What is the minimum mass of iron(III) chloride required to be added until the dark blue precipitate first appear?

Question

How do I solve this question?

The presence of iron(III) ion is confirmed in a qualitative analysis by adding potassium hexacyanoferrate(II) ($\ce{K4[Fe(CN)6]}$) solution.

A dark blue precipitate, iron(III) hexacyanoferrate(II) ($\ce{Fe4[Fe(CN)6]3}$), is formed.

What is the minimum mass of iron(III) chloride (in grams) required to add to a $\pu{200 cm3}$ of $\pu{0.0010 mol dm-3}$ potassium hexacyanoferrate(II) solution until the dark blue precipitate first appears?

• solubility product of $\ce{Fe4[Fe(CN)6]3} = \pu{3.30 x 10^-41 mol^7dm^-21}$
• relative molecular mass of $\ce{FeCl3} = \pu{162.3 g/mol}$

Choices:

1. $\pu{1.35 x 10^-8 g}$
2. $\pu{4.38 x 10^-7 g}$
3. $\pu{5.48 x 10^-7 g}$
4. $\pu{2.19 x 10^-6 g}$

This is what I came up with just now :

$K_\mathrm{sp} = \ce{[Fe^3+]^4 [Fe[CN]6^4-]^3}$

$\pu{3.3x10^-41}$ = $\ce{[Fe^3+]^4[0.001]^3}$

$\ce{[Fe^3+]^4}$ = $\pu{3.3 x 10^-32 mol dm^-3}$

$\ce{[Fe^3+]}$ = $\pu{1.35 x 10^-8 mol dm^-3}$

Mass of $\ce{FeCl3} = \pu{1.35 x 10^-8 mol dm^-3 \times 0.2 dm^3 \times 162.3 g/mol}$

= $\pu{4.38 x 10^-7 g}$

I am still relatively new to this topic so I am not sure if I am doing this correctly. Any help will be appreciated. Thank you !

Answer 1

Yes that is right.

However I would not round intermediate results to avoid round off errors. In fact I'd carry two extra significant figures for intermediate calculations and only round off final answer. (Easy now with a calculator, not so easy to do when I was doing these problems with a slide rule...) So I'd get Fe as $1.3478\times10^{-8}$ molar. Using that number I get 4.3749893530170255598967635197789e-7 as the final answer which rounds to 4.37e-7.

The problem's answer is really wrong since the concentration of the hexacyanoferrate(II) is only given to two significant figures. So the answer shouldn't have more than two either. If the problem wanted three significant figures then the hexacyanoferrate(II) should have been specified as 0.00100 molar.

You have to realize that significant figures are a crude way to do error propagation. So two significant figures for 0.0010 molar means +/- 0.00005 molar, so a +/-5% specification. However if the answer to two significant figures is 9.9 then that means +/- 0.05 which only has an error of +/-0.5%. So although crude using significant figures keeps you from stupidly assuming that all the figures in the calculator's answer, like 4.3749893530170255598967635197789e-7, are significant.

Answer 2

The only trouble is that by adding small amounts of a $\ce{FeCl3}$ solution in a $\ce{K4[Fe(CN)6]}$ solution, the rection does not produce a blue precipitate of $\ce{Fe3[Fe{CN}6]4}$ as given in the text. It produces a blue solution of "soluble Prussian blue" which is $\ce{KFe^{III}[Fe^{II}(CN)6]}$ by the reaction $$\ce{Fe^{III}Cl3 + K4[Fe^{II}(CN)6 -> KFe^{III}[Fe^{II}(CN)6 (aq) + 3 KCl}$$ and no precipitate does happen at all !! If insoluble Prussian blue is wanted, the contrary should be done : small amount of $\ce{K4[Fe(CN)6]}$ should be added to a $\ce{FeCl3}$ solution.

Ref.: 1) G. Fornosier, A. Bleuzen, Bleu de Prusse, Act. chim. 444, Oct. 2019, p. 16-21 (in French)
2) N. N. Greenwood, A. Earnshaw, Chemistry of the Elements, Pergamon, Oxford, 1984, p. 1271