Chemistry-Chemical analysis (anions)

Anions test

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  • Created by: chloefaun
  • Created on: 31-03-14 12:51

Titration

If you know the concentration of one of the reactants present in the titration, you can work out the concentration of the other reactant.

Step 1: Convert volumes to dm3

25 cm3 of HCl = 25 ÷ 1000 = 0.025 dm3

20 cm3 of NaOH = 20 ÷ 1000 = 0.020 dm3

Step 2: Determine the number of moles of sodium hydroxide

moles of NaOH = concentration × volume

moles of NaOH = 0.5 × 0.020 = 0.010 mol

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Titration con.

Step 3: Work out the number of moles of acid using the balanced equation HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) In this reaction, one mole of HCl reacts with one mole of NaOH. This is a 1:1 ratio. Therefore, in our titration, 0.010 mol of NaOH must neutralise 0.010 mol of HCl. Step 4: Calculate the concentration of the acid concentration of HCl = number of moles ÷ volume concentration of HCl = 0.010 ÷ 0.025 = 0.4 mol/dm3

Answer

The concentration of the HCl is 0.4 mol/dm3

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Other anion tests

Halide-To test for halide ions add nitric acid and sliver nitrate. If a halide is present it will form a coloured precipitate. KI-Yellow, BaCl-White, KBr-Cream.

Sulphate-Add a diltue HCl acid and a barium salt, it will form a white precipitate if sulphate ions are present.

Carbonate-Add dilute acid to the solution and collect any gas given off with a pipet. Place the pipet in the limewater, if it turns cloudy a carbonate is present.

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