VCE Chemistry Question Thread

[Adequace]

http://m.imgur.com/a/nogqP

Could I have some help with these 2 questions.

Q13) I'm not sure what I'm doing wrong but the answer is 0.3 mol/L

Q18) I don't know how to approach this question, I've tried finding the concentration of the diluted solution and then found the mole of that but I don't think this is correct.

Thanks

[Jakeybaby]

http://m.imgur.com/a/nogqP

Could I have some help with these 2 questions.

Q13) I'm not sure what I'm doing wrong but the answer is 0.3 mol/L

Q18) I don't know how to approach this question, I've tried finding the concentration of the diluted solution and then found the mole of that but I don't think this is correct.

Thanks

Hey,

 Here we have some concentration conversions, you may or may not have seen a diagram such as the following:
https://i.gyazo.com/3636474d3a9f8fd923cca3949658cae0.png

This will help you with both questions.

Question 13:

4.26%w/v (although weight is not mass, they are used interchangeably)

To convert to mol/L, firstly:
4.26 x 10 = 42.6
then: 42.6/M = 42.6/142.04 = 0.3mol/L

Question 18:

n=cv
n=0.0015x0.05
n=0.000075 mol prior to dilution.

And as the number of moles are kept constant through the dilution:
0.000075 = c x 10 (10L)
0.0000075 = c
Therefore after the dilution, 0.0000075mol/L was present.
Using the conversion table:
0.0000075 * 110.98 = 0.00083235 g/L
0.00083235 * 1000 = 0.83235 ppm

Jake

[Adequace]

Thanks Jake!!

[chemnerd345]

Can someone please tell me for the reaction:
Fe3+ (aq) +SCN- (aq) <=> Fe(SCN)2+ (aq)
will adding distilled water to the test tube move the equilibrium position to the left because its equivilant to an increase in pressure? Or will it have no effect apart from diluting the concentrations of the reactants and products?

[Jakeybaby]

Can someone please tell me for the reaction:
Fe3+ (aq) +SCN- (aq) <=> Fe(SCN)2+ (aq)
will adding distilled water to the test tube move the equilibrium position to the left because its equivilant to an increase in pressure? Or will it have no effect apart from diluting the concentrations of the reactants and products?

The equilibrium would shift to the left, but your reasoning is incorrect.

Upon the addition of H2O, the Fe3+ ions react with OH- and the SCN ions react with H+. This initially decreases the concentration of Fe3+ and SCN-.
The equilibrium then shifts to the left to produce more reactants to re-establish equilibrium.

[Swagadaktal]

The equilibrium would shift to the left, but your reasoning is incorrect.

Upon the addition of H2O, the Fe3+ ions react with OH- and the SCN ions react with H+. This initially decreases the concentration of Fe3+ and SCN-.
The equilibrium then shifts to the left to produce more reactants to re-establish equilibrium.

Um I think the other person's reasoning was correct (according to the VCE course)
A dilution = decrease in concentration. To partially oppose this change, the equilibrium will seek to increase concentration. There are 2 particles on the left side and 1 particle on the right. Therefore it will favour a net backwards reaction to increase the concentration of the equilibrium.

[chemnerd345]

The equilibrium would shift to the left, but your reasoning is incorrect.

Upon the addition of H2O, the Fe3+ ions react with OH- and the SCN ions react with H+. This initially decreases the concentration of Fe3+ and SCN-.
The equilibrium then shifts to the left to produce more reactants to re-establish equilibrium.

thankyou!

[keltingmeith]

Um I think the other person's reasoning was correct (according to the VCE course)
A dilution = decrease in concentration. To partially oppose this change, the equilibrium will seek to increase concentration. There are 2 particles on the left side and 1 particle on the right. Therefore it will favour a net backwards reaction to increase the concentration of the equilibrium.

You're both right, however chemnerd is technically wrong as s/he's made an incorrect reference to pressure. Remember, in VCE chemistry, the only things that exert a pressure are gases!

[Adequace]

Hey,

 Here we have some concentration conversions, you may or may not have seen a diagram such as the following:
https://i.gyazo.com/3636474d3a9f8fd923cca3949658cae0.png

This will help you with both questions.

Question 13:

4.26%w/v (although weight is not mass, they are used interchangeably)

To convert to mol/L, firstly:
4.26 x 10 = 42.6
then: 42.6/M = 42.6/142.04 = 0.3mol/L

Question 18:

n=cv
n=0.0015x0.05
n=0.000075 mol prior to dilution.

And as the number of moles are kept constant through the dilution:
0.000075 = c x 10 (10L)
0.0000075 = c
Therefore after the dilution, 0.0000075mol/L was present.
Using the conversion table:
0.0000075 * 110.98 = 0.00083235 g/L
0.00083235 * 1000 = 0.83235 ppm

Jake

I just checked for Q18 that the answer is 0.533, would this be a mistake?

Also, if any VCE chem people could answer this, would we need to memorise the conversions of concentration units? My teacher hasn't talked about at all, or is it something that should be obvious depending on the conversion that needs to be done?

[Elizawei]

would we need to memorise the conversions of concentration units?

My teacher tells us to memorise them, it'll definitely be useful in calculations. Although, VCAA usually makes it quite clear so interpreting which unit to use shouldn't be a problem on the real exams.

[jyce]

I just checked for Q18 that the answer is 0.533, would this be a mistake?

Also, if any VCE chem people could answer this, would we need to memorise the conversions of concentration units? My teacher hasn't talked about at all, or is it something that should be obvious depending on the conversion that needs to be done?

I agree with Jakeybaby's answer. EDIT: Actually, the question didn't ask for the concentration of calcium chloride; it asked for the concentration of chloride ions:

c(Cl^-)undiluted = 2 x c(CaCl₂)undiluted = 0.05 x 2 = 0.1 M
c(Cl^-)diluted = 0.1 x (1.5/10000) = 0.000015 M
To convert to g L^-1, multiply the molarity by the molar mass of Cl^-: 0.000015 x 35.5 = 0.0005325 g L^-1
0.0005325 g L^-1 = 0.5325 mg L^-1 = 0.533 ppm.


And no, I wouldn't say you need to "memorise" conversions of concentration units. Especially because there are so many different units of concentration (e.g., %w/w, %w/v, %v/v, ppm, ppb, mol L^-1, g L^-1, mg g^-1, and the list goes on) and VCAA doesn't usually just say "convert this to that"; instead, the conversions are woven into larger questions (e.g., you might be given data about the grams of a solute in a certain number of millilitres of solvent, but to answer a question you might need to have the concentration in molarity). What you should focus more on is understanding why the conversions make sense and knowing what some of the more seemingly complicated units of concentration actually mean (e.g., why are concentration and volume of solvent inversely proportional? what does ppm actually mean?). Unit manipulation is something that you practice and get better at over time. It's definitely one of the areas of the course that student's find difficult; just keep practicing and you'll find that it becomes easier 

[blacksanta62]

Hey, can anyone help with part B? I can't seem to rearrange it the same way they have 
Thank you

[zsteve]

Hey, can anyone help with part B? I can't seem to rearrange it the same way they have 
Thank you

At pH = 2, [H+] = 10^-2 M. Hence:


Hence

[Jakeybaby]

I agree with Jakeybaby's answer. EDIT: Actually, the question didn't ask for the concentration of calcium chloride; it asked for the concentration of chloride ions:

c(Cl^-)undiluted = 2 x c(CaCl₂)undiluted = 0.05 x 2 = 0.1 M
c(Cl^-)diluted = 0.1 x (1.5/10000) = 0.000015 M
To convert to g L^-1, multiply the molarity by the molar mass of Cl^-: 0.000015 x 35.5 = 0.0005325 g L^-1
0.0005325 g L^-1 = 0.5325 mg L^-1 = 0.533 ppm.

Sorry for that mistake, just misread the question

If I were you, I'd recommend that conversion graphic which I posted before, normally they are the units which are commonly used in questions.

[Adequace]

I agree with Jakeybaby's answer. EDIT: Actually, the question didn't ask for the concentration of calcium chloride; it asked for the concentration of chloride ions:

c(Cl^-)undiluted = 2 x c(CaCl₂)undiluted = 0.05 x 2 = 0.1 M
c(Cl^-)diluted = 0.1 x (1.5/10000) = 0.000015 M
To convert to g L^-1, multiply the molarity by the molar mass of Cl^-: 0.000015 x 35.5 = 0.0005325 g L^-1
0.0005325 g L^-1 = 0.5325 mg L^-1 = 0.533 ppm.


And no, I wouldn't say you need to "memorise" conversions of concentration units. Especially because there are so many different units of concentration (e.g., %w/w, %w/v, %v/v, ppm, ppb, mol L^-1, g L^-1, mg g^-1, and the list goes on) and VCAA doesn't usually just say "convert this to that"; instead, the conversions are woven into larger questions (e.g., you might be given data about the grams of a solute in a certain number of millilitres of solvent, but to answer a question you might need to have the concentration in molarity). What you should focus more on is understanding why the conversions make sense and knowing what some of the more seemingly complicated units of concentration actually mean (e.g., why are concentration and volume of solvent inversely proportional? what does ppm actually mean?). Unit manipulation is something that you practice and get better at over time. It's definitely one of the areas of the course that student's find difficult; just keep practicing and you'll find that it becomes easier 

Thanks jyce!

For concentration(ppm), my notes say that the solution has to be in kgs, so why don't we need to do anything to the denominator when going from g/L -> ppm?