Experiment 6 Determining Ka for pH Indicator

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Experiment 6 Determining Ka for pH Indicator Total Points: 67 Student’s Name: Adam Miller Lab Section: 804 NOTE: You must show your work for all calculations; no work, no credit. Data Table 1 (5 pts) Buret Reading (mL) pH Buret Reading (mL) pH 0 4.69 24 10.43 2 4.96 26 10.47 4 5.24 28 10.52 6 5.66 30 10.57 8 6.53 10 9.35 10.5 9.60 11 9.71 11.5 9.78 12 9.85 12.5 9.87 13 9.91 13.5 9.97 14 10.00 16 10.14 18 10.24 20 10.31 22 10.39

Experiment 6 Determining Ka for pH Indicator Total Points: 67 Data Table 2 (10 pts) pH Absorbance Determined K a Explanation if K a NOT Included in Average 7.28 .594 ------------------------ 6.38 .590 5.21E-6 5.49 .200 1.80E-6 4.62 .076 3.52E-6 4.09 .038 5.56E-6 3.83 .034 8.98E-6 3.49 .031 1.78E-5 3.20 .021 2.31E-5 2.86 .019 3.58E-5 Average K a (3 pts) → 1.27E-5 Show Q test calculation if any test performed and include a justification for why you used the Q test of why you did not use the Q test (5 pts): Q calculated = gap/range 3.58E-5 – 2.31E-5 = gap = 1.27E-5 3.58E-5 – 1.80E-6 = range = 3.40E-5 Q calculated = .374 < .468 (Q table ) No outliers

Experiment 6 Determining Ka for pH Indicator Total Points: 67 Section Results Table 1 – Titration Curve Method In the table below, enter at K a values determined by the titration method from all groups in your lab section. Calculate the overall average K a . Determined K a – Titration Method (5 pts) Explanation If Determined K a Excluded from Average 5.75E-6 7.94E-6 6.31E-6 5.13E-6 Average K a – Titration Method (3 pts) 6.28E-6 Show work if Q test performed and include a justification for why you used the Q test or why you did not use the Q test (5 pts) (5.75e-6 - 5.13e-6) / (7.94e-6 - 5.13e-6) = .22 Q value is less than .765, so all the values must be used in the average.

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Experiment 6 Determining Ka for pH Indicator Total Points: 67 Section Results Table 2 – Spectrophotometric Method In the table below, enter all average K a values determined by the spectrophotometric method from all groups in your lab section. Calculate the overall average K a . Determined Average Ka – Spectrophotometric Method (5 pts) 1.27E-5 1.27E-5 1.18E-5 1.18E-5 Average of Average K a – Spectrophotometric Method (3 pts) 1.23E-5

Experiment 6 Determining Ka for pH Indicator Total Points: 67 Post-lab Questions NOTE: You must show your work for all calculations; no work, no credit. 1. (5 pts) A commonly accepted value for the K a for bromocresol green is 2.0 × 10 -5 . Based on this value, which of the two methods is more accurate? Briefly justify your answer. Titration (6.28E-6 – 2.0E-5) / 2.0E-5 = .68 = 68% error Spectrophotometric (1.23E-5 – 2.0E-5) / 2.0E-5 = .38 = 38% error Spectrophotometric was more accurate because it had a lower percent error compared to titration. 2. (5 pts) Looking at the variation in the determined values for the two methods, which method is more precise? Briefly justify your answer. Spectrophotometric = 5.2E-7 Titration = 1.2E-6 Spectrophotometric was also more precise because it had a lower standard deviation compared to titration.

Experiment 6 Determining Ka for pH Indicator Total Points: 67 3. (5 pts) The spectrophotometric analysis done in this activity is measuring the absorbance of the base form (blue form) of the indicator. The blue form absorbs light best at 615 nm light; thus, the K a value was calculated used the equation K a = ( 10 − pH )( A ) A ref − A for absorbance values measured at 615 nm. The derivation for why this equation is equation to the K a is included in your Background Section – Spectrophotometric Method. Starting from the equilibrium constant expression for a weak acid K a = ¿¿ , derive the equation that would have to be used to determine the K a value based on pH and absorbance measures if 440 nm light were used. Measuring the absorbance at 440 nm would be measuring the absorbance of the acid form (yellow form) of the indicator (assume any absorbance of the blue form at 440 nm can be ignored). NOTE: “Derive” means show every step in the mathematical derivation of the equation. See pages 78-80 for the derivation of the equation when 615 nm light is used. Ka = [H 3 O + ] [A - ] / [HA] pH = -log[H 3 O + ] à [H 3 O + ] = 10 -pH A ref =Eb[HA] total (at pH <=7) [HA] total = A ref / Eb A = Eb[HA] remaining (at pH > 7.0) [HA] remaining = A/Eb [A - ] = [HA] total – [HA] remaining [A - ] / [HA] = [HA] total – [HA] remaining / [HA remaining] ( A ref /Eb-A/Eb)/(A/Eb) = (1/Eb(A ref – A))/1/Eb(A) = A ref – A/A Ka = 10 -pH (A ref – A)/A

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