OCR A 2016 Chemistry A* A-Level Resources

I came across a question in the June 2012 F321 paper, and I'm not entirely sure how to work it out.

Question:
A student is given a sample of an unknown Group 2 chloride.
- The student dissolves 2.86g of the chloride in water.
- The student adds excess aqueous silver nitrate.
- 8.604g of solid silver chloride, AgCl, forms.

In part i) I worked out that there are 0.06 moles of AgCl

Part ii) asks: to deduce the amount in moles of the group 2 chloride that the student dissolves. Hence deduce the relative atomic mass and identity of the group 2 metal. Give RAM to 1 d.p.


Help and explanation please?

What do you guys think will come up this year in the exam? for F324/F325

Guys with the copper iodine equation, it says two of the iodide ions do not change oxidation state?

I was wondering how this fits into the half equations?

So I thought it would be:

Cu2+ + e- --> Cu+

2I- --> I2 + 2e-

But I know 2 iodide ions are missing from the overall equation so where would we put them? Or is it a matter of just balancing each side?


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Guys with the copper iodine equation, it says two of the iodide ions do not change oxidation state?

I was wondering how this fits into the half equations?

So I thought it would be:

Cu2+ + e- --> Cu+

2I- --> I2 + 2e-

But I know 2 iodide ions are missing from the overall equation so where would we put them? Or is it a matter of just balancing each side?


Posted from TSR Mobile

I didn't really get this either when I looked at it but I think it might be because the iodide ions are in excess?? So two of them change oxidation state to become I2 and then two iodide ions are needed to react with the 2 Cu2+ to form 2 CuI2?? Sorry that was a really crap explanation but I think if you just remember that two extra Iodide ions are required to react with copper (II) ions to form copper iodide you'll be fine

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You need to balance both sides and then I think just add in the remaining iodide ions because they don't change oxidation state they are not involved in an ionic equation.
If someone could explain it better would be great!


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Yeah that's what I thought aha thanks, still not quite sure of the half equations we'd need but maybe they won't ask us for those...or they'll accept the ones I put? because technically that's what's happening right, if we ignore the Cu lol


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Thinking about it as I2 is a solid they will probably say that a precipitate is formed which will hunt at the l2.


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As part of an investigation, a student needed to prepare a buffer solution with a pH value of8.71. From the Ka value of phenol, the student thought that a mixture of phenol and sodiumphenoxide could be used to prepare this buffer solution.The student decided to use a 0.200 mol dm–3 solution of phenol, mixed with an equalvolume of sodium phenoxide.Use your knowledge of buffer solutions to determine the concentration of sodium phenoxidesolution that the student would need to mix with the 0.200 mol dm–3 phenol solution ka=1.3*10^-10

i keep getting 0.013 but answer is 0.13
1.3*10^-10 *0.2/10^-8.71=0.013

But what would that hint to exactly?


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Suggested question.
Copper and iodine are mixed forming a solution and a solid.
Work out the full equation.

That sort of hint ???


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is that the full question?

Good work helping each other guys! In most cases people seem to answer each other before I have a chance to even see the question! I'll get round to it eventually just post it again if I miss it.

I'm not going to work out what the question is, please post the full question if you want me to answer

Here's a video that explains it well I think
https://www.youtube.com/watch?v=OAkFX-7n8iU

Any further questions just ask.