Class 12 Exam > Class 12 Videos > Acid Base Titration : Solutions - Chemistry, Class 12
Acid Base Titration : Solutions - Chemistry, Class 12 Video Lecture
FAQs on Acid Base Titration : Solutions - Chemistry, Class 12 Video Lecture
| 1. What is acid-base titration? |  |
Ans. Acid-base titration is a technique used in chemistry to determine the concentration of an unknown acid or base solution by neutralizing it with a standardized solution of known concentration. This is done by adding the standardized solution, known as the titrant, to the unknown solution until the reaction between the acid and base is complete, indicated by a color change or pH indicator.
| 2. How does acid-base titration work? | |
Ans. Acid-base titration works by performing a reaction between the acid and base solutions, in which the number of moles of acid and base are equal. This is achieved by slowly adding the titrant to the unknown solution and monitoring the pH or using an indicator to detect the endpoint of the reaction. The volume of titrant required to reach the endpoint is used to calculate the concentration of the unknown solution using the stoichiometry of the balanced chemical equation.
| 3. What indicators can be used in acid-base titration? | |
Ans. There are several indicators that can be used in acid-base titration to detect the endpoint of the reaction. Some commonly used indicators include phenolphthalein, methyl orange, bromothymol blue, and bromocresol green. These indicators change color at different pH ranges, allowing for the determination of the endpoint based on the observed color change.
| 4. What factors can affect the accuracy of acid-base titration? | |
Ans. Several factors can affect the accuracy of acid-base titration. These include inaccuracies in measuring the volumes of solutions, presence of impurities or contaminants in the solutions, improper mixing of the solutions, and the choice of indicator. It is important to ensure proper technique, accurate measurements, and use of appropriate indicators to obtain reliable and accurate results.
| 5. What are some applications of acid-base titration? | |
Ans. Acid-base titration has various applications in chemistry. It is commonly used in pharmaceutical analysis to determine the concentration of active ingredients in medications. It is also used in environmental analysis for testing water quality by determining the concentration of pollutants. Additionally, acid-base titration is utilized in the food industry to determine the acidity or alkalinity of various food products.
Text Transcript from Video
I've taken this problem from
Chapter 4 of the Chemistry &
Chemical Reactivity book by
Kotz, Treichel and Townsend,
and I've done it with
their permission.
So let's do this example.
A 1.034 gram sample of impure
oxalic acid is dissolved in
water and an acid-base
indicator added.
The sample requires 34.47
milliliters of 0.485 molar
sodium hydroxide to reach
the equivalence point.
What is the mass of oxalic acid,
and what is its mass
percent in the sample?
So before we even break into the
math of this, let's just
think about what's happening.
We have some oxalic acid,
which looks like this.
It's really two carbolic acid
groups joined together, if
that means anything to you.
Watch the organic chemistry play
list if you want to learn
more about that.
So we have a double bond to one
oxygen, and then another
bond to a hydroxide.
We have that on the other
carbon as well.
This right here is what
oxalic acid is.
And it's an interesting acid,
because it can actually donate
two protons.
This proton can be nabbed off,
and this proton can also be
contributed.
And it's actually resident
stabalized.
If that doesn't mean anything
to you, don't worry.
You'll learn more about that
in organic chemistry.
But the important thing to
realize here is that there's
two protons to nab off of it.
Now each molecule of sodium
hydroxide-- remember when you
put it in the water it really
just dissolves, and you can
really just think of it as
hydroxide-- each molecule of
hydroxide can nab one of
the hydrogen protons.
So for every one molecule of
oxalic acid, you're going to
need two hydroxides-- one to nab
this hydrogen proton, and
then another one to nab
that hydrogen proton.
So let's write down the balanced
equation that we're
dealing with here.
So we're going to start off
with some oxalic acid.
So that has two hydrogens-- so
it's H2-- two carbons, and
then four oxygens-- O4.
It's dissolved in water, so
it's an aqueous solution.
And to that, we're going to
add sodium hydroxide.
Now I just told you that you're
going to need two of
the hydroxides to fully
neutralize the oxalic acid.
So you're going to
need two of them.
And this is also in our
aqueous solution.
And once the reaction happens,
this guy will have lost both
of the hydrogen protons,
so let me draw that.
So it will look like this.
No more hydrogen,
so it's C2O4.
It'll have a negative
2 charge.
And actually, you could imagine
that it might be
attracted to these positively
charged sodiums. And 2 sodiums
in particular.
So this has a negative
2 charge.
We could even write it there
if you want-- 2 minus.
And then you could have
the sodiums over here.
You have these two sodiums
that have two plus.
And this entire molecules
becomes neutral.
They are attracted
to each other.
They are still in an
aqueous solution.
And then, the hydroxide nabs the
protons, and then you are
left with just water.
So plus 2 moles-- or 2 molecules
depending on how
we're viewing this--
plus two waters.
I'll just use that same
orange color.
Plus two H2Os.
One of the hydrogens in each
of the water molecules are
coming from the oxalic acid, and
so two of these hydrogens
in these two moles of the water
are coming from one
entire molecule of
oxalic acid.
Now let's actually
do the math.
We have 34.47 milliliters of
the solution that has the
sodium hydroxide.
And I'm just going to convert
that to liters just so it's
easier to deal with the molarity
right over there.
So we have 34.47 milliliters--
we could write it of the
solution, but we understand
that, that's the case.
So let's just So this is times,
we have one liter for
every 1,000 milliliters.
And then this will give us-- the
milliliters cancel out--
34.47 divided by 1,000 is
0.03447 liters of this 0.485
molar sodium hydroxide
solution.
So let's figure out how many
actual molecules of sodium
hydroxide we have. This
is the solution.
And we know its concentration,
0.485 molar-- so let me do
that in a different color--
0.485 molar, this information
allows us to figure
out the actual
molecules of sodium hydroxide.
So we want to multiply this by--
we have 0.485 moles of
sodium hydroxide for every
1 liter of this solution.
That's what the molarity
tells us.
We have 0.485 moles per liter.
So the liters cancel out, and
then now we're going to
actually have to get
a calculator out.
And this'll tell us how many
moles of sodium hydroxide we
have in this solution.
So let me get my calculator.
There we go.
All right, let me just multiply
these two numbers.
So we have 0.03447 times 0.485
is equal to-- let me put this
down here-- 0.167.
And we only have three
significant digits here, so
we're going to round to three
significant digits.
So we'll just go with 0.0167.
So let me move that over
off the screen.
So this is going to be equal
to 0.0167, and all we have
left here are moles of
sodium hydroxide.
Now we know that this many moles
of sodium hydroxide are
going to completely react with
however many moles of oxalic
acid we have. Now we know that
we need two moles of this for
every mole of oxalic acid.
Or for every mole of oxalic acid
that completely reacts,
we need two moles of this.
So let's write that down.
And then you color.
So we need two moles of sodium
hydroxide, we got that from
our balanced equation right
there, and it's obvious it
needs one mole, or one molecule
will take this
proton, and then you need
another molecule
to take that proton.
So we need two moles of sodium
hydroxide for every one mole
of oxalic acid.
For every one mole of H2C2O4.
So essentially, we are
just going to divide
this number by 2.
Let me get the calculator
back.
So we're just going to divide
0.0167 divided by 2.
Once again, three significant
digits 0.00835.
So this is going to 0.00835
moles of oxalic acid, H2C2O4.
So we have the number
of moles, but we to
figure out the mass.
And we know the molar
mass of oxalic acid.
Let me write these down.
We know that hydrogen has a
molar mass-- let me write it
this way-- molar mass if you
have a mole of hydrogen, it
has a molar mass of one gram.
If you have-- and this comes
from its atomic weight-- if
you have carbon its molar mass
is 12 grams. And if you have
oxygen, its molar mass
is 16 grams.
So what's the molar mass
of oxalic acid?
Well we have two hydrogens,
so that's going to
be two grams, right?
2 times 1 gram.
That's the hydrogens there.
We have two carbons, so it's
going to be plus 24 grams. 12
grams for each of
these carbons.
And then we're going to have
four oxygens that weigh in, if
we have a mole of them,
at 16 grams. So that's
going to be plus 64.
So what does this come out to?
24 plus 64 is 88, right?
2 plus 6 is 88, right.
So it's 88 plus 2 more is 90
grams. So if you had a mole of
oxalic acid, it would be 90
grams. So we could say 90
grams per mole of H2C2O4.
So let's go back to
the math here.
I'll rewrite it over here.
We know we're dealing with
0.00835 moles of
oxalic acid, H2C2O4.
And now we know its
molar mass.
We know that there are 90
grams-- let me do this in a
different color, this color's
getting motonous-- we know
that there are 90 grams
of H2C2O4 for
every one mole of H2C2O4.
This is its molar mass.
And now we just multiply this
number, and we'll figure out
the grams of oxalic acid.
That and that cancels out.
and.
Then we just take the number
that we had and multiply it by
90, so times 90.
This just says the previous
answer, which is the number of
moles of oxalic acid times its
molar mass will tell us the
grams of oxalic acid.
So we get 0.75.
I'll just round it to
2 since we only have
three significant digits.
The 90 isn't exact, so it's a
little bit-- but I'll just
round it to three significant
digits.
So 7--.752.
This is equal to 0.752 grams
of oxalic acid: H2C2O4.
And I think we've answered
part of the question.
So the first question is, what
is the mass of oxalic acid?
We've just answered it.
That answer right there
is 0.752 grams.
Now the next part is,
what is its mass
percent in the sample?
Well the sample of impure oxalic
acid right over here,
was 1.034 grams. So we just have
to say, what percentage
is 0.752 of 1.034?
So let's get the calculator
back.
So we have the 0.752 divided
by 1.034 and we get 72.7%.
So the answer to the
second part right
over there is, 72.7%.
We were able to figure out that
this impure oxalic acid
sample is 72.7% actual
oxalic acid.
Chapter 4 of the Chemistry &
Chemical Reactivity book by
Kotz, Treichel and Townsend,
and I've done it with
their permission.
So let's do this example.
A 1.034 gram sample of impure
oxalic acid is dissolved in
water and an acid-base
indicator added.
The sample requires 34.47
milliliters of 0.485 molar
sodium hydroxide to reach
the equivalence point.
What is the mass of oxalic acid,
and what is its mass
percent in the sample?
So before we even break into the
math of this, let's just
think about what's happening.
We have some oxalic acid,
which looks like this.
It's really two carbolic acid
groups joined together, if
that means anything to you.
Watch the organic chemistry play
list if you want to learn
more about that.
So we have a double bond to one
oxygen, and then another
bond to a hydroxide.
We have that on the other
carbon as well.
This right here is what
oxalic acid is.
And it's an interesting acid,
because it can actually donate
two protons.
This proton can be nabbed off,
and this proton can also be
contributed.
And it's actually resident
stabalized.
If that doesn't mean anything
to you, don't worry.
You'll learn more about that
in organic chemistry.
But the important thing to
realize here is that there's
two protons to nab off of it.
Now each molecule of sodium
hydroxide-- remember when you
put it in the water it really
just dissolves, and you can
really just think of it as
hydroxide-- each molecule of
hydroxide can nab one of
the hydrogen protons.
So for every one molecule of
oxalic acid, you're going to
need two hydroxides-- one to nab
this hydrogen proton, and
then another one to nab
that hydrogen proton.
So let's write down the balanced
equation that we're
dealing with here.
So we're going to start off
with some oxalic acid.
So that has two hydrogens-- so
it's H2-- two carbons, and
then four oxygens-- O4.
It's dissolved in water, so
it's an aqueous solution.
And to that, we're going to
add sodium hydroxide.
Now I just told you that you're
going to need two of
the hydroxides to fully
neutralize the oxalic acid.
So you're going to
need two of them.
And this is also in our
aqueous solution.
And once the reaction happens,
this guy will have lost both
of the hydrogen protons,
so let me draw that.
So it will look like this.
No more hydrogen,
so it's C2O4.
It'll have a negative
2 charge.
And actually, you could imagine
that it might be
attracted to these positively
charged sodiums. And 2 sodiums
in particular.
So this has a negative
2 charge.
We could even write it there
if you want-- 2 minus.
And then you could have
the sodiums over here.
You have these two sodiums
that have two plus.
And this entire molecules
becomes neutral.
They are attracted
to each other.
They are still in an
aqueous solution.
And then, the hydroxide nabs the
protons, and then you are
left with just water.
So plus 2 moles-- or 2 molecules
depending on how
we're viewing this--
plus two waters.
I'll just use that same
orange color.
Plus two H2Os.
One of the hydrogens in each
of the water molecules are
coming from the oxalic acid, and
so two of these hydrogens
in these two moles of the water
are coming from one
entire molecule of
oxalic acid.
Now let's actually
do the math.
We have 34.47 milliliters of
the solution that has the
sodium hydroxide.
And I'm just going to convert
that to liters just so it's
easier to deal with the molarity
right over there.
So we have 34.47 milliliters--
we could write it of the
solution, but we understand
that, that's the case.
So let's just So this is times,
we have one liter for
every 1,000 milliliters.
And then this will give us-- the
milliliters cancel out--
34.47 divided by 1,000 is
0.03447 liters of this 0.485
molar sodium hydroxide
solution.
So let's figure out how many
actual molecules of sodium
hydroxide we have. This
is the solution.
And we know its concentration,
0.485 molar-- so let me do
that in a different color--
0.485 molar, this information
allows us to figure
out the actual
molecules of sodium hydroxide.
So we want to multiply this by--
we have 0.485 moles of
sodium hydroxide for every
1 liter of this solution.
That's what the molarity
tells us.
We have 0.485 moles per liter.
So the liters cancel out, and
then now we're going to
actually have to get
a calculator out.
And this'll tell us how many
moles of sodium hydroxide we
have in this solution.
So let me get my calculator.
There we go.
All right, let me just multiply
these two numbers.
So we have 0.03447 times 0.485
is equal to-- let me put this
down here-- 0.167.
And we only have three
significant digits here, so
we're going to round to three
significant digits.
So we'll just go with 0.0167.
So let me move that over
off the screen.
So this is going to be equal
to 0.0167, and all we have
left here are moles of
sodium hydroxide.
Now we know that this many moles
of sodium hydroxide are
going to completely react with
however many moles of oxalic
acid we have. Now we know that
we need two moles of this for
every mole of oxalic acid.
Or for every mole of oxalic acid
that completely reacts,
we need two moles of this.
So let's write that down.
And then you color.
So we need two moles of sodium
hydroxide, we got that from
our balanced equation right
there, and it's obvious it
needs one mole, or one molecule
will take this
proton, and then you need
another molecule
to take that proton.
So we need two moles of sodium
hydroxide for every one mole
of oxalic acid.
For every one mole of H2C2O4.
So essentially, we are
just going to divide
this number by 2.
Let me get the calculator
back.
So we're just going to divide
0.0167 divided by 2.
Once again, three significant
digits 0.00835.
So this is going to 0.00835
moles of oxalic acid, H2C2O4.
So we have the number
of moles, but we to
figure out the mass.
And we know the molar
mass of oxalic acid.
Let me write these down.
We know that hydrogen has a
molar mass-- let me write it
this way-- molar mass if you
have a mole of hydrogen, it
has a molar mass of one gram.
If you have-- and this comes
from its atomic weight-- if
you have carbon its molar mass
is 12 grams. And if you have
oxygen, its molar mass
is 16 grams.
So what's the molar mass
of oxalic acid?
Well we have two hydrogens,
so that's going to
be two grams, right?
2 times 1 gram.
That's the hydrogens there.
We have two carbons, so it's
going to be plus 24 grams. 12
grams for each of
these carbons.
And then we're going to have
four oxygens that weigh in, if
we have a mole of them,
at 16 grams. So that's
going to be plus 64.
So what does this come out to?
24 plus 64 is 88, right?
2 plus 6 is 88, right.
So it's 88 plus 2 more is 90
grams. So if you had a mole of
oxalic acid, it would be 90
grams. So we could say 90
grams per mole of H2C2O4.
So let's go back to
the math here.
I'll rewrite it over here.
We know we're dealing with
0.00835 moles of
oxalic acid, H2C2O4.
And now we know its
molar mass.
We know that there are 90
grams-- let me do this in a
different color, this color's
getting motonous-- we know
that there are 90 grams
of H2C2O4 for
every one mole of H2C2O4.
This is its molar mass.
And now we just multiply this
number, and we'll figure out
the grams of oxalic acid.
That and that cancels out.
and.
Then we just take the number
that we had and multiply it by
90, so times 90.
This just says the previous
answer, which is the number of
moles of oxalic acid times its
molar mass will tell us the
grams of oxalic acid.
So we get 0.75.
I'll just round it to
2 since we only have
three significant digits.
The 90 isn't exact, so it's a
little bit-- but I'll just
round it to three significant
digits.
So 7--.752.
This is equal to 0.752 grams
of oxalic acid: H2C2O4.
And I think we've answered
part of the question.
So the first question is, what
is the mass of oxalic acid?
We've just answered it.
That answer right there
is 0.752 grams.
Now the next part is,
what is its mass
percent in the sample?
Well the sample of impure oxalic
acid right over here,
was 1.034 grams. So we just have
to say, what percentage
is 0.752 of 1.034?
So let's get the calculator
back.
So we have the 0.752 divided
by 1.034 and we get 72.7%.
So the answer to the
second part right
over there is, 72.7%.
We were able to figure out that
this impure oxalic acid
sample is 72.7% actual
oxalic acid.
Related Exams
About this Video
|
Nov 15, 2024 Last updated |
Acid Base Titration : Solutions - Chemistry, Class 12 for Class 122024 is part of Class 12 preparation. The notes and questions for Acid Base Titration : Solutions - Chemistry, Class 12
have been prepared according to the Class 12 exam syllabus. Information about Acid Base Titration : Solutions - Chemistry, Class 12 covers all important topics
for Class 122024 Exam. Find important definitions, questions, notes, meanings, examples, exercises
and tests below for Acid Base Titration : Solutions - Chemistry, Class 12.
Introduction of Acid Base Titration : Solutions - Chemistry, Class 12 in English is available as part of our Class 12 preparation & Acid Base Titration : Solutions - Chemistry, Class 12 in Hindi
for Class 12 courses. Download more important topics, notes, lectures and mock test series for Class 12 Exam by
signing up for free.
Description
Video Lecture & Questions for Acid Base Titration : Solutions - Chemistry, Class 12 Video Lecture - Class 12 full syllabus preparation | Free video for Class 12 exam.
Information about Acid Base Titration : Solutions - Chemistry, Class 12
Here you can find the Acid Base Titration : Solutions - Chemistry, Class 12 defined & explained in the simplest way possible. Besides explaining types of Acid Base Titration : Solutions - Chemistry, Class 12
theory, EduRev gives you an ample number of questions to practice Acid Base Titration : Solutions - Chemistry, Class 12 tests, examples and also practice Class 12 tests.
|
Explore Courses for Class 12 exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
past year papers
,Sample Paper
,Viva Questions
,Class 12 Video Lecture
,video lectures
,Acid Base Titration : Solutions - Chemistry
,Free
,Exam
,Objective type Questions
,ppt
,Important questions
,MCQs
,Semester Notes
,Class 12 Video Lecture
,mock tests for examination
,Acid Base Titration : Solutions - Chemistry
,Previous Year Questions with Solutions
,practice quizzes
,Summary
,shortcuts and tricks
,study material
,Acid Base Titration : Solutions - Chemistry
,Class 12 Video Lecture
,Extra Questions
;
Study Acid Base Titration : Solutions - Chemistry, Class 12 on the App
Students of Class 12 can study Acid Base Titration : Solutions - Chemistry, Class 12 alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the Acid Base Titration : Solutions - Chemistry, Class 12,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of Acid Base Titration : Solutions - Chemistry, Class 12 is prepared as per the latest Class 12 syllabus.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup