Lab 7 Report

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Feb 20, 2024

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BAGHRAMIAN 1 Name: X Partner: X Date:10/12/2023 Lab 7 Report – How Do Buffers Work? Prelab Questions: 1. a. What is the pH of a solution formed by combining 50.00 mL of a 0.0855 M acetic acid solution, 25.00 mL of 0.1061 M NaOH solution and 25.00 mL of deionized water? Show your work. pKa of Acetic Acid (AA): 4.756 0.0855 M AA × 50 mL = 4.275 mmol AA 0.1061 M NaoH × 25 mL = 2.6525 mmol NaOH 4.275 − 2.6525 = 1.6225 mmol HA in excess pH = 4.756 + log ( 2.6525 ???? 1.6225 ???? ) = ?. ?? b. What is the pH when the solution in 1a is diluted 1 mL in 10 mL total volume? Show your work. 1 mL × 1.6225 mmol 9 ?? = .18027 mmol HA 1 ?? × 2.6525 ???? 9 ?? = .29472 mmol NaOH pH = 4.756 + log ( . 29472 ???? . 18027 ???? ) = ?. ?? 2. What is the pH when 25.00 mL of the solution in #1a has 5.00 mL of a 0.1056 M HCl solution added to it? Show your work. 25 mL 1a 100 mL × 1.6225 mmol HA = 0.4056 mmol HA 25 mL 1a 100 mL × 2.6525 mmol NaOH = 0.6631 mmol NaOH 0.1056 M × 0.005L = 0.528 mmol HCl mmols of HA = 0.4056 mmol HA + 0.528 mmol HCl = 0.9336 mmols HA mmols of NaOH = 0.6631 mmol NaOH − 0.528 mmol HCl = 0.1351 mmol pH = 4.756 + log ( 0.1351 0.9336 ) = ?. ?? 3. What is the pH when 25.00 mL of the solution in #1a has 5.00 mL of a 0.1056 M solution of NaOH has been added? 0.1056 M × 0.005L = 0.528 mmol NaOH 25 mL 1a 100 mL × 1.6225 mmol HA = 0.4056 mmol HA 25 mL 1a 100 mL × 2.6525 mmol NaOH = 0.6631 mmol NaOH mmols HA = 0.4056 − 0.528 = 0 mols HA mmols NaOH = 0.6631 mmols + (0.528 − 0.4056) = 0.7855 mmol NaOH

BAGHRAMIAN 2 0.7855 mmol 30 mL = 0.02618 M NaOH pOH = − log(0.02618) = 1.58 pH = 14 − 1.58 = ??. ?? 4. What volume of a 0.1048 M NaOH is needed to neutralize (get to the equivalence point) 50.00 mL of a 0.0876 M acetic acid solution? (0.05 L HA)(0.0876 M) × 1 mol NaOH 1 mol HA × 1 0.1048 M NaOH = ??. ?? ?? What is the pH at the equivalence point? Show your work. K b = K w K a K a = 10 −4.756 = 1.75 × 10 −5 K b = (1 × 10 −14 ) 1.75 × 10 −5 = 5.70 ∗ 10 −10 K b = [OH − ][HA] [A − ] [A − ] = (0.05L)(0.0876 M) = 0.00438 mol 5.70 × 10 −10 = x 2 0.00438 − x 2.50 × 10 −12 = x 2 → x = 1.58 × 10 −6 mols [OH − ] = 1.58 × 10 −6 mols (0.05? + 0.04179) ? = 1.72 × 10 −5 pOH = − log(1.72 × 10 −5 ) = 4.76 → pH = 14 − 4.76 = ?. ?? 5. In the titration of 15.00 mL of a 0.1000 M acetic acid solution by a 0.1000 M NaOH solution, find the pH of the solution being titrated when a) 0.00 mL of NaOH has been added K a = [H + ][A − ] [HA] K a = 10 −4.756 = 1.75 × 10 −5 mols of HA = (0.015 L) × (0.1000M) = 0.0015 mols HA 1.75 × 10 −5 = x 2 0.0015−x → 2.63 × 10 −8 = x 2 → x = 1.62 × 10 −4 mols H + [H + ] = 1.62×10 −4 mols 0.015 L = 0.01081 M H + pH = − log(0.01081) = ?.?? b) 5.00 mL of NaOH has been added 0.1000 M HA × 0.015 L HA = 0.0015 mols HA 0.1000 M NaOH × 0.005 L NaOH = 0.0005 mols NaOH ???? 𝐻𝐴 ?? 𝑒??𝑖?𝑖??𝑖??: (0.0015 mols HA − 0.0005 mols NaOH) = 0.0010 mols HA pH = 4.756 + log ( 0.0005 0.0010 ) = ?. ?? c) 7.50 mL of NaOH has been added Halfway to equilibrium point, meaning pH will equal pKa 0 .0075 L NaOH × 0 .1000 M NaOH = 0 .00075 mols NaOH 0 .015 L HA × 0 .1000 M HA = 0 .0015 mols HA mols HA at equilibrium : ( 0 .0015 mols HA − 0 .00075 mols NaOH) = 0 .00075 mols HA

BAGHRAMIAN 3 pH = 4 .756 + log ( 0 .00075 0 .00075 ) = ? . ??? d) 14.90 mL of NaOH has been added 0 .0149 L NaOH × 0 .1000 M NaOH = 0 .00149 mols NaOH 0 .015 L HA × 0 .1000 M HA = 0 .0015 mols HA mols HA at equilibrium : ( 0 .0015 mols HA − 0 .00149 mols NaOH ) = 0 .00001 mols HA pH = 4 .756 + log ( 0 .00149 0 .00001 ) = ? . ?? e) 15.00 mL of NaOH has been added 0 .015 L NaOH × 0 .1000 M NaOH = 0 .0015 mols NaOH = NaA K b = (1 × 10 −14 ) 1.75 × 10 −5 = 5.70 ∗ 10 −10 K b = 5 .70 × 10 −10 5 .70 × 10 −10 = x 2 0 .0015 − x → 8 .55 × 10 −13 = x 2 → x = 9 .25 × 10 −7 mols [ O H − ] [ O H − ] = 9 .25 × 10 −7 0 .030 L = 3 .08 × 10 −5 M [ O H − ] pOH = − log ( 3 .08 × 10 −5 ) = 4 .51 pH = 14 − 4 .51 = ? . ?? f) 30.00 mL of NaOH has been added 0 .03 L NaOH × 0 .1000 M NaOH = 0 .003 mols NaOH 0 .015 L HA × 0 .1000 M HA = 0 .0015 mols HA moles OH-: 0 .003 mols NaOH − 0 .0015 mols HA = 0 .0015 mols excess O H − [ O H − ] = 0 .0015 mols O H − 0 .045 L = 0 .03333 M O H − pOH = − log ( 0 .03333 M O H − ) = 1 .48 → pH = 14 − 1 .48 = ?? . ?? Report: Write the molarity and standard deviation of each reagent used. M of NaOH: .109 M ± .008051 M of HCl: .0998 M ± .00042 M of Vinegar: .083 M ± .0002 (Used Hunter’s and Jarod’s; our M was far too low) Step 1 – Effect of Dilution on an Acidic Buffer Solution OUR READINGS ARE 100% WRONG! I AM NOT SURE WHAT HAPPENED. Solution Measured pH Calculated pH AB1 12.10 5.04 AB2 11.19 5.04 AB3 10.11 5.04 AB4 9.18 5.04 AB5 8.36 5.04

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BAGHRAMIAN 4 Sample Calculation for AB1: 0.083 M HA × 0.05 L HA = 0.00415 mols HA 0.109 M NaOH × 0.025 L NaOH = 0.002725 mols NaOH ???? 𝐻𝐴 ?? 𝑒??𝑖?𝑖??𝑖??: (0.00415 mols HA − 0.002725 mols NaOH) = 0.001425 mols HA pH = 4.756 + log ( 0.002725 0.001425 ) = ?. ?? Step 2 – Effect of Acid & Base Addition on Acidic Buffer Solution: Solution Measured pH Calculated pH AB1 + HCl 5.52 3.53 AB1 + NaOH 12.30 10.64 Calculation for AB1 – AB1 with HCl: 0.005 ? × .0998 ? 𝐻𝐶? = 4.99 × 10 −4 ???? 𝐻𝐶? 0.025 L × 0.00415 mols AA = 0.000104 AA 0.025 L × 0.002725 mols NaOH = 0.000068 mols NaA 0.000104 AA – 0.000068 NaA = 0.000036 mols 0.000104 𝐴𝐴 + 4.99 × 10 −4 ???? 𝐻𝐶? = 0.000603 ?𝐻 = 4.756 + log [0.000036] [0.000603] = ?. ?? Calculation for AB2: 0.005 ? × .109 ? ???𝐻 = 5.54 × 10 −4 ???? ???𝐻 − 0.000104 ??? 𝐴𝐴 = 4.41 × 10 −4 ???𝐻 𝑖? 𝐸𝑥?𝑒?? 0.025 L × 0.00415 mols AA = 0.000104 AA 0.025 L × 0.002725 mols NaOH = 0.000068 mols NaA pOH = -log [?𝐻 − ] → ??𝐻 = − log( 4.41 × 10 −4 ) = 3.36 → ?𝐻 = 14 − 3.36 = ??. ?? Step 3 – The Buffer Region:

BAGHRAMIAN 5 Amount of NaOH to neutralize Acetic Acid: 50 mL AA × 0.83 M AA × 1 mol NaOH 1 mol AA × 1 .109 M NaOH = ??. ?? ?? ?𝐚?𝐇 Total Volume of Solution X: 50.00 ml AA + 38.07 mL NaOH = 88.07 mL Total Volume of Solution Y : 50.00 mL AA + 38.07 DI Water = 88.07 mL Beaker Volume “X” Volume “Y” Measured pH Calculated pH A 1.00 24.00 3.49 3.38 B 5.00 20.00 4.27 4.15 C 10.00 15.00 4.85 4.58 D 12.50 12.50 5.03 4.756 E 15.00 10.00 5.32 4.93 F 20.00 5.00 4.52 5.36 G 24.00 1.00 X 6.14 Ran out of solution by the time we got to G, but this was okay. Sample Calculation of pH for Beaker A: Solution X: 0.083 ? × 0.05 ? = 0.00415 ??? → 0.00415 ?𝑜? 0.08807 ? = 0.0471 ? 𝑆????𝑖?? “?” Solution Y: 0.083 ? × 0.05 ? = 0.00415 ??? → 0.00415 ?𝑜? 0.08807 ? = 0.0471 ? 𝑆????𝑖?? “?” 𝐴 − = 0.0471 ? × 0.001? = 4.71 × 10 −4 ???? 𝐻𝐴: 0.0471 ? × 0.024? = 0.0011304 ???? ?𝐻 = 4.756 + log ( 4.71 × 10 −4 𝐴 − 0.0011304 𝐻𝐴 ) = ?. ?? Graph pH (y-axis) versus volume (x-axis) in each of the solutions A-G. Calculate the pH of each solution. 0 1 2 3 4 5 6 7 0.00 5.00 10.00 15.00 20.00 25.00 30.00 pH Volume of Solution X (mL) Volume of Solution X vs pH meaured pH calculated pH

BAGHRAMIAN 6 Step 4 – Titration Curve: (Used Hunter’s and Jarod’s Data) Suggested Volume of NaOH Actual Volume of NaOH Measured pH Calculated pH 0.00 0.00 2.85 2.12 3.00 2.48 3.62 3.92 6.00 3.41 3.84 4.09 9.00 8.48 4.35 4.65 11.00 9.01 4.38 4.69 14.00 9.60 4.43 4.75 16.00 9.81 4.47 4.77 17.00 10.02 4.48 5.76 17.50 10.52 4.52 5.80 18.00 11.28 4.59 5.87 18.50 12.99 4.77 6.03 19.00 16.95 5.36 6.57 19.50 19.00 6.53 7.41 20.00 19.12 6.58 6.60 20.50 19.20 7.09 6.75 21.00 19.32 8.06 7.18 21.50 19.33 9.42 7.25 22.00 19.40 9.92 9.26 22.50 19.45 10.16 10.14 23.50 19.49 10.44 10.37 24.00 20.11 11.01 11.23 24.50 20.52 11.19 11.42 25.00 21.10 11.35 11.55 25.50 22.19 11.53 11.80 26.00 26.10 11.82 12.15 26.50 28.19 11.91 12.25 0 1 2 3 4 5 6 7 0.00 5.00 10.00 15.00 20.00 25.00 30.00 pH Volume of Solution X (mL) Volume of Solution X vs pH meaured pH calculated pH

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BAGHRAMIAN 7 27.00 33.01 12.01 12.40 27.50 40.02 12.11 12.53 Calculate the pH of your new vinegar (it was diluted with the water) and continue to calculate the pH after each addition of NaOH from the buret. 0.025 ? × 0.083 ? 𝐴𝐴 = 0.002075 ??? 𝐴𝐴 → 0.002075 ??? 𝐴𝐴 0.050 ?? ????? ?????𝑒 = 0.0415 ? ?𝐻 = − log(0.0415) = ?. ?? Show a sample calculation for each area of this table where the pH calculation changes (Vb=0mL, Vb before equivalence, Vb at equivalence, Vb after equivalence) (Where b is for base). Expected Equivalence Point: 0.083 ? 𝐴𝐴 × 25.00 ?? 𝐴𝐴 × 1 ?𝑜? ?𝑎?𝐻 1 ?𝑜? 𝐴𝐴 × 1 ?? ?𝑎?𝐻 .1070 ? ?𝑎?𝐻 = 19.39 ?? Vb = 0 mL ?? = 10 −4.756 = 1.75 × 10 −5 1.75 × 10 −5 = 𝑥 2 0.002075 − 𝑥 → 3.5393 × 10 −8 = 𝑥 2 → 𝑥 = 1.9077 × 10 −4 ???? [𝐻 + ] = 1.9077 × 10 −4 0.025 ? = 0.00763 ? 𝐻 + ?𝐻 = − log(0.00763) = ?. ?? Vb before equivalence 3.41 ?? → 0.00341 ? × 0.1070 ? = 3.6487 × 10 −4 ??? ???𝐻 𝐻𝐴 = 0.002075 ???𝑒? − 3.6487 × 10 −4 ??? ???𝐻 = 0.00171 ???𝑒? ?𝐻 = 4.756 + log ( 3.6487×10 −4 0.00171 ) = ?. ?? Vb at equivalence They did not record the exact calculated equivalence point, but within 0.01 mL of it. NaOH = 0.01912 L × 0.1070M NaOH = 0.002046 moles NaOH HA = 0.002075 − 0.002046 = 2.9 × 10 −5 pH = 4.756 + log ( 0.00206 2.9×10 −5 ) = ?. ?? Vb after equivalence NaOH = 0.02819 L × 0.1070 M = 0.00353 mol NaOH NaOH = 0.00301633 − 0.002075 = 9.41 × 10 −4 mols NaOH [OH − ] = 9.41×10 −4 moles [OH − ] 0.05319 L total solution = 0.017697 M NaOH pOH = − log(0.02512) = 1.75 → pH = 14 − 1.60 = ??. ??

BAGHRAMIAN 8 Graph the pH (calculated and measured) versus volume titrated on the same graph. Find the pKa of acetic acid from your data (lab measurement). At ½ V e the pH = pKa. 0.01939 L × 1 2 = .009625 L NaOH added. When 9.63 mL of NaOH has been added, the pH would ideally be equal to the pKa of 4.756. NaOH = 0.00963 L × 0.1070M NaOH = 0.00103041 moles NaOH HA = 0.002075 − 0.00103041 = 0.00104459 pH = 4.756 + log ( 0.00103041 0.00104459 ) = ?. ???, ???𝐜? ?? ?𝐞?𝐲 𝐜???𝐞 ?? ??𝐞 ??𝐚. Identify the buffer region in your graph. 0 2 4 6 8 10 12 14 0 5 10 15 20 25 30 35 40 45 pH Volume of NaOH (mL) Volume of NaOH vs pH Measured pH Calculated pH 0 2 4 6 8 10 12 14 0 5 10 15 20 25 30 35 40 45 pH Volume of NaOH (mL) Volume of NaOH vs pH Measured pH Calculated pH

BAGHRAMIAN 9 Post-Lab Questions: 1. For each step how do the calculated values compare with the measured values? The calculated pH for measurements before the equivalence point were somewhat close to the measure values until about 10.02 mL, when the difference was over 1. After this point, the measured pH were 1 or more away from the calculated pH. After the equivalence point, at 19.40 mL, the difference becomes very minimal again, with .20-.30 difference. 2. What happens to the pH upon dilution of buffered solutions? Based on the Henderson-Hasselbach equation, it can be understood that diluting a solution would affect the volume of the entire solution—equivocally affecting the volume of acid and base—and therefore has little effect on the pH of the solution. 3. Explain the buffer region. A buffer region is a region of a titration curve where the pH of a solution changes very little & remains relatively stable. This curve’s buffer region was from 0.00 mL to 19.40 mL (the equivalence point). Once the mL has surpassed the maximum buffering capacity, the pH will rise drastically. On the titration curve, this looks like the vertical line suddenly shooting up and exponentially increasing to a flat point off to the right. Harris, D. C., & Lucy, C. A. (2020). Quantitative chemical analysis . Macmillan Learning. 4. Where is the maximum buffering capacity in a buffer region? The maximum buffering capacity in a buffer region is best defined as the amount of acid or base that can be introduced into system before the pH experiences a drastic change. This usually occurs when the molar ratio of acid & conjugate base in solution are equal. This is also when pKa = pH. Harris, D. C., & Lucy, C. A. (2020). Quantitative chemical analysis . Macmillan Learning. 5. What happens when small amounts of acid or base are added to a buffer solution? What would happen if these solutions were not buffered? What conclusions can you draw about buffer solutions? Adding a small amount of acid or base to the buffer solution would not drastically affect the pH. A weak acid & conjugate base solution will react with additions of acid or base. In a buffered solution, the acid and base additions would maintain a relatively constant pH. The buffer would donate its own hydrogen ions to maintain the pH. However, without a buffer, adding acid would decrease pH and adding base would increase pH. In conclusion, buffers help to resist pH changes in solutions where a constant, stable pH is needed. Our body uses buffers to maintain the pH within us for a variety of purposes, such as keeping enzymes active and fighting infection.

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