You are Here: Home

# Could anyone help me with couple of these questions? watch

1. Both the questions have been uploaded to this post.

Attached Images

2. What have you tried?
3. I haven't done A level chemistry for over 5 years. From memory and some reasonable knowledge of the equilibrium constants, i worked out the answer for question 8 as Exothermic and the temperature increasing. Not sure if i am entirely correct?

I do not have an idea how to start about answering question 9.

Do you have an idea?
4. I think i have figured out the Ka value and i have got it as B (5 ka). Am i correct?
5. Kc increases with temperature. Kc = [products]/[reactants]. Increasing Kc means more products than before, i.e. equilibrium -> RHS. Therefore endothermic. Q9: I'd go with Ka = [H+] x [A-]/[HA]. When you add half the volume of OH- needed to fully neutralise (i.e. the volume that makes the graph go vertical) then [A-] = [HA], so Ka = [H+] and pKa = pH = 5.
6. (Original post by Saint132)
I think i have figured out the Ka value and i have got it as B (5 ka). Am i correct?
Yes, at the half-equivalence point. Since the equivalence point is , the half-equivalence point is which corresponds to so . Thus .
7. Yes. Thank you for confirming!

Question 8, i think i have made a mistake and the answer should be in fact be D (Increases and it should be endothermic as Kc has increased due to the temperature being increased)?

That is from what i have read. Is this correct also?
8. For question 8:

So as the value of has increased when the temperature has increased, it means the has increased in relation to the and . Therefore, the forward reaction has to be endothermic so equilibrium would shift to the right to oppose the increase the temperature thus increasing the concentration of .
9. One final question. I have worked out question 10 but question 11, i have no recollection of the theory.

Could you help with question 11 if possible.
Attached Images

10. (Original post by NeverLucky)
For question 8:

So as the value of has increased when the temperature has increased, it means the has increased in relation to the and . Therefore, the forward reaction has to be endothermic so equilibrium would shift to the right to oppose the increase the temperature thus increasing the concentration of .
Yeah. Thanks again. I think i should have done a chemistry degree rather than a physics related one as i had more intrigue in the last day working these out than any of my physics topics.
11. (Original post by Saint132)
One final question. I have worked out question 10 but question 11, i have no recollection of the theory.

Could you help with question 11 if possible.
The answer to this can simply be deduced from the first half of the hydrogen peroxide half-equation:

In order for hydrogen peroxide to react, ions have to be present and the only species out of the 4 that provide ions is .
12. (Original post by NeverLucky)
The answer to this can simply be deduced from the first half of the hydrogen peroxide half-equation:

In order for hydrogen peroxide to react, ions have to be present and the only species out of the 4 that provide ions is .
Nice and simple way to put it..
13. Thank you for your help. Understood!

### Related university courses

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: July 28, 2016
Today on TSR

### Edexcel GCSE Maths Unofficial Markscheme

Find out how you've done here

### AQA Maths Paper 1 exam discussion

Poll

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE