Titration Help!

[azngirl456]

This was done as orientation prac:

A vitamin C tablet (containing ascorbic acid) that weighs 0.689 grams is crushed up in deoinised water and diluted to 500 ml in a volumetric flask. 10ml aliquote of vitamin C solution was placed in a small conical flask. 10ml of 0.1M HCl and 3 drops of starch indicator was placed in conical flask as well. Iodine solution was placed in the burette. Iodine solution is titrated with idoine solution until a dark blue black colour appears.

My friends and I didnt get concordant results. We've asked around and no one else has concordant results. The closest someone got was 6.2, 5.3, 5.7.

Questions I have:

1) Why is HCl solution placed in the conical flask?
2) One of the questions asks: 'Use stoichiometric ratios from the overall redox equation to determine the ammount of ascorbic acid delivered in each 20ml aliquot'
But only 10ml of the vitamin solution was aliquot. Does 20ml refer to HCl+starch indicator and vitamin C solution? Also I don't really get how it's a redox reaction when it involves acids

[thushan]

Haha this was quite similar to the prac I did a few days ago at monash:

1) I'm not 100% sure on this, but ascorbic acid is a weak acid and would become deprotonated in solution of pH ~ 7. The HCl is to keep the pH low and keep the proton attached to the ascorbic acid, so that it can be oxidised properly.

2) I'm supposing they're using the term "aliquot" loosely here. They would be referring to the whole mixture of HCl and ascorbic acid + indicator.

Acids can be oxidised and/or reduced; eg. H2SO4, whilst a really strong acid, can be reduced to SO2. Just because something is an acid doesn't mean that it can act only as an acid; it can be a reductant or oxidant.

[azngirl456]

Haha this was quite similar to the prac I did a few days ago at monash:

1) I'm not 100% sure on this, but ascorbic acid is a weak acid and would become deprotonated in solution of pH ~ 7. The HCl is to keep the pH low and keep the proton attached to the ascorbic acid, so that it can be oxidised properly.

2) I'm supposing they're using the term "aliquot" loosely here. They would be referring to the whole mixture of HCl and ascorbic acid + indicator.

Acids can be oxidised and/or reduced; eg. H2SO4, whilst a really strong acid, can be reduced to SO2. Just because something is an acid doesn't mean that it can act only as an acid; it can be a reductant or oxidant.

Thank you Thushan for the help I know you said you weren't 100% on this, but could you explain more about weak acids being deprotonated at pH~7. do all weak acids behave like this? how does low pH keep the proton attached to ascorbic acid?

[thushan]

Suppose ascorbic acid has the formula HAsc, where Asc- is the rest of the molecule.

Say we dump this stuff in water (which has pH 7), and we get the equilibrium:
HAsc + H2O <-> Asc- + H3O+

But I think that the Asc- species doesn't oxidise, so we don't want this species in solution.

So what we do is dump a strong acid like HCl into the solution to lower the pH and increase [H+]. So we're effectively shifting the above equilibrium to the left (Le Chatelier's Principle). If you haven't done equilibria shifts at school, you can think of it as forcing protons onto the Asc- ion. So at low pH, pretty much all the ascorbic acid is present as the protonated form.

[azngirl456]

Suppose ascorbic acid has the formula HAsc, where Asc- is the rest of the molecule.

Say we dump this stuff in water (which has pH 7), and we get the equilibrium:
HAsc + H2O <-> Asc- + H3O+

But I think that the Asc- species doesn't oxidise, so we don't want this species in solution.

So what we do is dump a strong acid like HCl into the solution to lower the pH and increase [H+]. So we're effectively shifting the above equilibrium to the left (Le Chatelier's Principle). If you haven't done equilibria shifts at school, you can think of it as forcing protons onto the Asc- ion. So at low pH, pretty much all the ascorbic acid is present as the protonated form.

Oh I see what's going on. Thankyou again for your clarification Well appreciated