Neutralization Equivalent Determination (NE): NE - empirical approximation of the acid's molecular weight (MW) by neutralization with base. 2pop ● MW= NE for monocarboxylic acids, MW= 2XNE for dicarboxylic acids vedno of 15 fpol 1. Standardize the Conc NaOH by titration of known amount of benzoic acid: ● D NE = ● Where: macid (mg) Conc NaOH* Vol NaOH Dissolve 25 mg of benzoic acid in H₂O. Add 2 drops of phenolphthalein indicator. Add NaOH dropwise by the use of a volumetric burette until pink endpoint. Record volume of NaOH used. Repeat two more times. Average the Vol NaOH of the three trials, and plug into the above equation to determine Conc NaOH. 2. Repeat the above procedure using your unknown acid (3 trials). This time solve for NE. Based on the physical properties and NE, select all possible carboxylic acids within a ±7°C of the observed melting point and arrange the data for these acids and their derivatives into the lab report. Table 1 shown below (next page). Include this table in your Insigent

Chemistry & Chemical Reactivity
9th Edition
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Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
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Chapter16: Principles Of Chemical Reactivity: The Chemistry Of Acids And Bases
Section: Chapter Questions
Problem 116IL: Amino acids are an important group of compounds. At low pH, both the carboxylic acid group (CO2H)...
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I need help determining the concentration of NaOH and to solve the equation for both the monocarboxylic acid and the dicarboxylic acid for the unknown. I've attached the data obtained for the trials of benzoic, and for the unknown. 

Neutralization Equivalent Determination (NE):
MONE - empirical approximation of the acid's molecular weight (MW) by neutralization with base.
Yearm
2pot
NE=
macid (mg)
Conc NaOH * Vol NaOH
Where:
MW= NE for monocarboxylic acids,
MW= 2XNE for dicarboxylic acids
rux vebn
1. Standardize the Conc NaOH by titration of known amount of benzoic acid:
GO AVES)
Dissolve 25 mg of benzoic acid in H₂O.
Add 2 drops of phenolphthalein indicator.
Add NaOH dropwise by the use of a volumetric burette until pink endpoint.
Record volume of NaOH used.
Repeat two more times.
oliey)
Average the Vol NaOH of the three trials, and plug into the above equation to
determine Conc NaOH.
(1096
2. Repeat the above procedure using your unknown acid (3 trials). This time solve for NE. d
Based on the physical properties and NE, select all possible carboxylic acids within a 17°C of the
bilde
observed melting point and arrange the data for these acids and their derivatives into the
Table 1 shown below (next page). Include this table in your lab report.
18617
sbl
Transcribed Image Text:Neutralization Equivalent Determination (NE): MONE - empirical approximation of the acid's molecular weight (MW) by neutralization with base. Yearm 2pot NE= macid (mg) Conc NaOH * Vol NaOH Where: MW= NE for monocarboxylic acids, MW= 2XNE for dicarboxylic acids rux vebn 1. Standardize the Conc NaOH by titration of known amount of benzoic acid: GO AVES) Dissolve 25 mg of benzoic acid in H₂O. Add 2 drops of phenolphthalein indicator. Add NaOH dropwise by the use of a volumetric burette until pink endpoint. Record volume of NaOH used. Repeat two more times. oliey) Average the Vol NaOH of the three trials, and plug into the above equation to determine Conc NaOH. (1096 2. Repeat the above procedure using your unknown acid (3 trials). This time solve for NE. d Based on the physical properties and NE, select all possible carboxylic acids within a 17°C of the bilde observed melting point and arrange the data for these acids and their derivatives into the Table 1 shown below (next page). Include this table in your lab report. 18617 sbl
Neutralization
Benzoic Trial 3
initial. 15, 3mL
final: 20.2mL
Trial 1
initial: 4.3mL
final: 10.1mL = 5.8
Trial 2
initial: 10.1mL
final: 15.3mL = 5.2
Un known Trial 1
initial: 1.2
final: 10.8
Trial 2
initial; 11.2
final: 19.2
Equivalent Determination
=4.9
Trial 3
Initial: 4.8
finals 11.0
= 9.6
= 8
=6.2
7
> 5.3 average
7. ay average
Transcribed Image Text:Neutralization Benzoic Trial 3 initial. 15, 3mL final: 20.2mL Trial 1 initial: 4.3mL final: 10.1mL = 5.8 Trial 2 initial: 10.1mL final: 15.3mL = 5.2 Un known Trial 1 initial: 1.2 final: 10.8 Trial 2 initial; 11.2 final: 19.2 Equivalent Determination =4.9 Trial 3 Initial: 4.8 finals 11.0 = 9.6 = 8 =6.2 7 > 5.3 average 7. ay average
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