Experiment 14 Formal Report for pH and pH titration PDF

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Experiment 14: pH and pH Titration Dr. Jiang March 27, 2021 Chemistry 1110-04

Purpose In this experiment, the purpose is to see the different rates of pH changes at different stages of the titration of solutions of weak acids. Which gets used to help determine pKa of the benzoic acid by measuring pH of the solution at the half-equivalence point. Theories/Principles The purpose of this experiment is to see the different rates of pH changes at different stages of the titration of solutions of weak acid to help determine the pKa of our benzoic acid by measuring the pH of the solution at the half-equivalence point. We know that a weak monoprotic acid HA is partially ionized in solution in water, establishing the equilibrium HA + H2O → H3O+ + A-. To determine the pKa for a weak monoprotic acid we need to measure the pH of the solution at the half-equivalence point using the following equations: Ka = [H3O+][A-] / [HA] Take the logarithm: logKa = log(H3O+) + log(A-) - log(HA) Multiply equation by -1: -logKa = -log(H3O+) - log(A-) + log(HA) pKa = pH - log(A-) + log(HA) Materials and Methods The graphical method is able to help determine Ka because it shows a clear idea of how the pH is changing that is occurring with each mL increment during titration and can accurately predict the Ka for benzoic acid. For part A of the experiment we would measure and record the pH of the solutions given with a pH meter. With our collected pH measurements and calculated expected pH of each of these solutions, we then can compare our measured pH values with the calculated one. In this table for part A includes final calculations for the solutions with their pH measured with meter and the expected pH.

Table #2: Formatted Data for Part A of experiment and calculated expected pH. Solutions

pH measured with meter

Expected pH

0.015 M HCl

1.81

1.82

0.06 M HOAc

2.76

4.22

0.04 M NaOAc

8.97

4.22

0.022 M NaOH

12

12.34

0.1 M NH4Cl

5.31

5.13

0.1 M NaOAc

8.12

8.87

0.015 M NHACl with 0.02 M NH3

8.97

9.38

For Part B Table #3: Formatted Data for mass of beaker and mass of beaker + benzoic acid for Part B of the experiment. B #1

B #2

Mass of empty beaker

126.68 g

122.42 g

Mass of beaker + benzoic acid

126.97 g

122.72 g

Table #4: Formatted Data for Titration B#1 with increments of 1.0 mL along with its graph of pH vs. NaOH volume.

Table #5: Formatted Data for titration B#2 with increments of 0.2 mL, 1.0 mL, and 0.1 mL along with its graph of pH vs. NaOH volume.

Table #6: Equilibrium point, half-equivalence point, accepted benzoic acid Ka, actual benzoic

acid MM and calculated equivalent weight of benzoic acid along with its % error. Equilibrium Point:

Between 22.84 mL and 22.95 mL

Half-Equivalence Point:

11.52 mL

Ka of our Benzoic Acid:

1.91 x 10-5

Accepted Benzoic Acid Ka:

6.5 x 10-5

Actual Benzoic Acid MM:

122.12 g/mol

Equivalent Weight of Benzoic Acid:

130.43 g/mol

% Error of Molecular Weight Benzoic Acid:

6.80%

Results and Discussion Sample Calculations: Part A: 0.015 HCl, fully dissociates (strong acid) pH = -log(0.015 M) = 1.82 0.06 M HOAc, 0.04 M NaOAc Ka = 4x10-5 pH = -log(4x10-5) + log(0.04/0.06) = 4.22 0.022 M NaOH, fully dissociates (strong base) pOH = -log(0.022) = 1.66 pH = 14.00 - 1.66 = 12.34 0.1 M NH4Cl R

NH4 (aq) + H2O (l) ⇌ H3O+ (aq) + NH3 (aq)

I(M) 0.1

---

0

0

C(M) -x

---

0

0

E(M) 0.1-x

---

0

0

Kb NH3 = 1.8x10-5 Ka = kw/kb = 1.0x10-14 / 1.8x10-5 = 5.6x10-10 Ka = [H3O][NH3] / [NH4] = x2 / 0.1-x = 5.6x10-10 √x2 = √5.6x10-11 = 7.48x10-6 -log(7.48x10-6) =5.13 0.1 M NaOAc Just like above we set up the RICE table Ka acetic acid = 1.86x10-5 Kb = 1.0x10-14 / 1.86x10-5 = 5.38x10-10 Kb = x2 / 0.1-x = 5.38x10-10 √x2 = √5.38x10-11 = 7.34x10-6 -log(7.34x10-6 ) = 5.13 pH = 14.00 - 5.13 = 8.87 0.015 M NH4Cl with 0.02 M NH3 Kb = 1.8x10-5 -log(1.8x10-5) = 4.74 pOH = 4.74 + log(0.015/0.02) = 4.62 pH = 14.00 - 4.62 = 9.38 Part B: Equilibrium point: between 22.84 mL - 22.95 mL pH = pKa

Half-Equivalence point: 11.52 mL pH = pKa = 4.72 Ka = 10-4.72 = 1.91x10-5 Accepted benzoic acid Ka = 6.5x10-5 Actual benzoic acid Molar Mass: 122.12 g/mol MNaOH = 0.10 M VNaOH = 22.84 mL (closer to neutralization) V Benzoic Acid = 50 mL M Benzoic Acid = 0.10 x 22.84 / 50 = 0.046 M Moles of Benzoic Acid = 0.046 M x 0.050 L = 0.0023 moles Molecular Weight = 0.30 g/0.0023 moles = 130.43 g/mol (for B#2) % error of Molecular Weight Benzoic Acid = 130.43 - 122.12 / 122.12 x 100 = 6.8% Discussion: Based on our data and calculations in part A we notice that our measured pH values are pretty consistent with the expected pH values. However for 0.06 M HOAc and 0.04 M NaOAc we made the error of not mixing them together and measuring its pH value, instead we got the pH value for each. For part B, our benzoic acid #1 with increments of 1.0 mL, we notice that the pH slowly is rising with each mL. As the titration continues, there must have been a malfunction with our pH meter because the pH values aren’t showing much of a difference, which we reach a peak at 27.8 mL with a pH value of 11.18. In our next titration for benzoic acid #2 with increments of 0.2 mL, 1.0 mL and 0.1 mL, we see pH values slowly increasing, which helps in determining the equivalence point and half equivalence point by looking at our Table #5. The equivalence point in our data occurs between 22.84 mL - 22.95 mL, so given that 22.84 mL has the pH value closest to the neutralization of 7, I went ahead and used this measurement to

help determine the molar mass of benzoic acid. So when calculating the equivalent weight of benzoic acid we got 130.43 g/mol while the actual mass of benzoic acid is 122.12 g/mol, which isn’t much of a difference with its actual mass along with % error of 6.80%. Conclusion By carrying out the titrations of solutions of weak acids and noticing the different pH changes occurring at different rates, we were able to determine the pKa of our benzoic acid at the half-equivalence point to be 1.91 x 10-5. References 1. Lab Manual Answers to Questions from Lab Manual 1. We can not prepare a solution of pH=8 by diluting sufficiently a 0.01 M solution of HCl with water because with a higher acid or base concentration then water will be neglected unless the acid or base concentrations are lower. 2. The graphical method used in our experiment to determine Ka is a better way because we are able to see the pH changes that are occurring with each mL increment during titration. Which with the help of this method, we can accurately predict the Ka for benzoic acid. 3. Using methyl orange as an indicator would be better because we would be able to find a bigger value of molecular weight. In this occasion, I will be using my second titration data for calculation: (Using Methyl Orange) Moles NaOH: 0.10 M x 0.61 mL = 0.061 moles , Molecular Weight: 0.30 g / 0.061 moles = 4.91 g/mols (Using Phenolphthalein) Moles NaOH: 0.10 M x 23.01 mL = 2.301 moles Molecular Weight: 0.30 g / 2.301 moles = 0.13 g/mols

4. % unc. in g BA = 0.0001 g / 0.30 g x 100% = 0.03 % % unc. Veq. = ± 0.10 mL / 22.84 x 100% = 0.44% % unc. [NaOH] = 0.01% % unc. in Molecular Weight BA = 0.03% + 0.44% + 0.01% = 0.48%...