Volumetric Analysis

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Volumetric Analysis
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Volumetric Analysis

• Volumetric analysis is the method of finding out
  the concentration of a solution.
• For example
• Adding a solution of acid to a solution of base,
  in a measured way, until you have added enough to
  neutralise the base (which an indicator will tell
  you)
• Using an indicator like this to find exact
  amounts of solutions that react is called is
  called titration.

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Titration

• In this method, the concentration of one of the
  two solutions must be known this we call the
  standard solution (and it never changes with
  time).
• Knowing the concentration of one of the
  solutions, and the exact volumes of both, will
  allow you to work out the unknown concentration.
• E.g. looking back at the earlier example if we
  know the volume of acid needed to neutralise the
  base and the volume of the base the acid is
  being added to and the concentration of the base
  then we have all the info. we need to work out
  the acid concentration.
• The base solution is the standard solution
  because we know its volume and its concentration.

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Titration Method

1. Usually 25.0 cm3 of the standard solution is
   known. So for example, a pipette would be used to
   deliver this amount of the base (NaOH) into a
   clean conical flask.
2. Add a few drops of indicator (e.g.
   phenolphthalein)
3. Wash burette with a little bit of acid. Then fill
   the burette with acid, until meniscus is at zero.
4. Run the acid into the conical flask, until
   indicator changes colour to show neutralisation
   has been reached call this the end-point.

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Titration Method

1. First run is normally a rough one to see where
   end point will roughly be on burette. After this
   the titration is repeated more accurately, with
   acid being run into base quickly up to the near
   the end-point, and then drop by drop to find the
   exact volume of acid that causes neutralisation.
2. The titre is the volume of acid needed to
   neutralise 25cm3 of the base.
3. Use the average titre obtained from the repeat
   runs in calculation. First run ignored whhen
   working out average.

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Key Notes
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Calculation Example

• By titration you find that 15.0 cm3 of
  hydrochloric acid neutralise 25.0 cm3 of a 0.100
  mol dm-3 solution of sodium hydroxide. What is
  the concentration of hydrochloric acid?
• First you always need equation for reaction
  first!!!
• Hydrochloric acid sodium hydroxide ? sodium
  chloride water
• HCl(aq) NaOH(aq) ? NaCl(aq) H2O(l)
• This tell you that 1 mole of HCl neutralises 1
  mole of NaOH.
• Next work out the moles for either the base or
  the acid.
• You know the concentration for the base, so you
  start there

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• Amount (mol) volume (dm3) x concentration (mol
  dm-3)
• Amount (mol) NaOH 25.0 cm3 x 0.100 mol
  dm-3
• 2.50 x 10-3 mol.
• Now work out concentration of acid
• From equation amount (mol) of HCl amound (mol)
  of NaOH 2.50 x 10-3 mol.
• Amount (mol) volume (dm3) x concentration (mol
  dm-3)
• Re-arrange equation to give
• Concentration HCl amount (mol) / volume (dm3)
• Concentration HCl 2.50 x 10-3 mol / 15.0 cm3
• Concentration HCl 0.167 mol dm-3.

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Another Example
Volume of 250 cm3 of acid was 3.02g Answer next
slide
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REDOX titrations

• A titration, where the reaction is not a
  neutralisation, but a redox reaction.
• Major example is reaction between iodine and
  thiosulphate ions.
• Reaction is carried out the same way as a
  acid-base titration.

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REDOX titrations
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Finding the purity of potassium iodate(V)
The iodine can be generated by adding iodate(V)
ions to excess potassium iodide
solution. Iodate(V), IO3-, reacts as a strong
oxidising agent.
This mixture above can be titrated with standard
sodium thiosulphate solution. STARCH is the
indicator! (blue or blue-black to colourless)
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So 0.001 mol of I2 liberated by 25cm3 of impure
potassium iodate(V) solution
0.001 mol
Mol 0.000333 mol
V 25 cm3
V 20 cm3 M 0.1 mol dm-3
Start here
Mol 0.002 mol
0.001 mol
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0.001 mol
Mol 0.000333 mol
V 25 cm3
V 20 cm3 M 0.1 mol dm-3
Mol 0.002 mol
0.001 mol
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Finding the Percentage of copper in brass

• Redox titration can be used in the analysis of
  brass (an alloy of copper and zinc)

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Questions
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Answers
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Try this

• A standard solution is prepared by dissolving
  1.185g of potassium dichromate(VI) and making up
  to 250 cm3 of solution. This solution is used to
  find the concentration of a sodium thiosulphate
  solution. A 25 cm3 portion of the oxidant was
  acidified and added to an excess of potassium
  iodide to liberate iodine
• Cr2O72-(aq) 6I-(aq) 14H(aq) ? 3I2(aq)
  2Cr3(aq) 7H2O(l)
• When the solution was titrated against sodium
  thiosulphate solution, 17.5cm3 of thio was
  required. Find the concentration of the
  thiosulphate solution.

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Answer

• Need following two equations
• Cr2O72-(aq) 6I-(aq) 14H(aq) ? 3I2(aq)
  2Cr3(aq) 7H2O(l)
• Moles of K2Cr2O7 in standard solution mass/RMM
• 1.185g/294 0.00403 mol
• 0.00403 X 4 0.0161 mol dm-3
• Moles of K2Cr2O7 in 25cm3 (25/1000) x 0.0161
  4.025x10-4
• Moles of I2 liberated 4.025x10-4 x 3
  1.208x10-3
• Moles of thio in 17.5cm3 1.208x10-3 x 2
  2.415x10-3
• Conc. of thiosulphate 2.415x10-3 / (17.5/1000)
  0.138 mol dm-3