In: Chemistry
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.)
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).