A 0.130 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)₆²⁺ is 5.5 × 10⁸

Answer : The concentration of $Ni^{2+}$ ions at equilibrium is, $4.31\times 10^{-8}$

Explanation : Given,

Moles of $NiCl_2$ = 0.130 mol

Volume of solution = 1 L

$Concentration=\frac{Moles }{Volume}$

Concentration of $NiCl_2$ = Concentration of $Ni^{2+}$ = 0.130 M

Concentration of $NH_3$ = 1.20 M

$K_f=5.5\times 10^8$

The equilibrium reaction will be:

$Ni^{2+}(aq)+6NH_3(aq)\rightarrow [Ni(NH_3)_6]^{2+}$

Initial conc. 0.130 1.20 0

At eqm. x [1.20-6(0.130)] 0.130

= 0.42

The expression for equilibrium constant is:

$K_f=\frac{[Ni(NH_3)_6^{2+}]}{[Ni^{2+}][NH_3]^6}$

Now put all the given values in this expression, we get:

$5.5\times 10^8=\frac{(0.130)}{(x)\times (0.42)^6}$

$x=4.31\times 10^{-8}$

Thus, the concentration of $Ni^{2+}$ ions at equilibrium is, $4.31\times 10^{-8}$

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