The Student Room Group
If it is completely dissociated, i.e. it is a one-way reaction and not equilibrium, you just need to look at the molar ratio between the Sr(OH)2 and the OH ions (which I'm guessing is 1:2)

I can't see why the atomic masses are relevant, though
Reply 2
That is what I thought. I divided the concentration of the solution given (the 5x10^-3) and then times by two as there are two OH ions per Sr ion.

The RFM's may be unnecessary as this is only a part of a whole question.
RMorley
That is what I thought. I divided the concentration of the solution given (the 5x10^-3) and then times by two as there are two OH ions per Sr ion.

The RFM's may be unnecessary as this is only a part of a whole question.


So are you happy with this answer?

Lol I can't tell :P
I think the RAM's are there to confuse you.

For every mole of Sr there are 2 moles of OH there for if the solution is completely dissociated the OH conc will be 5x10^-3 * 2
Reply 5
Well yes that is what came to me when I saw the question, but it was really just a guess as there were no examples in the text books etc.

Thanks for your help
Reply 6
The Referee
I think the RAM's are there to confuse you.

For every mole of Sr there are 2 moles of OH there for if the solution is completely dissociated the OH conc will be 5x10^-3 * 2


Do you not have to divide the conc by 3 first and then times by 2?
RMorley
Do you not have to divide the conc by 3 first and then times by 2?


No, why would you do that?

Sr(OH)2 --> 2OH- + Sr2+

Right?

It's just 5x10^-3 * 2
Reply 8
oh because I thought if it is the ratio 1:2 you would divide by 3 to find the concentration of one part and then times by 2.

Thankyou then, that has been a great help.
I know what you mean, but that doesn't apply here.

You only do that with totals. For example, if 12 apples are split 1:2, I would divide 12 by the total of the ratios (2 + 1) and then multiply by the ratio for that person.

However, in this case, if you have 1 mole of something you just multiply it by 2 to get the amount on the right side.
Reply 10
Ah well thanks very much for your help, I think I understand concentrations quite a bit better now!

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