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Impurity??!!!!!

Heya!
i need help on a question it's based on titrations.

"Sodium hydroxide can absorb water and carbon dioxide from the atmosphere. 2.16g of impure sodium hydroxide was dissolved in 250 cm3 of distilled water, and 25 cm3 of this solution was titrated against 0.20 mol dm-3 hydrochloric acid. If the average titre was 22.20 cm3, calculate the percentage purity of the sodium hydroxide."

I looked at the answer (below) and it doesn't make any sense to me . They somehow worked out the moles of HCl first and then did mole ratio thing to work out the mole of NaOH. And from there worked out the pure NaOH in grams?

The ANSWER :
Amount of HCl = 4.44 x 10-3 mol,
ratio is 1:1, so amount NaOH in 25cm3 = 4.44 x 10-3 mol
hence amount NaOH in 250cm3 = 4.44 x 10-2 mol
mass pure NaOH =1.776 g, percentage purity =82.2%

:s-smilie: :s-smilie: :s-smilie:

Help me pls

I sure hope this isn't on the real exam!
Use number of moles=concentration x volume to work out the amount of HCl used. So that would be 0.2 x 0.0222

1 mol of hydrochloric acid is needed to neutralise 1 mol of sodium hydroxide so 4.44 x 10-3 is needed to neutralise 4.44 x 10-3.

4.44 x 10-3 mol was the amount in a 25cm3 sample so multiply by 10 to get the amount in your 250cm3 solution.

Your mass of NaOH is your number of moles x formula mass. Then just divide your masss of pure NaOH by your mass at the start and multiply your answer by 100 to get percentage purity.
:biggrin: Thank you!!!!

(so sorry for the late reply)

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