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Experiment 6: Acid – Base Titration Required reading: Ebbing, 11th Edition Chapters 15 and 16.  CH 15: Arrhenius definition for acids and bases.  CH 15: Brønsted-Lowry definition for acids and...

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Experiment 6: Acid – Base Titration
Required reading: E
ing, 11th Edition Chapters 15 and 16.
 CH 15: A
henius definition for acids and bases.
 CH 15: Brønsted-Lowry definition for acids and bases.
 CH 15: Autoionization of water and pH.
 CH 16: Solutions of weak acids or bases.
 CH 16: pH of salt solutions.
 CH 16: Buffer solutions.
 CH 4: See Figure 4.22 (section 4.10) in textbook for quick overview of a titration as a lab technique.
 CH 16: Acid – Base titration curves.
Learning Goals:
 To understand the concepts of pH and pKa.
 To create a titration curve
 To understand buffering action.
Background information and theory :Titration is a technique where a solution of known concentration (titrant) is carefully added to a solution of unknown concentration (analyte) until an endpoint is detected, which can be used to determine the analyte concentration. The endpoint can be any observable change in properties – pH, color, formation of a precipitate, conductivity, etc. – that indicates a very slight excess of titrant. With proper technique, the endpoint can be a very accurate approximation of the equivalence point, where the moles of titrant and analyte match the stoichiometry of the reaction. In an acid-base titration, pH can be monitored to determine the endpoint, as there is usually a sudden change in pH as the smallest excess of acid or base is added. While pH meters and pH indicators (compounds that change color as pH changes) can be used to observe the endpoint, pH meters offer the advantage of producing a titration curve when pH is plotted against the volume of titrant. Titration curves provide a more reliable measurement of the equivalence point and additional information about the analyte including the relative strength of the acid or base.
CH3CO2H (aq) + OH- (aq) ⇄ CH3CO2-(aq) + H2O (l)
Figure 1 Titration curve of an aqueous acetic acid solution with 0.1 M NaOH.
Consider the titration curve shown in Figure 1. An aqueous solution containing acetic acid (CH3CO2H) and a small amount of phenolphthalein starts off colorless. Initially there is a large excess of acetic acid in the flask which reacts with hydroxide to form the acetate ion (CH3CO2-). Since acetic acid and its conjugate base are both weak, the resulting mixture is a
uffer and the pH increases slowly. At the equivalence point, pH increases dramatically as acetic acid and hydroxide have reacted completely and the buffer capacity is exceeded. On any titration curve, the equivalence point will be the inflection point or halfway up on the S-shaped curve. Note the solution is slightly basic at the equivalence point because the acetate ion (CH3CO2-) is still present and is a weak base. If a strong acid such as HCl was titrated with NaOH, the solution would be neutral at the equivalence point as the product, NaCl, is a neutral salt. In either case, the slightest excess of titrant would make the solution basic enough to turn phenolphthalein pink, which starts around pH 8.5.The goals of this experiment are to determine the molar mass and pKa of an unknown weakacid. For the titration in Figure 1, we know the identity and formula of the acid, but let’s pretend we don’t. There are two additional pieces of information needed to find the molar mass of the acid, which you will have in your experiment: 1) the volume of the acid solutionand 2) the concentration in g acid/L solution For the sake of an example, assume a 25-mL sample of acetic acid was titrated and its concentration was 3.6 g/L. At the equivalence point,
moles base = moles acid
MbVb = MaVa
Based on Figure 1, the 25 mL of acetic acid solution required 15 mL of 0.10 M NaOH to reach the equivalence point. Solving for the unknown molarity gives a 0.060 M acetic acid solution
Ma =
(0.10M )(15mL)
(25mL) = 0.060 M CH3CO2H.Since each liter of solution contains 3.6 g CH3CO2H and 0.060 moles CH3CO2H, we can take the ratio to find the molar mass.
3.6 gC H 3CO 2H
0.060molesC H 3CO2H
=60. g/mol
The pKa of a weak acid is easy to determine from the titration curve if you know where to look. As the solution is a buffer, we can use the Henderson-Hasselbalch equation to determine its pH. At half the equivalence volume, half of the acid (HA) has been converted to its conjugate (A-), so [HA] = [A-].
pH = pKa + log¿¿
pH = pKa + log(1)
pH = pKaSo, one way of determining the pKa of a weak acid is to measure pH at half the equivalence point volume. Considering the titration curve in Figure 1, the half volume is 15 mL/2 = 7.5 mL. The pH of the solution at this point is just under 5, close to the actual pKa of 4.76. Figure2 shows titration curves for acids with a variety of pKa values for comparison. Note that the pH at the equivalence point is high enough in two of these curves that phenolphthalein would turn pink BEFORE the equivalence point.
Figure 2 Comparison of titration curves for weak acids with varying strength.
A second method to determine pKa is by measuring pH for a known concentration of the acid. Since molarity of the acid is determined during the titration, the pH of the solution at 0mL of titrant can be used to calculate Ka based on the equili
ium below:
CH3CO2H (aq) + H2O (l) ⇌ CH3CO2-(aq) + H3O+ (aq)
Ka = ¿¿For the titration in Figure 1, [CH3CO2H] = 0.060 M and pH = 2.97. Therefore, the equili
iumvalues are
[H3O+] = [CH3CO2-] = 10–2.97 = 0.00102M
[CH3CO2H] = (0.060 – XXXXXXXXXXM = 0.059 MPlugging into the Ka expression gives:
Ka = ¿¿ = ¿¿ = 1.76×10−5
pKa = – log ( 1.76×10−5) = 4.75
Procedure:
1. If this is your first time using a virtual web on the ChemCollective site, watch an introductory video here: http:
www.chemcollective.org/chem/common/vlab_walkthrouh_html5.php
2. Click on the following link to open the virtual lab for this experiment: http:
chemcollective.org/activities/autograded/ XXXXXXXXXXUsing a 100-mL volumetric flask, prepare an aqueous solution containing 15 g/L of the unknown solid acid. In another 100-mL volumetric flask, dilute the stock solution of 1.00M NaOH to 0.100M.4. Add 25.00 mL of the unknown acid solution and 0.5 mL of phenolphthalein indicatorto a 250-mL Erlenmeyer flask. Click on the flask and record the initial pH of the solution.5. Open the “Other” tab under glassware and add a 50-mL burette to the workbench. Fill the burette with 0.100M NaOH and begin adding it to the Erlenmeyer flask 1 mL at a time. 6. After each addition, record the total volume of titrant added and the pH of the unknown solution (2 decimal places each). After the first few additions, the pH should change slowly until close to the endpoint. If the pH changes by > 0.3, slow thetitration down by adding titrant in 0.10-mL increments.7. The endpoint is indicated by a change in color to light pink and a sudden increase in pH. Continue adding titrant and recording the volume and pH for a few mL after the endpoint. 8. Add a fresh 25-mL sample of unknown solution to an Erlenmeyer flask. Add exactly half the volume required to reach the endpoint and record the resulting pH.
http:
chemcollective.org/activities/autograded/143
http:
www.chemcollective.org/chem/common/vlab_walkthrouh_html5.php
Name: ____________________________________
Acid – Base Titration
Data Sheet:
Table 1. General titration data.
Volume of unknown acid solution, mL
Initial volume in burette, mL
Final volume in burette (endpoint), mL
Net volume of titrant, mL
pH halfway to endpoint
Table 2. Titration curve data.
VNaOH
(mL)
pH
VNaOH
(mL)
pH
VNaOH
(mL)
pH
VNaOH
(mL)
pH
Data Analysis:
1. Determine the molarity of the unknown solution and molar mass unknown compound. Show your work.
Table 3. Molar mass results.mmol NaOH at endpointmmol acidMolarity of acid solution, M
Molar mass of unknown, g/mol
Calculations:
2. Determine the Ka and pKa of the unknown acid based on the pH of the solution prior to adding titrant. Show your work.
Table 4. pKa results.
Molarity of acid solution, MpH at 0 mL titrant[H3O+]eq[HA]eq[A-]eqKa
pKa
Calculations:
Discussion:1. What is the pKa of your weak acid based on the pH of the half-neutralized solution? Howdoes this value compare to the one calculated above?
2. If the endpoint of your titration were 0.10 mL past the equivalence point, what would the percent e
or be for the volume measurement? Would the percent e
or for the molar mass be the same or different? Explain and/or show calculations to support your answer.
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    Experiment 6: Acid – Base Titration
    Background information and theory:
    Procedure:
    Data Sheet:
    Feedback Block
Answered Same Day Jul 30, 2021

Solution

Varun answered on Jul 31 2021
152 Votes
Volume of NaOH
    pH
    1
    2.38
    2
    2.52
    3
    2.65
    4
    2.76
    5
    2.86
    6
    2.95
    7
    3.03
    8
    3.11
    9
    3.19
    10
    3.26
    11
    3.33
    12
    3.4
    13
    3.46
    14
    3.53
    15
    3.61
    16
    3.68
    17
    3.776
    18
    3.84
    19
    3.94
    20
    4.04
    21
    4.17
    22
    4.32
    23
    4.54
    24
    4.93
    25
    10.71
Experiment 6: Acid – Base Titration
Table 1. General titration data.
    Volume of unknown acid solution, mL
    25 mL
    Initial volume in burette, mL
    50 mL
    Final volume in burette (endpoint), mL
    25 mL
    Net volume of titrant, mL
    25 mL
    pH halfway to endpoint
    3.5
Calculations:
1. mmol NaOH at endpoint
Solution:
100 mmol
moles base = moles acid
MbVb = MaVa
Mb = (0.1 M)(25mL)/(25mL)
= 0.1 moles
2. mmol acid
Solution:
100 mmol
moles base = moles acid
MbVb = MaVa
Ma = (0.1 M) (25mL)/(25mL)
= 0.1 moles
3. Molarity of acid solution,...
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