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### Course: Chemistry archive > Unit 8

Lesson 4: 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
- Introduction to titration

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# Titration of a weak acid with a strong base (continued)

Calculating the pH for titration of acetic acid with strong base NaOH at equivalence point and past the equivalence point. Created by Jay.

## Want to join the conversation?

- For the last part, how come you didn't have to solve for whatever OH- concentration the acetate then turned the water into like in part c?(27 votes)
- You're right. Hydrolysis of A⁻ does generate some OH⁻.

A⁻ + H₂O ⇌ HA + OH⁻

But the amount is so small compared to the excess NaOH that you have added that it is negligible.

It's like saying that x is negligible in all your other calculations.(39 votes)

- At5:40why did we further react acetate with H2O instead of just using the Molar concentration of 0.04 of acetate to calculate the pH?(28 votes)
- Since CH3COO- is a weak basic salt, it would not stay disassociated like Cl- salt. So some CH3COO- will be protonated to form CH3COOH. This reassociation to form the acid is important here since there is only the acetate present because as the acetate converts back to acetic acid, OH- is formed. And there is no CH3COOH present in the original solution to neutralize the OH- to form water, and will result in a pH change. Basically, the 0.04 M acetate solution is not a buffer solution, so protonation of the acetate will result in a pH change.(18 votes)

- How can you tell if X is going to be a very small number to count as negligible ?(6 votes)
- If the initial concentration is greater than 400×Ka or 400× Kb, x is negligible.(11 votes)

- If I have a solution which contains is affected by common ion effect (lets say 0.5 mol HF and 0.9 mol NaF) and then I apply titration using 2g of NaOH, how do i take account of both common ion effect equilibrium and the weak acid titration equilibrium?(5 votes)
- You are starting with 0.50 mol of HF and 0.90 mol of F⁻.

You are adding 0.05 mol of OH⁻.

This will react with 0.05 mol of HF and form 0.05 mol of F⁻.

You will then have 0.45 mol of HF and 0.95 mol of F⁻.

These are the numbers you use to calculate the new pH.(4 votes)

- For part D, why didnt you use the same steps for part C, given a second ice table?(4 votes)
- In part D, the acetic acid had all been neutralized, so all you had was excess NaOH.

There was effectively no equilibrium to calculate, so there was no need for an ICE table.(4 votes)

- I understand this completely, and this process. Will we have to memorize Ka values for acids and bases on the MCAT? I can't do these calculations without a calculator.(2 votes)
- No you will not have to memorize the Ka values. On the MCAT, the values they give you are easily doable without the calculator so it won't be a big deal.(8 votes)

- Why can't we use the Henderson-Hasselbach Equation for part c?(2 votes)
- The CH3COOH is all used up in the reaction with OH from NaOH. That means that it is no longer a buffer solution, and Henderson-Hasselbach is only for buffers.(8 votes)

- At9:00, are we ignoring Hydrolysis of CH3COO- due to common ion effect of OH- ?(5 votes)
- at 13.48, why can we calculate the POH from left-over OH- concentration directly instead of putting both CH3COO- and OH- together and re-calculate the [OH-]. though the increase amount is too small and the final result is the same. but, why we can do it? under what situation we can and cannot? Tks.(3 votes)
- I think you answered your own question ...

When dealing with a strong base and a weak acid/weak base buffer system, once you pass the equivalence point the contribution from the weak acid/weak base becomes trivial. This can easily be checked using the known excess of OH¯, the concentration of the weak conjugate base (CH₃OO¯) and the Kb.

We first assume that the OH¯ only comes from NaOH and calculate what happens to the acid-base equilibrium at this point.

Plug those values into the equation for Kb:`Kb = [OH¯] • [CH₃OOH] / [CH₃OO¯]`

5.6e-10 = 0.014 M • [CH₃OOH] / [CH₃OO¯]

[CH₃OOH] / [CH₃OO¯] = 5.6e-10 / 0.014 M = 4.0e-8

So, for every CH₃OO¯ that reacts with water to form CH₃OOH and OH¯, there will be 25 million CH₃OO¯ that don't react.(4 votes)

- The reactions vary a lot in different stages. How to know which one to use at which point?(4 votes)
- Trust the Henderson Hasselbach equation and know precisely what it is you are trying to do. If you want pKa or pH and have the other, use the Hend-Hass equation. If you want to know how much dissociate, use the other formula.(1 vote)

## Video transcript

- [Voiceover] We've been
looking at the titration curve for the titration of a
weak acid, acetic acid, with a strong base, sodium hydroxide. And in Part A, we found
the pH before we'd added any base at all. So, we found this point
on our titration curve. And we also found in Part B, the pH after you add 100 mL of base. So, we found the pH at this point. In Part C, our goal is to
find the pH after the addition 200 mL of 0.500 molar solution of sodium hydroxide. So, how many moles of hydroxide
ions are we adding here? So, if the concentration
of sodium hydroxide is 0.0500 M, that's the same concentration for hydroxide. So, for hydroxide, we
have the concentration of 0.0500 M, and molarity is moles over liters, right? So moles over liters. So, we have 200 mL, and if
we move our decimal place one, two, three, that's 0.2 liters. All right, so this is 0.2000 liters here. So, we solve for moles, so 0.05 times 0.2 is going to give us 0.0100, 0.0100 moles of hydroxide ions. So, that's how many moles of
hydroxide ions we're adding to our original acid solution. How many moles of acid did we start with? Well, we had 50 mL of a 0.200 molar solution of acetic acid. So, the concentration of acetic acid, the concentration of
acetic acid was equal to 0.200 M, and that's equal to moles over liters. So, 50.0 mL would be 0.0500 liters. So this would be 0.0500 liters. Once again, solve for moles. So, 0.200 times 0.0500 gives us 0.0100 moles of acetic acids. So notice we have the same number of moles of acid as we do of base. And the base is going
to neutralize the acid. Let's go ahead and write the reaction. Let's write the neutralization reaction. If we have, we start
with some acetic acid. And to our acidic solution we add our sodium hydroxide. So, we're adding some
sodium hydroxide here. The hydroxide ions are going
to take the acidic proton. So, a hydroxide ion takes
this acidic proton right here. H plus and OH minus give us H2O. If you take away the acidic
proton from acetic acid, you're left with acetate, CH 3 COO minus. And we're starting with .01 moles of acetic acid. So, let's color coordinate here. So, we're starting with 0.0100 moles. Over here we'll put moles. So, 0.0100 moles of acetic acid. And that's the same
number of moles of base. 0.0100, so we have 0.0100 moles of base. Notice our mol ratio is one to one, so we have one to one here. So, all of the base is going to react. It's going to completely
neutralize the acid that we originally had present. So, all of our base
reacts, and we end up with zero here, and all of our acid has been completely neutralized. We lose all of this. So, we lose all of that,
and so we've neutralized all of our acid too. So, this is our equivalence
point for this titration. And if we're losing acetic acid, we're converting acetic acid into acetate. So, if we think about
starting with zero moles of acetate, and we lose 0.0100 moles of acetic acid, that turns into acetate. So, we have to write
plus 0.0100 over here, so we're making that. So, we end up with 0.0100 moles of our acetate anion. All right, next. If we have moles of our acetate, and we have a concentration. So, if we find the total
volume of our solution, we can find the concentration
of acetate anions. So, let's do that next. We'll go back up here. What is the total volume now? We started with 50 mL,
and we added 200 more. So, 50.0 and 200.0 give us 250.0 mL. So, that's our total volume now. And 250.0 mL would be, move
our decimal place three, 0.25 liters. So, next we find our
concentration of acetate. So, what is our
concentration of acetate now? It would be moles over liters, so 0.0100. So we have 0.0100 moles over 0.2500 liters, over 0.2500 liters. So, we can go ahead and
find the concentration. You can do this in your
head or I'll just show you on the calculator, .01 divided by .25. It's going to give us a
concentration of 0.04. So, the concentration is 0.0400 M. So, we have acetate anions in solution at the equivalence point. And acetate reacts with water. So, let's go ahead and show that reaction. So, we have acetate, which
can react with water. And this reaction will eventually come to an equilibrium here. So the acetate anion acts as a base and takes a proton from water. So, if acetate picks up a proton, it turns into acetic acid. So, CH3COOH. If we take a proton away from water, we're left with hydroxide, so OH minus. So, now let's think about our initial concentration of acetate. So, our initial concentration
of acetate was 0.0400. So, we write here initial concentration is 0.0400 M. And we're assuming that we
don't have any products yet, so we write zero here. Next we think about our change. Well, a certain concentration
of acetate is going to react, so we're going to lose
a certain concentration, which we call X. Whatever we lose for acetate,
we gain for acetic acid. So if it's minus X for acetate, it must be plus X over here for acetic acid. And therefore, also, plus X for hydroxide. All right, so at equilibrium... At equilibrium, our concentration would be 0.0400 minus X for acetate. And then we'd have X over
here and X over here. So, if we write an equilibrium
expression for this, acetate is acting as a base. So, for our equilibrium
expression, we would write K b. And remember, that's concentration of products over reactants. So that would be X times X, so X times X, all over 0.0400 minus
X, leaving water out. So, this is over 0.0400 minus X. And we did all this in
weak base equilibrium, so make sure that you watched that video before you watch this one, because I have to go a
little bit faster here. All right, next we need
to find the Kb value. We can get the Kb by... Because we know the Ka. In the last video, the Ka for acetic acid was 1.8 times 10 to the negative 5. And we know Ka times Kb for a conjugate acid-based
pair is equal to 1.0 times 10 to the negative 14. Again, this is from an earlier video. So, if we plug in Ka into here, we can solve for Kb. And I won't do it, to save
time, on the calculator. I'll just give you the
answer, that Kb is equal to 5.6 times 10 to the negative 10. So, we plug this in to our
equilibrium expression, so we get 5.6 times 10 to the negative 10 is equal to X squared over... Here we make the assumption,
like we did in all the earlier videos, that
X is a very small number, it's a very small concentration. So, 0.0400 minus X is approximately the same thing as 0.0400. So, we write 0.0400 in here. And next we need to solve for X. So, let's take out the calculator. So we have 5.6 times 10 to the negative 10. We need to multiply that by .04. And then we need to take the
square root of our answer. And so we get X is equal to 4.7 times 10 to the negative 6. So, X is equal to 4.7 times 10 to the negative 6. What does X represent? We go up here, and we notice that X represents the concentration
of hydroxide ions. So this is equal to... This is equal to the
concentration of hydroxide ions. So 4.7 times 10 to the negative 6 molar. All right, our goal was to find the pH. So, at this point it makes
sense to find the pOH. pOH is equal to the negative
log of the concentration of hydroxide ions. So we plug that into here,
and we solve for the pOH. Negative log of 4.7 times 10 to the negative 6 give us a pOH of 5.33. So, the pOH is equal to 5.33. And one more step. pH plus pOH is equal to 14.00. So, if we plug in our pOH into here, pH is equal to 14.00 minus 5.33, which is 8.67. So, we're at the equivalence point, but this is a titration of a
weak acid with a strong base. And so, we have a basic salt solution at the equivalence point. So, our pH is in the basic range. It's above seven. So, let's find that point
on our titration curve. So, we've added here 200 mL of our base. And the pH was 8.67. So we've added 200.00 mL of our base, and the equivalence point
should be somewhere in there. So right about there, about 8.67. So, that's our equivalence point for a titration of a weak acid with a strong base for
this particular example. Finally we're on to Part D, which asks us, what is
the pH after the addition of 300.0 mL of a 0.0500 molar solution of sodium hydroxide? So, once again, we need
to find the moles of hydroxide ions that we are adding. The concentration would
be equal to 0.0500, so 0.0500 molar is our concentration of hydroxide ions. And that's equal to moles over liters. So, 300.0 mL would be 0.3000 liters. So, we have 0.3000 liters here. Multiply 0.0500 by
0.3000, and you get moles. So, 0.0500 times 0.300 is equal to 0.0150 moles of hydroxide ions. In Part C, we saw that
we needed 0.0100 moles of hydroxide ions to completely neutralize the acid that we originally had present. So, we're going to use up... We're going to use up 0.0100 moles of hydroxide. That's how much was necessary
to neutralize our acid. All right, so how many moles
of hydroxide are left over after the neutralization? Well, that would just be 0.0050. So, 0.0050 mol of hydroxide ions are left over after all the
acid has been neutralized. So, our goal is to find the pH. So, we could find the pOH if we found if we found the
concentration of hydroxide. So, what is the concentration
of hydroxide ions now after the neutralization has occurred? So, concentration is moles over liter, so it's 0.0050 divided by... What's our total volume? We started with 50.0, and
we have now added 300.0 mL more, so 300.0 plus 50.0 is 350.0 mL. Or 0.3500 liters. So, what is our concentration
of hydroxide ions? So, this is .005 divided by .35. So our concentration of hydroxide ions is 0.014. So, let's write 0.014 M here. Once we know that, we
can calculate the pOH. So the pOH is the negative log of the concentration of hydroxide ions. So, it's the negative log of 0.014. So, we can do that on our calculator. Negative log of .014. And we get a pOH of 1.85. All right, so our pOH. Let's get a little bit more room here. pOH is equal to 1.85. And finally, to get the pH, we know that pH plus pOH is once again equal to 14.00. So, we plug in the pOH into here. And the pH would be equal to 14.00 minus 1.85. So 14.00 minus 1.85 gives us 12.15. So, we're now past the equivalence point. All right, so we're past
the equivalence point here. So our pH is 12.15. After we've added 300.0 mL of our base. So, let's find the point
on our titration curve. We've added 300.0 mL of our base, so we're right here. So, we go up to our titration curve, and that would be right here. So, we just found the pH
is a little bit over 12. So approximately, we've got 12.15. So that's the pH right here
on our titration curve.