Acetic Acid Naoh Titration Equation

Determining the Concentration of Acetic Acid in Vinegar by Direct Titration

Key Concepts

Vinegar is produced past the oxidation of ethanol.
Table vinegar typically contains between 4 and 8 % v/v acetic acid (ethanoic acid).

Acetic acrid is a weak, monoprotic, organic Brønsted-Lowry acid:

acetic acid
(ethanoic acrid)

acetate ion
(ethanoate ion)

	H \|		O \|\|
H-	C	-	C	-O-H
	\| H

H⁺

	H \|		O \|\|
H-	C	-	C	-O^-
	\| H

acrid

conjugate base

Acetic acid (ethanoic acid) will react with a stiff base in neutralisation reaction:

acid

base

→

water

salt

	H \|		O \|\|
H-	C	-	C	-O-H
	\| H

NaOH(aq)

H-OH(l)

	H \|		O \|\|
H-	C	-	C	-O^- Na⁺
	\| H

acetic acrid
(ethanoic acrid)

sodium hydroxide

→

water

sodium acetate
( sodium ethanoate)

It is therefore possible to make up one's mind the concentration of acerb acid in vinegar by titrating the vinegar with a strong base such every bit aqueous sodium hydroxide solution.

moles of acetic acid in vinegar sample	=	moles of sodium hydroxide used in titration

n(CH₃COOH)	=	north(NaOH)

concentration of acetic acid	=	moles of acetic acid volume of vinegar
c(CH₃COOH)	=	n(CH₃COOH) V(vinegar)

Using the known density of acetic acid it is possible to calculate the concentration of acetic acid in vinegar equally a v/v%:

v/v % =	book of acetic acid book of vinegar	× 100

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Experiment: Determining the Acetic Acid in Vinegar past Titration

For this experiment it is best to choose a "white vinegar" (that is, a colourless vinegar) rather than a "dark-brown vinegar".
The colour of a "chocolate-brown vinegar" can mask the colour change at the stop point of the titration.

When a weak acid such every bit acetic acid is titrated with a strong base such as aqueous sodium hydroxide solution, the pH at the equivalence point will be greater than 7.
Suitable indicators for this experiment are phenolphthalein or thymol blue.

Process to titrate the acetic acid in vinegar:

Rinse a clean 250 mL conical (erlenmeyer) flask with h2o.
Rinse a make clean 25.00 mL pipette (pipet) with vinegar.
Pipette 25.00 mL of vinegar into the 250 mL conical (erlenmeyer) flask.
Add 2 drops of phenolphthalein indicator to the vinegar.
(The solution volition remain colourless)
Rinse a make clean 50.00 mL burette (buret) with standardised ane.00 mol L^-1 aqueous sodium hydroxide solution.
Fill the burette (buret) with this standardised i.00 mol Fifty^-1 NaOH(aq).
Set upward the equipment as in the diagram on the right.
Run NaOH(aq) from the burette (buret) into the conical (erlenmeyer) flask until the solution changes color from colourless to pink.
Repeat the titration carefully several times until concordant titres are achieved.

Sample Results

	Trial one / mL	Trial ii / mL	Trial 3 / mL
Terminal volume of NaOH(aq)	21.82	21.79	21.81
Initial volume of NaOH(aq)	0.00	0.00	0.00
Titre (volume of NaOH(aq) used)	21.82	21.79	21.81
Average Titre /mL	21.82 + 21.79 + 21.81 3		= 21.81

Computing the Concentration of Acetic Acid in Vingear in mol Fifty^-1

Write the balanced chemical equation for the neutralisation reaction:

word equation	acetic acid (ethanoic acid)	+	sodium hydroxide	→	sodium acetate (sodium ethanoate)	+	water
balanced chemical equation	CH_threeCOOH(aq)	+	NaOH(aq)	→	CH₃COO^-Na⁺(aq)	+	H₂O

Extract all the relevant data from the experiment.

counterbalanced chemical equation	CH_iiiCOOH(aq)	+	NaOH(aq)	→	CH₃COO^-Na⁺(aq)	+	H₂O
volume /mL	25.00		21.81
concentration /mol 50^-1	?		i.00

Bank check the data for consistency:
Concentrations are ordinarily given in M or mol L^-1 but volumes are often given in mL.
Y'all will need to convert the mL to L for consistency.
The easiest way to practise this is to multiply the book in mL × x^-3 (which is the aforementioned as dividing the volume in mL by 1000)

relevant species	acid	base
relevant species	CH₃COOH(aq)	NaOH(aq)
volume /mL	25.00	21.81
volume /L	25.00 ÷ 1000 = 0.02500	21.81 ÷ 1000 = 0.02181
concentration /mol 50^-1	?	ane.00

Summate the moles of NaOH(aq), n(NaOH)
moles = concentration in mol L^-one x book in L = due north = c ten 5
volume of NaOH(aq) = v(NaOH) = 0.02181 50
concentration of NaOH(aq) = c(NaOH) = 1.00 mol Fifty-1

moles NaOH(aq) = n(NaOH) = c(NaOH) × V(NaOH)
northward(NaOH) = 0.02181 × 1.00 = 0.02181 mol
Employ the counterbalanced chemic equation to make up one's mind the stoichiometric (mole) ratio of acid to base of operations:
n(acid):n(base)
due north(CH_threeCOOH):northward(NaOH)
1:i
Use the stoichiometric (mole) ratio to summate the moles of acerb acid
1 mole of NaOH neutralises 1 mole of CH₃COOH
therefore 0.02181 moles of NaOH neutralises 0.02181 moles of CH₃COOH
moles of acetic acid = n(CH₃COOH) = 0.02181 mol
From the volume of vinegar (acerb acid solution) and the moles of acetic acid, calculate its concentration (c) in mol Fifty^-1 :
concentration (mol L^-i) = moles ÷ volume (L)
concentration of acerb acrid = moles of acetic acid ÷ volume of acetic acid in L
moles of acetic acid = n(CH_iiiCOOH) = 0.02181 mol
volume of acerb acid solution (vinegar) = 5(CH₃COOH) = 0.02500 L
concentration of acetic acid solution (vinegar) = n(CH₃COOH) ÷ 5(CH_iiiCOOH)
c(CH₃COOH) = 0.02181 ÷ 0.02500 = 0.8724 mol 50^-1

Concentration of acetic acid in vinegar in mol L^-1 (molarity) is 0.8724 mol L^-1

Calculating the Concentration of Acetic Acid as a Percentage Past Volume (v/five %)

Write an expression for calculating v/v % concentration.
v/v % = book of solute ÷ volume of solution × 100
acetic acid in vinegar v/5 % = book of acerb acid ÷ volume of vinegar × 100
Excerpt the relevant data from the experiment
volume of acetic acid = ? (can be calculated using mass and density)
volume of vinegar = 25.00 mL
Calculate mass of acetic acid
mass = moles × tooth mass
moles acetic acid = n(CH₃COOH) = 0.02181 mol (come across calculations in previous section)
molar mass acetic acrid = 2 × 12.01 + 4 × one.008 + 2 × sixteen.00 = threescore.052 m mol^-1
mass acetic acid = moles acetic acid x molar mass acerb acid = 0.02181 × 60.052 = i.3097 g
Calculate volume of acerb acid using its mass and known density
density (g/mL) = mass (g) ÷ volume (mL)
book (mL) = mass (g) ÷ density (one thousand/mL)
density of acerb acid = one.049 thou mL^-ane (at 25°C)
mass acerb acid in vinegar = 1.3097 g
volume of acetic acrid = mass acetic acrid ÷ density acetic acid
volume acetic acid = 1.3097 ÷ 1.049 = 1.249 mL
Calculate concentration of acerb acid in vinegar as v/v %
acetic acid in vinegar v/v % = volume of acetic acid ÷ volume of vinegar × 100
volume of acetic acid = 1.249 mL
volume of vinegar = 25.00 mL
acerb acid in vinegar v/five % = ane.249 ÷ 25.00 × 100 = 4.996 %

5/v % acetic acid in vinegar is 4.996%