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Determination of the % by volume of ethanoic acid in 100cm^3 of water

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Introduction

I.B. CHEMISTRY: PRACTICAL 08-12-12. P2,P3,& P4-STOICHIOMETRY SULAIMAN JALLOH AIM: 1) To prepare a standard solution of potassium hydrogen phthalate by weighing and dissolving it water of correct volume. 2)To use the standard solution of 1) to standardize(find the concentration of) sodium hydroxide solution. 3)To determine the percentage of ethanoic acid,by volume,in vinegar,by titrating with standardized sodium hydroxide in 2) 1) MAKING A STANDARD SOLUTION OF POTASSIUM HYDROGEN PHTHALATE RAW DATA TABLE mass of weighing bottle,m+/-0.01g Volume of solution titrated,v+/-0.15 Concentration of sodium hydroxide given ,Cb/mol/ 33.88 250.00 0.1 The purity of the acid is given as 99.5+/-0.5% Calculations for the mass of the acid needed to make 250 of solution of the acid. Equation for the reaction: C6H5COOKCOOH+NaOH C6H5COOKCOONa+H2O so,the mole ratio of the acid to base is 1:1 This means 1 mole of acid gives 1 mole of the base. number of moles of sodium hydroxide,Nb=c v where c=concentration of sodium hydroxide v=volume of solution titrated Nb=0.1 250/1000) Nb=0.0250moles Hence the number of moles of the acid will be 0.0250 moles since the mole ratio is 1:1 number of moles ...read more.

Middle

trial 4 1.90 19.30 CALCULATION TABLE Initial volume +/-0.05 1.80 0.30 1.90 Final volume +/-0.05 19.20 17.80 19.30 Volume of acid used,/ 17.40 17.50 17.40 Average volume of acid used,Va=(17.40+17.50+17.40)/3 =17.43 Volume of acid used is given by Final volume-Initial volume =19.20-1.80 =17.40 Since the mole ratio of acid to base in 1) is 1:1 Concentration of the acid,Ca is calculated to be 0.100+/-0.001 Average volume of acid used,Va=17.43 Volume of sodium hydroxide used,Vb=25.00+/-0.03 Concentration of sodium hydroxide,Cb=? From Ca =Cb (mole ratio of acid to base is 1:1) Cb=(Ca )/Vb Cb=(0.100 )/25.00 Cb=0.06972 (3 S F +1) Error Propagation Sources of error % error Calculations concentration of the acid =1 volume of the acid-burette =0.10/17.43 =0.6 Pipette volume =0.03/25.00 =0.1 Hence, %error=1+0.6+0.1 =1.7% Error in the concentration of the base =1.7/100 =0.001 Therefore,the concentration of the base,Cb=0.070+/-0.001 3)Determination of Ethanoic acid content in vinegar by volumetric analysis Raw data Tests Observations Add three drops of phenolphthalein indicator in the base(sodium hydroxide) ...read more.

Conclusion

Volume of sodium hydroxide used,Vb=25.00+/-0.03 Since the mole ratio of the acid to base is 1:1 Ce Cb Ce=(Cb )/Va =(0.070 25.00)/29.30 =0.05973 (3 SF + 1) Error propagation Sources of error % error calculations Concentration of sodium hydroxide solution =1.7 Pipette for dilution =0.03/25.00 =0.1 Pipette for titration =0.03/25.00 =0.1 Burette =0.10/29.30 =0.3 Volumetric flask for dilution =0.15/250.00 =0.1 =2.3% The error in the concentration of the acid=2.3/100 =0.001 Concentration of ethanoic acid in 250 volumetric flask will be the dilution factor(250/25) concentration obtained in the titration f=10 =0.5973 Hence,the concentration of ethanoic acid in 250cm volumetric flask is: =0.597+/-0.001 To Calculate The Percentage By Volume of Ethanoic acid in the solution we know, molar mass,Mr(CH3COOH)=12.01+3(1.01)+12.01+2(16.00)+1.01 =60.06 Concentration of Ethanoic acid in =35.86 We also know that density =mass/volume density of ethanoic acid is given as 1.045 mass of the acid is calculated to be 35.86 Volume,V=35.86 /1.05 =34.15 / but 1 =1000 )/10 V=3.415 The error in the volume is the same as the percentage error in the concentration of the acid =2.3% of3.415 =2.3/100 =0.08 Hence,the percentage by volume of ethanoic acid in 250 of solution is =3.42+/-0.08 CONCLUSION ...read more.

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