Back titration and gravimetric analysis (1 Viewer)

[lanvins]

1. Lawn fertiliser contains nitrogen. A 1.3g sample of lawn fertiliser was weighed and carefully transferred to a 250ml volumetric flask. The weighing bottle was washed with a little de-ionised water and added to the content, until it reached the calibration line. A 20ml aliquot of this solution was added to a flask containing 20ml of .1M sodium hydroxide solution. 50ml of de-ionised water was than added. The flask was heated until the reaction NH4+ (aq) +OH-(aq) = NH3 (aq) +H20) (l) was complete. The burette was than titrated with .1M hydrochloric acid, using methyl red as indicator. The end point was reached when 44.3ml had been added.
Calculated the amount of NH4+ ions in the 1.3g fertiliser sample and the percentage by mass of nitrogen in the fertiliser, assuming all the nitrogen is present as ammonium ions.

2. 1.0 of ground fertiliser was placed in a 100ml beaker. 50ml of de-ionised water was added and filtered into a 600ml beaker. 3ml of 2M hydrochloric acid and water was added, until the total volume was 200ml. The solution was than boiled and 15ml of .5M barium chloride solution was added drop by drop. The precipate was colected filtrated, dried and weighed. A mass of 1.05g was obtained. How do u find the proportion of sulfate in the fertiliser?

(Can u plz show working?) thanks so much

 

[Undermyskin]

Ouch.

I'll try the 2nd one since it looks shorter...

Now I'm going to justify that 1.05g precipitate is mostly pure BaSO4. (assuming there's no presence of ions such as Ag or Pb that can lead to the precipitation of AgCl and PbCl2)

In the step of dissolving ground fertiliser in de-ionised water, only insoluble salts are filtered off. (assuming there's no Ca or Ba as well) So you have a solution of sulfate, probably carbonate and chloride.

When adding HCl and boil, sulfates and chlorides don't react but only carbonates. Boiling is the assuring any CO2 formed is released. This increases the purity of BaSO4 precipitate by preventing the formation of BaCO3. (What I'm saying can be marginal because even if BaCO3 is formed, excess HCl converts it immediately to CO2 gas anyway).

Chlorides don't react with BaCl2 (of course) so it doesn't affect.

After filtering, only BaSO4 is presence to a high purity (demonstrated above that no other precipitate can be formed).

number of mol of BaSO4: 1.05/(137.3+32.1+64)=4.4987...*10^-3 (app.)

mass of SO4 2-: n (above)*(32.1+64)= 0.432...g (app.)

%: 41.17% (w/w) (app.)

Feel free to correct.

 

[Undermyskin]

I think I'll try the 1st one as well...

n NaOH: 0.02*0.1=2*10^-3 mol

n HCl: 44.3*0.1*10^-3=4.43*10^-3 mol.

Na can't be excess because twice n NaOH < HCl which is expected to react with both NH3 to give NH4Cl and some Na salts to give off CO2 or st else.

The only way is NH4+ is in excess.

Assuming all the rest NH4+ reacts with HCl --> n NH4+=(4.43-2)*10^-3=2.43*10^-3 mol

mass of N= 2.43*10^-3*14.01=0.034g

%= 2.62% (w/w) (app.)

You can ask what is the significance of adding NaOH. It's to convert NH4+ to NH3 since if (NH4)2SO4 is present, adding HCl doesn't convert this into NH4Cl at all--> can't carry out titration with high accuracy.

By the way, I don't expect my answers to be totally correct because the questions don't give any specific impurities so you can see I assume a few things. Hope it helps anyway.

Which school do you go to?