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If I have the following equilibrium: $$\ce{Cu^2+ + 4NH3 <=> Cu(NH3)4^2+}$$ Given an initial concentration of $\ce{Cu^2+}$ ($C_\ce{Cu}$) and an initial concentration of ammonia $C_{NH_3}$, the equilibrium constant $\beta$ can be expressed: $$\beta=\frac{x}{(C_\ce{Cu}-x)(C_{NH_3}-x)^4}$$
My question is: is there any approximation I can use in order to solve this expression to know the value of $x$? $\beta$ is known.
Thanks
There is some chemistry behind this which you must understand to recognize the approximation. First the copper ammine complex doesn't really form in the manner indicated rather: $$\ce{Cu^2+ + 4H2O <=>T[acidic solution] Cu(H2O)4^2+}$$
But in basic solution copper will form a gelatinous precipitate which can perhaps be represented as:
$$\ce{Cu(H2O)4^2+ + 2OH^{-} <=>T[basic solution] Cu(H2O)2(OH)2}$$
With ammonia in the aqueous solution various ammine complexes will be formed such that the soluble tetra-ammine complex is formed in a series of reactions something like:
$$\ce{Cu(H2O)2(OH)2(s) + NH3 <=> Cu(H2O)(OH)2(NH3)} $$
$$\ce{Cu(H2O)(OH)2(NH3) + NH3 <=> Cu(OH)2(NH3)2} $$
$$\ce{Cu(OH)2(NH3)2 + NH3 <=> Cu(OH)(NH3)3^+} $$
$$\ce{Cu(OH)(NH3)3^+ + NH3 <=> Cu(NH3)4^2+(aq)} $$
Ammonia is typically present in excess so the overall reaction is written "in shorthand" as the following equation (I highly doubt that the exact sequence of chemical reactions between the gelatinous precipitate and ammonia is fully understood):
$$\ce{Cu^2+ + 4NH3 <=> Cu(NH3)4^2+} \quad \text{K}_\text{eq}$$
and the equilibrium for this reaction is greatly favored to the right (the ammine complex form).
If you let $[\ce{Cu2+}]_i$ be the initial concentration of $\ce{Cu^{2+}}$, let $[\ce{NH3}]_i$ be the initial concentration of $\ce{NH3}$ and the final concentration of the ammine complex be $[\ce{Cu(NH3)4^2+}]_f$ then in excess ammonia a simplfying assumption would be: $$ [\ce{Cu(NH3)4^2+}]_f \approx [\ce{Cu2+}]_i$$ $$[\ce{NH3}]_f = [\ce{NH3}]_i - 4[\ce{Cu(NH3)4^2+}]_f \approx [\ce{NH3}]_i - 4[\ce{Cu2+}]_i$$
the equations could of course be solved exactly, but you end up with a quadratic expression which really isn't a difficult mathematical problem to solve. It just takes a wee bit more calculating.
