Kate Collier
Lab 10
5th Hour
Titration Lab
Introduction:
Confirmation is commonly used every day. Repairmen usually know what the problem is before they even get to it to run tests. Before they start to order parts or fix what they think is the problem, they run a test or tests to confirm that they are about to fix the right problem instead of something that didn’t really need to be fixed because the problem was elsewhere. The reason for doing a titration is to find the equivalence point, which is found when the base and acid neutralize. By taking the equivalence point and dividing it by 2, the equilibrium constant can be determined. If a titration is performed, then the equilibrium constant, Ka or Kb can be determined.
Materials and Methods:
Labs were set up with burettes; this was taken place at a lab with vinegar and NaOH. A calculator and pH reader was obtained before beginning lab. 27.6 mL of vinegar was added to a beaker, the initial pH was recorded. NaOH was added .5 mL at a time, recording the new pH every time. This continued until the amount of NaOH doubled the initial amount of vinegar that was placed in the beaker. Lab was not completed in first day; saran wrap was placed around top of beaker and taped to keep shut, so as to not change the pH level too much. The lab was completed on the second day, the solution was dumped and beaker cleaned, materials put away except for the burets, which were left for the other class.
Results:
Table 2: Strong Acid vs. Strong BaseTrial / Volume at
Equivalence Pt. (mL) / pH at
Equivalence Pt. / Initial Volume
of HCl (mL) / Molarity
1 / 30.75 / 7.85 / 25 / .31
2 / 13.75 / 9.4 / 10 / .34
3 / 25.75 / 9.5 / 20 / .32
4 / 43.25 / 6.25 / 35 / .31
5 / 38.25 / 7.8 / 30 / .32
6 / 18.25 / 6.65 / 15 / .30
Average / 7.9 / .32
Standard Deviation / 1.3 / .01
Without Outliers / Avg.=7.1
Stan Dev. =.81
Table 4: Weak Base vs. Strong Acid
Trial / Volume at Equivalence (mL) / pH at Equivalence / Initial Volume
Ammonia (mL) / M Ammonia / 1/2 Equivalence Pt. / pKb / Kb
1 / 11.75 / 4.45 / 12 / .31 / 5.88 / 9.2 / 6.3*10-10
2 / 21.75 / 3.47 / 20 / .34 / 10.88 / 9.6 / 2.5*10-10
3 / 6.75 / 7.05 / 10 / .21 / 3.38 / 13.3 / 5.0*10-10
4 / 35.75 / 3.5 / 35 / .33 / 17.88 / 9.7 / 1.9*10-10
5 / 26.75 / 3.5 / 30 / .29 / 13.38 / 9.6 / 2.5*10-10
6 / 22.25 / 3.4 / 25 / .28 / 11.13 / 9.4 / 3.9*10-10
Standard Deviation / .44 / .025 / .2 / 1.77*10-10
Average / 3.67 / .31 / 9.5 / 3.42*10-10
True Value / 1.8*10-5
Dropping Trial 1 / StDev. = .047
Avg. =3.47 / StDev
=8.49*10-11
Avg. = 2.7*10-10
Table 6: Weak Acid vs. Strong Base
Trial / Volume at
Equivalence / pH at
Equivalence / Initial Volume
of Vinegar / M Vinegar / 1/2 Equ. Pt. / pKa / Ka
1 / 7.75mL / 8 / 15 / .13 / 30875 / 4.21 / 6.17*10-5
2 / 10.75mL / 8.25 / 15 / .18 / 5.375 / 4.6 / 2.51*10-5
3 / 12.25mL / 8.8 / 27.6 / .11 / 6.125 / 4.205 / 6.24*10-5
4 / 7.75mL / 8 / 10 / .19 / 3.875 / 4.7 / 1.995*10-5
5 / 15.75mL / 9.15 / 20 / .20 / 7.875 / 4.615 / 2.43*10-5
6 / 22.7 mL / 7.95 / 25 / .23 / 11.375 / 4.658 / 2.20*10-5
Standard Deviation / .501 / .045 / 2.17*10-5
Average / 8.36 / .17 / 3.68*10-5
True Value / .69 / 1.7*10-5
Table 7: Calculations
M of NaOH
5g NaOH / 1 mol / = .25M NaOH
40g NaOH / .5 L NaOH
M of HCl
30.75mL NaOH / .25M NaOH / 1 mol HCl / = .31M HCl
1L NaOH / 1 mol NaOH / .25mL HCl
Ka from pKa
Ka= 10 ^-pKa / = 10^-4.21 / = 6.17*10-5
Conclusion:
The average pH of HCl and NaOH at equilibrium was found to be 7.9, when it should be 7.0, and the standard deviation is 1.3, when the closer it is to zero, the more accurate the data is. In trials 2 and 3, the pH at equilibrium was around 9, which could play into effect for the average pH being greater than 7. When you take out trials 2 and 3, the average pH becomes 7.1 and the standard deviation becomes .81.Dropping these trials brings the pH closer to 7, which is where it should be, along with the standard deviation coming closer to zero.
The pH of ammonia, at equilibrium is 3.67, and the standard deviation is .44. Because the titration was a mixture of a strong acid and weak base (NH3), the pH should be below 7, which for these trials, where the equilibrium should be exactly, is unknown. The standard deviation is less than 1, which is also good because it is closer to zero. Trial 3 (green) had already been dropped during for the results because it wasn’t done correctly, by dropping trial 1, the standard deviation becomes .047 which significantly changes the data and brings it much closer to zero. When dropping trial 1, the average didn’t change much, but went down to 3.47. The average value of Kb was found to be 3.42*10-10. The true value of Kb is 1.8*10-5. The accuracy of the Kb calculated is 99.9%, which is very good. The precision of Kb is determined by the standard deviation which is 1.77*10-10, the smaller the number, the more precise, this number is very close to zero; therefore the calculations made were precise.
The average pH of vinegar at equilibrium is 8.36; it should be greater than 7, because it is vinegar (weak acid) mixed with a strong base. The standard deviation is .501. Dropping any trials isn’t necessary because all pH levels are close to each other. The average Ka value calculated was 3.68*10-5, the true value of Ka should be 1.7*10-5. The precision of the trials was fairly good; however, something went wrong because the accuracy is 116%.
According to the data if Kb is known than the identity of an unknown weak base can be found. Because the accuracy of Ka was over 100%, it is hard to tell if unknown identities of weak acids can be found by a known Ka. This lab could have been done better if every group had started and finished the same day, so levels couldn’t be tampered with, as well as pressure and temperature. It also would have helped if the same amounts were used for all trials.