Main content

## Chemistry library

### Unit 14: 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
- Introduction to titration

© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice

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

Calculating the pH for titration of weak base, ammonia, with strong acid, HCl, at the equivalence point and past the equivalence point. Created by Jay.

## Want to join the conversation?

- In the last part of this problem (d), there will also be formation of 0.004 moles of ammonium ion ( after the equivalence point ) like in part (c).Why didn't we consider the reaction of ammonium ion with water, that I think will afect the pH ??(32 votes)
- You're right. The reason you can ignore the fact that the ammonium exists in solution is that it is relatively a much weaker acid than the HCl. The HCl will completely dissociate, compared to only a very small portion of the ammonium, so the HCl will be much more responsible for the pH of the solution.(28 votes)

- In part a) we only did the ICE and find pH

In part b) we did the ICE and then did the HH because it is a buffer solution

In part c) we did the ICE and then did it again for NH4 to find the pH using Ka

In part d) we only found the mol of HCl and subtracted NH3 from HCl and divided it by .1 liters to find the M of H30+ to calculate pH.

Are the procedure all different because of the difference in liters compared to NH3? If so do we use d) to solve 80ml?(7 votes)- The procedures are different because they are all different chemical situations.

In part a), you have a solution of a**weak base**, so you must use an ICE table.

In part b), you use HH because you have a buffer, a solution of a**weak base and its conjugate acid**.

In part c), you use ICE again because you have a solution of a**weak acid**.

In part d) you have a solution of a**strong acid**.

You must use the**total**volume in each case to calculate the concentrations of the components.

After adding 80.0 mL of 0.100 mol/L HCl, you have in solution (0.00800 -0.00400) mol = 0.00400 mol of H₃O⁺.

The total volume of the solution is (40.0 + 80.0) mL = 120.0 mL = 0.1200 L.

[H₃O⁺] = 0.00400 mol/0.1200 L = 0.0333 mol/L

pH = -log [H₃O⁺] = -log(0.0333) = 1.48(21 votes)

- at the beginning of all these problems you equation HCL to H30. how and why are we able to do this? my biggest problem with titrations is just setting up the equation.(2 votes)
- The reaction is hapenning in water so:

HCl + nH2O -» (H3O+) + Cl-

According to Arhenius definition of acids and bases, HCl is a proton giver. So, we'Re left with Cl-. And H+ has to go somewhere so it sticks to H2O. because it has a free pair of electons.

Also, you can watch

https://www.khanacademy.org/science/chemistry/acids-and-bases-topic/acids-and-bases/v/acid-base-definitions-1(2 votes)

- Why do we go through so much work in the previous questions b and c and then only need to do so little in the last one?(2 votes)
- Because it's an easy question, particularly if you did all the work in b and c, and know that the acid dissociates completely.(1 vote)

- Inception like question here...

The first part of the video shows how NH3 + HCL -> NH4+ + Cl- in aqueous solution

After this reaction, 0.002 moles of 0.1M HCl (H3O+) remains, as well as 0.004 moles of NH4+

In previous videos we saw how the product from a similar reaction, NH4+, itself reacts with H2O to also form H3O+ ions... If we follow this same logic through, when solving for this reaction by using a previously given Ka value, we solve for incremental H3O+ moles = 4.73*10^-7

So is it fair to say that the total number of H3O+ moles with which we solve for pH are in reality 0.002 mol + 4.73*10^-7 mol which in effect equals ~0.002 mol given how small the latter value is?

The pH of 0.002 mol of H3O+ ions = 1.699 (rough rounding)

The pH of ~0.002 mol of H3O+ ions = 1.698 (rough rounding, but lower pH value...)

Is it correct to say then that because the incremental H3O+ mols are so negligent, we overlook these moles as well as the (ever smaller) products from continued reactions that will lead to additional H3O+ moles?

(NH3 reacts with a strong acid that produces NH4+ which in turn reacts with H2O weak base that produces NH3, that itself will react with weak H2O acid that will produce NH4+, etc., etc. etc... approaching 0) ?

If so, does a titration curve behave as if it were approaching an asymptote as the reactions within a solution never fully cease to occur?

I really hope this makes sense! I'm going nuts over here.(2 votes)- Don't worry about that last bit too much. The Ka of NH4+ and Kb of NH3 take what you're talking about into account.(1 vote)

- At4:47, we calculate the total moles of NH4 given off by the reaction assuming it starts with 0 moles of NH4. Although, wouldn't we start with more than 0 moles because the NH3 molecule would have been disassociating before we even added the HCl?(2 votes)
- You start with NH3 and you have 0.004mol of NH3. Afterward, it reacts with HCl to form NH4. Before the reaction, NH3 remains as NH3 and does not start dissociating.(1 vote)

- At10:08But x also represents NH3 why should we not take it as the concentration of NH3 ions ?(2 votes)
- We only care about the concentration of H3O+ because that's what we use to calculate the pH, which was what the question was asking for. x does represent the concentration of NH3 as well, but we don't need to use its value for anything.(1 vote)

- Can something that is an acid in one solution behave as a base in another solution (and vice verse for bases)?(1 vote)
- Yep! Water is a perfect example. (may be the only one, I'm not 100% sure) Depending on the other reactant, water can act as a base or as an acid!(2 votes)

- since this is a titration of a weak base and a strong acid, I assume ammonia is a weak base; doesn't that mean that its conjugate acid, NH4+, is a strong acid, therefore having a complete dissociation in part c?(1 vote)
- Nope!

While a strong acid always has a (very) weak conjugate base, moderately weak acids have, moderately weak conjugate bases (and*vice versa*). This is true for NH4+ and its conjugate base NH3 and will always be true for an acid-base pair that constitutes a buffer.

Remember that a strong acid (or base) more-or-less completely dissociates in water. This implies that its conjugate base (or acid) more-or-less completely**fails**to go in the reverse direction. In practice the conjugate base (or acid) is not really acting as a base (or acid) at all!

In contrast, moderately weak acids and bases undergo partial dissociation in water and by necessity their conjugate base or acid**must**also undergo partial dissociation.(2 votes)

- Why is the value of X negligible?(1 vote)
- x can be negligible only in comparison with something else.

It is not negligible by itself.

For example, if you had $100, you might consider $5 to be a negligible amount.

If you had only $5, $5 would not be negligible.

A common criterion is that a quantity is negligible if it is less than 5 % of something else.

Most weak bases dissociate by only less than 5%, so the initial concentration usually decreases by less than 5 %.

x is negligible only**in comparison with**the initial concentration.(2 votes)

## Video transcript

- [Voiceover] In the
last video, we looked at the titration curve for the titration of a weak base with a strong acid. And we found the pH at two
points on our titration curve. We found the pH before
we'd added any of our acid. So that was right here. We did that in part a of our question. And part b of our
question, we found the pH after we'd added 20
milliliters of our acid. So that point is right here
on our titration curve. And we found the pH in part b. In part c our goal is to find the pH after we've aded 40 milliliters of our 0.1 molar solution of HCl. So how many moles of acid
did we add at this time? So the concentration of HCl is equal to 0.1 molar, so that's equal to 0.1 molar, molarity is moles over liters. So how many moles of acid did we add? We added 40 milliliters which, if we move our decimal
place one, two, three, that's 0.04 liters. So we added 0.04 liters and we can just solve for moles here. So the moles of HCl will be equal to 0.1 times 0.04. So 0.1 times 0.04 is equal to 0.004. So that's how many moles
of HCl we have added. So you can think about
this as H plus and Cl minus and, if you have 0.004 moles of HCl, that's how many moles
of H plus you have too. Or you can think about H plus and H two O giving you H three O plus, or hydronium. So you can also say, 0.004 moles of H three O plus. Alright, how many moles of
ammonia did we start with? Right, well our concentration was 0.1, so the concentration of
ammonia that we started with was equal to 0.1 molar, and once again, molarity
is moles over liters, we started with 40 milliliters of ammonia, which is 0.04 liters. So we started with 40 milliliters, or 0.04 liters, so to find moles of ammonia, it's the same calculation we just did. 0.1 times 0.04 gives us 0.004 moles of ammonia, moles of NH three. So the acid that we
added, in our titration, is going to react with the
base that we started with. So, let's go ahead and write out the neutralization reaction. So one way you can write that would be, NH three plus H three O plus. So, ammonia, NH three, plus
hydronium, H three O plus, so H three O plus is our acid, right? It donates a proton to ammonia, so if NH three picks up a proton, it turns into NH four plus and if H three O plus donates a proton, we're left with H two O. So that's one way to
represent our reaction. The other way we could write it would be, NH three plus HCl. So let's write it that way too. So ammonia plus HCl, HCl donates a proton to NH three giving us NH four plus and once HCl donates a proton, you have Cl minus, the chloride
anion, the conjugate base. So this would be NH four plus Cl minus. Alright, let's write down how many moles of acid that we have. So we added 0.004 moles of our acid. So let's write down 0.004 here, for moles of acid that we added. And for moles of base
that we started with, that's also equal to 0.004. Let's get some more room down here. So we started with
0.004 moles of our base. If you look at your molar ratios there, one to one mole ratio. So we have now added enough acid to completely react with the base that we originally had present. So this represents the equivalence point of our titration. This is the equivalence
point for our titration, because we've now added enough acid to completely react with our base. So all of this acid is
going to react, right? So we use up all of that and it uses up all of the
base that we had present. So all this base is going to react with the acid we're gonna lose, all 0.004 moles of NH three. So we're left with zero of that. Ammonia turns into ammonium, NH four plus. And so, if we start with
zero for NH four plus, whatever we lose for ammonia, was what we gain for NH four plus. So for losing 0.004 moles of ammonia, that's how many moles of
ammonium we are gaining. So we gain 0.004 moles of NH four plus. So, after we add the acid, we're left with 0.004 moles of NH four plus. So this is at our equivalence point. So we have some ammonium present. What's the concentration of
ammonium that we have present? Well, the concentration, once again, is moles over liters. So we have 0.004 moles of NH four plus, so let's write 0.004
moles of NH four plus. What is the volume? What's the total volume now? Well let's go back up to the problem. We started with 40 milliliters of our... We started with 40 milliliters and we added another 40. So 40 plus 40 is 80 milliliters. Or one, two, three, 0.08 liters. So our total volume now is 0.08 liters. So let's go ahead and write that in here. So 0.08 liters. So 0.004 divided by 0.08 is just equal to 0.05. So 0.05 molar is the concentration of NH four plus that is present. Alright, ammonium is a weak acid. Ammonium can react with
the water that is present. So let's now write out that reaction. So NH four plus is going
to react with the water. And ammonium is going
to function as an acid. Ammonium donates a proton to water. So if NH four plus donates
a proton to H two O, H two O turns into H three
O plus, or hydronium. And if you take away a
proton for NH four plus, you're left with NH three or ammonia. So ammonium reacts with water and that's going to affect your pH. If I go back up here, so NH four plus does react with water,
but the chloride anion does not react appreciably with water. So you don't have to worry
about this being present, but the NH four plus will affect the pH, which our goal, of course,
is to figure out the pH. Alright, so let's get some more room and let's write in our
initial concentration of NH four plus, which is 0.05. So the initial concentration
of NH four plus is 0.05. And, once again, for these problems, these weak acid equilibrium problems, we assume that we're starting with zero for the concentration of
our products and our change, we're going to lose a certain
concentration of ammonium. A certain concentration of ammonium is going to react with water. So we're gonna lose x. And then whatever we lose for
ammonium turns into ammonia. So if we lose x for the
concentration of ammonium, that's the same
concentration that we gain, for NH three, for ammonia. And therefore, that's also
the same concentration of NH three O plus. So at equilibrium, we're gonna have 0.05 minus x for the concentration of NH four plus. For the concentration on
H three O plus, we have x. And for the concentration
of NH thee, we also have x. Next, this is functioning
as an acid, alright? NH four plus is donating
a proton to water, so for the equilibrium expression, we have to write Ka here. And Ka is equal to concentration
of products over reactants, so we have concentration
of H three O plus, which is x times
concentration of NH three, which is x all over the
concentration of NH four plus, which is 0.05 minus x. So 0.05 minus x. Ka for NH four plus, Ka for ammonium, we calculated in an earlier video and we found it to be 5.6 times ten to the negative ten. So this is equal to... This is equal to, x times x is x squared and 0.05 minus x, if x is a very small concentration, 0.05 minus x is approximately
the same as 0.05. So we write 0.05 in here, to make our lives easier
for the calculation and we can get out the
calculator and solve for x. 5.6 times ten to the negative ten. Alright, we're gonna multiply that by 0.05 and then we're gonna take
the square root of our answer and solve for x. X is equal to 5.3 times ten to the negative six. So let's go ahead and write that down. X is equal to 5.3 times ten to the negative six and we go back up here
to see what x represents. X represents the concentration of hydronium ions at equilibrium. So this would be the
concentration of H three O plus. We've done a lot of work and the last thing we
have to do is find the pH. And now we've found the
concentration of hydronium ions, the pH is just the negative log of what we just found. So negative log of 5.3 times
ten to the negative six Let's get out the calculator and do it. So we have the negative log of 5.3 times ten to the negative six and we get a pH of 5.28 if we round that. So our pH is equal to 5.28. So at the equivalence point,
our solution is acidic, right? A pH of 5.28. Let's find that on our
titration curve here. So we've added 40 milliliters of our acid and our pH is 5.28. So. let's say that is 5.28 here. So this is part c) of our question. The pH is equal to 5.28 at the equivalence point. Finally in part d), our
goal is to find the pH after we've added 60 milliliters of our 0.1 molar solution of HCl. So this time, how many moles of acid have we added? We can find that from the molarity since our concentration
is 0.1 molar, of HCl. And molarity is moles over liters. So we're adding 60 milliliters,
which is 0.06 liters. So that's 0.06 liters, we solve for moles. 0.1 times 0.06 is equal to 0.006. So that's how many moles
of HCl we've added now. HCl is H plus and CL minus, so you can think about this
and being 0.006 moles of H plus or add your proton onto water to form hydronium, H three O plus. So 0.006 moles of acid and remember from part
c), we needed 0.004 moles to completely react with the ammonia. So, at this point, we're
gonna lose 0.004 moles, because that much is necessary
to react with our ammonia. So how much is left over? So after our acid has
reacted with the ammonia, we're left with 0.002 moles of acid. So that's how much we
have present in solution. So, next, since our
goal is to find the pH, we need to find the
concentration of hydronium ions. So the concentration of hydronium ions will be equal to the moles, which is 0.002 and we need to divide that by our volume. Well, our volume will be 40
milliliters plus 60 milliliters, which will be 100 millililters. So that's our total volume here. 100 milliliters will be 0.1 liters, so we have 0.1, we have 0.1 liters here. and we can solve for the
concentration of hydronium ions. So 0.002 divided by 0.1 gives us a concentration of hydronium equal to, this is 0.0200 molar. So now we're ready to find the pH since the pH is equal to the negative log of the concentration of hydronium. So the negative log of 0.0200, let's get out the calculator, is equal to, so the negative log of 0.0200 is equal to 1.699, so our pH here, our pH is equal to 1.699, or however you want to think
about rounding these things. You can just say 1.70 just to make things a little bit easier, a little bit easier to
think about that number. So approximately 1.7. After that it's 60
milliliters of our acid, so on our titration curve, we've added 60 milliliters of acid, so here we are right here and our pH is pretty close to 1.7, so we're past the equivalent point on our titration curve now.