Retha Puvogel and Mya Gardner
02 April 2018
Potentiometric Analyses
Introduction:
The value that represents the equilibrium constant for the titration of a weak acid into a strong base is the Ka value. The reason that we determine this value is because the Ka value is used to determine the acid dissociation in a solution. The molar mass is found by dividing mass over moles of a substance – i.e. grams/moles. Molar mass is an important aspect in acid titration in order to determine the exact amount of acid that was put into the beakers for pH testing. In potentiometric titration, it is just like regular titration, except instead of using a Universal Indicator – like phenolphthalein from the titration lab in Chemistry-151 – in this kind of titration, a pH meter is inside of the solution as it is being stirred by a stir-top. In order to determine the Ka value from the potentiometric titration, it is found by using the Henderson – Hasselbalch Equation which uses pKa and the log of the concentrations of the acid over the base to determine the pH. From this specific experiment, the concentrations of the acid and the base were equal to one (1) which made the log of one (1) be zero (0) which was insignificant to this experiment. This made the pH value equal to the pKa value. To determine the Ka value, you had to take the pKa value from the pH and raise it negatively to ten (10) to get the Ka value. Molar mass is found for each sample by using the sample size – of zero point two five zero (0.250g) grams – over the moles of the acid – which was determined from how much acid was dispensed from the burette into the solution. If our hypothesis is correct, then we will have a perfect titration curve graph with an equivalence point in the middle with points leading to and from it. And also, there will be a derivative graph that has many plot points and an obvious equivalence point.
Results:
Unknown 7: Trial 1 Unknown 7: Trial 2 Unknown 7: Trial 2.2 Unknown 7: Trial 3
Mass of Unknown Acid 0.250 g 0.250 g 0.250 g 0.250 g
[NaOH] 0.0498 M 0.0498 M 0.0498 M 0.0498 M
Volume of NaOH dispensed 45.07 mL 37.00 mL 29.46 mL 30.62 mL
Volume of NaOH @ Equiv. Point 27.25 mL 26.00 mL 26.10 mL 25.50 mL
pH @ ½ Equiv. Point 5.69 5.46 5.40 5.20
pKa @ ½ Equiv. Point 5.69 5.46 5.40 5.20
Ka=10^-pKa 2.042 * 10 ^-6 3.467 * 10^-6 3.981 * 10^-6 6.310 * 10^-6
pH initial of NaOH 12.94 12.09 12.94
pH initial of Unknown 4.29 4.35 4.31 4.40
Sample One:
Solid Weight: 0.250 g
Initial Volume NaOH: 50.00 mL
Initial pH Unknown: 4.34
End Volume: 4.93
Initial pH NaOH: 12.94 mL
Volume Base(ml) pH
1.05 4.43
3.21 4.67
5.7 4.86
7.84 5.04
9.92 5.21
12.05 5.36
14.3 5.48
16.53 5.64
18.71 5.77
20.71 5.95
22.95 6.17
25.09 6.47
27.25 11.37
29.56 11.77
31.58 11.97
33.62 12.1
35.87 12.19
38.01 12.27
40.1 12.32
42.34 12.37
45.07 12.4
0 4.34
Sample Two:
Solid weight 0.250 g
Initial Volume NaOH 50.00 mL
Initial pH Unknown 4.35
End Volume 13.00 mL
mL NaOH pH
25.01 9.81
26 10.92
26.5 11.13
27.01 11.3
28.01 11.48
29.02 11.62
30.01 11.72
31.2 11.82
32 11.86
33.21 11.94
34.05 11.97
35.12 12.01
36.1 12.05
37 12.09
0 4.35
Sample Two.Two:
Solid Weight 0.250 g
Initial Volume NaOH 50.00 mL
Initial pH Unknown 4.31
End Volume 20.54 mL
mL NaOH pH
2.1 4.55
5 4.88
10.21 5.29
15.01 5.61
20.01 6.13
22.11 6.34
24 6.8
25.12 7.55
26.1 10.8
26.51 11.06
27.1 11.26
27.62 11.42
28.12 11.51
28.5 11.59
29.1 11.66
29.46 11.7
0 4.31
Sample Three:
Solid Weight 0.250 g
Initial Volume NaOH 50.00 mL
Initial pH Unknown 4.40
End Volume 19.38 mL
mL NaOH pH
3.1 4.71
6.22 5.01
9.26 5.24
13.3 5.52
17.01 5.8
21 6.2
22.21 6.37
22.69 6.48
23.51 6.7
24.1 6.94
24.54 7.14
25.2 8.44
25.5 10.4
26.12 10.85
26.6 11.16
27.13 11.32
27.51 11.42
28.6 11.61
29.48 11.71
30.62 11.83
0 4.4
Discussion:
Our equivalence point for Sample One ended up being at 27.25 mL that had a pH of 11.37. At that ½ equivalence point for Sample One was 13.63 mL with a pH of 5.69. Because the pH is equal to pKa value, in Sample One the pKa value for Sample One was 5.69. From the pKa value, the Ka value was calculated to be 2.042 * 10^-6. The molar mass that was determined for Sample One was 183.8 g/mol of acid.
In this lab, Sample Two had to be repeated because there was a lack of data points from the start to the equivalence point. When we repeated Sample Two we ended up with a derivative graph and titration curve graph that looked much better than the original data from Sample Two gave to us. From the original data of Sample Two, we had an equivalence point at 26.00 mL with a pH of 10.92. For the ½ equivalence point values, they were: 13.00 mL with a pH of 5.46. And because pH equals the pKa, the pKa value was 5.46 as well. After calculations including the pKa value, the Ka was determined to be 3.476 * 10^-6. The molar mass value for this sample was 193.1 g/mol of acid.
After repeating Sample Two, we obtained data for Sample Two.Two. The equivalence point for Sample Two.Two was at 26.10 mL with a pH of 10.80. This means that the ½ equivalence point was at 13.05 mL with a pH of 5.40. Which from this ½ pH value, we get the pKa of 5.40. Using this pKa value to calculate the Ka value, the Ka values is determined to be 3.981 * 10^-6. This sample of data had a molar mass of 192.3 g/mol of acid.
For the final set of data for Sample Three, the equivalence point was found at 25.50 mL with a pH of 10.40. At the ½ equivalence value, the data points were: 12.75 mL and a pH of 5.20. From this pH at the ½ equivalence point, we get the pKa value of 5.20. After calculations occurred, the Ka value was determined to be 6.310 * 10^-6. The molar mass was calculated out to be 196.9 g/mol of acid.
For the molar mass, the average from all four (4) samples of data was determined to be 191.5 g/mol of acid. Taking each value and subtracting out the average, then squaring that number gets you halfway to determining the standard deviation. You have to average out those value you got from squaring and divide by the number of data points you had and then take the square root of that value. Our standard deviation for the molar mass was calculated to be 4.8. and from there, the rsd percentage turned out to be 2.5%.
Finding the average for the pKa values, was adding all of the values together and dividing by four (4) because that was how many samples of data we had. The average pKa value turned out to be 5.44. From doing the calculations for the standard deviation, we got a value of 0.1747 for the standard deviation. To calculate the rsd percentage, you take the standard deviation and divide it by the average value and multiply by one hundred (100) to get the percentage. Our rsd value was determined to be 3.212% for the pKa values.
Conclusion:
We did not prove our hypothesis entirely. Our data values and from the multiple graphs, we did not obtain perfect graphs and values. According to our RSD percentages, our data was precise for both the molar masses and as well for the pKa values. From the standard deviation values, they prove how accurate our data was. Based off of our data value for each of the standard deviation numbers – the pKa and the molar mass values – our data was accurate. This lab taught us how to use the Henderson – Hasselbalch equation with the pH value to find the pKa value. And to then in turn use the pKa value to find the Ka values. This lab also taught us the difference between potentiometric titration and acid-base titration.