Using p H Electrode for Acid Base Titration post lab PDF

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Using pH Electrode for Acid Base Titration Introduction The purpose of the Using pH electrode for Acid Base Titration lab was to analyze the equivalence point using potentiometric titrations. Another purpose was to see how well the electrode picked up the exact equivalence point compared to the indicator. For this lab potentiometric titrations were used so that we could get familiar with the methods, to find the equivalence point, and to determine the molecular mass of the unknown compound. This experiment was carried out by getting a sample of unknown and dissolving it in water to create an aliquot. Then figure out if it was acid or base so that the proper indicator could be used. Then do a fast run on the titration to find the end point, and a careful run where the aliquot is titrated with titrant slowly so that the pH electrode could find the end point and so that we can see when the indicator changed color. Then using this data create a graph to find the difference between the readings on the pH electrode and the indicator.

Ag (s)∨AgCl (s )∨Cl −( aq )∨¿ H +( aq , outside)⋮ H +(aq , inside), Cl −( aq )∨ AgCl (s )∨ Ag(s ) Methodology Gather around 2.00g of unknown and dissolve in deionized water in a 250mL volumetric flask. After all of the unknown is dissolved add enough water to the dilution mark. This is the aliquot for the experiment. Then gather a pH meter, a glass electrode, and a sample of pH buffer 4 and 7. Using the material gathered calibrate the pH meter and glass electrode using the pH buffers. Also find the pH of the unknown solution by using pH paper. If the sample is acidic use phenolphthalein, and if basic use bromocresol green. This experiment dealt with an acid so phenolphthalein. The first titration is a rough titration to find an estimated volume of titrant that gives the end point. Place in a 125mL flask 25mL of unknown and add 3 drops of phenolphthalein. Titrate this solution with 0.1M NaOH until the endpoint. Then doing a rough titration with the electrode, and 100mL of unknown solution titrate the solution again but more carefully. Add 1drop of indicator, and start adding 2mL of NaOH, then as the solution gets closer to the end point start adding increments of 1mL and then 0.5mL of NaOH. Then as the equivalence point is passed start increasing the amount of titrant. Data, Calculations and Error Analysis Mass of unknown Titrant Titrant concentration

2.0197 NaOH 0.1025

On the fast run the value was when the indicator changed color

Fast run Volume

pH 10.27

10.55

Actual Titration Volume 0.000 5.000 10.00 15.00 20.00 25.00 27.00 29.00 31.00 32.00 33.00 34.00 35.00 36.00 37.00 37.50 38.00 38.25 38.50 38.60 38.70 38.80 38.90 39.00 39.25 39.50 40.00 41.00 42.00 43.00 44.00 45.00

pH 4.430 4.720 4.930 5.100 5.260 5.440 5.530 5.610 5.730 5.790 5.850 5.960 6.070 6.140 6.460 6.610 6.920 7.190 8.720 9.310 9.630 9.840 10.060 10.110 10.320 10.450 10.590 10.670 10.790 10.870 10.910 11.010

Δph/ΔV 0.000 0.058 0.042 0.034 0.032 0.036 0.045 0.040 0.060 0.060 0.060 0.110 0.110 0.070 0.320 0.300 0.620 1.080 6.120 5.900 3.200 2.100 2.200 0.500 0.840 0.520 0.280 0.080 0.120 0.080 0.040 0.100

(Δph/ΔV)/Δ V 0.000 0.012 -0.003 -0.002 0.000 0.001 0.004 -0.002 0.010 0.000 0.000 0.050 0.000 -0.040 0.250 -0.040 0.640 1.840 20.160 -2.200 -27.000 -11.000 1.000 -17.000 1.360 -1.280 -0.480 -0.200 0.040 -0.040 -0.040 0.060

47.00 49.00

11.170 11.310

0.080 0.070

-0.010 -0.005

Actual Titration pH 12.000 10.000 8.000 6.000 4.000 2.000 0.000 0.000

10.000

20.000

30.000

40.000

50.000

60.000

1st derivative 7.000 6.000 5.000 4.000 3.000 2.000 1.000 0.000 30.000

32.000

34.000

36.000

38.000

40.000

42.000

44.000

2nd derivative 30.000 20.000 10.000 0.000 35.000 -10.000

36.000

37.000

38.000

39.000

40.000

41.000

42.000

-20.000 -30.000

End point volume based on graphs is 38.50mL and the pH at 38.50mL is 8.720 moles of unknown acid

Titration concentration∗Equivalence point volume(L)=moles of NaOH L=0.003946 moles ( 38.50 1000 )

0.1025 M∗

1 molesof NaOH=1 moles of unknown acid at the equivalence point 0.003946 moles of NaOH =0.003946 moles of unknown acid Molecular mass of unknown acid

(0.003946/100 ml) *1000mL=0.3946M Total moles of unknown acid

Total moles of acid∗.25 L of solution 0.009866= 0.3946∗.25 L

mass of unknown=moles of unknown∗molecular mass of unknown

mass of unknown =molecular mass of unknown moles of unknown 2.0197 =204.7 0.009865625 The indicator changed at 38.60 mL and around a pH of 9.31 The actual equivalence point was 38.50mL and around a pH of 8.72 Indicator error

( Indicator volume−actual volume ) ∗100 % actual volume

((38.60-38.50)/38.50)*100=0.2597% Results and Discussion In the experiment we used an unknown acid . We know that the acid monoprotic and not diprotic or polyprotic because on the graphs that I created there were only one maximum on the 1st derivative graph, and also because the titration curve only had 1 inflection point. Diprotic, and polyprotic acids will have more than one inflection point and also more than 1 maximum on the first derivative graph. Also we know that the acid used was a weak acid because at the equivalence the pH was >7, and since we used a strong base it makes the acid weak. If the unknown acid was a strong acid then the pH would =7 at the equivalence point. It is important to control the addition of titrant added to a few milliliters before and after the endpoint is so that when graphing you can graph a cleaner graph that shows the equivalence point better so that the volume of titrant and pH can be found. Also you could over titrate the solution causing you to miss the equivalence point and not be able to record the correct values. Potentiometric titrations are more advantageous when the availability of an indicator is limited, analyte solution is colored or cloudy, and because it works well with diluted solutions. Also the potentiometric titrations are able to more accurately achieve the equivalence point of a titration than indicators. Some disadvantages that this method has is that you have to be very careful with how much you titrate at a time, which causes it to become more time consuming. We used a machine in this experiment so there could have been error in the machine, and also the magnetic stir bar kept being

attracted to the electrode which caused some problem in the experiment. Overall the experiment seemed to go well because the titration was with a weak monoprotic acid and the pH was over 7 and had 1 inflection point. The molecular mass seems a little high but 200 is a reasonable molecular mass. Conclusion In this experiment an unknown acid was analyzed to find the equivalence point with potentiometic titrations with a strong base. This means that using potentiometric titrations allowed us to more accurately calculate the equivalence point and the molecular mass of the unknown....