Question

One millimole of Ni(NO3)2 dissolves in 240.0 mL of a solution that is 0.500 M in ammonia. The formation constant of Ni(NH3)62+ is 5.5×108. What is the initial concentration of Ni(NO3)2 in the solution? What is the equilibrium concentration of Ni2+(aq ) in the solution?

Answer 1

Answer:

$[Ni^{2+}]_0=0.0042M$

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

Explanation:

Hello,

In this case, we consider the formation reaction of such cationic complex as:

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

Thus, the law of mass action is:

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

So, we can compute the initial concentration of nickel ions which matches with the nickel (II) nitrate concentration:

$[Ni^{2+}]_0=\frac{0.001mol}{0.240L} =0.0042M$

Next, by using the formation constant, we write the law of mass action in terms of the change $x$ due to the reaction extent:

$5.5x10^8=\frac{x}{(0.500-6x)^6(0.0042-x)}$

Thus, solving for $x$ we obtain:

$x=0.0042M$

Which means all the nickel ions were consumed:

$[Ni^{2+}]_{eq}=0.0042M-x=0.0042M-0.0042M=0$

This has sense since the formation constant is large enough.

Best regards.