Chapter 16, Problem 72E

A solution is prepared by adding 0.10 mole of Ni(NH₃)₆ Cl₂ to 0.50 L of 3.0 M NH₃. Calculate [Ni(NH₃)₆²⁺] and [Ni²⁺] in this solution. K_overall for Ni(NH₃)₆²⁺ is 5.5 × 10⁸. That is,

$5.5\times {10}^8=\frac{[Ni{(N{H}_3)}_6{}^{2+}]}{[N{i}^{2+}]{[N{H}_3]}^6}$

for the overall reaction

${\text{Ni}}^{\text{2+}}(aq)+6N{H}_3(aq)\rightleftharpoons {\text{Ni(NH}}_{\text{3}}{\text{)}}_{\text{6}}{}^{\text{2+}}(aq)$