# Question 7b75d

May 23, 2016

Here's my take on this.

#### Explanation:

For starters, I assume that you're actually looking for the molar concentration of the ferrocyanide anion, "Fe"("Cn")_6^(4-), in a saturated solution of zinc ferrocyanide, "Zn"_2"Fe"("CN")_6, that contains $\text{0.04 M}$ zinc sulfate, ${\text{ZnSO}}_{4}$.

Zinc ferrocyanide, also called zinc hexacyanoferrate(II), is actually insoluble in water, which means that when this coordination compound is placed in aqueous solution, an equilibrium will be established between the undissociated solid and the dissociated ions

"Zn"_ color(red)(2)"Fe"("CN")_ (6(s)) rightleftharpoons color(red)(2)"Zn"_ ((aq))^(2+) + "Fe"("CN")_ (6(aq))^(4-)

Notice that every mole of zinc ferrocyanide will produce $\textcolor{red}{2}$ moles of zinc cations and $1$ mole of ferrocyanide anions in solution.

Now, zinc sulfate is soluble in water, which is why it completely dissociates into zinc cations and sulfate anions

${\text{ZnSO"_ (4(aq)) -> "Zn"_ ((aq))^(2+) + "SO}}_{4 \left(a q\right)}^{2 -}$

Here every mole of zinc sulfate will produce $1$ mole of zinc cations in solution.

This means that you will have

["Zn"^(2+)]_("coming from ZnSO"_4) = ["ZnSO"_4] = "0.04 M"

By definition, the solubility product constant, ${K}_{s p}$, for zinc ferrocyanide will be equal to

K_(sp) = ["Zn"^(2+)]^color(red)(2) * ["Fe"("CN")_6^(4-)]

You can find the value of ${K}_{s p}$ listed here

${K}_{s p} = 2.1 \cdot {10}^{- 16} {\text{M}}^{3}$

http://www2.ucdsb.on.ca/tiss/stretton/database/Solubility_Products.htm

Rearrange the above equation to solve for ["Fe"("CN")_6^(4-)], the molar concentration of the ferrocyanide anions in this solution

["Fe"("CN")_ 6^(4-)] = K_(sp)/(["Zn"^(2+)]^color(red)(2))

Plug in your values to get

["Fe"("CN")_ 6^(4-)] = (2.1 * 10^(-16) "M"^color(red)(cancel(color(black)(3))))/( (0.04)^color(red)(2) color(red)(cancel(color(black)("M"^2)))) = color(green)(|bar(ul(color(white)(a/a)1.3 * 10^(-13)"M"color(white)(a/a)|)))#

I'll leave the answer rounded to two sig figs.

SIDE NOTE You can compare this value to the molar solubility of zinc ferrocyanide in pure water. If you take $s$ to be the molar solubility of the salt, you will have

${K}_{s p} = {\left(\textcolor{red}{2} s\right)}^{\textcolor{red}{2}} \cdot s = 4 {s}^{3}$

Rearrange to get

$s = \sqrt[3]{{K}_{s p} / 4} = \sqrt[3]{\frac{2.1 \cdot {10}^{- 16}}{4}} = 3.7 \cdot {10}^{- 6} \text{M}$

Notice that the molar solubility of the salt decreases significantly in your solution because of the presence of the zinc cations $\to$ think common-ion effect here.