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The Relative Formula Mass of an Unknown Acid.

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The Relative Formula Mass of an Unknown Acid Introduction I have been provided with a solution of unknown monobasic acid. By titration with a standard solution of sodium hydroxide I am to calculate its molarity and hence the relative formula mass of this unknown acid. (We can assume the concentration of the sodium hydroxide is 0.1M) Procedure 1. Using the funnel, rinse the burette with the acid solution and fill it accurately with the same solution until the 0cm3 mark (Do not forget to rinse and fill the tip first though). 2. Using the pipette filler, rinse the pipette with some of the sodium hydroxide solution and carefully transfer 25.0cm3 of the solution to a clean 250cm3 conical flask (Remembering to touch the bottom of the flask with the tip of the pipette for ten seconds to fully empty the pipette). ...read more.


8. Repeat steps 5,6 and 7 at least until there are three concurrent results. 9. Empty the burette and wash it carefully immediately after the titration, especially if it has a ground glass tap. Results Titration no. 1 2 3 4 5 Titre value 21.7cm3 21.2cm3 21.2cm3 21.3cm3 21.2cm3 To calculate the mean titration value I added together the last four results and divided by four. I opted to leave out the first result as this was my trial run and wasn't concurrent with the others. *MEAN TITRE ? (21.2 + 21.2 + 21.3 + 21.2) ? 4 = 21.23cm3 Analysis As we are told that the acid is monobasic the equation for the reaction can then be written as " HA + NaOH ? NaA + H2O " (Where HA is the unknown acid) Following on from the fact that the acid is monobasic, and therefore reacts with the sodium hydroxide on a one to one ratio, it is then possible to use the formula 'V1C1=V2C2' (V = volume in dm3 and C = conc. ...read more.


7.5g ? (21.23 ? 1000) = 0.159225g Now we can simply use the formula below: "Relative formula mass = Mass ? No. of moles" = (0.159225) ? (0.002505) = 63.56g From the acids given as possibilities it is clear that the unknown acid is "HNO3" which has a relative molecular mass of 63g. This is a difference of 0.88% from my result (calculated as shown below). [ (63 ? 63.56) ? 100 = 99.12% ] This percentage difference may be due to slight inaccuracies of the equipment used for the experiment. Below I have calculated the maximum percentage error for each piece of apparatus and also the total percentage error. * 250cm3 volumetric flask ? 0.5cm3 ? (0.5 ? 250) ? 100 = 0.2? * 25cm3 pipette ? 0.05cm3 ? (0.05 ? 25) ? 100 = 0.2? * Burette ? 0.15cm3 ? (0.15 ? 21.23) ? 100 = 0.71? TOTAL MAXIMUM PERCENTAGE ERROR ? 1.11% This helps explain the slight differences in answers. ...read more.

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