Main content

### Course: AP®︎/College Chemistry > Unit 13

Lesson 2: Titrations- Acid–base titrations
- Worked example: Determining solute concentration by acid–base titration
- Titration of a strong acid with a strong base
- Titration of a strong acid with a strong base (continued)
- Titration of a weak acid with a strong base
- Titration of a weak acid with a strong base (continued)
- Titration of a weak base with a strong acid
- Titration of a weak base with a strong acid (continued)
- 2015 AP Chemistry free response 3b
- 2015 AP Chemistry free response 3c
- 2015 AP Chemistry free response 3d
- 2015 AP Chemistry free response 3e
- 2015 AP Chemistry free response 3f
- Titration curves and acid-base indicators
- Redox titrations

© 2024 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Worked example: Determining solute concentration by acid–base titration

The concentration of an acid solution can be determined by titration with a strong base. First, calculate the number of moles of strong base required to reach the equivalence point of the titration. Then, using the mole ratio from the balanced neutralization equation, convert from moles of strong base to moles of acid. Finally, divide the number of moles of acid by the given volume of the acid solution to find the concentration. Created by Jay.

## Want to join the conversation?

- When using the shortcut MV = MV for the acid and the base, shouldn't the unit of the volume be in litres (the same way that molarity uses litres)?(11 votes)
- When you divide both sides by 20 mL, the units cancel out so it does not matter what you choose. All that matters is that the unit you choose is the same throughout the equation. Dividing 27.4mL by 20mL is the same as dividing 0.0274L by 0.02L. You will get the same number and there will be no units! So, we can say that mL were used simply because the information was given in mL and it would have been unecessary to change.(48 votes)

- why are we not dividing the molarity by total volume for both parts when solving? Or in general, why are we not adding both volumes to make one value for a total volume when solving for this question?(4 votes)
- You have a base reacting with an acid. The question asks how much acid you need to react with base so that they neutralize each other (and form a salt with water, but no floating acids or bases). So, when are MV(basic)=MV(acidic). The greater volume that is made will not influence the equilibrium point because water is at pH 7 (neutral) so the ratio to total volume is irrelevant.(7 votes)

- what is the molar solution?(2 votes)
- Molar solution is actually a concentration term. This means when a given solution contains one mole of atoms , ions, molecules, or any chemcial compound in its one litre solution, that solution is called molar solution(4 votes)

- how did you predict the product of the acid-base reaction at1:57? I never learned that(3 votes)
- Acids (using the Arrhenius definition) are chemicals which produce H+ ions, while bases are chemicals which produce OH- ions. So when barium hydroxide (Ba(OH)2) and hydrochloric acid (HCl) enter a water solution they will dissociate, or split up. The barium hydroxide will split into one barium ion, Ba^(2+), and two hydroxide ions, OH-. And the hydrochloric acid will split into one hydrogen ion, H+, and one chloride ion, Cl-.

At this point the reaction becomes a double replacement reaction essentially where the hydroxide and hydrogen ions combine to form water, OH- + H+ = H2O. And the barium and chloride ions form barium chloride, BaCl2. For the barium chloride we need that 1:2 ratio of barium to chloride ions because it is an ionic compound and the charges must combine and cancel each other to make a neutral compound.

Hope that helps.(3 votes)

- at2:03how did we know we had Ba2+ and cl-?(2 votes)
- We work with ions. The barium ion has a charge of 2+ and chlorine ion has a cahrge of 1-.(5 votes)

- what does titration and finding the equilibrium help us with(2 votes)
- That simply tells you how much acid (or base) you need to add to a solution to neutralize it. This becomes particularly important when you do research in a lab because neutralization will become very important.(4 votes)

- What was the pH indicator used for this titration?

Would this influence the final result?

***(2 votes)- The ph indicator used here is phenolphthalein. these indicators are used not more than 1 to 2 drops, so they influence very less. To know more how the structure of Hph (phenolphthalein) changes after reacting with a base. hope it helps.(3 votes)

- Is it necessary to include all the zeros after the number ? ( 0.02*00*)(2 votes)
- They are significant figures so yes they’re quite important, they show the accuracy of the measurement.(2 votes)

- So when using the shortcut equation, you wouldn't have to convert volumes into liters? That made no sense to me as molarity is in moles per liter, but his volumes were in milliliters?(2 votes)
- You're right, you wouldn't need to convert the volumes to Liters, because the units are in milliliters on both sides of the equation. When you divide both sides by 20.00 mL, the units cancel just as they would if they were written in Liters, and the ratio would still be the same. So, 27.4mL/20.0mL = 0.0274 L/0.020 L = 1.370(unitless).(2 votes)

- but what would be the conc of oh- ions

is is twice that of Ba(OH)2(2 votes)

## Video transcript

- [Voiceover] Let's do
another titration problem, and once again, our goal is to find the concentration
of an acidic solution. So we have 20.0 milliliters
of HCl, and this time, instead of using sodium
hydroxide, we're going to use barium hydroxide, and it
takes 27.4 milliliters of a 0.0154 molar solution
of barium hydroxide to completely neutralize
the acid that's present. All right, so let's
start with what we know. We know the concentration
of barium hydroxide. It's 0.0154 molar, and we also know that molarity is equal to moles over liters. All right, so we have 0.0154 as equal to, let's make moles X, over liters. 27.4 milliliters is 0.0274 liters, right? So that's 0.0274 liters. We solve for X, and X of course represents the moles of barium hydroxide. So let's get out the calculator
here, and let's do that. So let's get some room over here. So we take 0.0154 and we multiply that by 0.0274, and that gives us, this will be 4.22 times 10 to
the negative fourth, right? So that's equal to 0.000422 moles of barium hydroxide. All right, next, let's write
the neutralization reaction. So we have barium
hydroxide reacts with HCl. So barium hydroxide plus HCl
gives us, for our products, we have H plus and OH
minus, so that's H20. And then our other product,
this is barium two plus, right? This is BA2 plus, and over
here we have Cl minus 1. So we have BA2 plus and CL minus 1, so you could cross those over. So BACl2, right? So BACl2, barium chloride,
as our other product here. All right, next we need to
balance our equation, right? We need to balance the
neutralization reaction here. So let's start by
looking at the chlorines. So over here on the left,
we have one chlorine. On the right, we have two. So we need to put a two right here, and now we have two
chlorines on both sides. Next, let's look at hydrogens. So on the left side, we
have two hydrogens here, and then we have two over here. So we have four hydrogens on the left. On the right, we have only two hydrogens. So we need to put a two
here for this coefficient to give us four hydrogens on the right. So now we have four, and we
should be balanced, right? Everything else should be balanced. Let's look at the mole ratio
for barium hydroxide to HCl. For every, right, there's a one here. So for every one mole of barium hydroxide, we have two moles of HCl. So we already calculated how
many moles of barium hydroxide that we used in our titration, right? That's 0.000422. So therefore, we had twice as many of HCl. So we can multiply this number by two, and we'd figure out how
many moles of HCL we have. Or you could set up a proportion. Right, so if we're
talking about a ratio of barium hydroxide to HCl, our mole ratio is one to two. Right, and our moles of barium hydroxide, let me go ahead and use
a different color here. That's up here, that's 0.000422
moles of barium hydroxide. Our goal is to find how many
moles of HCl were present. And so obviously you just need
to multiply 0.000422 by two. And so we get X is equal to 0.000844, right? That's how many moles of HCl we have at our equivalence points. All right, so finally,
we just have to calculate the concentration of our
acid solution, right? Let's go back up here so we
can see what we started with. Right, so we started with
20 milliliters of HCl. Right, and 20 milliliters would be, move our decimal place, 0.0200 liters. So now we have moles, right? We have moles, and we have liters. So we can calculate the concentration. All right, so the concentration of HCl in our original solution would be, we had 0.000844 moles. All right, divide that by liters. That was 0.0200 liters, right? 20 milliliters is equal to 0.0200 liters. And so we can do our calculation here. So we can take 0.000844, and we can divide that by 0.0200, and we get for our answer here 0.0422 molar. All right, so the concentration
of HCl is equal to 0.0422 molar, and we are finally done, right? That's our concentration
of our acid solution. Let's see what happens if you try to use MV is equal to MV, that shortcut that we learned about in the last video. So this would be MV is
equal to MV, and let's do the molarity of the base
times the volume of the base is equal to the molarity of the acid times the volume of the acid. So for our base, the
concentration was 0.0154 molar, and the volume of base that we used was 27.4 milliliters in our titration. For the acid, we don't
know what the molarity is. That's what we're trying
to find in the problem, and the volume was 20.0
milliliters, right? So let's do that calculation. So trying to use the shortcut way, 0.0154 times 27.4 gives us that number, divide by 20, right? So we get 0.0211, right? So for our answer, for X, we get 0.0211 molar. And so you can see that's not
the correct answer, right? Here we've got a
concentration that's half of the concentration that we got
when we did it the longer way. And so if you want to use the
shortcut way for this problem, you would have to multiply by two, right? So if you multiply your answer by two, then you get the correct answer, 0.0422 molar. And a lot of the time,
students have a hard time figuring out what you do, right? So where do you mulitply by two? How do you do that? Of course, you can figure
it out by looking at your balanced equation up here, right? But it's tricky for a lot of students. And so that's why the shortcut way isn't always the best way. You can still use it if you
understand how to use it, right? But it's a little bit
better for these problems, when your mole ratio is not one to one, to go through the longer way and do it, if you want to be absolutely sure that you're getting it correct.