Chemistry 3/4 2013 Thread

[Mr Keshy]

Dude you were practically right!

Yay!

I thought it had something to do with that.

[Edward21]

HELP, I've never seen a question like this   
A bottle of ‘Wunda White’ bleach lists the following as its active ingredients: sodium hypochlorite, NaOCl, 40 g L−1; sodium hydroxide, NaOH, 9 g L−1. A chemist wishes to check both these claims using volumetric analysis.

To check the sodium hydroxide claim, an analyst had at his disposal some 0.100 mol L−1 hydrochloric acid, a 20.00 mL pipette, a burette and a 250.0 mL volumetric flask. For various reasons, a titre of about 20.00 mL is preferable.

       What is the approximate volume of bleach that should be added to the volumetric flask for dilution before the analysis?

A    10 mL

B    50 mL

C    100 mL

D    200 mL

[Limista]

HELP, I've never seen a question like this   
A bottle of ‘Wunda White’ bleach lists the following as its active ingredients: sodium hypochlorite, NaOCl, 40 g L−1; sodium hydroxide, NaOH, 9 g L−1. A chemist wishes to check both these claims using volumetric analysis.

To check the sodium hydroxide claim, an analyst had at his disposal some 0.100 mol L−1 hydrochloric acid, a 20.00 mL pipette, a burette and a 250.0 mL volumetric flask. For various reasons, a titre of about 20.00 mL is preferable.

       What is the approximate volume of bleach that should be added to the volumetric flask for dilution before the analysis?

A    10 mL

B    50 mL

C    100 mL

D    200 mL

Is the answer C?

[Limista]

Also the mass of solute required? 400 mL of a solution that is 1.00 mol L−1 with respect to Fe3+ ions, using Fe2(SO4)3.9H2O? Again, I haven't seen this type of question in these circumstances!

I got 224840g (this is deliberately not in sig figs by the way). However, I don't really understand the question which you typed. 

[Edward21]

Is the answer C?

It's A, and I got B!

[Edward21]

I got 224840g (this is deliberately not in sig figs by the way). However, I don't really understand the question which you typed. 

And I copied and pasted directly from Chemistry 2. The answer for that one is 112g and I don't know how to get there, if m=nM, then logically divide 112g/55.8g/mol for approx 2mol... where did that come from!?

[Limista]

And I copied and pasted directly from Chemistry 2. The answer for that one is 112g and I don't know how to get there, if m=nM, then logically divide 112g/55.8g/mol for approx 2mol... where did that come from!?

It's A, and I got B!

Oh god. Well I gave them both a shot & honestly I've never been faced with questions like these as yet (which is not saying much, since I only just started year 12 haha). Hmm - you might have to ask Aurelian or TRex or someone... I'm sure they'll be able to provide an answer! 

[Edward21]

Oh god. Well I gave them both a shot & honestly I've never been faced with questions like these as yet (which is not saying much, since I only just started year 12 haha). Hmm - you might have to ask Aurelian or TRex or someone... I'm sure they'll be able to provide an answer! 

Yeah, and at this point I'm now a student that has no life, other than to study chemistry on a saturday night #vcelyf don't worry, hope when someone (hint hint) answers these we'll both learn what we got wrong 

[Chazef]

I have no idea what to do but I tried this:
1. assume it's 20.00mL of HCl you want, meaning the n(HCl) = 0.1 x .020 = 0.002 mol
2. n(NaOH) = n(HCl) = 0.002 mol
3. 9 g/L NaOH, convert to mol/L by dividing by molar mass, = 0.229 mol/L
4. find reciprocal of above value to get the amount of litres per mol of NaOH (1/0.229 = 4.367 L/mol)
5. you want 0.002 mol so multiply by above value to get the amount of litres you need, 4.367 x 0.002 mol = 0.00873 L = 8.73 mL which is around about 10mL
am I even close?

[MonsieurHulot]

I have no idea what to do but I tried this:
1. assume it's 20.00mL of HCl you want, meaning the n(HCl) = 0.1 x .020 = 0.002 mol
2. n(NaOH) = n(HCl) = 0.002 mol
3. 9 g/L NaOH, convert to mol/L by dividing by molar mass, = 0.229 mol/L
4. find reciprocal of above value to get the amount of litres per mol of NaOH (1/0.229 = 4.367 L/mol)
5. you want 0.002 mol so multiply by above value to get the amount of litres you need, 4.367 x 0.002 mol = 0.00873 L = 8.73 mL which is around about 10mL
am I even close?

Yup, that's what I did, albeit in a slightly different way.
I worked out that you need about 1.86mL of bleach for the appropriate amount of sodium hypcohlorite, so 10mL is the most reasonable answer.

[Reckoner]

This is what I did, got 8.89ml. Hope you can read it, and follow it. I didn't set it out well either.

[Mr Keshy]

Gosh, keen fellows discussing VCE Chemistry first thing in the morning (literally).

Interesting question this, haven't seen anything similar to it (at least I can't remember)

[Edward21]

This is what I did, got 8.89ml. Hope you can read it, and follow it. I didn't set it out well either.

Oh wow, thanks; I totally understand, and the setting out was fine 

[Chazef]

I'm still confused because surely you would need more than 0.002 mol of NaOH in the volumetric flask as you would be doing multiple titrations wouldn't you? I don't see why the concentration of the NaOH in the volumetric flask would be .002/.250 because wouldn't that suggest that you are just going to use the entire volumetric flask for one titration?

[Reckoner]

^yeah, thats what I thought. I don't think its a very good question. Best to see what the experts think I reckon.

EDIT: He only has a 20.00 mL pipette, a burette and a 250.0 mL volumetric flask. If he did multiple titres, he would have to use 20ml of the NaOH as thats the only amount he could accurately measure. If this is the case, he would need 100ml of the original solution (see attached).

With the glassware he has, can he do multiple titres? I've forgotten most of this lab technique stuff lol! I'd better relearn it for my student haha

EDIT 2 ignore my question haha, I decided to actually think about it rather than remember. And what do you know I figured it out. Always think!