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

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:

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