HSC 2012-2015 Chemistry Marathon (archive) (6 Viewers)

someth1ng · Feb 21, 2012

Re: HSC 2012 Chemistry Marathon

nightweaver066 said:
Good answer. In a test, hopefully you'll draw a nice structural formula of ethanol showing both the hydroxide functional group and the alkyl chain.

Post a new questions please!

Yeah, I normally draw pictures/diagrams of some sort when I can.

nightweaver066 said:
Notice how i wrote E(ox). That means i already had reversed the reduction potential given on the data sheet to obtain the oxidation potential.

If you look on your sheet, it tells you that the reduction potential is 0.34V.

Yeah, so it's ALWAYS:
Ecell=higher number-lower number ??

Next question:
A student, attempting to obtain the equivalence point of a titration between acetic acid and sodium hydroxide, was unable to reach a consistent result using methyl orange as an indicator. Suggest one modification to the procedure which would give an acceptable result. Explain why sodium acetate solution is not pH neutral.

nightweaver066 · Feb 21, 2012

Re: HSC 2012 Chemistry Marathon

someth1ng said:
Yeah, I normally draw pictures/diagrams of some sort when I can.

Yeah, so it's ALWAYS:
Ecell=higher number-lower number ??

Yep.. The Fe3+ to Fe2- reduction potential is 0.77V.
The CU2+ to Cu reduction potential is 0.34V.

By following what you said, Ecell = 0.77 - 0.34 = 0.43V.

nightweaver066 · Feb 21, 2012

Re: HSC 2012 Chemistry Marathon

someth1ng said:
Yeah, I normally draw pictures/diagrams of some sort when I can.

Yeah, so it's ALWAYS:
Ecell=higher number-lower number ??

Next question:
A student, attempting to obtain the equivalence point of a titration between acetic acid and sodium hydroxide, was unable to reach a consistent result using methyl orange as an indicator. Suggest one modification to the procedure which would give an acceptable result. Explain why sodium acetate solution is not pH neutral.

Use phenolphthalein instead of methyl orange as the indicator.

$CH_3COONa_{(aq)} \rightarrow CH_3COO^-_{(aq)} + Na^+_{(aq)}$

$CH_3COO^-_{(aq)} + H_2O_{(l)} \rightarrow CH_3COOH_{(aq)} + OH^-_{(aq)}$

Therefore sodium acetate solution is not pH neutral but is basic, i.e. pH > 7.

someth1ng · Feb 21, 2012

Re: HSC 2012 Chemistry Marathon

nightweaver066 said:
Use phenolphthalein instead of methyl orange as the indicator.

$CH_3COONa_{(aq)} \rightarrow CH_3COO^-_{(aq)} + Na^+_{(aq)}$

$CH_3COO^-_{(aq)} + H_2O_{(l)} \rightarrow CH_3COOH_{(aq)} + OH^-_{(aq)}$

Therefore sodium acetate solution is not pH neutral but is basic, i.e. pH > 7.

Oh, you use a basic indicator (phen...) when you are testing for basic substances and acidic indicators (methyl orange) for acidic substances?

Also, is the anion of the weak acid make basic solutions? eg CH3COO-

I don't know how to do that question:
How do you do this question?
A standard solution was prepared by dissolving 1.432g of sodium carbonate in water. The solution was made up to 250mL in a volumetric flask. Calculate the concentration of the sodium carbonate solution.The solution was used to determine the concentration of a solution of hydrochloric acid. Four 25.00mL samples of acid were titrated with the sodium carbonate solution. The average titration volume was 22.35mL. Why was it necessary to conduct four titrations? Calculate the concentration of the hydrochloric acid solution.

Dubble-u25 · Feb 22, 2012

Re: HSC 2012 Chemistry Marathon

someth1ng said:
I don't know how to do that question:
How do you do this question?
A standard solution was prepared by dissolving 1.432g of sodium carbonate in water. The solution was made up to 250mL in a volumetric flask. Calculate the concentration of the sodium carbonate solution.The solution was used to determine the concentration of a solution of hydrochloric acid. Four 25.00mL samples of acid were titrated with the sodium carbonate solution. The average titration volume was 22.35mL. Why was it necessary to conduct four titrations? Calculate the concentration of the hydrochloric acid solution.

Firstly, to calculate the concentration of Na₂CO₃ we will need to use the formula c=n/v.
Therefore from the given information v= 0.25L, and n will need to be calculated using n=m/M.
n(Na₂CO₃)= 1.432/2(22.99)+12.01+3(16) = 0.0135....
c(Na₂CO₃)= 0.0135...../0.25 = 0.0540......
Therefore the concentration of Na₂CO₃= 0.05404 (4 sig.figs)

Now, write a balanced equation:
Na₂CO₃_(aq) + 2HCl_(aq) → 2NaCl_(aq) + H₂O_(l) + CO₂_(g)
Again, we need to use c=n/v to calculate the concentration of HCl but we are only give v=0.025L.
So using the concentration of Na₂CO₃ we found earlier and v=0.02235L we can calculate the number of mols that reacted.
c=n/v
n= c x v
n(Na₂CO₃)= 0.05404..... x 0.02235 = 0.001207......
From the equation we can see that there is a 1:2 ratio (Na₂CO₃: HCl)
Therefore HCl will have the twice the number of mols as Na₂CO₃.
Now, c=n/v
c(HCl)=0.002414..../0.025 = 0.09656 (4sig.figs.)

So the concentration of HCl is 0.09656

It is necessary to conduct the titration several times to obtain an average to ensure that your results are accurate.

Now, write a balanced equation:
Na₂CO₃_(aq) + HCl_(aq) → NaCl_(aq) + H₂O_(l) + CO₂_(g)

*facepalm*

nightweaver066 · Feb 22, 2012

Re: HSC 2012 Chemistry Marathon

Solution:
$n(BaSO_4) = \frac{1.8}{137.3+32.07+16 \times 4}$

$= 7.71 \times 10^{-3} moles$

$n(BaSO_4) = n(SO_4^{2-}) = 7.71 \times 10^{-3} moles$

$m(SO_4^{2-}) = n(SO_4^{2-}) \times (32.07+16 \times 4)$

$= 0.741g$

$(SO_4^{2-})\%(mass) = \frac{0.741}{2}$

$= 37\%$

Qn:

someth1ng · Feb 22, 2012

Re: HSC 2012 Chemistry Marathon

(a)
H2O(l)+CO2(g) <--> H2CO3(aq)
- Increasing the pressure within the system will favour the forward reaction as it would decrease pressure within the system.
- If the forward reaction is favoured, an increased amount gas is dissolved into solution.

(b)
The forward reaction of "H2O(l)+CO2(g) <--> H2CO3(aq)" is exothermic so a lower temperatures, the reaction favours the direction that increases temperature - the forward reaction and so more gas will be dissolved.

Consider the following equation for the reaction of potassium dichromate with an acidified solution of sodium iodide.
Cr2O7-2(aq) + 14H+(aq) +6I-(aq) 2Cr3+(aq)+ 3I2(aq) + 7H2O(l)
What is the oxidation state of chromium in the dichromate ion?
Identify which products have been reduced and which have been oxidised.
Justify the omission of sodium and potassium from the ionic equation above in terms of movements of electrons.

someth1ng · Feb 22, 2012

Re: HSC 2012 Chemistry Marathon

What exactly is "titration volume"?

Also, I don't see why this line, "n(Na2CO3)= 0.05404..... x 0.02235 = 0.001207......" works. How does that show the number of moles that reacted.

Dylanamali · Feb 22, 2012

Re: HSC 2012 Chemistry Marathon

someth1ng said:
What exactly is "titration volume"?

Also, I don't see why this line, "n(Na2CO3)= 0.05404..... x 0.02235 = 0.001207......" works. How does that show the number of moles that reacted.

titration volume in that question refers to the amount of sodium carbonate needed to neutralise/reach the end point the HCl.
So thus 22.35mL of sodium carbonate was needed to neutralise 25mL of HCl.

Using that knowledge, we can then determine the no. of mols of sodium carbonate needed to neutralise the HCl, which is found by n = C x V = 0.05404 (conc. from part 1) x 0.02235 (volume) = 0.001207 mols.

someth1ng · Feb 22, 2012

Re: HSC 2012 Chemistry Marathon

Dylanamali said:
titration volume in that question refers to the amount of sodium carbonate needed to neutralise/reach the end point the HCl.
So thus 22.35mL of sodium carbonate was needed to neutralise 25mL of HCl.

I was probably confused since I didn't have the definition of "titration volume" in my mind.

Dylanamali said:
titration volume in that question refers to the amount of sodium carbonate needed to neutralise/reach the end point the HCl.
So thus 22.35mL of sodium carbonate was needed to neutralise 25mL of HCl.

Why is it 22.35mL Na2CO3:25mL HCl and not the 250mL Na2CO3:22.35mL HCl?

Dubble-u25 said:
c(HCl)=0.001207...../0.025 = 0.04831 (4sig.figs.)

Why is it divided by 0.025L if only 0.02235L reacted?

To clear things up:
I was thinking, why isn't it how much HCl is needed to neutralise the Na2CO3?

After reading the question again:
"Four 25.00mL samples of acid were titrated with the sodium carbonate solution. The average titration volume was 22.35mL. "

Does this mean: there were four flasks of 25mL of HCl and in each, they put 22.35mL of the Sodium Carbonate solution?

Dylanamali · Feb 22, 2012

Re: HSC 2012 Chemistry Marathon

Have you executed the titration prac yet at school? If not, then it might be difficult to understand.

But yeah, in that question - a 250mL solution of sodium carbonate was made with a concentration of 0.05404 M as easily determined in part 1.
This solution is then poured into the burette. Whilst 25mL of the acid is pipetted into the conical flask.
The burette is then used to pour the sodium carbonate solution into the conical flask containing the acid (it also contains an indicator - most likely methyl orange in this case - ask if you want to know why its methyl orange). The methyl orange indicator changed colour once 22.35mL of the sodium carbonate base had been added to the acid - indicating the equivalence point and point of neutralisation. That is why its 22.35mL of sodium carbonate not 250mL - because only 22.35mL of sodium carbonate was buretted into the flask not 250.

Firstly, we used the titration volume and multiplied that by the concentration of the sodium carbonate to find the no. of mols of sodium carbonate. We then apply molar ratios, in this case 1:2 sodium carbonate:HCl, to find the no. of mols of HCl that were in the flask. We found this to be 0.001207x2 mols. After this to find the concentration of HCl we use C = n/V.
In this case the volume is 0.025 and not 0.02235 because the 22.35mL was indicative of the sodium carbonate needed, but in this calculation we are determining the conc of the HCl - and we know from the question that 25mL of HCl was pipetted into the flasks.

someth1ng · Feb 22, 2012

Re: HSC 2012 Chemistry Marathon

Yeah, I read your response then read the question then realised that I didn't read it properly. Four 25.00mL samples of acid were titrated with the sodium carbonate solution.
I thought it meant the Sodium Carbonate solution was titrated with HCl so I was so lost! haha

It all makes perfect sense now - thanks guys.

In the HSC, will they follow you through solving it? Like, can it ask you somthing like:

"A standard solution was prepared by dissolving 1.432g of sodium carbonate in water. The solution was made up to 250mL in a volumetric flask. Four 25.00mL samples of acid were titrated with the sodium carbonate solution. The average titration volume was 22.35mL.

Calculate the concentration of the hydrochloric acid solution."

Dylanamali · Feb 22, 2012

Re: HSC 2012 Chemistry Marathon

someth1ng said:
Does this mean: there were four flasks of 25mL of HCl and in each, they put 22.35mL of the Sodium Carbonate solution?

Not neccessarily.. it probably means that they had four flasks of 25mL of HCl and performed 4 titrations.
The first being a rough titration to gain an approximate indication of where the eq. point is and next 3 being more accurate. The would then discard the rough titration value and use the other 3 values to determine a titration volume average - which was 22.35mL of Na2CO3.

Dylanamali · Feb 22, 2012

Re: HSC 2012 Chemistry Marathon

someth1ng said:
In the HSC, will they follow you through solving it? Like, can it ask you somthing like:

"A standard solution was prepared by dissolving 1.432g of sodium carbonate in water. The solution was made up to 250mL in a volumetric flask. Four 25.00mL samples of acid were titrated with the sodium carbonate solution. The average titration volume was 22.35mL.

Calculate the concentration of the hydrochloric acid solution."

Yes - generally with titration questions in the HSC they are split up into 2 or 3 parts. So they get you to break down each component of the question.
a) find the concentration of the sodium carbonate solution (1 mark)
b) find the concentration of the HCl solution (2 marks)

SanjoyM · Feb 22, 2012

Re: HSC 2012 Chemistry Marathon

Explain why Sodium Hydorxide is not suitable to be used as a primary standard and justify your answer (3 marks)
Define Le Chatlier's principle and outline the factors it effects.
Analyse how neutralisation reactions can be used to minimise damages during chemical spills.
What is condensation polymerisation?
Describe the process for the production of Hight density and Low density Polyethylene, and compare and contrast its uses with its properties (6 marks)

More questions coming up

LOL

SanjoyM · Feb 23, 2012

Re: HSC 2012 Chemistry Marathon

You have performed a practical on Esterfication. Outline and steps your undertook and assess the reliability, validity and accuracy of the experiment.
Define pH? (Simple q

)
Discuss why the pipette and burette should be rinsed with the solution they will contain before conducting the titration.

Dylanamali · Feb 23, 2012

Re: HSC 2012 Chemistry Marathon

btw - just out of curiosity where is everyone at in the syllabus in HSC chem atm? how far into acidic are you guys?

^ sanjoy you already done esterification at school?? =O

SanjoyM · Feb 23, 2012

Re: HSC 2012 Chemistry Marathon

Dylanamali said:
btw - just out of curiosity where is everyone at in the syllabus in HSC chem atm? how far into acidic are you guys?

^ sanjoy you already done esterification at school?? =O

We started with the acidic environment, and we are up to the 3rd context point of Production of materials. We'll probably finish POM soon, and start Monitoring chemical... Or revise for Half yearlies

Dylanamali · Feb 23, 2012

Re: HSC 2012 Chemistry Marathon

ohh right, no wonder.

ah k fair enough.. yeah I wouldn't think that you'd begin monitoring before half yearlies, you'll probs do revision.

someth1ng · Feb 23, 2012

Re: HSC 2012 Chemistry Marathon

1 more thing about the titration, isn't there 2 moles of HCl per 1 mole Na2CO3? Hence, you need to double that final concentration?

After doing that, I got 0.09662 mol/L

I'm only early into the Acidic Environment - end of PFA 2 and start of PFA 3.

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