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Topic: titration problem with no given volume  (Read 4499 times)

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Offline peptea

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titration problem with no given volume
« on: March 13, 2009, 02:41:27 AM »
so most titration problems have a given volume, whether it's the strong acid/base or the weak acid/base. mine for some reason doesnt give any volume.

example

a 0.12 M sample of sulferous acid was titrated with 0.10 M of KOH. at what added volume of base and pH does each equivalence point occur?

i just dont understand how to get around this problem! please help...

Offline AWK

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Re: titration problem with no given volume
« Reply #1 on: March 13, 2009, 03:06:06 AM »
In any titration you should solve an equation with one variable. Your problem is solvable if it is 0.12 mole of H2SO3 (for compounds use always formulas, not names, because non-standard names may be ambigous) not 0.12 molar solution (then volume is needed). pH you can calculate from basic protolysis of SO32- anion  (salt hydrolysis of Na2SO3).
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Offline Borek

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Re: titration problem with no given volume
« Reply #2 on: March 13, 2009, 03:58:51 AM »
You may express your result in mL titrant added per mL of original sample (something like "5 mL of 0.2 HCl needed to neutralise 10 mL 0.1M NaOH") or you may assume - and state it clearly - you have started with some known volume of titrated substance ("Assuming we started with 25 mL of 0.1M HCl, 20 mL of 0.125 M NaOH is required to reach the equivalence point"). Both answers will show you know how to deal with the calculations.
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Offline peptea

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Re: titration problem with no given volume
« Reply #3 on: March 13, 2009, 04:10:02 AM »
it is given in molar, not moles. and that's almost exactly how the problem was phrased, though i changed the numbers a little

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