Data Sheet with Sample Results 1. Estimated equivalence point of HCl titration pH = 2. Estimated equivalence point of acetic acid titration PH = 3. Estimated pk. of acetic acid PK. = Vol NAOH (mL) pH 0.10 M HC1 Vol NaOH (mL) рн .10 M Ac Асid 0.00 1.21 0.00 2.88 2.00 1.23 2.00 3.38 4.00 1.31 4.00 3.81 6.00 1.44 6.00 4.24 8.00 1.59 8.00 4.51 10.00 1.82 10.00 4.77 12.00 2.03 12.00 5.01 14.00 2.34 14.00 5.30 16.00 2.59 16.00 5.66 18.00 2.89 18.00 6.02 20.00 10.50 20.00 9.04 22.00 11.20 22.00 11.40 24.00 11.50 24.00 11.80 Titration Curve of 0.10 M HCI with 0.10 M NaOH Titration Curve of 0.10 M Acetic Acid with O.10 M NaOH 200 200 Questions to consider when preparing the laboratory report: 1. Compare the estimated equivalence points to their theoretical values 2. Compare the estimated pKa of acetic acid to its theoretical value

Data Sheet with Sample Results 1. Estimated equivalence point of HCl titration pH = 2. Estimated equivalence point of acetic acid titration PH = 3. Estimated pk. of acetic acid PK. = Vol NAOH (mL) pH 0.10 M HC1 Vol NaOH (mL) рн .10 M Ac Асid 0.00 1.21 0.00 2.88 2.00 1.23 2.00 3.38 4.00 1.31 4.00 3.81 6.00 1.44 6.00 4.24 8.00 1.59 8.00 4.51 10.00 1.82 10.00 4.77 12.00 2.03 12.00 5.01 14.00 2.34 14.00 5.30 16.00 2.59 16.00 5.66 18.00 2.89 18.00 6.02 20.00 10.50 20.00 9.04 22.00 11.20 22.00 11.40 24.00 11.50 24.00 11.80 Titration Curve of 0.10 M HCI with 0.10 M NaOH Titration Curve of 0.10 M Acetic Acid with O.10 M NaOH 200 200 Questions to consider when preparing the laboratory report: 1. Compare the estimated equivalence points to their theoretical values 2. Compare the estimated pKa of acetic acid to its theoretical value

Chemistry: The Molecular Science

5th Edition

ISBN:9781285199047

Author:John W. Moore, Conrad L. Stanitski

Publisher:John W. Moore, Conrad L. Stanitski

Chapter15: Additional Aqueous Equilibria

Section: Chapter Questions

Problem 34QRT

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Fifty cm3 of 1.000 M nitrous acid is titrated with 0.850 M NaOH. What is the pH of the solution (a) before any NaOH is added? (b) at half-neutralization? (c) at the equivalence point? (d) when 0.10 mL less than the volume of NaOH to reach the equivalence point is added? (e) when 0.10 mL more than the volume of NaOH to reach the equivalence point is added? (f) Use your data to construct a plot similar to that shown in Figure 14.10 (pH versus volume NaOH added).
n 0 The following pH curve is for the titration of 20.0 mL of a 0.10 M solution of a weak monoprotic acid with a 0.10 M solution of a strong monoprotic base. Data points at 1 mL increments. 14.0 12.0 10.0 Hd 8.0 6.0 4.0 2.0 0.0 0.0 10.0 20.0 ession# 188... 30.0 Volume of titrant (mL) 40.0 50.0
A The graph shows the titration curves of a strong acid and a weak acid. Estimate the pk, of the weak acid from its titration curve V 14 12 10 pH 12086420 Volume of strong base (ml) 1st attempt Write the estimated pK, to one decimal place 1 OF 8 QUESTIONS COMPLETED 82 W S # 3 Z X E 80 FI D $ 4 C R F K5 V 4 FS T G A 6 B S Y 27 & H 8 U N 8 00 DHI 14 M 9 K TV O hi See Periodic Table O L SUBMIT ANSWER P ** FIL
2. Below is a titration curve generated by a similar titration to the ones simulated, Vinegar Titration Curve 14.00 13.00 12.00 11.00 10.00 9.00 8.00 * 7.00 6.00 5.00 4.00 3.00 2.00 1.00 0.00 0.00 5.00 10.00 15.00 20.00 25.00 30.00 35.00 Volume NaOH (mL) We can use this curve to estimate the pka of our weak acid by what is called the "half equivalence point method". The equivalence point is the middle of the steepest part of this graph. Estimate the volume of NaOH added to the titration at that point by using a rule or other straight edge to follow that point down to the x-axis. Divide that volume by 2 (half the equivalence point) and estimate the pH at that volume of NaOH using your straight edge again. pKa = pH at the half equivalence point. What do you estimate the pka to be? 3. The pka of acetic acid is 4.76. Calculate percent error between the true value and your estimated value from question 2.
Determine the pH at the point in the titration of 40.0 mL of 0.200 M HNNH2 with 0.100 M HNO: after 10.0 mL of the strong acid has been added. The value of Kb for H2NNH2 is 3.0 x 10-6. PREV 1 2 4 NEXT > Based on your ICE table and definition of Kb, set up the expression for Kb in order to determine the unknown. Do not combine or simplify terms. = 3.0 x 10-6 RESET [0] [0.200] [0.0200] [0.100] [0.140] [0.175] [x] [2x] [0.200 + x] [0.200 - x] [0.0200 + x] [0.0200 - x] [0.100 + x] [0.100 - x] [0.140 + x] [0.140 - x] [0.175 + x] [0.175 - x]
In the titration of 50.00 mL 0.0200M weak base B using 0.0150 M HCI, which indicator is the best one to be used? (pK, of the weak base B is 10.50). Indicator 4: Thymolthalein: pH transition = 8.5-10.5 OIndicator 5: Indigo Carmine: pH transition= 11.6- 14.0 Indicator 3: Cresol red: pH transition 7.2-8.8 Indicator 1: Cresol purple: pH transition = 1.2-2.8 Indicator 2: Methyl red: pH transition = 4.8-6.0
1) The equivalence points of the two titrations curves were not in the same pH range explain why. 2) Using the experimental data at 1/2 the equivalence point volume, calculate the Ka of acetic acid. Volume of NaOH at Equivalence point for HCl Trial = 5.083ml Equivalence point pH for HCl Trial = 6.57 pH 1/2 Equivalence point for HCl Trial = 3.285 Volume of NaOH at Equivalence point for HC2H3O2 = 4.025ml Equivalence point pH for HC2H3O2 Trial = 4.13 pH 1/2 Equivalence point for HC2H3O2 Trial = 2.065
Determine the pH at the point in the titration of 40.0 mL of 0.200 M H2NNH2 with 0.100 M HNO3 after 80.0 mL of the strong acid has been added. The value of Kb for H2NNH, is 3.0 x 10-6. ( PREV 1 2 3 4 NEXT > Based on the result of the acid-base reaction, set up the ICE table in order to determine the unknown. H2NNH3*(aq) + H20(1) - H;O*(aq) + H,NNH2(aq) Initial (M) Change (M) Equilibriu m (M). 5 RESET 0.100 0.200 0.0333 0.0667 +x -x 0.100 + x 0.100 - x 0.200 + x 0.200 - x 0.0333 + x 0.0333 - x 0.0667 + x 0.0667 - x
[Review Topics] Calculate the pH at the halfway point and at the equivalence point for each of the following titrations. a 100.0 mL of 0.30M HC,H,O2 (K. = 6.4 x 10) titrated by 0.10 M NaOH %3D pH at the halfway point= %3D pH at the equivalence point b. 100.0 ml. of 0.30 M C2H; NH, (K, = 5.6 x 10) titrated by 0.50 M HNO, %3D pH at the halfway point = pH at the equivalence point c. 100.0 mL of 0.80 M HCl titrated by 0.25 M NaOH pH at the halfway point= pH at the equivalence point %3D
In the titration of 50.00 mL 0.0200M weak base B using 0.0150 M HCI, which indicator is the best one to be used? (pK, of the weak base B is 10.50). Indicator 2: Methyl red: pH transition = 4.8-6.0 Indicator 1: Cresol purple: pH transition = 1.2-2.8 Indicator 4: Thymolthalein: pH transition = 8.5-10.5 O Indicator 5: Indigo Carmine: pH transition= 11.6- 14.0 O Indicator 3: Cresol red: pH transition = 7.2-8.8
5. A 50.00 mL 0.1000 M HNO2 (Ka = 7.2 x 10-+) is being titrated with 0.1000 M NaOH. Calculate the pH of the solution before any amount of titrant is added. (Hint: Construct an ICE table) а. b. Calculate the volume of NAQH required to reach the equivalence point. Calculate the pH after the addition of 25.00 mL of NaOH. с. d. Calculate the pH at equivalence point. (Hint: Determine first the concentration of the conjugate base that was produced. You will construct an ICE table for the conjugate base.) e. Calculate the pH after adding 75.00 mL of NAQH. f. Using your answers from ato f, roughly generate a titration curve.
e The following pH curve is for the titration of 20.0 mL of a 0.10 M solution of a weak monoprotic acid with a 0.10 M solution of a strong monoprotic base. entation ok F 14.01 12.0 10.0 рн Data points at 1 mL increments. 8.0 6.0 4.0 2.0 44444 0.0 0.0 (a) What is the pH before any titrant has been added? 10.0 1444444 20.0 30.0 Volume of titrant (mL) 40.0 444444 50.0 CO G +