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    4. 2.5 g of a sample of impure ethanedioic acid, H2C2O4.2H2O, was dissolved in water and the solution made up to 250 cm3. This solution was placed in a burette and 21.3 cm3 were required to neutralise 25 cm3 of 0.1 moldm-3 NaOH. Given that ethanedioic acid reacts with NaOH in a 1:2 ratio, calculate the percentage purity of the sample.
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    (Original post by m.leeda.n)
    calculate the percentage purity of the sample.
    Which bit are you stuck on?
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    Calculate the moles of NaOH. Use the given ratio to work out the moles of ethanedioic acid. Calculate the corresponding mass based on the Mr. Look at that number as a proportion of 2.5 g.
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    (Original post by m.leeda.n)
    4. 2.5 g of a sample of impure ethanedioic acid, H2C2O4.2H2O, was dissolved in water and the solution made up to 250 cm3. This solution was placed in a burette and 21.3 cm3 were required to neutralise 25 cm3 of 0.1 moldm-3 NaOH. Given that ethanedioic acid reacts with NaOH in a 1:2 ratio, calculate the percentage purity of the sample.
    you see, the original acid is impure.

    so when you write an equation involving the "pure acid", when we worked backward from what actually reacted, it would give you only the mass of the pure acid.

    hence this mass of pure acid in an original impure sample should be lesser - the first hint of your answer being sensible or not sensible.

    the other mistake often overlooked is that the actual titration is with 25 cm3 of the acid, whereas the original mass of pure acid is way more than just in the 25 cm3.

    working backward,
    1) mole of NaOH reacted = (1)

    2) mole of acid in 25 cm3 / mole of NaOH reacted = 1 / 2

    so mole of acid in 25 cm3 = 1/2 * (1) = (2)

    3) in 25 cm3, you have (2) mole of acid
    so in 250 cm3 of the original acid, you have ??? mole of acid = (3)

    4) the total acid has

    mole = mass/Mr
    mass of original acid (pure) = (3) * Mr (acid)

    5) percentage purity = ???
 
 
 
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