44-0134!00!01 rpt using buffers

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Using BuffersGino A. Romeo, Jr., Ph.D. Version 42-0134-00-01

L ab Rep o R t a s s i s t ant This document is not meant to be a substitute for a formal laboratory report. The Lab Report Assistant is simply a summary of the experiment’s questions, diagrams if needed, and data tables that should be addressed in a formal lab report. The intent is to facilitate students’ writing of lab reports by providing this information in an editable file which can be sent to an instructor.

o b s e R v a t i o ns

Data Table 1: Add 0.1M HCl

Drops pH Paper Color0123456789101112131415

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Data Table 2: Add 0.1M NaOH


Data Table 3: Add 6M HCl


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Data Table 4: Add 6M NaOH


Data Table 5: Add 0.1 M NaOH


Data Table 6: 0.1M NaO


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Que st ions A. What are the similarities and differences between the data in Tables 1 and 2? Explain how the data supports the role of a buffer.

B. What are the similarities and differences between the data in Tables 3 and 4? Explain how the data supports the role of a buffer.

C. Explain how the data from Tables 1 and 2 are similar and different from the data in Tables 3 and 4. What does this indicate about a buffers ability to maintain pH when an acid or a base is added?

D. What do the data in Tables 5 and 6 show about the ability of water to behave as a buffer? Considering the definition of a buffer, elaborate on your answer.

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E. In the procedures for Part C, “Adding Dilute Concentrations of Acid and Base to Distilled Water:” Do you think it would be useful to repeat procedures 2 and 3 but instead use concentrated NaOH and HCl? Why or why not?

Optional Question for Science Majors – or Extra Credit

A. One beaker contains 10.0 mL of acetic acid/sodium acetate buffer at maximum buffer capacity (equal concentrations of acetic acid and sodium acetate), and another contains 10.0 mL of pure water. Calculate the hydronium ion concentration and the pH after the addition of 0.25 mL of0.10 M HCl to each one. What accounts for the difference in the hydronium ion concentrations? Explain this based on equilibrium concepts; in other words, saying that “one solution is a buffer” is not sufficient.

Ka

of acetic acid = 1.8 x 10

Acetic acid/sodium acetate buffer

Acetic acid .1M

Sodium acetate .1M

C2H

3O

2+ + → HC

2H

3O

2Before After Reaction:

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H O+ + C H O - ⇄ HC H O +H O3 (aq) 2 3 2 (aq) 2 3 2(aq) 21) Convert 0.25 mL of 0.10M HCl → mols HCl

2) Convert 10 mL of 0.1M HC H O2 3 2→ mols HC H O2 3 2

3) Convert 10 mL of 0.1M C H O → mols C H O2 3 2 2 3 2

4) Stoichiometry CalculationsH

5) Convert 0.00975 mol C H O after reaction to concentration:2 3 2

HC2H

3O

2→ + + C

2H

3O

2

Initial: 0Change: +xEquilibrium x

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6) Convert 0.001025 mol HC H O after reaction to concentration:2 3 2

7) Concentration Calculations:H

3O

8) Solve for [H O+]

9) Solve for new pH:

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Alternatively, calculate pH using the following method:

The addition of the acid changed the pH by 0.02 units

Water: Add 0.25 mL of 0.1M HCl to 10 mL H2O1) Convert 0.25 mL of 0.10M HCl → mols HCl

2) Concentration of H O+ in 10.25 mL:3) Calculate new pH:

44-0134!00!01 rpt using buffers

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