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7. Rinse the burette and pipette with water and put the burette upside down with the tap open. Data Table Exp Rough titration Vi(mL) Vf(mL) Volume of NaOH used (mL) 1 2 3 0 29.8 29.8 0 30.0 30.0 30.2 30.2 Average volume of NaOH used = pH of vinegar (ethanoic acid solution) = 2.8 [NaOH)= 0.40 mol/L Volume of ethanoic acid used in the titration 10.00 mL 29.8+30.0+30.2 = 30.0 (mL) 3 1. Write the balanced equation for the neutralization reaction between ethanoic acid and sodium hydroxide solutions. Include states. CH3COOH(aq) + NaOH(aq) CHǝ CooNa (aq) + H2O (e) 2. Calculate the concentration of the ethanoic acid. Use the ratio in which the acid and base react according to the balanced equation. 30mL = 0.03L 144 C₁ V₁ = C2 V₂ = CH3COOH NaoH (0.4 mol/L) (0.03L)= C (0.01L) C = 1.2 mol/L 100mL=0.01 L molar ratio: 二十 CIVI C2V2 = + => CIV₁ = C₂ V2 3. Calculate the [H+] (or [H3O+]) from the pH measurement. PH = -log [H+] 2.8 = -109, CH+] 1028 = [H+] CH+] 1.58x10 mol/L 4. Write the Ka expression for ethanoic acid in water.[K/U 1 mark] CH3COOH (99) + H2O (l) → CH3 C00¯ +H30+ H2O(CH3 [CH3 C00][H3O+] ka: = [CH3COOH] 5. Set up into ep Keep out of ch Do not get in o leugh c of adım so 5. Set up an ICE table for the ionization of acetic acid in water. Then substitute the equilibrium concentrations into the Ka expression and calculate the value of Ka. CH3COOH caq) + H2O (e) CH 3 COO (aq) + H30+ (ag). I 1.2 -x ICE 1.2-X 0 X 0 X X X 6. Calculate the % ionization of the ethanoic acid in the vinegar. Write the equation and show the substitution.

7. Rinse the burette and pipette with water and put the burette upside down with the tap open. Data Table Exp Rough titration Vi(mL) Vf(mL) Volume of NaOH used (mL) 1 2 3 0 29.8 29.8 0 30.0 30.0 30.2 30.2 Average volume of NaOH used = pH of vinegar (ethanoic acid solution) = 2.8 [NaOH)= 0.40 mol/L Volume of ethanoic acid used in the titration 10.00 mL 29.8+30.0+30.2 = 30.0 (mL) 3 1. Write the balanced equation for the neutralization reaction between ethanoic acid and sodium hydroxide solutions. Include states. CH3COOH(aq) + NaOH(aq) CHǝ CooNa (aq) + H2O (e) 2. Calculate the concentration of the ethanoic acid. Use the ratio in which the acid and base react according to the balanced equation. 30mL = 0.03L 144 C₁ V₁ = C2 V₂ = CH3COOH NaoH (0.4 mol/L) (0.03L)= C (0.01L) C = 1.2 mol/L 100mL=0.01 L molar ratio: 二十 CIVI C2V2 = + => CIV₁ = C₂ V2 3. Calculate the [H+] (or [H3O+]) from the pH measurement. PH = -log [H+] 2.8 = -109, CH+] 1028 = [H+] CH+] 1.58x10 mol/L 4. Write the Ka expression for ethanoic acid in water.[K/U 1 mark] CH3COOH (99) + H2O (l) → CH3 C00¯ +H30+ H2O(CH3 [CH3 C00][H3O+] ka: = [CH3COOH] 5. Set up into ep Keep out of ch Do not get in o leugh c of adım so 5. Set up an ICE table for the ionization of acetic acid in water. Then substitute the equilibrium concentrations into the Ka expression and calculate the value of Ka. CH3COOH caq) + H2O (e) CH 3 COO (aq) + H30+ (ag). I 1.2 -x ICE 1.2-X 0 X 0 X X X 6. Calculate the % ionization of the ethanoic acid in the vinegar. Write the equation and show the substitution.

Fundamentals Of Analytical Chemistry
Fundamentals Of Analytical Chemistry
9th Edition
ISBN:9781285640686
Author:Skoog
Publisher:Skoog
Chapter14: Principles Of Neutralization Titrations
Section: Chapter Questions
Problem 14.9QAP
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7. Rinse the burette and pipette with water and put the burette upside down with the tap open. Data Table Exp Rough titration Vi(mL) Vf(mL) Volume of NaOH used (mL) 1 2 3 0 29.8 29.8 0 30.0 30.0 30.2 30.2 Average volume of NaOH used = pH of vinegar (ethanoic acid solution) = 2.8 [NaOH)= 0.40 mol/L Volume of ethanoic acid used in the titration 10.00 mL 29.8+30.0+30.2 = 30.0 (mL) 3 1. Write the balanced equation for the neutralization reaction between ethanoic acid and sodium hydroxide solutions. Include states. CH3COOH(aq) + NaOH(aq) CHǝ CooNa (aq) + H2O (e) 2. Calculate the concentration of the ethanoic acid. Use the ratio in which the acid and base react according to the balanced equation. 30mL = 0.03L 144 C₁ V₁ = C2 V₂ = CH3COOH NaoH (0.4 mol/L) (0.03L)= C (0.01L) C = 1.2 mol/L 100mL=0.01 L molar ratio: 二十 CIVI C2V2 = + => CIV₁ = C₂ V2 3. Calculate the [H+] (or [H3O+]) from the pH measurement. PH = -log [H+] 2.8 = -109, CH+] 1028 = [H+] CH+] 1.58x10 mol/L 4. Write the Ka expression for ethanoic acid in water.[K/U 1 mark] CH3COOH (99) + H2O (l) → CH3 C00¯ +H30+ H2O(CH3 [CH3 C00][H3O+] ka: = [CH3COOH] 5. Set up into
Transcribed Image Text:7. Rinse the burette and pipette with water and put the burette upside down with the tap open. Data Table Exp Rough titration Vi(mL) Vf(mL) Volume of NaOH used (mL) 1 2 3 0 29.8 29.8 0 30.0 30.0 30.2 30.2 Average volume of NaOH used = pH of vinegar (ethanoic acid solution) = 2.8 [NaOH)= 0.40 mol/L Volume of ethanoic acid used in the titration 10.00 mL 29.8+30.0+30.2 = 30.0 (mL) 3 1. Write the balanced equation for the neutralization reaction between ethanoic acid and sodium hydroxide solutions. Include states. CH3COOH(aq) + NaOH(aq) CHǝ CooNa (aq) + H2O (e) 2. Calculate the concentration of the ethanoic acid. Use the ratio in which the acid and base react according to the balanced equation. 30mL = 0.03L 144 C₁ V₁ = C2 V₂ = CH3COOH NaoH (0.4 mol/L) (0.03L)= C (0.01L) C = 1.2 mol/L 100mL=0.01 L molar ratio: 二十 CIVI C2V2 = + => CIV₁ = C₂ V2 3. Calculate the [H+] (or [H3O+]) from the pH measurement. PH = -log [H+] 2.8 = -109, CH+] 1028 = [H+] CH+] 1.58x10 mol/L 4. Write the Ka expression for ethanoic acid in water.[K/U 1 mark] CH3COOH (99) + H2O (l) → CH3 C00¯ +H30+ H2O(CH3 [CH3 C00][H3O+] ka: = [CH3COOH] 5. Set up into
ep Keep out of ch Do not get in o leugh c of adım so 5. Set up an ICE table for the ionization of acetic acid in water. Then substitute the equilibrium concentrations into the Ka expression and calculate the value of Ka. CH3COOH caq) + H2O (e) CH 3 COO (aq) + H30+ (ag). I 1.2 -x ICE 1.2-X 0 X 0 X X X 6. Calculate the % ionization of the ethanoic acid in the vinegar. Write the equation and show the substitution.
Transcribed Image Text:ep Keep out of ch Do not get in o leugh c of adım so 5. Set up an ICE table for the ionization of acetic acid in water. Then substitute the equilibrium concentrations into the Ka expression and calculate the value of Ka. CH3COOH caq) + H2O (e) CH 3 COO (aq) + H30+ (ag). I 1.2 -x ICE 1.2-X 0 X 0 X X X 6. Calculate the % ionization of the ethanoic acid in the vinegar. Write the equation and show the substitution.
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