Question

(a) From the solubility product of zinc ferrocyanide, ${\mathbf{Zn}}_{\mathbf{2}}\mathbf{Fe}{\left(\mathrm{CN}\right)}_{\mathbf{6}}$ , calculate the concentration of $\mathbf{Fe}{\left(\mathrm{CN}\right)}_{\mathbf{6}}^{\mathbf{4}\mathbf{-}}$in 0.1 0m M ${\mathbf{ZnSO}}_{\mathbf{4}}$saturated with ${\mathbf{Zn}}_{\mathbf{2}}\mathbf{Fe}{\left(\mathrm{CN}\right)}_{\mathbf{6}}$ . Assume that ${\mathbf{Zn}}_{\mathbf{2}}\mathbf{Fe}{\mathbf{\left(}\mathbf{CN}\mathbf{\right)}}_{\mathbf{6}}$is a negligible source of ${\mathbf{Zn}}^{\mathbf{2}\mathbf{+}}$ .

(b) What concentration of ${\mathbf{K}}_{\mathbf{4}}\mathbf{Fe}{\mathbf{\left(}\mathbf{CN}\mathbf{\right)}}_{\mathbf{6}}$should be in a suspension of solid ${\mathbf{Zn}}_{\mathbf{2}}\mathbf{Fe}{\left(\mathrm{CN}\right)}_{\mathbf{6}}$ in water to give$\left[{\mathrm{Zn}}^{2+}\right]\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathbf{}\mathbf{M}$?

Short answer

a)

The concentration of $\mathrm{Fe}{\left(\mathrm{CN}\right)}_6^{4-}$ is $2.1\times {10}^{-8}M$

b)

The concentration of ${\mathrm{K}}_4\mathrm{Fe}{\left(\mathrm{CN}\right)}_6$ is $8.4\times {10}^{-4}M$

Step-by-step solution

Concept used

Solubility product $\left(K_{\mathrm{sp}}\right)$:

It is defined as the product formed from the mathematical product of its concentration of dissolved ion raised to the power of its stoichiometric coefficient.

Calculate the concentration of  Fe(CN)64-

a)

${\mathrm{Zn}}_2\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\leftrightharpoons {\mathrm{Zn}}^{2+}+\mathrm{Fe}{\left(\mathrm{CN}\right)}_6$

The concentration is:

$$\begin{gathered} {\mathrm{K}}_{\mathrm{sp}}={\left[{\mathrm{Zn}}^{2+}\right]}^2\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6^{4-}\right] \\ 2.1\times {10}^{-16}={\left(0.00010\right)}^2\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6^{4-}\right] \\ \left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6^{4-}\right]=\frac{2.1\times {10}^{-16}}{{10}^{-8}} \\ \left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6^{4-}\right]=2.1\times {10}^{-8}\mathrm{M} \end{gathered}$$

Calculate the concentration of  K4Fe(CN)6

b)

$$\begin{gathered} {\mathrm{K}}_{\mathrm{sp}}={\left[{\mathrm{Zn}}^{2+}\right]}^2\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6^{4-}\right] \\ 2.1\times {10}^{-16}={\left(5.0\times {10}^{-7}\right)}^2\left[{\mathrm{K}}_4\mathrm{Fe}{\left(\mathrm{CN}\right)}_6^{4-}\right] \\ \left[{\mathrm{K}}_4\mathrm{Fe}{\left(\mathrm{CN}\right)}_6^{4-}\right]=\frac{2.1\times {10}^{-16}}{{\left(5.0\times {10}^{-7}\right)}^2} \\ \left[{\mathrm{K}}_4\mathrm{Fe}{\left(\mathrm{CN}\right)}_6^{4-}\right]=8.4\times {10}^{-4}\mathrm{M} \end{gathered}$$