Trial 1 Volume of acid 30 ml Volume NaOH ADDED TO HALF NEUTRALIZE ACID 6.5ML Ph...
|
Trial 1 |
|
|
Volume of acid |
30 ml |
|
Volume NaOH ADDED TO HALF NEUTRALIZE ACID |
6.5ML |
|
Ph of Half neutralized solution |
4.34 |
|
Ka |
? |
|
PKa of acetic acid |
? |
Homework Answers
We know that, pH corresponding to half equivalence point = pKa
In this case, pH corresponding to half equivalence point = 4.34
Therefore, pKa of acetic acid = 4.34
We have relation, pKa = - log 10 Ka
Ka = 10 - pKa = 10 -4.34 = 4.57 
10 -05
ANSWER : pKa of acetic acid = 4.34 & Ka of acetic
acid = 4.57
10 -05
Add Answer to:
Trial 1
Volume of acid
30 ml
Volume NaOH ADDED TO HALF NEUTRALIZE ACID
6.5ML
Ph...
How do I calculate the ionization constant for acetic acid from the measured pH of the acedic aci...
How do I calculate the ionization
constant for acetic acid from the measured pH of the acedic acid
samples?
Data and Calculations 1. Measurement of pH and fitration of acetfic acid solution Concentration of standardized NaOH titrant Mass concentration of acetic acid-aao Trial 3 Trial 2 Trial 1 Measured pht of the acetic acid solution Mass of acetic acid solution taken for titration. Initial buret reading of NaOH titrant Final buret reading of NaOH titrant Net volume of NaOH Millimoles...
1. The equivalence point of a weak, monoprotic acid with a volume of 22.00 mL was...
1. The equivalence point of a weak, monoprotic acid with a volume of 22.00 mL was reached after adding 22.10 mL of 0.1025 M NaOH(aq) and the pH at this volume was 8.91. The pH was 3.37 when the volume of NaOH(aq) added was 11.05 mL. What is the value of Ka for this unknown acid? 0.1025 0.1030 3.37 8.91 1.23 x 10-9 4.27 x 10-4 2. A solution of acetic acid, HC2H3O2, a weak monoprotic acid, was standardized by...
Exp 7 TRIAL 1 ation Results TRIAL 1 NaOH volume at equivalencepH 3.02 volume OmL pH 4.41 vol. 3mL...
please help determine the circled area from the data recorded
during titration
Exp 7 TRIAL 1 ation Results TRIAL 1 NaOH volume at equivalencepH 3.02 volume OmL pH 4.41 vol. 3mL point NaOH volume at half H 3.14 volume .5mL pH 4.50 vol. 3.5mL equivalence point pH at the equivalence point pH 3.73 volume 1mL pH at half equivalence pointpH 3.96 volume 1.5mL pKa of acetic acid Ka of acetic acid pH 4.63 pH 4.74 pH 4.88 pH 5.14 vol....
millimoles of NaOH titrant Chemical Equilibria: K, of a Weak Monoprotic Acid Report Form Name: Partner's...
millimoles of NaOH titrant
Chemical Equilibria: K, of a Weak Monoprotic Acid Report Form Name: Partner's Name: (if any) Lab Section MWITTHM-TH (Circle) Data and Calculations 1. Measurement of pH and titration of acetic acid solution Concentration of standardized NaOH titrant 0.1054 Mass concentration of acetic acid 2.40 - mol/L Trial 3 Measured pH of the acetic acid solution Mass of acetic acid solution taken for titration Trial 1 3.09 20.16 0 11.4 Trial 2 3.32 30.06 Initial buret reading...
1. Calculate the volume (in mL) of the amount of 0.200 M NaOH required to neutralize...
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Trial 1 Acetic acid / L solution 2.402g/L Volume of acetic acid (mL) 30 mL Initial...
Trial 1 Acetic acid / L solution 2.402g/L Volume of acetic acid (mL) 30 mL Initial buret volume (mL) 0.2 mL Final buret volume at end point (mL) 18.2 mL Total volume of NaOH titrated (mol) 18.0 mL Moles of NaOH titrated (mol) Molarity of acetic acid (mol/L) Molar mass of acetic acid Average volume of NaOH titrated mL Average molar concentration of acid Average molar mass of acetic acid
1. Calculate the volume (in mL) of the amount of 0.200 M NaOH required to neutralize...
1. Calculate the volume (in mL) of the amount of 0.200 M NaOH required to neutralize a monoprotic weak acid solution made by 2.00 g of potassium hydrogen phthalate (KHP) dissolved in water. Hint: the complete neutralization occurs at the equivalence point, where the number of moles of the analyte (in this case, the weak acid) equal to the titrant (in this case, the strong base). 2. Identify the equivalence point, the half-equivalence point on the titration curve below and...
1. Calculate the volume (in mL) of the amount of 0.200 M NaOH required to neutralize...
1. Calculate the volume (in mL) of the amount of 0.200 M NaOH required to neutralize a monoprotic weak acid solution made by 2.00 g of potassium hydrogen phthalate (KHP) dissolved in water. 2. Identify the equivalence point, the half-equivalence point on the titration curve below and determine the pKa of the acid. 12 10- DH 8 6 4 N 0 0 40 20 Titrant Volume (mL)
1. What volume of 0.10 M NaOH is needed to neutralize 100 mL of 0.050 M...
1. What volume of 0.10 M NaOH is needed to neutralize 100 mL of 0.050 M HCl? 2a. When 100 mL of 0.01 M NaOH solution is mixed with 100 mL of 0.01 M of HNO3 solution, what is the concentration of H+ and the pH in the resulting mixture? 2b. When 100 mL of 0.01 M NaOH solution is mixed with 100 mL of 0.02 M of HCl solution, what is the concentration of H+ and the pH in...
2) 19.63 mL of 0.100 M NaOH is required to neutralize 2.00 mL of a solution containing acetic acid according to th...
2) 19.63 mL of 0.100 M NaOH is required to neutralize 2.00 mL of a solution containing acetic acid according to the following reaction: HC2H302(aq) + NaOH(aq) → NaC2H3O2(aq) + H2O(1). How many grams of acetic acid are present in 2 mL solution? Report the correct number of significant figures, and report the units. 3) 19.63 mL of 0.100 M NaOH is required to neutralize 2.00 mL of a solution containing acetic acid according to the following reaction: HC2H3O2(aq) +...
How do I calculate the ionization
constant for acetic acid from the measured pH of the acedic acid
samples?
Data and Calculations 1. Measurement of pH and fitration of acetfic acid solution Concentration of standardized NaOH titrant Mass concentration of acetic acid-aao Trial 3 Trial 2 Trial 1 Measured pht of the acetic acid solution Mass of acetic acid solution taken for titration. Initial buret reading of NaOH titrant Final buret reading of NaOH titrant Net volume of NaOH Millimoles...
please help determine the circled area from the data recorded
during titration
Exp 7 TRIAL 1 ation Results TRIAL 1 NaOH volume at equivalencepH 3.02 volume OmL pH 4.41 vol. 3mL point NaOH volume at half H 3.14 volume .5mL pH 4.50 vol. 3.5mL equivalence point pH at the equivalence point pH 3.73 volume 1mL pH at half equivalence pointpH 3.96 volume 1.5mL pKa of acetic acid Ka of acetic acid pH 4.63 pH 4.74 pH 4.88 pH 5.14 vol....
millimoles of NaOH titrant
Chemical Equilibria: K, of a Weak Monoprotic Acid Report Form Name: Partner's Name: (if any) Lab Section MWITTHM-TH (Circle) Data and Calculations 1. Measurement of pH and titration of acetic acid solution Concentration of standardized NaOH titrant 0.1054 Mass concentration of acetic acid 2.40 - mol/L Trial 3 Measured pH of the acetic acid solution Mass of acetic acid solution taken for titration Trial 1 3.09 20.16 0 11.4 Trial 2 3.32 30.06 Initial buret reading...
1. Calculate the volume (in mL) of the amount of 0.200 M NaOH required to neutralize a monoprotic weak acid solution made by 2.00 g of potassium hydrogen phthalate (KHP) dissolved in water. 2. Identify the equivalence point, the half-equivalence point on the titration curve below and determine the pKa of the acid. 12- 10 PH 8-1 6 N 0+ 20 Titrant Volume (ML) 40
1. Calculate the volume (in mL) of the amount of 0.200 M NaOH required to neutralize a monoprotic weak acid solution made by 2.00 g of potassium hydrogen phthalate (KHP) dissolved in water. Hint: the complete neutralization occurs at the equivalence point, where the number of moles of the analyte (in this case, the weak acid) equal to the titrant (in this case, the strong base). 2. Identify the equivalence point, the half-equivalence point on the titration curve below and...
1. Calculate the volume (in mL) of the amount of 0.200 M NaOH required to neutralize a monoprotic weak acid solution made by 2.00 g of potassium hydrogen phthalate (KHP) dissolved in water. 2. Identify the equivalence point, the half-equivalence point on the titration curve below and determine the pKa of the acid. 12 10- DH 8 6 4 N 0 0 40 20 Titrant Volume (mL)
2) 19.63 mL of 0.100 M NaOH is required to neutralize 2.00 mL of a solution containing acetic acid according to the following reaction: HC2H302(aq) + NaOH(aq) → NaC2H3O2(aq) + H2O(1). How many grams of acetic acid are present in 2 mL solution? Report the correct number of significant figures, and report the units. 3) 19.63 mL of 0.100 M NaOH is required to neutralize 2.00 mL of a solution containing acetic acid according to the following reaction: HC2H3O2(aq) +...
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