Edexcel A2 Chemistry Exams -6CH04 (14th June) and 6CH05 (22nd June) Discussion Thread

Can anyone help?
Find the pH of the mixture formed when 25 cm3
of 2 mol dm–3 sodium hydroxide solution is
added to 50 cm3
of 2 mol dm–3 ethanoic acid, for which Ka
= 1.7 × 10–5 mol dm–3.
pH=
A 2.2
B 2.5
C 4.5
D 4.8

Thanks

Hey, there's something I don't get, how the Sodium thiosulphate is used to find the concentration of Iodine of it reduces it back continuously, like when it is used up, there would still be Iodide ions that would oxidise to Iodine (hope you understand what I mean )

Just to check first, is the answer D?

Okay I'm kinda of going to answer this all together instead of in the separate parts because I think it'll make more sense that way.

Basically, the job of sodium thiosulfate is to reduce any iodine formed back into iodide ions:

2S₂O₃^2- + I₂ --> S₄O₆^2- + 2I^-

So, the solution initially will be colourless, since the persulfate ions oxidise iodide ions to iodine, but the sodium thiosulfate reduces the iodine immediately back to iodide ions, so a black colour doesn't form straight away (which is what would happen if sodium thiosulfate wasn't used --> answer to part ii). The final colour change occurs when all of the sodium thiosulfate has reacted, so any iodine formed will no longer be reduced back down to iodide ions. Therefore, the final colour change is colourless to black (answer to part i). In relation to part iii), the iodide concentration initially remains constant due to persulfate ions oxidising iodide ions to iodine but then sodium thiosulfate reducing iodine immediately back down to iodide ions again.

Hope that makes da sense

Yes it is.

Thank you very much! It makes perfect sense now, I did not know I had to know Persulphate reaction.

Okay well, this is a sneaky one...

First, we know Ka = [H+][CH3COO-] / [CH3COOH]

Firstly, calculating the moles of ethanoic acid and sodium hydroxide shows you which one is in excess. In this case, you get 0.1 moles and 0.05 moles respectively. The ethanoic acid is therefore in excess.

You then calculate the moles of ethanoic acid left over after the reaction has occurred; 0.1 - 0.05 = 0.05 moles of ethanoic acid left over.

The number of moles of ethanoate ions formed is worked out using the number of moles of the thing that isn't in excess - in this case the sodium hydroxide. The number of moles of ethanoate ions is therefore 0.05 moles also.

Now here's the twist: in this question, the number of moles of ethanoic acid left over happens to be the same as the number of moles of ethanoate ions formed. Their concentrations will also be the same since they are in the same total volume.

Conc. of ethanoic acid will = 0.05 * 1000 / 75 = 2/3.
Conc. of ethanoate ions will = 0.05 * 1000 / 75 = 2/3.

So, subbing in the values into Ka:

Ka = [H+] * (2/3) / (2/3).

As you can see, the 2/3 cancels out on the top and bottom, leaving:

Ka = [H+]. So [H+] = 1.7x10-5.

From here, we can work out pH:

pH = -log(1.7x10-5) = 4.769.... = 4.8.

Hence, the answer is D. This question is an example of a half-equivalence point question, where the pH = the pKa value of the acid (pKa = -log(Ka)).

Thank you so much!!

@Sandy_Vega30 This video should help explain half-equivalence point if you're unsure I'm watching it right now ^_^ https://www.youtube.com/watch?v=ezbFkhMFoCo

I'm not sure what you mean, sorry !

Okay well, this is a sneaky one...

First, we know Ka = [H+][CH3COO-] / [CH3COOH]

Firstly, calculating the moles of ethanoic acid and sodium hydroxide shows you which one is in excess. In this case, you get 0.1 moles and 0.05 moles respectively. The ethanoic acid is therefore in excess.

You then calculate the moles of ethanoic acid left over after the reaction has occurred; 0.1 - 0.05 = 0.05 moles of ethanoic acid left over.

The number of moles of ethanoate ions formed is worked out using the number of moles of the thing that isn't in excess - in this case the sodium hydroxide. The number of moles of ethanoate ions is therefore 0.05 moles also.

Now here's the twist: in this question, the number of moles of ethanoic acid left over happens to be the same as the number of moles of ethanoate ions formed. Their concentrations will also be the same since they are in the same total volume.

Conc. of ethanoic acid will = 0.05 * 1000 / 75 = 2/3.
Conc. of ethanoate ions will = 0.05 * 1000 / 75 = 2/3.

So, subbing in the values into Ka:

Ka = [H+] * (2/3) / (2/3).

As you can see, the 2/3 cancels out on the top and bottom, leaving:

Ka = [H+]. So [H+] = 1.7x10-5.

From here, we can work out pH:

pH = -log(1.7x10-5) = 4.769.... = 4.8.

Hence, the answer is D. This question is an example of a half-equivalence point question, where the pH = the pKa value of the acid (pKa = -log(Ka)).

may I ask where 75 came from?

Like, if the sodium thiosulphate job is to reduce the iodide ions back to iodine, then when it finishes, how can we know the concentration of iodide ions that was in the reaction ?

I'm not sure, sorry xD

Help needed!

The pH of a sample of water is 7.
10cm3 of 1.0mol dm-3 hydrochloric acid is added to 990cm3 of this water.
What is the pH of the solution formed?
A 0
B 1
C 2
D 3 Correct answer is C but can anyone please provide an explanation?

Doesn't the sodium thiosulphate react with iodine to produce iodide ions and not the other way round?

Posted from TSR Mobile