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A1 A-Level Chemistry Ratio/concentration question

3.88 g of a monoprotic acid, HA, was dissolved in water and made up to 250 cm3.
25.0 cm3 of this solution was titrated with 0.095 mol dm–3 KOH solution, requiring
49.5 cm3.
Calculate the relative molecular mass of the acid.
Original post by ARMY101
3.88 g of a monoprotic acid, HA, was dissolved in water and made up to 250 cm3.
25.0 cm3 of this solution was titrated with 0.095 mol dm–3 KOH solution, requiring
49.5 cm3.
Calculate the relative molecular mass of the acid.

Monoprotic (i.e. acids of the formula HA - diprotic acids have the formula H2A) acids react with KOH in a 1;1 ratio as shown by the equation below:

HA + KOH--> KA + H2O

Find moles of KOH required for neutralisation (moles KOH= conc x vol in dm3) (= 0.095 x 0.0495)

This is the same as the no of moles of HA in 25 cm3

Then work out moles HA in 250 cm3 (x previous answer by 10)

Then work out the Mr of the acid, rearrange moles=Mass/Mr to get:

Mr= mass/moles.... you know the mass of HA in 250 cm3 (it is given as 3.88g) and you have just worked out the moles.
Reply 2
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ARMY101
OP
10
Thank you @Davies Chemistry!
In case another student wants to compare my working I first had to divide 49.5cm3 by 1000 to convert it into dm3. Find moles of KOH: moles KOH= c x v in dm3= 0.095 x 0.0495=4.7025x10^-3mol. This is the same as the no of moles of HA in 25 cm3 because of the 1:1 ratio.
Then work out moles HA in 250 cm3 (x previous answer by 10) =0.047025mol
Then work out the Mr of the acid, rearrange moles=Mass/Mr to get: Mr= mass/moles.... you know the mass of HA in 250 cm3 (it is given as 3.88g): Mr=3.88/0.047025= 82.5mol (3sf)

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