Lab Report Ex 8 Acids and Bases I PDF

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Lab report for experiment 8...

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1 General Chemistry 1 Laboratory Report Name: Ivan Chan Date of experiment: 4/5/19

Lab day: Friday TA: Fang

8. Acids and Bases - I Purposes of the experiment and brief overview of experimental method(s) The purpose of this experiment is to analyze acid-base reactions, notably those between both strong and weak acids and a strong base. In this experiment, the pH values at the equivalence point were determined for both titrations, while in the weak acid-strong base solution the pH value of the half-equivalence point was also determined in order to calculate the equilibrium constant, Ka of the reaction. In part 1, the strong acid-strong base titration, HCl(aq) + NaOH(aq) → NaCl(aq) + H20(l), was analyzed and graphed to determine an accurate value of the [HCl]stock and pH value at equilibrium. This was accomplished by using a pH electrode and a bromothymol blue indicator to measure pH change over volume of NaOH added. Using the concept of complete dissociation, the concentration of the HCl stock solution could be determined from the known concentration and volume of NaOH added. In part 2, the weak monoprotic acid-strong base titration, CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l), was used to both determine the accurate value of the [CH3COOH]stock and the acid dissociation constant for the reaction by using the half equivalence point of the titration and the Henderson-Hasselbach equation. This was accomplished by using a pH electrode and phenolphthalein indicator to measure the pH change over volume of NaOH added. In both parts 1 and 2, the equivalence point was determined at the greatest value of first derivative of the titration curve, at which the rate of pH change over volume NaOH is the highest. The purpose of part 3 was to study the ability of a buffer to resist change in pH. The buffer solution, made from CH3COOH(aq) and CH3COONa(aq), underwent additions of water, strong acid, and strong base to examine this property. The pH after each of these additions was calculated by using the value of the [CH3COOH]stock determined in part 2. In each of the experiments, the pH electrode was calibrated using a buffer solution with pH = 4 and a buffer solution with pH = 7 to obtain an accurate reading. The titrant was added using a buret for the highest significant figure accuracy and appropriate glassware was used in the rest of the experiment.

2 Results & calculations Part 1 – HCl versus NaOH titration Plot pH on the vertical axis versus volume of NaOH added (not the buret reading). Connect the points by a smooth curve. On the same graph, plot the first derivative (pH/V); use the vertical axis on the right hand side for this plot, and use a different label for these points. Draw a vertical line to the horizontal axis thru the point where pH/V is a maximum. The point of intersection of this vertical line and the titration curve is the equivalence point; label this point on the graph. In addition, mark and label the end point (where the indicator changed color) on the graph. Make sure your graphs conforms to all requirements. 3.00

18.00

16.00 2.50 14.00

12.00 2.00

1.50

8.00

7.29 6.00 1.00 4.00

2.00 0.50 0.00

0.00 0.00

2.00

4.00

6.00

8.00

10.00

12.00

-2.00 14.00

Volume of NaOH added / mL

Figure 1. The titration of a strong acid, HCl, with a strong base, NaOH(aq) using bromothymol blue indicator is shown above on the green curve. The first derivative of this relationship is

ΔpH/ΔV /mL-

pH

10.00

3 shown on the red curve. Both the equivalence point (pH=7.29), at which moles of HCl is equal to moles of NaOH, and the endpoint (pH=7.31), at which the color change occurred, are labeled. The maximum value of this first derivative, at which the rate of pH increase per volume added is the highest, is synonymous with the equivalence point volume. Using this and the experimental graph, the pH at the equivalence point can be estimated.

Volume of HCl stock solution used: 10.00 mL Indicator used: bromothymol blue [NaOH] used: 0.1000 mol L-1 Eq. point volume of NaOH: 10.90 mL (from the graph) Calculate [HCl] in the stock bottle (show work below) : (Note that [HCl] written on the stock bottle has only one SF) [HCl] = (Concentration of NaOH) x (Volume of NaOH at equivalence point) x (1molHCl/1molNaOH) / (Volume of HCl stock solution used) = (0.1000M) x (0.01090L) x (1molHCl/1molNaOH) / (0.01000L) = .1090M = 0.1090M

pH at the equivalence point of the titration: 7.29 (from the graph) pH at the end point of the titration (where the indicator changed color): 7.31

Part 2 – CH3COOH versus NaOH titration Plot your titration data similar to what you did in part 1 above. As you did for HCl, determine the equivalence point and label it on the graph. In addition, mark and label the following: The end point (where the indicator changed color), the equivalence point volume on the horizontal axis, the half equivalence point volume on the horizontal axis, the half equivalence point on the titration curve, and the pH at the half equivalence point on the pH axis.

4 20.00

14.00

18.00 12.00 16.00

13.8

14.00

12.00

11.8

8.00

pH

10.00

6.00 8.00

6.00

5.76

4.00

4.00 2.00 2.00

0

0 0.00 0.00

0.00 5.00

10.00

15.00

20.00

25.00

30.00

35.00

40.00

45.00

50.00

Volume of NaOH added / mL

Figure 2. The titration of a weak acid, CH3COOH(aq) with a strong base, NaOH(aq) using phenolphthalein indicator is shown above on the blue curve. The first derivative of this relationship is shown on the red curve. The half equivalence point (pH=4.03), at which the concentration of weak acid is equal to the concentration of its conjugate base, the equivalence point (pH=8.2), the point at which all of the weak acid has been neutralized and converted to conjugate base, and the endpoint (pH=7.31), at which the color change occurred, are labeled. The maximum value of the first derivative, at which the rate of pH increase per volume added is the highest, is synonymous with the equivalence point volume. Using this and the experimental graph, the pH at the equivalence point can be estimated. The half equivalence point volume is equal to half of the volume of the equivalence point. Volume of acetic acid solution used: 10.00 mL Indicator used: Phenolphthalein [NaOH] used: 0.1000 mol L-1 Eq. point volume of NaOH: 12.40 mL (from the graph)

ΔpH/ΔV / mL-

10.00

5 Calculate [CH3COOH] in the stock bottle (show work below): .1240M [CH3COOH] = (Concentration of NaOH) x (Volume of NaOH at equivalence point) x (1mol CH3COOH /1molNaOH) / (Volume of CH3COOH stock solution used) = (0.1M) x (0.01240L) x (1molCH3COOH /1molNaOH) / (0.01000L) = 0.1240M = .1240M pH at the end point of titration (where the indicator changed color): pH at the equivalence point of the titration: 8.20 Equivalence point volume (of NaOH): 12.40 mL (same as above) Half-equivalence point volume : 6.20 mL pH at half-equivalence point volume: 4.03 Ka of acetic acid (show work to the right): Ka of acetic acid (show work to the right): 9.33 x 10-5 At the half equivalence point volume, p(Ka) = pH so p(Ka) = 4.03, 10-pH =10-4.03 = Ka = 9.3 x 10-5

6 Part 3 - Acetic acid – acetate buffer. Molarity of acetic acid solution used: 0.1240 mol L-1 Molarity of sodium acetate solution used: 0.10 mol L-1 Describe below how you prepared the buffer. Be brief and to the point. 30.00mL of 0.1240M CH3COOH(aq) was added to 30.00mL of 0.10M CH3COONa(aq) in 100mL beaker.

Table 1 – Measured pH values Solution tested Measured pH Original acetic acid 4.29 + acetate buffer 20.0 mL of the 4.29 buffer above 20.0 mL of the 4.29 buffer above 20.0 mL of the 4.29 buffer above 20.0 mL of acetic 2.82 acid 20.0 mL of acetic 2.68 acid 20.0 mL of 6.66 sodium acetate 20.0 mL of 6.88 sodium acetate

Reagent added None

pH after addition Not applicable

Calculated pH 3.94

2.00 mL 0.10 M HCl

4.06

3.75

2.00 mL 0.1000 M NaOH 20.0 mL water

4.48

4.09

4.31

3.94

2.00 mL 0.10 M HCl

2.11

Not needed

2.00 mL 0.1000 M NaOH 2.00 mL 0.10 M HCl

3.67

Not needed

5.06

Not needed

2.00 mL 0.1000 M NaOH

11.76

Not needed

Using the value of [CH 3COOH]stock, and the value of K a you calculated in part 2, calculate the expected pH, to two digits to the right of the decimal point, of the first four solutions listed above, and enter the values in the table. For each of the four situations, show the following: Moles of acetic acid to begin with, moles of acetate to begin with, moles of HCl or NaOH added as appropriate, moles of acetic acid and acetate after the addition if appropriate, and finally the pH. Show work. Refer to appropriate worked out examples in the manual or ask Dr. G if you need help.

7 Questions. Be sure to first list each question, followed by your answer to that question. Be brief and to the point. Brevity and relevance are more important than volume. Use separate paragraphs if and when appropriate.

Solution 1: Moles of CH3COOH to begin: ([CH3COOH]stock) x (Vol. CH3COOH in buffer / Total Vol. Buffer) = (0.1240M) x (0.03000L / 0.06000L) x .03000L = 1.860 x 10-3 moles Molarity of acetic acid = 1.86010-3mol/ 0.0600L = 0.03100mol/L Moles of CH3COONa to begin: ([CH3COONa]stock) x (Vol. CH3COONa in buffer) / (Total Vol. Buffer) = (0.1M) x (0.03000L / 0.06000L) x .03000L = 1.5 x 10-3 moles Molarity of sodium acetate = 1.5 x 10-3 / 0.06000L = 0.025mol/L pH = pKa because the initial concentration of both the acid and conjugate base is the same. Henderson Hesselbach equation: pH = pKa +log([base]/[acid]) = -log(K ) +log (0.025/0.03100) = -log( 9.3 x 10-5) +log (0.050/0.06200) = 3.94 a

Solution 2: [CH3COOH] to begin: ([CH3COOH]stock) x (Vol. CH3COOH in buffer) / (Total Vol. Buffer) x (Vol. buffer added) = (0.1240M x 0.03000L) / (0.06000L x 0.02000L) = 0.001240 moles Molarity of acetic acid = 1.240 x 10-3mol / 0.02200L = 0.05636mol/L Moles of CH3COONa to begin: ([CH3COONa]stock) x (Vol. CH3COONa in buffer) / (Total Vol. Buffer) x (Vol. buffer added) = (0.10M x 0.03000L) / (0.06000L x 0.02000L) = 0.0010moles Molarity of sodium acetate = 1.0 x 10-3mol / 0.02200L = 0.045mol/L Moles of HCl to begin: ([HCl] x Vol HCl added) = (0.1090M x 0.00200L) = 2.20x10-4moles Molarity of HCl = 2.20 x 10-4mol /0.02200L = 0.0100mol/L Since HCl is a strong acid, it dissociates completely, so the moles of HCl is the amount of H+ that react with the conjugate base.

Initial Molarity: Change: Final Molarity:

CH3COOH(aq) 0.05636mol/L +0.0100mol/L 0.06636 mol/L

↔

CH3COO-(aq) 0.045mol/L -0.0100mol/L 0.035 mol/L

pH = pKa+log[base/acid] pH = -log(9.3 x 10-5) + log(0.035/0.06636) = 3.75 Solution 3: Moles of CH3COOH to begin: ([CH3COOH]stock) x (Vol. CH3COOH in buffer) / (Total Vol. Buffer) x (Vol. buffer added) = (0.1240M x 0.03000L) / (0.06000L x 0.02000L) = 0.001240 moles Molarity of acetic acid = 1.240 x 10-3mol/0.02200L = 0.05636mol/L Moles of CH3COONa to begin: ([CH3COONa]stock) x (Vol. CH3COONa in buffer) / (Total Vol. Buffer) x (Vol. buffer added) = (0.10M x 0.03000L) / (0.06000L x 0.02000L) = 0.001moles Molarity of sodium acetate = 1.0 x 10-3mol/0.02200L = 0.045mo./L

8 Moles of NaOH to begin: ([NaOH]) x (Vol NaOH added) = (0.1000M x 0.00200L) = 2.00 x 10-4 moles Molarity of NaOH = 2.00 x 10-4mol / 0.02200L = 0.009091mol/L

Initial Molarity: Change: Final Molarity:

CH3COOH(aq) 0.05636 M -0.00909 M 0.0473 M

↔

CH3COO-(aq) 0.045 M +0.00909 M 0.054M

pH = pKa+log[base/acid] pH = -log(9.3 x 10-5) + log(0.054/0.0473) pH = 4.09

Solution 4: pH = pKa + log([base/acid]) Since the solution is already in water, the addition of 20.0mL of water is only doubling the total volume. However, the moles of acid and base are still going to remain the same. Moles of CH3COOH to begin: ([CH3COOH]stock) x (Vol. CH3COOH in buffer / Total Vol. Buffer) x (Volume of CH3COOH in solution) = (0.1240M) x (0.03000L / 0.06000L) x (0.02000L)= .00124 moles Molarity of acetic acid = 0.001240 mol / 0.04000L = 0.03100M Moles of CH3COONa to begin: ([CH3COONa]stock) x (Vol. CH3COONa in buffer / Total Vol. Buffer) x (Volume of CH 3COOHNa in solution) = (0.10M) x (0.03000L) / (0.06000L) x (0.02000L) = 0.0010 moles Molarity of sodium acetate = 0.0010 mol / 0.04000L = 0.025 mol / L pH = pKa + log ( [base]/[acid]) = -log(9.3 x 10-5) + log(0.025/0.03100) = 3.94

1. Is the pH at the equivalence point in the titration of a monoprotic weak acid with a strong base expected to be less than or equal to or greater than 7? Explain why. Be brief and to the point. The pH at the equivalence point in the titration of a monoprotic weak acid with a strong base should be greater than 7, because at the equivalence point, all of the weak acid has been neutralized and converted to its conjugate base, making the solution more basic. 2. You were asked to determine and annotate the half equivalence point in the acetic acid titration, but not in the HCl titration. Explain why and why not. Be brief and to the point. Since in a strong base – strong acid titration both substances dissociate completely, the pH of the solution being titrated (in this case HCl) at a certain point can simply be calculated by determining the number of moles of H + neutralized by the OH - in the NaOH and subtracting the calculated concentration from the original concentration of H + in the solution. On the other hand, in the acetic acid titration, only the strong base dissociates completely, and the weak acid is instead in equilibrium with its conjugate base. This merits the use of K, the equilibrium constant,

9 and a different method for determining pH: the Henderson-Hasselbach equation. The half equivalence point is useful in this case because it allows the determination of the K a value, since at the half equivalence point, the concentration of the weak acid is equal to the concentration of its conjugate base, so pH = pKa. After determining Ka for the solution, the pH at any point of the titration can be calculated. This method would not work for strong acid--strong base titrations since they react completely and do not provide an equilibrium. 3. In a succinct paragraph or two, summarize your results in part 3, focusing on buffer action. Of the three ‘solutions’ tested (the buffer, weak acid alone, and conjugate base alone), which one exhibited satisfactory buffer behavior experimentally? On what basis were you able to conclude so? Is this what was expected? Why or why not? Be brief and to the point. The use of the CH3COOH (aq) and CH3COO-(aq) buffer kept the pH relatively stable when strong acids and bases were added to it. The differences in pH after the addition of several different substances is much greater in the weak acid or conjugate base solutions in comparison to the buffer. This is due to the fact that a buffer, with both a weak acid and conjugate base, will neutralize both strong acids and strong bases that are added to the solution, using up H + and OHin the reaction rather than allowing them to change the pH of the solution. In doing so, the weak acid will react to form its conjugate base, shifting the equilibrium to one side, and the conjugate base will thus naturally react to form weak acid to return the reaction to equilibrium, as postulated by Le Chatelier’s principle. As a result, the pH will stay relatively stable .On the other hand, a weak acid or base solution alone is only able to neutralize either one of a strong base or strong acid; the low amount of the conjugate acid or base makes returning to equilibrium difficult to achieve once concentrated strong acid or base has been added. The solution that exhibited satisfactory buffer behavior was, unsurprisingly, the buffer solution that we prepared, which had an average shift of less than 0.3 pH points when strong acid or strong base was added. In comparison, the weak acid alone and conjugate base alone solution shifted approximately 1 or more pH points.

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