[Ah Beng]

Hi, I need some help with this question :




I have several ideas for these questions :


(b) (ii)
While the graph has a gradient (ie. before it levels off), the limiting reactant is NaOH. And since the mole ratio of OH- to Fe(OH)3 is 3 : 1 (for original reaction), whilist the mole ratio of OH- to Fe(OH)2 is 2 : 1 (for new reaction), hence the gradient of the new graph should be steeper (that is to say, for the same volume of NaOH used, we get more of the ppt), and the new graph will hence be above the original graph.

However, where would it level off? The original graph levels off at 6mm (height of ppt).

If the height of ppt (after centrifugation) is directly proportional to number of ppt particles, then because the mole ratio of Fe ion (whether Fe2+ or Fe3+) to the ppt (whether Fe(OH)2 or Fe(OH)3 respectively) is 1 : 1 in both reactions, hence it should level off at the same height of 6mm.

However, if the height of the ppt (after centrifugation) is dependent on the molar mass, then the greater the molar mass, the more space each formula unit takes up (here, conceivably due to the greater number of ions per formula unit of Fe(OH)3 compared to Fe(OH)2). In which case, the height of the ppt for the new reaction (ie. using Fe2+ instead of Fe3+) should be lower, in other words, the graph levels off *below* 6mm.

Using calculations of molar masses, where molar mass of Fe(OH)2 is 90g, Fe(OH)3 is 107g, taking proportion of 6mm for 107g, we have 5mm for 90g.

So by this idea, the new graph levels off at 5.0 mm.

But exactly how much steeper will the gradient be? For the same height of ppt at say, 5mm, instead of 10cm3 of NaOH, only 2/3 of the no. of moles (hence volume) of NaOH needs to be used, because the mole ratio of OH- to the ppt product is now 2:1 (instead of 3:1), which works out to be 6.67 cm3 of NaOH.

In summary, the new graph will have level off after reaching (from the origin to) 6.67 cm3, 5.0 mm.

Is this correct?


b) (iii)
If Al3+ ions were used instead, the mole ratio (of metal ion to ppt product) would be the same as if Fe3+ ions were used.

However, because the molar mass of Al is much less (or 'lighter') than the molar mass of Fe, hence for any or each volume of aqueous NaOH used, the height of the ppt would be less. If this idea is true, is it possibly due to the fact that the number of electron shells for Fe (period 4) is 1 more than the number of electron shells for Al (period 3), and hence occupy more 'space' per formula unit (comparing Al(OH)3 to Fe(OH)3).

If the above is correct, then the new Al3+ graph will have a gentler gradient than the Fe3+ graph.

However, where will it level off? Assuming our earlier idea about the height of the ppt being proportional to molar mass, the new max height of ppt would be (6mm/107g)x(78g) = 4.37mm.

And to answer the "difference in shape?" and "reason for difference?" questions of (b)(iii), my idea is :

Difference in shape - The graph will have a gentler gradient, and level off at a lower maximum ppt height.

Reason for difference - The molar mass of Al(OH)3 is less than the molar mass of Fe(OH)3, hence the height of the ppt will be lower for each volume of NaOH.


Is this correct?

Thank you in advance for any replies.


Ah Beng

[Borek]

While the graph as a gradient (ie. before it levels off), the limiting reactant is NaOH. And since the mole ratio of OH- to Fe(OH)3 is 3 : 1 (for original reaction), whilist the mole ratio of OH- to Fe(OH)2 is 2 : 1 (for new reaction), hence the gradient of the new graph should be steeper (that is to say, for the same volume of NaOH used, we get more of the ppt), and the new graph will hence be above the original graph.

Assuming both ppts have identical density, which I doubt. Seems like you have no choice but to make this assumption.