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# For my Chem lab on buffers, I was wondering if the Ka and Kb value would be different for each person who

conducted the experiment differently. And if so, how would you calculate those values?

NAME: Frances Kato QUIZ SECTION: AB STUDENT ID: 0851379 LAB PARTNER: Tim Iwuoha Total Points = 60 pts (10 notebook, 50 template) Deprotonation of Acetic Acid 3 pt rxn, 2 pt for Ka, 2 pt for Kb (correct & entered) Provide the acid dissociation reaction for proton removal in water: DATA AND CALCULATIONS Parts A - D: Preparation of Acetic Acid/Acetate Solutions 5 pts: ALL solution prep data and expected initial pH values entered (M) 0.10 Molar Mass Mass of Solid Moles of Solid Volume of Water *Molarity (g/mol) Used (g) (mol) (mL) (M) 136.08 1.36 0.01 100.00 0.10 *(from calculation on pg 2) Solution ID mL Water Total Volume (mL) #1 20.00 0.00 30.00 50.00 2.87 *(based on example #2 15.00 5.00 30.00 50.00 4.27 calculations on pg 2) #3 10.00 10.00 30.00 50.00 4.74 #4 5.00 15.00 30.00 50.00 5.22 #5 0.00 20.00 30.00 50.00 8.87 DI Water -- -- 50.00 50.00 5.50 Part E: Preparation of Aspirin and Buffered Aspirin Solutions (g) (g) Aspirin 0.213 0.207 larity (M) of HCl for acid test: 0.500 Molarity (M) of NaOH for base test: 0.500 Buffered Aspirin 0.430 0.434 Volume (mL) of HCl added: 0.250 Volume (mL) of NaOH added: 0.250 Millimoles of HCl added: 0.125 Millimoles of NaOH added: 0.125 Parts C-E: pH Measurements and Determination of Buffer Capacity 5 pts: ALL pH values and calculated buffer capacities entered Solution ID Solution ID #1 2.96 2.09 0.14 #1 2.82 3.66 0.15 #2 3.98 3.28 0.18 #2 3.92 4.29 0.34 #3 4.53 4.12 0.30 #3 4.55 4.77 0.57 #4 5.11 4.67 0.28 #4 5.13 5.43 0.42 #5 7.54 5.27 0.06 #5 7.65 12.19 0.03 DI Water 7.32 2.07 0.02 DI Water 9.44 12.20 0.05 Aspirin (acid) 2.53 2.01 0.24 Aspirin (base) 2.62 3.07 0.28 Buffered Aspirin (acid) 5.93 4.76 0.11 Buffered Aspirin (base) 5.86 8.49 0.05 *example calculation for Solution #2 buffer capacity is on pg 3 Provide the K a value for the acid Provide the K b value for the conjugate base 1.8 x 10 -5 5.6 x 10 -10 Molarity of Stock Soln CH 3 COOH CH 3 COONa·3H 2 O mL of CH 3 COOH mL of CH 3 COONa *Summary of Expected Initial pH Mass of Solid (for acid test) Mass of Solid (for base test) Initial Measured pH pH Measured After H+ Addition *Buffer Capacity (mmol H+ added / D pH) Initial Measured pH pH Measured After OH- Addition Buffer Capacity (mmol OH- added / D pH) Chem 152 Lab #2: Buffers By signing below, you certify that you have not falsified data, that you have not plagiarized any part of this lab report, and that all calculations and responses other than the reporting of raw data are your own independent work. Failure to sign this declaration will result in 5 points being deducted from your lab score. Signature: Frances Kato (signature below) Note: All sections of this report must be typed CH_3 COOH(aq)+H_2 O(l)→H_3 O^+ (aq)+CH_3 COO^− (aq) Provide the calculation for the concentration of the stock CH 3 COONa solution you prepared (actual, not theoretical). (3 pts) moles of CH 3 COONa: 1.360g/(136.08g/mol)=0.01 moles "CH 3 COONa" mL to L: 100mL∗1L/1000mL=0.1L molarity (concentration): (0.01 moles CH_3 OONa" " )/(0.1 L)= 0.1 M stock CH_3 COONa Provide the calculation for the expected initial pH of solution #5. (3 pts) CH_3 COO^− (aq)+H_2 O(l)⟶OH^− (aq)+CH_3 COOH(aq) 0.1 0 0 −x +x +x 0.1−x x x K_b=[BH^+ ][OH^− ]/[B] =([CH_3 COOH][OH^−])/([CH_3 COO^−]) x^2/(0.1−x)≈x^2/0.1 K_b=K_w/K_a =(1.0 × 10^(−14))/(1.8 × 10^(−5) )=5.6 × 10^(−10) 5.6 × 10^(−10)=x^2/0.1⟹(0.1)(5.6 × 10^(−10) )=x^2⟹√((0.1)(5.6 × 10^(−10) ) )=x⟹x=7.48 × 10^(−6), x=[OH^− ] Give the predominant chemical reaction that occurs when CH 3 COONa is dissolved in water (omit spectator ions). (3 pts) CH_3 COONa (aq)→CH_3 COO^− (aq)+Na^+ (aq) **Na^+not included in predominant chemical reaction because is a spectator ion. CH_3 COO^− (aq)+H_2 O(l)→CH_3 COOH(aq)+OH^− (aq) Provide the calculation for the expected initial pH of solution #2 (as an example of the calculation for solutions #2-4). (3 pts) (0.1M CH_3 COOH)(0.015L)=0.0015mol CH_3 COOH (0.0015 mol CH_3 COOH)/(0.05 L (total vol. of solution))={0.03 M CH_3 COOH} (0.1M CH_3 COONa)(0.005L)=0.0005mol CH_3 COONa (0.0005 mol CH_3 COOH)/(0.05 L (total vol. of solution))={0.01 M CH_3 COONa} pH=pK_a+log (([A^−])/([HA]))=pK_a+log (([CH_3 COONa])/([CH_3 COOH]))=−log (1.8 × 10 ^(−5) ) +log (0.01/0.03)= expected initial pH soln. #2=4.27 Provide the calculation for the expected initial pH of solution #1. (3 pts) CH_3 COOH(aq)+H_2 O(l)⟶H_3 O^+ (aq)+CH_3 COO^− (aq) 0.1 0 0 -x +x +x 0.1−x x x K_a=[H^+ ][A^− ]/[HA] =([H_3 O^+ ][CH_3 COO^−])/([CH_3 COOH]) x^2/(0.1−x)≈x^2/0.1 K_a=1.8 × 10 ^(−5) 1.8 × 10 ^(−5)=x^2/0.1⟹(0.1)(1.8 × 10 ^(−5) )=x^2⟹√((0.1)(1.8 × 10 ^(−5) ) ) =x⟹x=1.3416 × 10^(−3), x=[H_3 O^+ ] 〖〖 pH=−log 〗〖 ([H_3 O^+ ])= log 〗〖 (1.3416 × 10^(−3) )= expected initial pH solution #1=2.87 Show entire document

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