Question

In: Chemistry

You mix a 131.0 mL sample of a solution that is 0.0115 M in NiCl2 with...

You mix a 131.0 mL sample of a solution that is 0.0115 M in NiCl2 with a 173.5 mL sample of a solution that is 0.256 M in NH3.

After the solution reaches equilibrium, what concentration of Ni2+(aq) remains? (Formation constant is Kf=2.0×108.)

Solutions

Expert Solution

The reaction between Ni (II) and ammonia can be shown as

Ni2+ (aq) + 6 NH3 (aq) -------> [Ni(NH3)6]2+ (aq)

Kf = [Ni(NH3)62+]/[Ni2+][NH3]6 = 2.0*108 ……(1)

Since the formation constant of the Ni-amine complex is very high, we can assume that the entire amount of Ni present in the solution has been complexed as Ni(II)-amine complex. Consequently, the molar concentration of Ni(II)-amine complex at equilibrium is the same as the initial concentration of Ni(II), i.e, 0.0115 M.

Moles of NH3 added = (volume of NH3 added)*(concentration of NH3) = (173.5 mL)*(0.256 mol/L) = 44.416 mmole (1 M = 1 mol/L).

Moles of Ni2+ added (as NiCl2) = (131.0 mL)*(0.0115 mol/L) = 1.5065 mmole.

As per the balanced stoichiometric equation,

1 mole Ni2+ = 6 moles NH3.

Therefore,

1.5065 mmole Ni2+ = (1.5065 mmole Ni2+)*(6 moles NH3/1 mole Ni2+) = 9.039 mmole NH3.

Mmoles of NH3 retained at equilibrium = (44.416 – 9.039) mmole = 35.377 mmole.

Total volume of the solution = (131.0 + 173.5) mL = 304.5 mL.

Equilibrium concentration of NH3 = (35.377 mmole)/(304.5 mL) = 0.1162 M.

Let x be the concentration of Ni2+ at equilibrium. We can write

Kf = [Ni(NH3)62+]/[Ni2+][NH3]6 = 2.0*108

===> 2.0*108 = (0.0115 M)/(x).(0.1162)6

===> 2.0*108 = (0.0115 M)/x.(2.4617*10-6)

===> x = (0.0115 M)/(2.0*108).(2.4617*10-6) = 2.3358*10-5 M

The concentration of free Ni2+ in the solution is 2.3358*10-5 M (ans).


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