Question

What concentration of $\mathbf{Fe}{\left(\mathrm{CN}\right)}_{\mathbf{6}}^{\mathbf{4}\mathbf{-}}$(ferrocyanide) is in equilibrium with$\mathbf{1}\mathbf{.}\mathbf{0}\mathit{\mu }\mathit{M}\mathit{A}{\mathit{g}}^{\mathbf{+}}$ androle="math" localid="1663331502441" $\mathit{A}{\mathit{g}}_{\mathbf{4}}\mathit{F}\mathit{e}{\mathbf{\left(}\mathbf{C}\mathbf{N}\mathbf{\right)}}_{\mathbf{6}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}$? Express your answer with a prefix from Table 1-3.

Short answer

$\left[Fe{\left(CN\right)}_8^{4-}\right]$ was calculated as $8.5\times {10}^{-21}\mathrm{M}$.

Step-by-step solution

Concept used

Solubility product :

It is defined as the mathematical product of its concentration of the dissolved ions raised up to the power of their stoichiometric coefficient.

$$\begin{gathered} M_y{X}_{z\left(s\right)}\leftrightharpoons y{M}^{z+}\left(aq\right)+z{X}^{y-}\left(aq\right) \\ {K}_{sp}={\left[M^{z+}\right]}^y{\left[X^{y-}\right]}^z \end{gathered}$$

Calculate the concentration of Ferrocyanide

$A{g}_{4}Fe{\left(CN\right)}_6\leftrightharpoons 4A{g}^{+}+Fe{\left(CN\right)}_6^{-4}$

Ferrocyanide is at equilibrium with $0.1\mu MA{g}^{+}$and$A{g}_{4}Fe{\left(CN\right)}_6^{4-}$

$$\begin{gathered} \left[{\mathrm{Ag}}^{+}\right]\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6^{4-}\right]={\mathrm{K}}_{\mathrm{sp}} \\ {\left[1.0\times {10}^{-6}\right]}^4\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6^{4-}\right]=8.5\times {10}^{-45}\mathrm{M} \\ \left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6^{4-}\right]=8.5\times {10}^{-21}\mathrm{M} \\ =8.5\mathrm{zM} \end{gathered}$$