VCE Chemistry Question Thread

[orgekas]

Hi guys,

I didn't know where to post this or if it's already answered but I'm going to ask here.
Where can I find solutions for Heinemann chemistry 2 2nd edition student workbook?
I've been googling for the past few days without finding anything.
Please help.

Thank you.

[wunderkind52]

Hi guys,

I didn't know where to post this or if it's already answered but I'm going to ask here.
Where can I find solutions for Heinemann chemistry 2 2nd edition student workbook?
I've been googling for the past few days without finding anything.
Please help.

Thank you.

They should be in the Student CD rom - Chem2-->Docs i THINK

[orgekas]

They should be in the Student CD rom - Chem2-->Docs i THINK

It only contains answers for the textbook question not the student workbook

[tashhhaaa]

Hi guys,
I'm having a bit of trouble with this question from Ch 4 of the Heinemann book:

Lawn fertiliser contains ammonium ions (NH +). A 1.234 g
sample of lawn fertiliser was dissolved in water to make a 250.0 mL solution. A 20.00 mL aliquot of this solution was added to a flask containing 20.00 mL of 0.1022 M sodium hydroxide solution. The flask was heated until the reaction:
NH +(aq) + OH−(aq) → NH (aq) + H2O(l)
was complete. Excess sodium hydroxide in the resulting solution was titrated with 0.1132 M hydrochloric acid, using phenolphthalein as indicator. The end point was reached when 9.97 mL had been added. Calculate:

a) the amount, in mol, of HCl used in the titration
b) the amount of NaOH in excess after reaction with the
fertiliser
c) the amount of NaOH that reacted with the NH + ions 4
d) the amount of NH + ions in the 1.234 g fertiliser sample 4
e) the percentage by mass of nitrogen in the fertiliser, assuming all the nitrogen is present as ammonium ions.

I'm not expecting anyone to work out the whole question but help with a) and b) to start me off would be great

[wunderkind52]

Hi guys,
I'm having a bit of trouble with this question from Ch 4 of the Heinemann book:

Lawn fertiliser contains ammonium ions (NH +). A 1.234 g
sample of lawn fertiliser was dissolved in water to make a 250.0 mL solution. A 20.00 mL aliquot of this solution was added to a flask containing 20.00 mL of 0.1022 M sodium hydroxide solution. The flask was heated until the reaction:
NH +(aq) + OH−(aq) → NH (aq) + H2O(l)
was complete. Excess sodium hydroxide in the resulting solution was titrated with 0.1132 M hydrochloric acid, using phenolphthalein as indicator. The end point was reached when 9.97 mL had been added. Calculate:

a) the amount, in mol, of HCl used in the titration
b) the amount of NaOH in excess after reaction with the
fertiliser
c) the amount of NaOH that reacted with the NH + ions 4
d) the amount of NH + ions in the 1.234 g fertiliser sample 4
e) the percentage by mass of nitrogen in the fertiliser, assuming all the nitrogen is present as ammonium ions.

I'm not expecting anyone to work out the whole question but help with a) and b) to start me off would be great

This is a back titration. What has happened here, is you want to find the amount of ammonium ions. To do this, you have put in an excess amount of sodium hydroxide. You put in lots, to make sure that all of the ammonium reacts. Since you have added an excess, some will react with ammonium, but some will remain unreacted yes? So what you do, is you titrate the unreacted NaOH with HCl. So, you know how many moles of HCl you added, which corresponds to the amount of mol of EXCESS NaOH. To find the amount of NaOH reacted, you subtract this amount from the initial amount. Ie. you started off with n(NaOH initial) and that split into n(NaOH reacted with nH4) and n(NaOH excess). So by doing n(NaOH initial) - n(NaOH excess), you can find the amount that reacted with the NH4.

Spoiler

a)n(HCl)=cV=0.1132*0.00997
b)NaOH + HCl --> NaCl + H2O => n(NaOH excess)=n(HCl)
c)n(NaOH reacted)=n(NaOH initial)-n(NaOH excess)=(0.1022*0.0200)*n(NaOH excess)
d) n(NH4+ ions in sample)=n(NH4+)*(250/20)
e)n(N)=n(NH4+)
m(N)=n(N)*M=n(N)*14.0
%mass=(m(N)/m(sample))*100=(m(n)/1.234)*100

[acinod]

Hi guys,
I'm having a bit of trouble with this question from Ch 4 of the Heinemann book:

Lawn fertiliser contains ammonium ions (NH +). A 1.234 g
sample of lawn fertiliser was dissolved in water to make a 250.0 mL solution. A 20.00 mL aliquot of this solution was added to a flask containing 20.00 mL of 0.1022 M sodium hydroxide solution. The flask was heated until the reaction:
NH +(aq) + OH−(aq) → NH (aq) + H2O(l)
was complete. Excess sodium hydroxide in the resulting solution was titrated with 0.1132 M hydrochloric acid, using phenolphthalein as indicator. The end point was reached when 9.97 mL had been added. Calculate:

a) the amount, in mol, of HCl used in the titration
b) the amount of NaOH in excess after reaction with the
fertiliser
c) the amount of NaOH that reacted with the NH + ions 4
d) the amount of NH + ions in the 1.234 g fertiliser sample 4
e) the percentage by mass of nitrogen in the fertiliser, assuming all the nitrogen is present as ammonium ions.

I'm not expecting anyone to work out the whole question but help with a) and b) to start me off would be great

This is a great question to understand for back titration. The above comment helps explain a bit more about the ideal of back titration. But you can see below to help you work it out if you still get stuck.

a) To calculate the amount of HCL, we use the formula n=cV.
n(HCl)=c(HCl)*V(HCl)=0.1132*0.00997=1.13*10^-3 mol

b) The amount of NaOH in excess is the amount that reacted with the HCl. From our previous answer and noting the equation: HCl+NaOH=H2O+NaCl. The amount of NaOH is the same amount as the amount of HCl as the ratio is 1:1.
n(NaOH excess)=1.13*10^-3 mol.

c) n(NaOH reacted) = n(NaOH initial) - n(NaOH excess) = 0.1022*0.02000 - 1.13*10^-3 = 9.14*10^-4 mol

d) The ratio of NH4+ and OH- is 1:1, so n(NH4+ in aliquot)=n(NaOH reacted)=9.14*10^-4 mol. To find the amount in the 1.234g fertiliser sample, we need to use a ratio. Since the aliquot was 20.00 mL from 250.0 mL and the 250.0 mL contains all the NH4+ ions from the fertiliser sample, n(NH4+ in fertliser) = (9.14*10^-4)*(250/20) = 0.0114 mol.
e) n(N)=n(NH4+ in fertliser) = 0.0114 mol
m(N) = n(N)*M(N) = 0.0114*14.0 = 0.156g
%(N) = 0.156/1.234 * 100% = 12.6%

[tashhhaaa]

This is a back titration. What has happened here, is you want to find the amount of ammonium ions. To do this, you have put in an excess amount of sodium hydroxide. You put in lots, to make sure that all of the ammonium reacts. Since you have added an excess, some will react with ammonium, but some will remain unreacted yes? So what you do, is you titrate the unreacted NaOH with HCl. So, you know how many moles of HCl you added, which corresponds to the amount of mol of EXCESS NaOH. To find the amount of NaOH reacted, you subtract this amount from the initial amount. Ie. you started off with n(NaOH initial) and that split into n(NaOH reacted with nH4) and n(NaOH excess). So by doing n(NaOH initial) - n(NaOH excess), you can find the amount that reacted with the NH4.

Spoiler

a)n(HCl)=cV=0.1132*0.00997
b)NaOH + HCl --> NaCl + H2O => n(NaOH excess)=n(HCl)
c)n(NaOH reacted)=n(NaOH initial)-n(NaOH excess)=(0.1022*0.0200)*n(NaOH excess)
d) n(NH4+ ions in sample)=n(NH4+)*(250/20)
e)n(N)=n(NH4+)
m(N)=n(N)*M=n(N)*14.0
%mass=(m(N)/m(sample))*100=(m(n)/1.234)*100

This is a great question to understand for back titration. The above comment helps explain a bit more about the ideal of back titration. But you can see below to help you work it out if you still get stuck.

a) To calculate the amount of HCL, we use the formula n=cV.
n(HCl)=c(HCl)*V(HCl)=0.1132*0.00997=1.13*10^-3 mol

b) The amount of NaOH in excess is the amount that reacted with the HCl. From our previous answer and noting the equation: HCl+NaOH=H2O+NaCl. The amount of NaOH is the same amount as the amount of HCl as the ratio is 1:1.
n(NaOH excess)=1.13*10^-3 mol.

c) n(NaOH reacted) = n(NaOH initial) - n(NaOH excess) = 0.1022*0.02000 - 1.13*10^-3 = 9.14*10^-4 mol

d) The ratio of NH4+ and OH- is 1:1, so n(NH4+ in aliquot)=n(NaOH reacted)=9.14*10^-4 mol. To find the amount in the 1.234g fertiliser sample, we need to use a ratio. Since the aliquot was 20.00 mL from 250.0 mL and the 250.0 mL contains all the NH4+ ions from the fertiliser sample, n(NH4+ in fertliser) = (9.14*10^-4)*(250/20) = 0.0114 mol.
e) n(N)=n(NH4+ in fertliser) = 0.0114 mol
m(N) = n(N)*M(N) = 0.0114*14.0 = 0.156g
%(N) = 0.156/1.234 * 100% = 12.6%

Thank you so much guys!!! Life savers!

[Eiffel]

Hey, you know for the second last spoiler in the second post. Brick cleaner Questions

why does the number of mols of HCl increase? (i understand the calculation) but are they not just diluting it to 250ml?


also, with sig figs, is it (the answer) supposed to be expressed to the least number from the actual question, or least from the values you used directly in that part of the equation ?


Thanks

[warya]

Does anyone know why the plunger of a gas syringe must be rotated in gravimetric analysis

[wunderkind52]

Hey, you know for the second last spoiler in the second post. Brick cleaner Questions

why does the number of mols of HCl increase? (i understand the calculation) but are they not just diluting it to 250ml?


also, with sig figs, is it (the answer) supposed to be expressed to the least number from the actual question, or least from the values you used directly in that part of the equation ?


Thanks

Its a bit early for me so i havent properly read the question, but as i take it, the sample has been diluted to form a 250mL solution, then 20mL sample of that has been taken. Now you appreciate that both solutions are mixed thoroughly, so they have the same concentration? its like cordial - if you pour some into a glass, your glass and the jug have exactly the same concentration! So, if you say let the concentration be the same, you can use the formula of n1/v1=n2/v2. It makes sense i think. If you have 5 mols of sample in 100 mL of cordial, then if you pour out 20mL of that without changing the concentration, you should have 1mol.

As for yor second question, you only apply significant figures to your actual calculations. If you don't use a number given in your working, then it is not affecting the accuracy of your solution.

[Spina]

Hi guys,

I just have a quick question regarding a chem practical. Could someone please answer the following question.

Suggest why the experimental procedure specified that all masses should be measured on the same balance.

Thanks guys, any response would be great.

[wunderkind52]

Hi guys,

I just have a quick question regarding a chem practical. Could someone please answer the following question.

Suggest why the experimental procedure specified that all masses should be measured on the same balance.

Thanks guys, any response would be great.

This answer doesn't sound too pretty, but I think it has something to do with how scales may sometimes be calibrated differently or just have discrepancies? If you weigh something on one scale, then weigh it on another scale, you may not get the same mass. This can affect your calculations, if you've got differences in mass. However, if you use the same balance, even if it's not accurate, all your records will be consistent.

[Fusuy]

Hey Guys, just a question. We did this gravimetric analysis prac:

So we had to find the % of sulfur in the fertilzer, which i know how to get. What I dont get is why did we have to add the HCL in step 3? Thanks

1) Finely grind a small quantity of fertiliser using a motar and pestle. Weigh out accurately about 1.0 g of the ground fertiliser into a 100 ml beaker. Record the mass and the brand of fertiliser.

2) Add 50ml of distilled water and stir to dissolve as much of the fertiliser sample as possible. Filter the mixture into a 600ml beaker, washing the residue several times using distilled water.

3) Add about 3 mL of 2 M hydrochloric acid to the filtrate and add more water so that the total volume is about 200ml. Boil the solution.

4) Slowly add 15 mL of 0.5 M barium chloride solution drop by drop from a burette to the hot solution. A white precipitate of barium sulfate will form. Stir continuously throughout this process.

5) Boil the mixture for a further minute. Remove it form the heat and allow the precipitate to settle. Ensure that no sulfate ions remain in the solution by adding several drops of barium chloride solution. If more precipitate forms, add a further 3 mL of barium chloride solution and test again for unreacted sulfate ions.

6) Weigh a Gooch crucible fitted with filter paper.

7) Collect the precipitate in the Gooch crucible using gentle vacuum filtration. Use about 10 mL of warm distilled water to wash any precipitate remaining in the beaker into the crucible.

Collect the last drops of filtrate in a 100 mL beaker and test for the presence of chloride ions by adding a few drops of silver nitrate solution to the filtrate. If the solution becomes cloudy, wash the precipitate further with a further 10 mL of warm water and repeat the test.

9) Place the crucible and contents in an oven heated to 100 -120°C and leave overnight.

10) Weight the crucible and contents and record the mass.

So we had to find the % of sulfur in the fertilzer, which i know how to get. What I dont get is why did we have to add the HCL in step 3? Thanks

[wunderkind52]

Hey Guys, just a question. We did this gravimetric analysis prac:

So we had to find the % of sulfur in the fertilzer, which i know how to get. What I dont get is why did we have to add the HCL in step 3? Thanks

1) Finely grind a small quantity of fertiliser using a motar and pestle. Weigh out accurately about 1.0 g of the ground fertiliser into a 100 ml beaker. Record the mass and the brand of fertiliser.

2) Add 50ml of distilled water and stir to dissolve as much of the fertiliser sample as possible. Filter the mixture into a 600ml beaker, washing the residue several times using distilled water.

3) Add about 3 mL of 2 M hydrochloric acid to the filtrate and add more water so that the total volume is about 200ml. Boil the solution.

4) Slowly add 15 mL of 0.5 M barium chloride solution drop by drop from a burette to the hot solution. A white precipitate of barium sulfate will form. Stir continuously throughout this process.

5) Boil the mixture for a further minute. Remove it form the heat and allow the precipitate to settle. Ensure that no sulfate ions remain in the solution by adding several drops of barium chloride solution. If more precipitate forms, add a further 3 mL of barium chloride solution and test again for unreacted sulfate ions.

6) Weigh a Gooch crucible fitted with filter paper.

7) Collect the precipitate in the Gooch crucible using gentle vacuum filtration. Use about 10 mL of warm distilled water to wash any precipitate remaining in the beaker into the crucible.

Collect the last drops of filtrate in a 100 mL beaker and test for the presence of chloride ions by adding a few drops of silver nitrate solution to the filtrate. If the solution becomes cloudy, wash the precipitate further with a further 10 mL of warm water and repeat the test.

9) Place the crucible and contents in an oven heated to 100 -120°C and leave overnight.

10) Weight the crucible and contents and record the mass.

So we had to find the % of sulfur in the fertilzer, which i know how to get. What I dont get is why did we have to add the HCL in step 3? Thanks

You are analysing for the sulfur content. I think fertilisers commonly contain phosphate ions, and referring to the solubility rules, barium phosphate is insoluble. So thats a problem because if you add barium chloride, you will get baroum sulphate and barium phosphate precipitate, and you cant distinguish between them leading to a greater percentage. The HCl should react with PO4 ions to form HPO4, and BaHPO4 is soluble, so it shouldnt affect your mass of precipitate.

[Fusuy]

You are analysing for the sulfur content. I think fertilisers commonly contain phosphate ions, and referring to the solubility rules, barium phosphate is insoluble. So thats a problem because if you add barium chloride, you will get baroum sulphate and barium phosphate precipitate, and you cant distinguish between them leading to a greater percentage. The HCl should react with PO4 ions to form HPO4, and BaHPO4 is soluble, so it shouldnt affect your mass of precipitate.

Thanks, that seems right