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# &amp;#145;The Relative Strength of an Unknown Acid&amp;#146;.

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Introduction

'The Relative Strength of an Unknown Acid' The aim of this experiment is to determine the relative strength of an unknown acid whose relative formula mass is 135. I am provided with the acid in as a white crystalline solid which is very soluble in water. The unknown acid monoprotic, which means for every one mole of acid, one mole of hydrogen is needed. The unknown acid can be completely neutralised by sodium hydroxide and the reaction is exothermic. The enthalpy change depends on the strength of the acid, so the stronger the acid, and the larger the enthalpy change. Therefore the weaker the acid, the smaller the enthalpy change. Below is a table with some typical values obtained by experiment. Acid Enthalpy change (kJ per mole of acid) HCl -57.9 HNO3 -57.6 CH2ClOOH -53.4 CH3COOH -50.1 HCN -38.2 To begin working out the unknown acids strength, I am going to use the enthalpy change equation (below) to work out the mass needed to make up a standard solution of the unknown acid. For this equation to work I am going to use 13?C as my hypothetical temperature rise and I am also going to choose an enthalpy change from the table above. I am going to use the enthalpy change -50.1, CH3COOH. ...read more.

Middle

Next I need to clean out a conical flask; I do this by again rinsing with distilled water. I will also need to swill out a 25cm3 pipette, again using distilled water. After this is completed I have to shake the volumetric flask of succinic acid, I do this by turning it upside down and shaking the contents, returning it upright, repeating this ten times, I must do this each time before I use the acid. After that I will use the 25cm3 pipette with a pipette filler to suck up some succinic acid. I will fill it up over the 25cm3 mark and then drain it out under gravity into a beaker, making sure the tip does not touch the liquid as it will not run out as good. Next I need to again shake the succinic acid. I will then fill the pipette up again to the 25.0cm3 mark, making sure the meniscus of the acid touches the 25.0cm3 mark. I will then transfer this to the conical flask. I will then add 4 drops of phenolphthalein indicator to the acid, which will turn the solution pink. To begin with the tap should be opened quite slowly allowing the sodium hydroxide to run into the conical flask slowly. ...read more.

Conclusion

It is important that only a few drops of indicator are added because excess indicator will result in a different volume of solution to be added from a burette before the required colour change is obtained. Now to combine all the errors worked out so I can get the total percentage of errors of the whole experiment. 0.04 + 2 + 2 + 0.2 + 0.15 = 4.39% From combining all the error rates I estimate the overall error rate for the experiment to be �4.39%. To make this clearer I will turn the percentage into a measurement: 25.2 x 4.39 = 1.10628cm3 Therefore the observed result was 25.2cm3 � 1.12cm3. 100 From my hypothetical results I am going to use the mean titre of 25.2cm3 of sodium hydroxide and the 25cm3 of succinic acid to work out the relative formula mass of succinic acid. (Concentration x Volume) NaOH = 2 (CV) NaOH = 2 = 0.1 x 25.2 (Concentration x Volume) H2A 1 (CV) H2A 1 2 x 25 x C H2A = 0.0504 Number of moles = Concentration x Volume n = 0.0504 x 250 n = 0.0126 1000 1000 Number of moles = mass taken 0.0126 = 1.32 Mr = 1.32 Mr Mr 0.0126 Mr = 104.7619046 Mr = 104.8 Looking back at the relative molecular mass of succinic acid I can see that when the Mr is 104, n = 1. Skill 1 - Planning Charlotte Nellist Page 1 ...read more.

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22.40 22.45 22.42 22.42 Initial Volume ( cm3) 0.00 0.00 0.00 0.00 Volume of titre ( cm3) 22.40 22.45 22.42 22.42 Discussion: The gram concentration of the acid solution =1.65/250cm3 =6.6g.dm-3 Assume that n=3, Mr of succinic acid will be 132.

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