Chemistry 3/4 2013 Thread

[Limista]

Assuming it's all multiplication and division, yes.

But I recommend you use 3 throughout (whilst keeping calculator values). This way, to determine the accuracy of your answer, you need only look at your previous line. This is a better way, because if you are doing a combination of additions and subtractions with multiplications and divisions, looking at the start of the qn for accuracy can be misleading.

Thanks!

[Patches]

To what extent are we going to need to memorise compounds for the exams? For instance, are there going to be questions that require you to know the formula for something like thiosulfate, or will S2O3^2- be listed? My school seems to be putting a lot of emphasis on rote memorisation. Thanks

[thushan]

To what extent are we going to need to memorise compounds for the exams? For instance, are there going to be questions that require you to know the formula for something like thiosulfate, or will S2O3^2- be listed? My school seems to be putting a lot of emphasis on rote memorisation. Thanks

Generally I think they give the formula in the exam, but best to know common ions like thiosulfate and permanganate and that.
But you can derive from the name:

sulfate: -ate = polyatomic ion containing maximum possible number of oxygens; sulf - = contains sulfur. Therefore, sulfate is SO4 (2-)
thiosulfate: thio - = one of the oxygen atoms is replaced by a sulfur atom. So instead of SO4 (2-) you have S2O3 (2-).

For these names, it's really good to look for patterns like this.

[LOLs99]

Ah yes. You are adding a 3 dp number to a 2 dp number in your final step; your answer is to 2 dp (in this case, it happens to be in 3 sf).

Do I use the number on the calc(i.e.1.02278703g & 0.82125264g) in the final plus step ?

[thushan]

yes

[xvt2101]

Hi
I'm have trouble with an acid-base titration activity. The experiment used 50ml of 0.1M NaOH and 50ml of unknown concentration HCl.
The titrant was NaOH (sorry, not sure if correct use of term), or was put into the burette. The HCl was diulted with with 200ml of de-ionised water.
These are the results after completing the experiment:
      Inital Reading    Final Reading                 Titre
1           5.59 ml            7.89 ml                 2.30 ml
2           7.89 ml          10.35 ml                 2.46 ml
3         10.35 ml          12.68 ml                 2.33 ml
                           Average Titre                 2.37 ml

I'm not really sure what to do for the following calculations:
1. The amount of sodium hydroxide (in moles), present in the average titre of sodium hydroxide solution.
2. The amount of hydrochloric acid present in each 20ml aliquot of diluted hydrochloric acid.
3. The concentration of hydrochloric acid, in moles per litre, in the undiluted hydrochloric acid.

I'm sorry if this wasn't the right place to post, I wasn't sure if I was supposed to make a new thread. If there's something I might have left out could you please tell me? Again, sorry if this wasn't correct and thanks in advance for any help.

[Bad Student]

Yo xvt2101. I'm not sure if my answer is correct so if someone finds a mistake, please tell me.

1. Since the concentration and volume of the NaOH is provided (0.1 M), we can use the formula n=cV.



2. We can use the balanced formula of the reaction to determine the amount of hydrochloric acid present in each 20 mL aliquot of diluted hydrochloric acid.





3. In total, there is 250 mL of HCl solution because 50mL of HCl was diluted with 200mL of deionised water. If 20mL of that solution contains  0.000237 mol of HCl, then 250mL will contain



Then we can use the formula n=cV to determine the concentration of the original solution.



EDIT: attempted to fix mistake

[xvt2101]

Thank you for the answer. 

I'm not sure where the 20 came from, could you please explain? Should I change it to 0.020 L or leave it as 20 mL. I'm grateful for the response, it's just that I'm really confused. 



EDIT: Thanks for the answer. I was really really confused, thanks for clarifying.

[Bad Student]

I think that's a mistake, I'll try to fix it.

[Reckoner]

^^Nearly.
For step 3, we have a 20ml aliquot with 0.000237 mol of HCl. We need to find out n(HCl) in 250 ml of the same concentration solution. So we use ratios. To convert the 20ml to 250ml, we need to multiply by 250/20. As as they are the same concentration, we need to multiply 0.000237 by 250/20 in order the find n(HCl) (not 20 as you put)
The second part of step 3 is correct though.


EDIT: you noticed, good job

[teletubbies_95]

average titre is wrong ! :0  there arent three concordant results due to +- 0.1 ml. So only, 2.30ml and 2.33 ml are concordant results !

[thushan]

In other words - you should only average concordant titres, which are titres within 0.10 mL of each other.

[Reckoner]

average titre is wrong ! :0  there arent three concordant results due to +- 0.1 ml. So only, 2.30ml and 2.33 ml are concordant results !

Good catch! This is a perfect example of why checking your work and actually reading the question properly are vital steps! 

[Limista]

For paper and thin-layer chromatography - are they methods that can only be used if there is polar solute that has to dissolve in polar solvent?

So, paper and thin-layer chromatography do not work for non-polar solutes and solvents?

Thanks

[thushan]

For paper and thin-layer chromatography - are they methods that can only be used if there is polar solute that has to dissolve in polar solvent?

So, paper and thin-layer chromatography do not work for non-polar solutes and solvents?

Thanks

TLC defs works for non-polar solvents. Dw about paper chrom, not on study design.