Titrations
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A-level chemistry

A-level chemistry

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Titrations Titrations Presentation Transcript

  • Titrations Aim: To understand how to use volumetric analysis to determine the concentration of acids or bases
  • Moles in solution No. of moles in solution = M x V ______ 1000
  • Moles in solution
    • How many moles of NaCl are there in each of these solutions:
    • 0.500mol dm -3 .
    • 0.750mol dm -3 ; 500cm 3
    • 0.650mol dm -3 ; 750cm 3
    • 1.5mol dm 3 ; 1.5dm -3
  • Calculations – Solution moles
    • Calculate the number of moles of calcium chloride needed to make a 2 Mol dm -3 solution, volume 250cm 3.
    • Calculate the mass of Calcium Chloride needed to make up this solution.
  • Number 2
    • Calculate the number of moles of sulphuric acid needed to make a 5 mol dm -3 solution; volume 100cm 3 .
    • What mass of sulphuric acid would be needed to make up the solution?
  • Solution calculations
    • If 10.3g of sodium bromide was dissolved in 250cm 3 of solution what would its concentration in mol dm -3 be?
  • Solution calculation 2
    • If 8g of copper sulphate was dissolved to make up a solution of volume 500cm 3 , what would be its concentration in mol dm -3 ?
  • Using titration to find out the exact concentration of NaOH.
    • Titration involves neutralisation – using an acid of known concentration to find out the concentration of NaOH
    • Equation for the reaction:
    • HCl + NaOH NaCl + H 2 O
  • Titration to find out the concentration of a solution of NaOH.
    • If we know the volume and concentration of HCl needed to neutralise 25cm 3 NaOH we can use this and the mole ratio from balanced equation to work out:
    • The number of moles of HCl that have reacted with the NaOH.
    • The balanced equation will than tell us the number of moles of NaOH that reacted with the HCl
    • Because we know the volume of NaOH and the number of moles we can then calculate the concentration.
  • Example
    • In a titration 21.45cm 3 of 0.500mol dm -3 HCl was needed to neutralise 25cm 3 NaOH. What is the concentration of the NaOH?
    • Step 1 – Calculate the moles of HCl needed to neutralise the NaOH.
    • Number of moles HCl= 21.45 x 0.5/1000 = 0.01072moles
    • From the balanced equation work out the mole ratio of NaOH to HCl
    • 1 mole HCl reacts with 1 mole NaOH
    • 0.01072 moles react with 0.01072 moles NaOH.
  • Part 2
    • Therefore, in 25cm 3 of solution there are 0.01072moles of NaOH……. What is its concentration?
    • No of moles = MV/1000
    • 0.01072 = M x 25/1000
    • M = 0.01072 x 1000/25
    • = 0.4288mol dm -3
  • An Experiment to find out the concentration of a solution of Sodium Hydroxide
    • Half fill the 25ml pipette with your solution of sodium hydroxide with a pipette filler.
    • Remove the filler and place your thumb or first finger over the round end of the pipette to stop the liquid escaping.
    • Invert the pipette a couple of times to rinse the pipette.
    • Drain the liquid into an empty beaker – DO NOT BLOW OUT THE LAST DROP OF LIQUID FROM THE PIPETTE.
    • Measure out accurately 25ml of the sodium hydroxide solution by using the pipette and filler and in a empty into a small conical flask. Add a few drops of the indicator.
  • Experiment continued
    • Fill the burette with the HCl using a funnel with enough acid to rinse the burette out – about 10ml. Place your thumb or first finger over the round end of the burette and invert the burette 2/3 times.
    • Drain the burette into a beaker using the tap so that the acid also drains through the jet.
    • Fill the burette ensuring that the jet is filled, and note the volume accurately – read to 2 decimal places.
  • Experiment (3)
    • Add the solution to the flask slowly and continually swirl the flask.
    • As the indicator begins to change colour add the acid drop by drop to the flask.
    • When the indicator has turned colourless and the end point has been reached record the volume to decimal places.
    • Repeat the titration until 3 CONSISTENT volumes of solution have been obtained – 3 readings within + or – 0.1ml.
    • These 3 consistent results should then be averaged to give the mean titre for the experiment.
  • Results table Volume delivered in cm 3 Final volume in cm 3 Start volume in cm 3 4 3 2 1 Titre – Volume of solution added from burette
  • Calculation
    • Calculate the average volume of HCl that reacted with 25cm 3 of NaOH Show your working.
    • Calculate the number of moles of HCl that reacted with the NaOH.
    • Using the balanced equation work out the number of moles of NaOH that reacts with this number of moles of HCl.
    • Using this number of moles of NaOH and the volume of NaOH used in the experiment calculate its concentration.
  • Confirming the concentration of your standard solution of Sodium carbonate
    • Carry out a titration to work out the concentration of your solution of sodium carbonate.
    • The acid you are provided with is 0.1M HCl.
    • The indicator you will use for this titration is screened methyl orange. The end point is reached when it grey/purple from green
  • Experiment write up
    • Diagram and method and explanation for each stage of the practical method.
    • A balanced symbol equation for the reaction
    • A results table identical to the HCl NaOH practical.
    • The calculation to find out:
    • The moles of HCl that reacted.
    • The moles of Na 2 CO 3 in the solution.
    • The concentration of the Na 2 CO 3.
    • A comparison of the calculated value v your stated value on the flask. Sources of experimental error?
  • Question
    • 25.0 cm 3 of a 0.0504 mol dm -3 solution of sulphuric acid was titrated with a solution of sodium hydroxide. The mean titre was 27.8cm 3.
    • Write the balanced equation for this reaction.
    • Calculate the number of moles of sulphuric acid that reacted with the NaOH.
    • Use the balanced equation to calculate the number of moles of NaOH that react with the acid.
    • Claculate the concentration of the NaOH.
  • Question.
    • Calculate the volume of 0.100 mol dm -3 hydrochloric acid solution required to neutralise 25 cm 3 of 0.0567 mol dm -3 sodium carbonate solution .
    • Write the balanced symbol equation for the reaction.
    • Calculate the number of moles of sodium carbonate solution.
    • Using the balanced equation calculate the number of moles of HCl that reacts with the sodium carbonate.
    • Calculate the volume of HCl used in the reaction.