OCR Chemistry A F325 Equilibria, Energetics and Elements Wed 13 June 2012

yes i used the root Ka x OH-
formula?
i get confused like which one we use sometimes Kw Ka Kc?

ah no wonder you used Ka.

KA is always there for weak acids and buffers

KW is always there for Bases

and KC is well you'll know when you need to use that the question will say

Do you understand how I worked it out? Thats also another trick they throw in sometimes

oh my god how many tricks do these sneaky OCR PEOPLE PUT
yes i think so , can you give me another just to see if i get it cause this is my weakest part i think

x

x

Guys sorry to but in..

Is this a mistake on p215 (the last question=Q3 )..

they say its exo but should be endo

yeah sure

Work out the pH of H2SO4 at a concentration of 0.200mol dm-3
Work out the pH of H3PO4 at a concentration of 0.400mol dm-3
Work out the H+ Conc of H2SO4 at pH 1.45
Work out the H+ conc of H2SO4 at pH 2.31
Work out the H+ conc of H3PO4 at pH 3.44

I'll just do the acid ones first I'll ask you some tricky base ones after you've done the above Calculations were my weakest point too but had a lot of practice for the Jan paper and now there just simple to do

1) 13.3
2)13.6
Are these two correct first of alll?

Your pH's are too high for them to be considered Acids.

Mind showing me your working on what your doing?

The answer says endothermic in the back of the book? Which is correct

oh god! i am using the Kw formula?

ah that explains it.

H2SO4 and H3PO4 are both strong acids. Kw is used just to calculate pH of Bases.

Only two formulas you need are pH = -log(H+) and H+ = 10 to the power of pH (put it as a minus though so if its pH 3 its 10 to the power of -3)

Just remember if your working out pH for an acid which has two H+ like H2SO4 then you times the Concentration by 2.
If your working out the Concentration for an acid which has 2 H+ like H2SO4 then you divide your answer by 2 after you've done 10 to the power of -pH

Understand a bit better? you probably know how to do them all you just select the wrong one to use

Really!

Let's break it down

1) Increased temp.. the reaction moves towards endothermic direction
2) It is stated in the question that the solution turns blue
3)we know The complex ion on the LHS is blue..therefore it only turns blue if t moves to Left ..hence forward reaction is exothermic!

pls correct me if am wrong

yes oh dear so sorry mate i am embarrased getting them wrong oh dear !
just is it possible mate if you can write me what each of them is used to calculate that would be really helpful sorrryyyyyyy

Sure can Right lemme just paste the questions;
Work out the pH of H2SO4 at a concentration of 0.200mol dm-3
Work out the pH of H3PO4 at a concentration of 0.400mol dm-3
Work out the H+ Conc of H2SO4 at pH 1.45
Work out the H+ conc of H2SO4 at pH 2.31
Work out the H+ conc of H3PO4 at pH 3.44

Right no 1.

So our concentration is 0.200 of H2SO4 to work out pH we need to multiply that concentration twice as there's two H+. So we do pH = -log(H+) which is -log=(0.400) which equals the answer of pH = 0.40

no 2. This is the same but we have 3 H+ in H3PO4. So we multiply the concentration by 3. So we have 1.20 as the conc for all 3 H+. So we -log(1.20) which equals a pH of -0.08. (sorry about that crap choice of numbers by me there)

So you see when working out the pH we times the concentration by the number of H's the Acid has. So as you can see H2SO4 meant we multiplied the conc by 2 and H3PO4 meant we multiplied the conc by 3.

no 3

With concentration its a little different but pretty much the same as how we work out a normal concentration. So to work out H+ we do 10 to the power of -pH.
So our pH is 1.45 we do 10 to the power of -1.45 which equals 0.0354. This is the concentration for 2 H's in H2SO4 so to get it for one we need to divide by 2. So the answer is 0.0177.

No 4
pH is 2.31 so we calculate H+ conc. 10 to the power of -2.31 = 4.90x10-3
We need to divide by 2 so we can get one H in H2SO4. So the answer is 2.45x10-3

no 5
Ph is 3.44 So first step as always work out the conc of H+ using 10 to the power of -pH. So 10 to the power of -3.44 = 3.63x10-4 Now this time instead of dividing by 2 we divide by 3 as the acid is H3PO4. It has 3 H's. So the Answer is 1.21x10-4.


That help? Just say if not

oh i get it now
Can you mate like tell me Ka,Kw,Kc,buffers what they each calculate so i can now and when to use this 1.00x10power-14 ?

Ka can be used for several things;
most commonly Weak acids. Formula for kA is:

Ka = [H+] [A-] Over [HA]

HA = the whole acid
H+ = well just H+
A- = the negative ion leftover
H+ and A- are usually classed as the same. So the formula of kA can be simplified to:

Ka = [H+] squared over HA

To work out weak acids with Ka we have to rearrange the second equation above.
So Ka = [H+] squared over HA
So to get H+ we have to do Ka x HA (conc of the whole acid)
So we Ka X HA = answer. This answer is H+ squared.
We square root the answer to get H+.
We then -log that answer to get the pH.

You can also use Ka to work out pKa by doing -log(Ka)
And work out Ka from pKa by doing 10 to the power of minus pKa.

Kw = [H+] x [OH-]
at room temperature that value is always 1.0x10-14
Kw is only used to work out Bases.
e.g. Work out pH of KOH at concentration of 0.200 mol dm-3
We have to rearrange Kw = [H+] x [OH-] so we can calculate H+

So we do H+ = kW over [OH-] The value for OH- is the one given in the question which is 0.200. And the kW value is always 1.0x10-14.

So we do 1.0x10-14 divided by 0.200 that equals the H+.
Now we simply use -log(Answer) to get the pH

(answer i got was like 14.70)

With Kc you don't need to worry about that with the acids and bases. its one value and you can't work out pH or anything using it. Kc is simply just products divided by reactants.

Need a bit more explaining let me know I know I haven't talk about the other base question I gave earlier I'll get to that once you've understood Kw!

i will revise this but thank you so much +++++reps

Your welcome, have a good crack at learning how to do them inside out after that its all pretty much easy

thanks and can we continue tomorrow asking questions?