A titration is carried out and 0.04dm^3 of sulphuric acid neutralises 0.08dm^3 sodium hydroxide of concentration 1mol/dm^3. Calculate the concentration of the sulphuric acid.

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Answers / Chemistry / GCSE / A titration is carried out and 0.04dm^3 of sulphuric acid neutralises 0.08dm^3 sodium hydroxide of concentration 1mol/dm^3. Calculate the concentration of the sulphuric acid.

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A titration is carried out and 0.04dm^3 of sulphuric acid neutralises 0.08dm^3 sodium hydroxide of concentration 1mol/dm^3. Calculate the concentration of the sulphuric acid.

The first thing to do with this titration calculation is write the balanced symbol equation for the reaction. Using the knowlege that when an acid and base react a salt and water is produced we can write:

H₂SO₄ _(aq)+ 2NaOH _(aq) ---> Na₂SO₄ _(aq)+ 2H₂O _(l)

From the balanced symbol equation we can see that 1 mole of H₂SO₄ neutralises 2 moles of NaOH

The next thing to do is calculate the number of moles of sodium hydroxide that were used up in the reaction.

We use the equation: number of moles = concentration x volume

N = 1 mol/dm³ x 0.08dm³ = 0.08 moles of NaOH

The number of moles of H₂SO₄ used up in the reaction is therefore half of this (0.04 moles) since we established from the symbol equation that 1 mole of H₂SO₄ neutralises 2 moles of NaOH.

Finally we can calculate the concentration of H₂SO₄ by using the equation :

concentration = number of moles/ volume

c = 0.04mol/ 0.04dm³ = 1mol/dm³

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A titration is carried out and 0.04dm^3 of sulphuric acid neutralises 0.08dm^3 sodium hydroxide of concentration 1mol/dm^3. Calculate the concentration of the sulphuric acid.

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