Question #7b75d
1 Answer
Here's my take on this.
Explanation:
For starters, I assume that you're actually looking for the molar concentration of the ferrocyanide anion,
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
Now, zinc sulfate is soluble in water, which is why it completely dissociates into zinc cations and sulfate anions
#"ZnSO"_ (4(aq)) -> "Zn"_ ((aq))^(2+) + "SO"_ (4(aq))^(2-)#
Here every mole of zinc sulfate will produce
This means that you will have
#["Zn"^(2+)]_("coming from ZnSO"_4) = ["ZnSO"_4] = "0.04 M"#
By definition, the solubility product constant,
#K_(sp) = ["Zn"^(2+)]^color(red)(2) * ["Fe"("CN")_6^(4-)]#
You can find the value of
#K_(sp) = 2.1 * 10^(-16)"M"^(3)#
http://www2.ucdsb.on.ca/tiss/stretton/database/Solubility_Products.htm
Rearrange the above equation to solve for
#["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
#K_(sp) = (color(red)(2)s)^color(red)(2) * s = 4s^3#
Rearrange to get
#s = root(3)(K_(sp)/4) = root(3)((2.1 * 10^(-16))/4) = 3.7 * 10^(-6)"M"#
Notice that the molar solubility of the salt decreases significantly in your solution because of the presence of the zinc cations