Unit 4 Questions MEGATHREAD!

[banditkeith]

Hi guys (and gals!)
Just going over my practice papers again and realised a confusion in my understanding of weak acids.
This was on the Neap 09 paper by the way..

2cii) 10.0mL of the 4.0M HOBr solution containing bromophenol blue indicartor was diluted to a total volume of 1.0L
What colour change, if any, would be expected? Explain.

I wrote that there would be a lighter yellow as dilution has occurred and since it's a weak acid, dilution would not change
the pH by much and hence not change the indicator colour.
Apparently it changes to a blue colour since the pH would rise above 4.6 (The initial pH was found to be 4.0)
Could someone explain how much effect dilution actually has on weak acids?
Thanks in advance!

[vea]

Hi guys (and gals!)
Just going over my practice papers again and realised a confusion in my understanding of weak acids.
This was on the Neap 09 paper by the way..

2cii) 10.0mL of the 4.0M HOBr solution containing bromophenol blue indicartor was diluted to a total volume of 1.0L
What colour change, if any, would be expected? Explain.

I wrote that there would be a lighter yellow as dilution has occurred and since it's a weak acid, dilution would not change
the pH by much and hence not change the indicator colour.
Apparently it changes to a blue colour since the pH would rise above 4.6 (The initial pH was found to be 4.0)
Could someone explain how much effect dilution actually has on weak acids?
Thanks in advance!

I see where you're trying to go by saying that dilution of a weak acid will not change the pH much but this is compared to if you dilute a same concentration and volume of a strong acid by the same factor.
Here you can actually calculate the pH of the HOBr using the Ka values from the data booklet.

BEFORE DILUTION
2.4 x 10^-9=[H3O+]^2/4.0
[H3O+]=9.798 x 10^-5
pH=-log(9.798 x 10^-5)=4.01

When it has been diluted from 10mL to 1.0L, the concentration changes by a factor of 1/0.01=100
4.0/100=0.04M

AFTER DILUTION
2.4 x 10^-9=[H3O+]^2/0.04
[H3O+]=9.798 x 10^-6
pH=-log(9.798 x 10^-6)=5.01
This pH is greater than 4.6 and so therefore the colour will change to blue!

[banditkeith]

Yeah.. I guess it was wrong of me to assume the pH doesn't change by that much. Shouldn't have been lazy and actually worked out the new pH. Thanks for your time

I feel so ready for chem but not ready at the same time. Tis stupid

[generalkorn12]

What's the equation to find the mole/mass of substance, if you only know the compounds
heat of combustion, and the energy released during the reaction.
I've noticed, when doing past papers, that they seem to divide Energy by (Delta Hc), and can
find the mole of an element...

[Panicmode]

What's the equation to find the mole/mass of substance, if you only know the compounds
heat of combustion, and the energy released during the reaction.
I've noticed, when doing past papers, that they seem to divide Energy by (Delta Hc), and can
find the mole of an element...

See attached scan = D

[DisaFear]

So, after so much methods and english cram, I've forgotten how to do these (done so many before)


Can't be

A. Where Pb is involved, it's negative during discharge, so positive during recharge
B. pH decreases as more H+ is made
C. PbSO4 is not produced during recharge
D. Those changes are for the discharge, not recharge?

What am I doing wrong? Probably interpreting D wrong

[Mao]

During discharge, Pb + SO42- --> PbSO4 + 2e is the anode equation (-ve)
During recharge, this equation is the reduction equation PbSO4 + 2e --> Pb + SO42-.
Reduction at cathode. During electrolysis, cathode is (-ve).

:. Pb is produced at the negative electrode.

You will find that the +/- will not change going from electrochemical to electrolytic. The only thing that changes is whether they are the cathode or anode. (Definition of cathode/anode depend on the flow of electrons, definition of +/- depends on the electrochemical potential, which is static)

(your other assessments are correct)

[Hamdog17]

Is the minimum voltage required to recharge a secondary cell equal to the operating voltage or the highest half-cell voltage?

Chemology '11 says half-cell, I think it's operating

[Panicmode]

Is the minimum voltage required to recharge a secondary cell equal to the operating voltage or the highest half-cell voltage?

Chemology '11 says half-cell, I think it's operating

Heinemann says, To recharge a cell, the  products of the reaction must be converted back into the original reactants: the reaction must occur in reverse. This is done by connecting the cell to a 'charger', a source of electrical energy, which has a potential difference greater than the potential difference of the cell

ie, You're right. It's the potential difference of the cell (operating voltage) not the highest half cell E naught voltage.


However... reading further into it it appears that electrolysis of water can occur when a minimum of only 1.23 volts is applied (the E naught of the higher half cell, not the potential difference) so I dunno.

[nos488]

With fuel cell questions that I've seen in a few practice exams i have trouble approaching it because i have no idea what the reactants are going to form, anyone have tips on these sort of questions?

[DisaFear]

With fuel cell questions that I've seen in a few practice exams i have trouble approaching it because i have no idea what the reactants are going to form, anyone have tips on these sort of questions?

Give us an example. I think I know which type of questions you are on about, I am not good at them as well.

[Hamdog17]

With fuel cell questions that I've seen in a few practice exams i have trouble approaching it because i have no idea what the reactants are going to form, anyone have tips on these sort of questions?

Most questions should tell you. Remember for any fuel cell using a hydrocarbon as fuel the overall cell reaction is the same as the combustion of that cell. i.e. any hydrocarbon or organic (eg. glucose) fuel cell produces CO2. Also remember a fuel cell using oxygen always reduces oxygen. Use you're electro chem series for the rest

[Christiano]

VCAA 2008 Question 19

Fuel cells are being developed that use fuels other than hydrogen as their energy sources. The table below shows
four potential fuels and their reactions in the fuel cell. (For simplicity, symbols of state have been omitted from
these reaction equations.)
Fuel Reaction in the fuel cell
methanol CH3OH CH3OH + H2O → CO2 + 6H+ + 6e–
ethanol C2H5OH C2H5OH + 3H2O → 2CO2 + 12H+ + 12e–
ethane C2H6 C2H6 + 4H2O → 2CO2 + 14H+ + 14e–
ethane-1, 2-diol C2H4(OH)2 C2H4(OH)2 + 2H2O → 2CO2 + 10H+ + 10e–

Which one of the fuels would produce the greatest amount of CO2 per coulomb of electrical charge
generated?
A. methanol
B. ethanol
C. ethane
D. ethane-1, 2-diol

I thought it would be C but it's D. I don't understand the assessor's report. Could someone elaborate?

[vea]

You look at the ratio of n(CO2)/n(e-)

methanol: 1/6
ethanol: 2/12=1/6
ethane: 2/14=1/7
ethane-1,2-diol 2/10=1/5

1/5 is the largest number there so the answer is D

[mickeymouse]

Alright I have a question:

Why does the concentration of OH- decrease as the concentration of lactic acid increases?

So I understand that as a result [H+] would increase, pH would decrease, but not too sure where the hydroxide ions come in ...