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 101119-74 Unfold and gently clean the surface use: Lightly brush away dust and an scratch. Not for use of test a small area first. Single Keep out of reach n. Do not get in eyes to be jou pue -----Open -- s Wipe ndiflzoq bat Tavo ma daw.als wornet beadlo stint 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. нош → CH₂ Cool caq) + H2O (e) CH 3 COO (aq) + H30+ (ag). 1.2 -x 1.2-X H₂Oce) 0 X と X X 6. Calculate the % ionization of the ethanoic acid in the vinegar. Write the equation and show the substitution. 7. a) Calculate your % error. The accepted Ka for acetic acid is 1.8 x 10-5 b) Give two sources of error(measurement/calculation errors are not acceptable) and how they affect your calculated Ka

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 101119-74 Unfold and gently clean the surface use: Lightly brush away dust and an scratch. Not for use of test a small area first. Single Keep out of reach n. Do not get in eyes to be jou pue -----Open -- s Wipe ndiflzoq bat Tavo ma daw.als wornet beadlo stint 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. нош → CH₂ Cool caq) + H2O (e) CH 3 COO (aq) + H30+ (ag). 1.2 -x 1.2-X H₂Oce) 0 X と X X 6. Calculate the % ionization of the ethanoic acid in the vinegar. Write the equation and show the substitution. 7. a) Calculate your % error. The accepted Ka for acetic acid is 1.8 x 10-5 b) Give two sources of error(measurement/calculation errors are not acceptable) and how they affect your calculated Ka

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 101119-74 Unfold and gently clean the surface use: Lightly brush away dust and an scratch. Not for use of test a small area first. Single Keep out of reach n. Do not get in eyes to be jou pue -----Open -- s Wipe ndiflzoq bat Tavo ma daw.als wornet beadlo stint 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. нош → CH₂ Cool caq) + H2O (e) CH 3 COO (aq) + H30+ (ag). 1.2 -x 1.2-X H₂Oce) 0 X と X X 6. Calculate the % ionization of the ethanoic acid in the vinegar. Write the equation and show the substitution. 7. a) Calculate your % error. The accepted Ka for acetic acid is 1.8 x 10-5 b) Give two sources of error(measurement/calculation errors are not acceptable) and how they affect your calculated Ka

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Chapter15: Acid-base Equilibria

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Problem 118CP: Malonic acid (HO2CCH2CO2H) is a diprotic acid. In the titration of malonic acid w ith NaOH,...

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please answer 7 ，thanks！！

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

101119-74
Unfold and gently clean the surface
use: Lightly brush away dust and
an scratch. Not for use of
test a small area first. Single
Keep out of reach
n. Do not get in eyes
to be jou pue
-----Open --
s Wipe
ndiflzoq bat
Tavo ma daw.als
wornet beadlo stint
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.
нош
→
CH₂ Cool caq) + H2O (e) CH 3 COO (aq) + H30+ (ag).
1.2
-x
1.2-X
H₂Oce)
0
X
と
X X
6. Calculate the % ionization of the ethanoic acid in the vinegar. Write the equation and show the substitution.
7. a) Calculate your % error. The accepted Ka for acetic acid is 1.8 x 10-5
b) Give two sources of error(measurement/calculation errors are not acceptable) and how they affect your
calculated Ka

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