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

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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.

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