Question: Using the provided materials, determine the buffer capacity, concentrations of the conjuga — Acid-Base Buffers and Titration Chemistry Question

Procedure should include:
1. Method for determining volume of a drop. This may be omitted if the student already has the volume memorized. Note: The student may elect to use the graduations on the side of the pipet. However, this is less desirable (and accurate) than what is described in the first two sentences above.
2. Volume of buffer used. Probably best to use a smaller volume (10mL) and titrate with 1.0M HCl or NaOH, then repeat with the other. If done dropwise, larger buffer volumes would take too long. Students may also do a survey trial to find an approximate volume of acid/base that destroys the buffer.
3. Adding a drop of designated reagent (NaOH or HCl) with stirring. The pH should be determined after each drop’s addition.
4. The addition of drops should be continued until the pH stabilizes at a higher or lower value depending on the reagent (NaOH or HCl) added.
5. A data table should be constructed that shows the pH after the addition of each drop.
6. If a student chooses to continue with the same buffer solution rather than a fresh sample: After concluding a titration, an equivalent volume of the complementary solution (NaOH if HCl was used in the titration or HCl if NaOH was used in the titration) to return to the starting point of the titration.
7. Repeat steps 2-5 above but for the complementary titrant (NaOH if HCl was used in the first titration or HCl if NaOH was used in the first titration).
8. Repeat steps 2-7 for multiple trials.

Data should include:
1. Volume (mL) of buffer used in each titration.
2. Data table with number of drops of titrant (NaOH or HCl) added with corresponding pH. Note: Students may elect to use the volume markings on their graduated pipets instead. This should not receive as much credit as it is less precise.
3. Sample Data - the values found below will be used in subsequent sample calculations and represents a titration where 10.0 mL of buffer is titrated. The values that each student obtains will undoubtedly vary slightly.
Number of Drops NaOH pH Number of Drops HCl pH
0 9 0 9
1 9 1 9
2 9 2 9
3 9 3 8.5
4 9 4 8
5 9 5 6.5
6 9 6 4
7 9.5 7 3
8 10 8 3
9 11 9 3
10 11.5
11 12
12 12
13 12

Plots will vary based on methods, but have some resemblance to the following: Multiple plots for multiple trials. Students may have marked equivalence and ½ equivalence points.
pH vs. Number of Drops of Titrant

a. Students should identify that buffer capacity is calculated by mol/L buffer divided by the change in pH (∆pH) in the buffering region (part of graph shown in #1 above before a dramatic change in pH).
b. Students should identify 4-10 drops of acid/base are added in the buffering region between pH values of ~10.0 and ~8.0.
c. Students should identify that approximately 4-10 drops of acid/base added corresponds to 0.0005 moles of acid/base or buffer being neutralized. This results in a concentration of 0.005 mol/L.
10 drops(0.05 mL/drop)(1 liter/1,000 mL)(1.00 mol/L)=0.0005 mol of acid/base or buffer
0.0005 mol / 0.0100 L = 0.05 mol/L
d. Students should recognize that the ∆pH = 1.0 from pKa in the buffering region on the graph. 8-10=1.0
e. Students should calculate a buffering capacity equal to ~0.005 (no units) when “c” above is divided by “d” above.
0.05 mol/L / 1.0 = 0.050

a. Students should identify the equivalence point for the NaOH titration (occurs at ~8.5 drops of base added to buffer).
b. Students should identify that 8.5 drops of base added corresponds to 0.0004 moles of NaOH being added or HA being neutralized.
8.5 drops(0.05 mL/drop)(1 liter/1,000 mL)(1.00 mol/L)=0.0004 mol of NaOH (or HA)
c. Students should calculate [HA]=0.040.
0.0004 mol / 0.0100 L = 0.040 mol/L
d. Students should identify the equivalence point for the HCl titration (occurs at ~5 drops of acid added to buffer).
e. Students should identify that 5 drops of acid added corresponds to 0.00025 moles of HCl being added or A^- being neutralized.
5 drops(0.05 mL/drop)(1 liter/1,000 mL)(1.00 mol/L)=0.00025 mol HCl (or A^-)
f. Students should calculate [A^-]=0.025.
0.00025 mol / 0.0100 L = 0.025 mol/L

a. Students may elect to use the Henderson/Hasselbach equation (see “b” below) or use an approach not utilizing this tool (see “c” below).
b. Use the initial pH (pH=9) from the data table, initial [HA] calculated in 3c above, the initial [A^-] calculated in 4c above and the Henderson^-Hasselbach equation to calculate the Ka. The value should be ~6 x 10^-10.
9 = -log Ka + log[0.025] / [0.04]
Ka = 6 x 10^-10
c. Determine the initial [H⁺] which should be 1 x 10^-9. Using this and the answers from 3c and 4c along with the acid ionization constant expression to calculate the Ka. The value should be ~6 x 10^-10.
10^-9 = [H⁺] = 1 x 10^-9
Ka = ([H⁺][A^-])/[HA] = ([1 x 10^-9][0.025])/[0.04]
Ka = 6 x 10^-10

Range of pH 8-10.
Around 4.0*10^-3 M
Around 2.5*10^-3 M
Around 6.0*10^-10