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# pH question... watch

1. Hi all,

I have this unusual pH question, but I don't know if I'm on the right lines:

Solution A contains 'n' moles of a weak acid, HX. The addition of some NaOH neutralises one third of the HX present in solution B.

i) In terms of the amount, n, how many moles of HX are present in Solution B?

ii) Determine the ratio [HX]/[X-] in Solution B

i) 1/3 n

ii) presumably 1/3n: 2/3n so 1:2?

Slightly confused. The part of my brain that really enjoys chemistry is throwing a tantrum because there's basic maths involved.

Any help is much appreciated,

Luke
2. (Original post by labt)
I have this unusual pH question, but I don't know if I'm on the right lines
You're on the right lines, but you've been tricked by the wording of the question. If a third of the HX has been neutralized, 2/3 of it is left in solution B.
3. (Original post by BJack)
You're on the right lines, but you've been tricked by the wording of the question. If a third of the HX has been neutralized, 2/3 of it is left in solution B.
I'm not entirely sure why, but I find your avatar incredibly entertaining.
4. (Original post by labt)
Hi all,

I have this unusual pH question, but I don't know if I'm on the right lines:

Solution A contains 'n' moles of a weak acid, HX. The addition of some NaOH neutralises one third of the HX present in solution B.

i) In terms of the amount, n, how many moles of HX are present in Solution B?

ii) Determine the ratio [HX]/[X-] in Solution B

i) 1/3 n

ii) presumably 1/3n: 2/3n so 1:2?

Slightly confused. The part of my brain that really enjoys chemistry is throwing a tantrum because there's basic maths involved.

Any help is much appreciated,

Luke

You are told that only (1/3) of n have been neutralised therefore there are (2/3)n remaining.

Only (1/3)n have reacted to form salt therefore there is exactly the same number of moles of salt = (1/3)n

So the ratio of acid/salt = (2/3)n/(1/3)n = 2
5. Indeed - answer is upside down, silly me!
Thanks for the help!

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