# Solving Acid-Base Titration Problems

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We're going to work on solving acid base titration problems.
To begin with we review the mole ratio.
Mole ratios are going to be used to determine the number of moles needed
to neutralize a solution.
Neutralization is a key word here.
So when we combine an acid which has the generalized formula of HA
(so we can say this is our acid) with a base
with a generalized formula of BOH (here's a base).
An acid plus a base combine together
to create a salt; where the cation came from the base
and the anion came from the acid. Salt plus water.
Well where did the water come from?
The water came from hydrogen and hydroxide
reacting together. So what's important
is that we have hydrogen ions coming off the acid.
The acid donates the hydrogen ion and the hydroxide ion is coming off the base.
And when those species get together - wooo - we make water.
And when we've neutralized all of the hydrogen ions with all the hydroxide ions
necessary to create water, and not have any extra hydrogens or hydroxides
we're done. Okay, and this is what we'll be solving for in the titration problem.
But these ratios are going to be really important for us
in figuring out how many hydrogens do we need, how many hydroxides do we need.
So the ratios are going to be important so we need a correct balance equation
with the correct salt product in order to solve the problem.
Additionally, we're going to go back to using the mole map.
So in the mole map that we started with in the beginning of the year
we had 3 arms - the volume of the gas, mass, and representative particles.
And we added into this an extension - molarity - or we could say 'concentration'.
And we said that molarity is equal to
moles of solute over liters of solution.
So I'm going to add molarity to both sides of my mole map here.
So we may not necessarily be solving for molarity
but if we're working with concentrations we want to use this molarity equation.
And we said we could connect the double mole map
by using the mole ratios from the balanced chemical equation.
So when we solve problems
involving acid base titrations we're typically going to start out
right over here. So we're going to start out
with a known amount of a known concentration.
Okay, then we're going to have to go through several steps.
We start with what we know, we'll use the molarity equation in some form
to get this to moles, then we'll go from
moles of our known, to moles of our unknown
using the mole ratio. And then we'll go from moles
of our unknown to molarity or volume in the next step.
So three main steps. They all stem off of
the first step which is actually the balanced chemical equation.
Now typically this is the hardest part for students so you want to make sure that you
remember when you're balancing chemical equations you first have to balance
each individual substance your reactants and your products species.
So don't forget to look for charges and have you written the right subscripts and is a
subscript supposed to be there?
Once you have a balanced chemical equation, we're going to follow the mole map to help us out.
We'll first identify the number of malls of known in our standard solution
and since we're typically going to be working with molarity
we know that molarity equals moles over
volume, so what we want to do is rearrange this
and we know that moles will be equal to molarity times volume.
Okay, molarity times liters.
Then the next step is going to be to convert to moles
of our unknown using the mole ratio. And then finally we'll solve for
our unknown. And when I say we're going to solve for our unknown, we could be solving
for molarity, or we could solve for volume;
we could solve for liters. Once again molarity equals moles
over liters. If I'm solving for
liters, then liters will be equal to moles over molarity.
Hey, so being able to rearrange equations is important.
Let's take a look at a sample problem, here we go.
How many milliliters of .45 molar hydrochloric acid
must be addd to 25 milliliters of 1.00 mole potassium hydroxide
to make a neutral solution?
So step number one is
write the balanced equation. Hydrochloric acid
plus potassium hydroxide.
Alright, rearrange, we're going to form a salt - potassium chloride
and water, H20.
And I can look and see that I have one potassium on each side,
one chlorine on each side, 2 hydrogens on each side,
one hydroxide on each side - so this one is nice and balanced.
Now when we did our stoichiometry problems earlier in the year,
I said to use the equation to help you organize the information.
So let's do that again. So underneath hydrochloric acid
I'm going to write the concentration which is .45 molar.
Hey, and I don't know what volume.
That's what we're trying to solve for.
Under potassium hydroxide I'd have a 1.00 molar solution,
and I have 25 milliliters.
So this allows me to see all of the information
in a simple format. so now I can get into solving the mathematics,
the stoichiometry - the titration kind of stoichiometry for this problem.
So first I want to find moles of my known solution, well what is my known solution?
My known solution is what I know the most about, so in this case
I know the most about potassium hydroxide - so that's going to be my known
solution. Okay, reviewing once again
if molarity equal moles over volume
then moles is equal to
molarity times volume.
So that's what I'm going to do here alright? I have
1.00 molar times .025 liters
okay, remember we want to use liters, so that gives me
.025 moles of my potassium hydroxide solution.
Hey, so I found moles of my known.
Now I'm going to use the mole ratio
to find moles of the unknown.
So this time I have .25 -oops sorry about that -
I have .025 moles of potassium hydroxide
my mole ratio is one to one but let's write it out for
ourselves because not every problem is going to be on to one so get in the habit of
doing this now. Moles of what I'm going to on top, moles of what I'm going to on the bottom
remember the ones come from the balanced chemical equation
and so this time I know that I still have .0 25
moles of HCl.
HCl is my unknown, my unknown is HCl. That's what I"m trying to solve for.
In this particular problem I'm being asked to solve for volume.
Once again, if molarity equals moles over liters
rearranging this equation I get
liters - or volume - equals moles over molarity.
So, I found that I have .025 moles HCl
.45 molar is the concentration the problem gave me,
Solving this, I get an answer of .056 liters
and if you move the decimal, one, two, three places...
then you get the answer of 56 milliliters.
So take a moment to look at all the steps.
We started with the balanced chemical equation - check!
We found the moles of the known solution - check!
We used the mole ratio to find moles of the unknown - check!
And then we solved for the unknown - and in this case we were solving
for volume. So let's see another problem.
In this problem we're being asked to solve for concentration.
Notice the steps are exactly the same.
We'll just have a rearrangement of the molarity equation in number four,
which is actually a little bit easier because for solving for molarity and not
volume this time. So once again start with the balanced chemical equation
You've got a lot of different species to deal with here, so it can be challenging.
Calcium hydroxide (CaOH2) plus
acetic acid (HC2H3O2)
goes to salt - calcium acetate
balanced for all the species and water
going back through and balancing them
Sorry about that.
I'm going to need 2 acetates
and I'm going to need 2 waters
So, balance chemical equation. Once again organize the information that was given
to you in the problem.
So if I take a look at the problem I'm told I have 24.6 milliliters
and I want to find molarity.
And then I have 14.2 milliliters and the known molarity are ready.
So remember, the moles of known solution is the one that you know the most
information about. So I know the most information about acedic acid.
So for acedic acid, I'm going to find moles of my known,
so molarity equals moles over liters
so moles equals molarity time liters
In this case I have .0140 molar
times .0142 liters
(remember I've converted milliliters to liters)
and I get a complicated looking number
1.988 times 10 to the -4th moles.
Very, very small number of moles.
Now I'm going to use the mole ratio to find moles of my unknown
1.988 times 10 to the -4th moles
- this is of the acedic acid -
Alright, so H2 (long pause)
HC2H3O2 (bell ringing in background)
You can tell I'm in my classroom doing this.
1 mole of my unknown
which is what I'm solving for on top
2 moles from the balanced chemical equation
of acedic acid.
And I get 9.94 times 10 to the -5th moles
of my calcium hydroxide. Now I must solve for the unknown. In this problem I want
to know moles. Moles is just - Molaritty.
Molarity is just volume so I have
moles....
volume comes all the way from up here
Alright, so I've got to cover that into volume in liters
so .0246 liters. And my molarity here is a small molarity
4.04 times 10 to the -3rd molar.
So I know it's a lot of information to look through but you've seen this before
we've done these types of problems before
using the double mole map; it's just using the molarity
component as part of the double mole map. Write a balanced chemical equation.
whichever one you know the most information about find the moles
that substance. Then use mole ratio to find that your other substance.
And then solve for the unknown; whether you're solving for molarity,
solving for volume, is just going to use the same equation.
So keep all those steps in mind as you're solving titration problems