Question

Using the Kf value of 1.2×109 calculate the concentration of Ni2+(aq) and Ni(NH3)62+ that are present at equilibrium after dissolving 1.47 gNiCl2 in 100.0 mL of NH3(aq) solution such that the equilibrium concentration of NH3 is equal to 0.20 M.
Express your answers in moles per liter to two significant figures separated by a comma.

Answer 1

This problem is describing the equilibrium whereby the formartion of a complex is attained when 1.47 g of nickel(II) chloride is dissolved in 100.0 mL of ammonia so that the latter's equilibium concentration is 0.20 M. Thus, it is asked to calculate the equilibrium concentrations of both nickel(II) ions and that of the complex.

Firstly, we can write out the chemical equation to be considered:

$Ni^{2+}+6NH_3\rightleftharpoons Ni(NH_3)_6^{2+}$

Next, we can calculate the initial concentration of nickel(II) ions by using the concept of molarity:

$[Ni^{2+}]=\frac{1.47gNiCl_2*\frac{1molNiCl_2}{129.6g}*\frac{1molNi^{2+}}{1molNiCl_2} }{0.1000L}=0.113M$

Afterwards, we set up an equilibrium expression for this chemical reaction:

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

Which can be written in terms of the reaction extent, $x$:

$Kf=\frac{x}{(0.113-x)(0.2)^6}$

Now, for the calculation of $x$, we plug in Kf, and solve for it:

$1.2x10^9=\frac{x}{(0.113-x)(0.2)^6}\\\\1.2x10^9=\frac{x}{(0.113-x)(6.4x10^{-5})}\\\\7.68x10^4(0.113-x)=x\\\\x=0.112999 M$

Which is about the same to the initial concentration of nickel(II) ions because the Kf is too large.

Thus, the required concentrations at equilibrium are about:

$[Ni(NH_3)_6^{2+}]=0.113M$

$[Ni^{2+}]=0M$

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