• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

Our titration was to find out how much 0.10 molar alkali neutralises 25cm3 of acid, using methyl orange as a indicator.

Extracts from this document...

Introduction

Chemistry coursework Analysing Our titration was to find out how much 0.10 molar alkali neutralises 25cm3 of acid, using methyl orange as a indicator. Here are the results. 1st Result 2nd Result 3rd Result 4th Result 5th Result Acid Amount 25cm3 25cm3 25cm3 25cm3 25cm3 Alkali amount 29.20cm3 28.80cm3 28.60cm3 28.85cm3 28.75cm3 Average (not including 1st Result) 28.75cm3 As you can see there is an anomalous result (1st Result) which I will not use in my calculations. The equation for the titration can be sown as: H2SO4 + Na2CO3 ------- H2O + Na2SO4 + CO2 This has a ratio of: H2SO4 + Na2CO3 ------- ...read more.

Middle

Na2CO3, has a relative mass of: 23+23+12+16+16+16 = 106g/mol This total if for 1.0 mol, I want 0.10mol. So 1M solution= 106g dissolved in 1 dm3 = 1mol/dm3 = 1M But the volumetric flask = 250cm3 So we need 1/4 mole in 1/4 of a litre = 1mol dm-3 = 106g mol-1 -------------- = 26.5g required 4 So for 0.10M of solution you need to dissolve 26.5 ------ = 2.65g in 250cm3 = 0.1M 10 Now that we have made a accurate solution we can find out the molarity of the acid by using the average result of alkali used. ...read more.

Conclusion

in 25.00cm3 So molarity of acid = 0.0028 moles in 25.00cm3 = 1000 ----- = 40 25 to get the amount of times 25.00cm3 goes into a litre. = 40 x 0.0028 moles = 0.112 mol dm-3 This gives you the exact molarity of the acid in dm3. From this we can see that the acid was a slightly stronger molarity. We used 25cm3 of acid : 28.75cm3 of alkali. It is clearly obvious that the acid is stronger because we used less of it to neutralise more alkali. So overall out results make sense because the acid was a stronger molarity than the acid. ...read more.

The above preview is unformatted text

This student written piece of work is one of many that can be found in our GCSE Aqueous Chemistry section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related GCSE Aqueous Chemistry essays

  1. How much Iron (II) in 100 grams of Spinach Oleracea?

    0.00004316 x 1000 = 0.04316mols dm-3 At a temperature of 70 oc Firstly work out the moles of Potassium Manganate (VII) (aq) in the titration. Moles = Concentration x Volume Moles = 0.01 mol dm -3 x 20.70 cm3

  2. Finding out how much acid there is in a solution.

    Something that I cannot stop from happening is human error, which an experiment is always affected by. If someone else were to follow my planning and method, they would easily be able to understand and follow the instructions, as I have clearly set out the method step by step.

  1. The aim of this experiment is to find out how much diluteHydrochloric acid is ...

    * Be careful not to split any acid or anything else if this happens then call a teacher * Clean the apparatus so the next person that uses it won't get acid on them. Prediction Before I have done the experiment, I have predicted that the sodium hydroxide will neutralise when about 25cm of acid is put in.

  2. Finding out how much acid there is in a solution.

    Conical Flask 5. Burette 6. Distilled Water 7. Dropping Pipette 8. Sulphuric Acid 9. Filter Funnel 10. Set up the stand and burette which should be held firmly in positions by the clamp. 11. A funnel will then be placed at the top of the burette and the valve set so that no solution can flow through.

  1. Finding out how much Acid there is in a Solution and the Molarity of ...

    The amount needed is therefore 2.65g of solid anhydrous sodium carbonate. Measure out 25cm3 of 0.1mol/dm3 sodium carbonate solution using a 25cm3 pipette and pour into a 250cm3 conical flask. Fill the burette with 50cm3 of sulphuric acid so that the level reads 0 at the top of the burette.

  2. To find out the molarity of the acid using 25cm3 of acid to neutralise ...

    So the average of these is 28.80cm3. We now use 28.80cm3 to work out the molarity of the acid. The equation for the titration can be sown as: H2SO4 + Na2CO3 H2O + Na2SO4 + CO2 This has a ratio of: H2SO4 + Na2CO3 H2O + Na2SO4 + CO2 1

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work