Unit 4 Questions MEGATHREAD!

[Panicmode]

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 ...

Remember the equation;

2H2O <--> OH- + H30+

As the concentration of H3O+ ions is increased, the system partially opposes this with an equilibrium shift to the left. Therefore, as the [H3O+], the [OH-] decreases and vice versa

[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 ...

Remember the equation;

2H2O <--> OH- + H30+

As the concentration of H3O+ ions is increased, the system partially opposes this with an equilibrium shift to the left. Therefore, as the [H3O+], the [OH-] decreases and vice versa

mmm oh yeah, so if a question involves a weak acid, should we always consider the self-ionisation constant of water? since all acids are aqueous?

[Aurelian]

If anyone's done CSE 2010 CSE -

Q2e) "Write the name of a suitable indicator to monitor the pH of a titration between ethanoic acid and sodium hydroxide."

Honestly no idea how to do this. I feel it may be something from Unit 3 stuff, but even then I only recall indicator selection based on titration curves.

Anyone able to help?

[Panicmode]

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 ...

Remember the equation;

2H2O <--> OH- + H30+

As the concentration of H3O+ ions is increased, the system partially opposes this with an equilibrium shift to the left. Therefore, as the [H3O+], the [OH-] decreases and vice versa

mmm oh yeah, so if a question involves a weak acid, should we always consider the self-ionisation constant of water? since all acids are aqueous?

If a question involves a weak acid, you should use the Ka value to determine the [H30+]. Yes, all acids are aqueous, but remember that the self ionisaiton constant o water only holds true at 25 degrees Celsius.

Q2e) "Write the name of a suitable indicator to monitor the pH of a titration between ethanoic acid and sodium hydroxide."

Honestly no idea how to do this. I feel it may be something from Unit 3 stuff, but even then I only recall indicator selection based on titration curves.

Anyone able to help?

In your data booklet, it lists the Ka values of most of the indicators. Use this to figure out the [H30+] at which they will change and help you to determine a suitable indicator.

[Aurelian]

In your data booklet, it lists the Ka values of most of the indicators. Use this to figure out the [H30+] at which they will change and help you to determine a suitable indicator.

Would you be able to elaborate? I can determine the [H3O+] at which they will change (assuming a 1:1 ratio of acid and conjugate base), but I don't know how I would use this to then determine a suitable indicator...

[Christiano]

This is a TSFX question (based off an article from The Age)

Effect on the c(Fe3+) and m(Fe2+) of adding water to the Fe3+(aq) + SCN-(aq) Fe(NCS)2+(aq) equilibrium at constant temperature.
Adding water instantaneously decreases all concentrations, and increases the concentration fraction. This pushes the system out of equilibrium. To return to equilibrium the concentration fraction must decrease back equal to K, so the reverse reaction dominates and the m(Fe3+) increases. Whilst this increases the c(Fe3+), equilibrium is reached again before it returns to its initial value. Overall the c(Fe3+) decreases.

Can you explain why the equilibrium 2Fe3+(aq) + Sn2+(aq) 2Fe2+(aq) + Sn4+(aq) is not pushed out of equilibrium by adding water?

[cltf]

This is a TSFX question (based off an article from The Age)

Effect on the c(Fe3+) and m(Fe2+) of adding water to the Fe3+(aq) + SCN-(aq) Fe(NCS)2+(aq) equilibrium at constant temperature.
Adding water instantaneously decreases all concentrations, and increases the concentration fraction. This pushes the system out of equilibrium. To return to equilibrium the concentration fraction must decrease back equal to K, so the reverse reaction dominates and the m(Fe3+) increases. Whilst this increases the c(Fe3+), equilibrium is reached again before it returns to its initial value. Overall the c(Fe3+) decreases.

Can you explain why the equilibrium 2Fe3+(aq) + Sn2+(aq) 2Fe2+(aq) + Sn4+(aq) is not pushed out of equilibrium by adding water?

Dilution decreases the concentration of solution, same amount of particles in a larger area. Now look at how decreasing pressure, particles are more spread out. Simialr theory/outcome. Therefore in an equation where there are equal particles or either side, any uniform changes to the concentrations would have no real net effect, and this applies to gas and solutions.

At least that's how I interpret it.

[DavidSheena]

CSE 2009
Question three:
 
Consider the equilibrium:
CO(g) + 2H2(g) <-> CH3OH(g)   Change in H = -92 kJ per mol.

Change to the System            Net Shift         Effec on no. of mol of H2.           Effect on [H2]
Decrease of pressure                Left                   Increase                                    Increase
Volume Halved                          Right                  Decrease                                   Decrease

Those were my answers; you had to fill in the table for each change.

The solutions say that the [H2] would be opposite for each one; decrease for the first and increase for the second.

Could anyone explain?

[Aurelian]

CSE 2009
Question three:
 
Consider the equilibrium:
CO(g) + 2H2(g) <-> CH3OH(g)   Change in H = -92 kJ per mol.

Change to the System            Net Shift         Effec on no. of mol of H2.           Effect on [H2]
Decrease of pressure                Left                   Increase                                    Increase
Volume Halved                          Right                  Decrease                                   Decrease

Those were my answers; you had to fill in the table for each change.

The solutions say that the [H2] would be opposite for each one; decrease for the first and increase for the second.

Could anyone explain?

You are right in your assessment of the direction in which the position of equilibrium shifts, but you must remember that Le Chatelier's principle relates to the system *partially* counteracting the imposed change. As a result, a halving of volume, for instance, would trigger a forward reaction and consume some H2, but the overall concentration is still higher than before the volume halving. Looking at a concentration/time graph, [H2] spikes up and then goes back down a bit (as the system responds), but it doesn't go down to where it was beforehand, letalone lower. Does that help?

[DavidSheena]

Yes! Thank you so much. I totally forgot about how everything spikes up before it goes down when you increase pressure. Thank you

[nos488]

oxidant, oxidation, reducant, reduction, reduced and oxidised what are the differences i always get confused

[REBORN]

A reductant is a substance that allows another substance to be reduced but itself is always oxidised.

[DisaFear]

How do?

[tony3272]

No idea if this is right, but this is how i went about it:



Therefore

Therefore

Therefore

[mickeymouse]

How do?
(Image removed from quote.)

oh I just did this question lol

times it by x100 and algebra

so just make that ratio in your ka equal 1/100

if that makes sense ... o_O