Answer to Question #111151 in General Chemistry for mya 20

Question

A 0.170 mole quantity of NiCl2 is added to a liter of 1.20 M NH3 solution. What is the concentration of Ni2+ ions at equilibrium? Assume the formation constant of Ni(NH3)2+6 is 5.5×108 .

Answer 1

Solution:

1) 0.170 mole of NiCl₂ is added to a liter of 1.20 M NH₃ solution.

Therefore, the concentration of NiCl₂ is 0.170 mol / 1L = 0.170 M.

C_o(Ni²⁺) = C(NiCl₂) = 0.170 M.

2) Concentration of the NH₃ solution is 1.20 M.

The number of moles of ammonia is 1.20 mol.

The balanced chemical equation is:

Ni²⁺+6NH₃ ⇒ [Ni(NH₃)₆]²⁺

The formation constant of the reaction is K_f = 5.5×10⁸.

The formation constant of the reaction is very high so we can say that almost the reactants are consumed up to the maximum level.

According to the chemical equation: 1 mole of Ni²⁺ reacts with 6 moles of NH₃.

Thus, we have 0.170 mol of Ni²⁺ reacts with 6 × 0.170 = 1.02 moles of NH3.

Amount of NH₃ remain in the solution is n = (1.20 − 1.02) = 0.18 mol.

Let's write the ICE (Initial, Change, Equilibrioum) table:

5.5×10⁸ * x * 0.18⁶ = 0.17;

x = [Ni²⁺] = 9.09×10^-6 M.

[Ni²⁺] = 9.1×10^-6 M.

Answer: [Ni²⁺] = 9.1×10^-6 M.