• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month
Page
1. 1
1
2. 2
2
3. 3
3

# 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

# Related GCSE Aqueous Chemistry essays

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

If it has been in contact with the skin, wash off using soapsuds. Bibliography 1) Understanding Biology for advanced level (Third Edition) Glenn and Susan Toole Stanley Thornes Publishers -Cheltenham (1995) 2) www.nms.on.ca/Secondary/iron_is_good_for_you.htm Canadian Health Authority 3) www.chemistry.eku.edu/GODBEy/Homepage_files/CHE_112.../oxred_oxalate.htm 4) Chemical Ideas (Second Edition) G. Bulon, J. Holman, J. Lazonby, G.

2. ## 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.

1. ## Finding out how much acid there is in a solution.

This is the end point of your titration. * Keep on repeating the titration until you have achieved a good set of results. Your values should be within 0.1cm of each other. * Remember to record the burette reading. Make sure that the reading is taken at eye level for accuracy, and is taken at the bottom of the meniscus.

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

The conical flask should be placed on a white tile. This makes it clearer to see the colour change of the indicator as the colour of the laboratory bench may interfere with the colour of the indicator and the exact end-point would be difficult to determine.

1. ## 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

2. ## Find out how much acid there is in a solution

Adding Sulphuric Acid Solution into the Burette * In a 100ml beaker add some Sulphuric Acid solution. * Place a small funnel on the burette. * Make sure the burette is below your head and you must not pour the solution if the burette is above your head.

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