0
$\begingroup$

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 !

|improve this question
$\endgroup$
  • $\begingroup$ The chemical formula in the line with the solubility product is wrong. $\endgroup$ – MaxW Jun 9 '18 at 15:21
  • $\begingroup$ Yes, I changed it just now. Any idea how to do the question ? $\endgroup$ – Gordon Jun 9 '18 at 16:14
  • $\begingroup$ Start with setting up the $K_{\text{sp}}$ equation. $\endgroup$ – MaxW Jun 9 '18 at 16:20
  • $\begingroup$ Ah, that was my fault, sorry. $\endgroup$ – pentavalentcarbon Jun 9 '18 at 17:01
0
$\begingroup$

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.

|improve this answer
$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.