. of mol of NH₃ is

Mol of NH₃ that remain is

From the balance reaction, the no. of mol of the complex will be equal to the no. of mol of NiCl₂ taken which is equal to 1.196 x 10^-3mol.

Total volume = 115.0mL + 195.0mL = 310.0mL = 310.0 x 10^-3L

Concentration of [Ni(NH₃)₆]²⁺ is

Therefore, the concentration of Ni²⁺ at equilibrium is 9.4 x 10^-7 mol/L.