Question

Using (Kf=1.2×109) calculate the concentration of Ni2+(aq), assuming that it is present as Ni(NH3)2+6, at equilibrium after dissolving 1.28 gNiCl2 in 100.0 mL of 0.20 MNH3(aq).

Answer 1

Concentration of Ni²⁺ = 0.06984M

Explanation

i) Given moles of NH3 = (0.20mol/1000ml)×100ml = 0.020mol

Given moles of NiCl2 = 1.28g/129.593g/mol = 0.009877mol

moles of Ni²⁺ = 0.009877mol

ii) Assume the complete reaction between Ni²⁺ and NH3

Ni²⁺(aq) + 6NH3(aq) --------> Ni(NH3)6²⁺(aq)

0.009877moles of Ni²⁺ should  react with 6×0.009877 = 0.059262moles of NH3 but available moles of NH3 is 0.02moles. So, limiting reagent is NH3

0.02 moles of NH3 react with 0.02mol/6 = 0.003333moles of NH3

moles of Ni(NH3)6²⁺ formed = 0.003333

remaining moles of Ni²⁺= 0.009877mol - 0.003333mol = 0.006544

[Ni(NH3)²⁺] = (0.003333mol/100ml)×1000ml =0.03333M

[Ni²⁺] = (0.006544mol/100ml) ×1000ml = 0.06544M

iii) Consider the dissociation equillibrium of Ni(NH3)6²⁺

Ni(NH3)6²⁺ <------> Ni²⁺(aq) + 6NH3(aq)

K_d = [Ni²⁺][NH3]⁶/[Ni(NH3)6²⁺]

K_f = 1/K_d

K_d = 1/ 1.2×10⁹ = 8.33×10^-10

iv) at equillibrium

[Ni²⁺] = 0.06544 + x

[NH3] = 6x

[Ni(NH3)6²⁺] = 0.03333 - x

Therefore,

(0.06544 + x)(6x)⁶/(0.0333 -x) = 8.33×10^-10

solving for x

x = 0.004400

Therefore,

[Ni²⁺] = 0.06544 + 0.004400= 0.06984M