A solution is prepared by adding (0.090 mathrm{~mol}) of (mathrm{K}_{3}left[mathrm{Fe}(mathrm{CN})_{6}ight]) to (0.60 mathrm{~L}) of (2.0 mathrm{M} mathrm{NaCN}).

Question:

A solution is prepared by adding $0.090 \mathrm{~mol}$ of $\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}ight]$ to $0.60 \mathrm{~L}$ of $2.0 \mathrm{M} \mathrm{NaCN}$. Assuming no volume change, calculate the concentrations of $\mathrm{Fe}(\mathrm{CN})_{6}{ }^{3-}$ and $\mathrm{Fe}^{3+}$ in this solution. The $K$ (overall) for the formation of $\mathrm{Fe}(\mathrm{CN})_{6}{ }^{3-}$ is $1 \times 10^{42}$.

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