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1412

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Chemistry

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Feb 20, 2024

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LAB REPORT Professor Nguyen Thanh Tin Lam Chem 1412 20 November 2022 GENERAL CHEMISTRY II Experiment #7: Acid – Base Titrations & Titration Curves Objective: To demonstrate the titration of acid with a base in different combinations (strong and weak) To prepare titration curves for four different combinations of acid-base titration. I. Introduction: From the lecture series, a titration curve is prepared by plotting the pH of the titration mixture versus the volume of the titrant added. When the points are traced properly, the resulting curve will look like the letter “S” or the mirror image. The equivalence point corresponds to the midpoint of the steep portion of the curve. This can be graphically located by drawing a vertical line dividing the curve into equal parts. The pH that corresponds to this point (where the vertical line and the curve intersect) is the pH at equivalence point. II. Procedure: Part C: Titration of NH 4 OH with CH 3 COOH 1. Calibrate the pH meter by placing the electrode into a buffer solution (of known pH) and adjusting the knob to the pH of the given buffer. Then lift the electrodes and rinse with DI water before using in the succeeding steps. 2. Obtain a clean buret and rinse with DI water. Then clamp the buret to a ring stand. 3. Obtain a stock solution of standard NH 4 OH solution (approximately 100 mL placed in an Erlenmeyer flask). Using a beaker, fill the buret with the NH 4 OH solution, just above the calibration mark. Drain a small amount into a beaker until the meniscus is at a point within the zero mark (or any chosen calibration mark for initial reading. 4. Obtain 15 mL of approximately 0.10 M CH 3 COOH solution in a clean, dry200-mL beaker. 5. Position the beaker with the acid on top of a magnetic agitator so that the pH electrode is immersed in the solution. Note the initial pH of the acid. THANH TIN LAM 1
LAB REPORT 6. Then, place the tip of the buret with the base titrant just inside the beaker, with the tip below the rim. 7. Titrate the acid by adding the base from the buret in 2-4-mL increments, stirring continuously and, noting the pH after each addition. (Note: When a weak base is involved, use your discretion on the volume of titrant to be added… sometimes 6-8 mL increments may be necessary) 8. When successive pH readings begin to increase rapidly, add the titrant in smaller 9. increments (1 mL or in drops) until successive pH readings are increasing only slightly. Gradually increase the amount of base added (not to exceed 2 mL at a time), until the pH reaches about pH = 10 (or over), or when the letter “S” is formed by the points. 10. Prepare a titration curve by plotting the volume of base added as the abscissa and the corresponding pH as the ordinate. Determine the equivalence point for each titration. Data & Result NH4OH with CH 3 COOH (Identify specific samples used) Trial mL Base pH Trial mL Base pH Trial mL Base pH 1 0.00 2.76 12 16.50 4.62 23 33.00 8.17 2 1.50 3.22 13 18.00 4.70 24 34.50 8.45 3 3.00 3.55 14 19.50 4.78 25 36.00 8.63 4 4.50 3.77 15 21.00 4.85 26 37.50 8.76 5 6.00 3.91 16 22.50 4.95 27 39.00 8.86 6 7.50 4.03 17 24.00 5.05 28 40.50 8.94 7 9.00 4.16 18 25.50 5.18 29 42.00 9.00 8 10.50 4.26 19 27.00 5.34 30 43.50 9.06 9 12.00 4.34 20 28.50 5.56 31 45.00 9.12 10 13.50 4.45 21 30.00 5.94 32 46.50 9.16 11 15.00 4.51 22 31.50 7.45 33 48.00 9.2 THANH TIN LAM 2
LAB REPORT Calculation based on theory B (aq) + CH 3 COOH (aq) HB (aq) + CH 3 COO - In order to reach the equivalence point the moles of CH 3 COOH added must equal the moles of the weak base NH 4 OH and since it is a 1:1 ratio. 0.1 moleCH 3 COOH L x 0.015 L 1 = 0.0015 moleCH 3 COOH Mole of CH 3 COOH = mole of CH 4 OH = 0.0015 mole The initial data we have [CH 4 OH] = 0.1 M 0.0015 moleCH 4 OH 1 x 1 L 0.1 mole = 0.0015 LCH 4 OH = 15 mL CH 4 OH Volume total = 15 mL + 15 mL = 30 mL = 0.03 L THANH TIN LAM 3
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LAB REPORT [ HB ] = 0.0015 mole HB formed 0.03 L = 0.05 M CH 4 OH : K b = 1.8 x 10 -5 Ka = Kw Kb = 1.00 x 10 14 1.8 x 10 5 = 5.55 x 10 10 HB (aq) + H 2 O B- (aq) + H 3 O + HB H 2 O B H 3 O + Initial 0.05 - 0.00 0.00 Change -x - +x +x Expression 0.05- x - x x 5.55 x 10 10 = x. x 0.05 x x 2 = 5.55 x 10 10 .0.05 x = 5.2678 x 10 6 pH = - log (5.2678x10 -6 ) = 5.278 The equivalence point between strong acid and weak base will be less than 7. III. Discussion In the above experiment, setting up the meter is an important job that needs to be done carefully. Because the meter plays almost the most important role in terms of data in performing the experiment. When the result is complete, the curve is not as expected. Therefore, if we want the curve to become smoother, we have to subdivide the amount of solution added. There is quite a clear difference between the calculated results and the experimental results. At the moment, I still can't explain clearly where this error comes from. The experiment needs to be repeated in order to compare the results. IV. Conclusion Plots of acid–base titrations generate titration curves that can be used to calculate the pH, the pOH, the pKa , and the pKb of the system. The shape of a titration curve, a plot of pH versus the amount of acid or base added, provides important information about what is occurring in solution during a titration. The shapes of titration curves for weak acids and bases depend dramatically on the identity of the compound. THANH TIN LAM 4