Chemical quantities (AS)
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Hi,
I don't understand how to solve this question possibly because I don't know the stoichiometric equation for it, and cannot find in my textbooks. Please help me with it.
25.0 cm3 of ethanedioic acid required 22.5 cm3 of 0.100 mol dm–3 potassium hydroxide solution for complete neutralisation.
The concentration of ethanedioic acid is
A 0.0225 mol dm–3
B 0.0450 mol dm–3
C 0.0560 mol dm–3
D 0.0900 mol dm–3
Thanks a lot.
Corrie writes ...
Ethanedioic acid, (COOH)2, as its name suggests, contains two carboxylic acid groups. It will therefore react with KOH in a 1:2 ratio, as follows:
(COOH)2 + 2KOH -> (COO-K+)2 + 2H2O
Quick answer: as two moles of alkali are needed to neutralise one mole of the acid, and the volumes are more or less the same, the acid must be approximately half the concentration of the alkali. Do you agree? This means B or C is the correct answer. As slightly less alkali than 25 cm3 is used, the acid must be slightly less concentrated than (0.1)/2, so the answer B must be the correct one.
Unless you are happy with the above 'quick' answer, it is probably safer to do the calculation:
Moles of KOH used = (22.5/1000) x 0.100 = 2.25 x 10-3
Therefore moles of ethanedioic acid in 25.0 cm3 taken = (2.25 x 10-3)/2 = 1.125 x 10-4
Conc. of acid = moles/volume (dm3) = 1.125 x 10-3/0.025 = 0.045 mol dm-3 - - Note the bits in bold!
Therefore answer is B, as above.
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updated: 17 June 2007
