Respuesta :
Explanation:
It is known that the coefficients change in concentration and in the exponents. Hence, the reaction equation will be as follows.
[tex]Cu^{2+}(aq) + 4NH_{3}(aq) \rightarrow Cu(NH_{3})^{2}_{4}(aq)[/tex]
According to the ICE table,
[tex]Cu^{2+}(aq) + 4NH_{3}(aq) \rightarrow Cu(NH_{3})^{2}_{4}(aq)[/tex]
Initial : 0.10 1.50 0
Change : -x -4x +x
Equilibrium: 0.10 - x 1.50 - 4x x
Hence, the mass action expression is as follows.
[tex]K_{f} = \frac{[Cu(NH3)^{2+}_{4}]}{[Cu^{2+}][NH_{3}]_{4}}[/tex]
= [tex]\frac{x}{(0.10 - x)(1.50 - 4x)^{4}}[/tex]
As, the value of is huge, it means that the reaction is very product favored. Hence, we need to find the limiting reactant first and then we get to know what x should be.
In the given reaction ammonia is the limiting reactant, because there is less than 4 times the ammonia as the copper cation. Thus, we expect it to run out first, and so, x is approximately equal to 0.25 M.
So, putting the given values into the above equation as follows.
[tex]1.03 \times 10^{13} = \frac{0.25}{(0.10 - 0.25)(1.50 - 4x)}^{4}[/tex]
=
From here
[tex][NH_{3}] = 1.50 - 4x = (\frac{2.33}{1.03 \times 10^{13}})^{\frac{1}{4}[/tex]
= M
Therefore, we can "re-solve" for x to get and verify that it is still ≈0.250 M.
x = [tex][Cu(NH_{3})^{2+}_{4}][/tex]
= [tex]\frac{1.50 - 2.31284 \times 10{-4}}{4}][/tex]
= 0.37491425 M
Thus, we can conclude that concentration of ([tex]Cu^{2+}[/tex]) is 0.37491425 M.
