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Concentration of Unreacted Metal Ion

Calculate the concentration of free calcium(II) ion in a solution prepared with initial concentrations of calcium of 0.020 M and EDTA4- of 0.10 M.

In solving this problem, we will start under very naïve circumstances.  We will not consider any other complexation of the calcium ion (for example, by something like hydroxide), and we have been given a concentration of E4– and do not need to worry about the pH of this solution, how we got a concentration of 0.10 M, or whether protonation occurs.

The first thing to do is to write the reaction and look up the relevant formation constant.

\[\ce{Ca^2+  +  E^4- \leftrightarrow  CaE^2-} \hspace{60px} \mathrm{K_f = 5.0\times 10^{10}}\]

What we see is that has a very large formation constant.  That means that this reaction will go to completion.  The approach to solving this problem is to allow it to go to completion, realizing that one of the reagents will limit the amount of product that forms.  Then we need to allow a small amount of back reaction to occur.  We can construct the following chart.

\[\begin{align}
& &&\ce{Ca^2+}\hspace{25px} + &&\ce{E^4-} \hspace{25px}\leftrightarrow &&\ce{CaE^2-} \\       
&\ce{Initial} &&\mathrm{0.020\: M} &&\mathrm{0.10\: M} &&0  \\
&\ce{Complete\: Reaction}  &&0 &&0.08 &&0.02  \\
&\ce{Back\: Reaction}  &&\ce{x} &&0.08 +\ce{x} &&0.02 - \ce{x}  \\
&\ce{Assumption}  && &&0.08 \gg\ce{x} &&0.02 \gg\ce{x}  \\
&\ce{Approximation}  &&\ce{x} &&0.08 &&0.02
\end{align}\]

The approximation that 0.08 \(\gg\) x and 0.02 \(\gg\) x is reasonable since the Kf value is so large, therefore the amount of back reaction will be excessively small.

\[\mathrm{K_f =\dfrac{[CaE^{2-}]}{[Ca^{2+}][E^{4-}]} =\dfrac{0.02}{(x)(0.08)} = 5.0\times10^{10}}\]

\[\mathrm{x = [Ca^{2+}] = 5\times 10^{-12}}\]

The amount of unreacted calcium ion is incredibly small.  Obviously the approximations we made were justified and we see how far these reactions can go toward completion.