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

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 !

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.