Question

You mix a 131.0 mL sample of a solution that is 0.0118 M in NiCl2 with a 171.0 mL sample of a solution that is 0.264 M in NH3. After the solution reaches equilibrium, what concentration of Ni2+(aq) remains? (Formation constant is Kf=2.0×108.)

Answer 1

moles of NiCl2 = 0.0118*0.131 = 0.0015458 moles NiCl2
moles of NH3 = 0.264*0.171 = 0.045144 moles NH3
Assuming the formula of the complex ion is: [Ni(NH3)6]^+2
NiCl2 + 6NH3 ==> [Ni(NH3)₆]^+2 + 2Cl-
0.0015458 moles NiCl2 * 6 moles NH3/1mole NiCl2 = 0.0092748 moles of NH3 are consumed.
Moles of NH3 that remain = 0.045144 – 0.0092748 = 0.0358692
Moles of [Ni(NH3)6]+2 that are formed =
(0.0015458 moles NiCl2 * 1mole [Ni(NH3)6]+2)/ 1 mole NiCl2 = 0.0015458 moles ]Ni(NH3)6]+2.
Final volume = 131.0mL + 171.0 mL = 302 mL = 0.302 Liters
[NH3] = 0.0358692 moles/0.302 liters = 0.1187 M
[Ni(NH3)6]+2 = 0.0015458 moles/0.302 liters = 0.005118 M
Kf = [Ni(NH3)₆]^+2/ [Ni+2][NH3]^6
Kf = 2.0*10^8
2.0*10^8 = (0.005118)/[Ni+2](0.1187)^6
[Ni+2] = 0.005118/(0.1187)^6*(2.0*10^8)
[Ni+2] = 9.1488*10^-6 M