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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 strong acid with a strong base

In a strong acid-strong base titration, the acid and base will react to form a neutral solution. At the equivalence point of the reaction, hydronium (H+) and hydroxide (OH-) ions will react to form water, leading to a pH of 7. This is true of all strong acid-strong base titrations. An example of this would be the titration of hydrochloric acid (strong acid) and sodium hydroxide (strong base) to form sodium chloride and water. Created by Jay.

## Want to join the conversation?

- I don't know if it's a stupid question but, why, at the beginning of the titration, we "add" water, or where the water come from?(28 votes)
- If you don't add water then HCl does not dissociate into H+ and Cl- ions. Without them there is no concentration of H+ ions and therefore no pH.(30 votes)

- Why don't you take into account the milliliters of water?(17 votes)
- I believe that this is because volumes are considered to be additive; in other words, the reaction does not create new volume– it merely alters the concentrations of the species (substances) within the overall volume. When you add 10.00 mL of NaOH solution (which includes the water that the NaOH is dissolved in) to 20.00 mL of HCl (also including the aqueous portion of the solution), you end up with a solution that is 30.00 mL. If you've ever watched a video of someone doing one of these titrations, you'll notice that as the reaction progresses, the volume doesn't change appreciably from the measurement taken just after the base is added– that is, the reaction itself does not really alter the volume of the overall solution. Remember that the reactants are consumed as the products (water and NaCl) are produced, so there is no noticeable net change in volume during the reaction, if that makes any sense.(13 votes)

- I'm having a problem about this acid and base topic. Can someone link me another source of this to learn? :((13 votes)
- Why is it important to say 0.500 M (Zero point five zero zero) Why are the two zeros after the 5 also important?(4 votes)
- It tells us they're significant figures and tells the precision of that number. Essentially we have weighed out exactly 0.500 moles of whatever, and we measured it accurately to the third decimal place. It wasn't 0.499 or 0.501.(11 votes)

- At1:36he says, that pH= -log[H3O+]. Can someone explain why?(3 votes)
- pH refers to the proton concentration, so, both -log[H30+] and -log[H+] are the same thing, and refer to pH.(7 votes)

- What is the point of titrating a strong acid or a strong base? Can the concentration of the acid/base not be calculated from a pH value since their protonation reactions with water are essentially unidirectional so that [H3O+] is equal to the total concentration.(2 votes)
- You’ve got it the wrong way around. A pH value is calculated AFTER titration, using the experimental concentration. Titration is generally done on a completely unknown solution whose
*approximate*pH is found using various indicators, mainly litmus in school laboratories.(3 votes)

- can a higher concentration of HCl affect a Ksp value?(3 votes)
- K is a constant that only changes with temperature. See it's derivation using rate laws.(1 vote)

- At6:28, notice the equation that corresponds with question "b". Because this is a 1:1 mole ratio, can't we just use the mv=mv equation to determine the new concentration for H3O+ and then take the -log[H3O+] to determine the new pH?(2 votes)
- Does it give you the same answer when you try it?(2 votes)

- If the concentration of the strong base is 0.100M and the strong acid is 0.120M, is the equivalence point still at 7?(2 votes)
- Yes, the concentrations of the acid and base need not be the same. If we have a more concentrated acid, that just means we’ll need more base to reach the equivalence point and so the volumes at the equivalence point will be different. But the pH will still be 7 because the conjugate acid (of the strong base) and base (of the strong acid) don’t appreciable react further in acid-base reactions to change the pH beyond neutral conditions. So it’s the identity of the acid and base in the titration whether the pH will be 7 or not, not the concentrations.

Hope that helps.(2 votes)

- At8:27, Why did he add the volume of the base 10 ml when he is trying to get the concentration of only the acid and not the total volume? Thanks in advance(1 vote)
- The question he is answering at that point in the video is: “what is the pH after the addition of 10.00mL of 0.500M NaOH”

Working out the concentration of H3O+ and then pH depends on the total volume of the solution.(3 votes)

## Video transcript

- [Voiceover] Let's say we're
doing a titration and we start with 20 mL of .500 molar HCl. So we're starting with a strong acid, and to the strong acid, we're going to add a solution of a strong base. We're going to add a .500
molar solution of NaOH, and as we add the base, the
pH is going to increase, and we can show this
on our titration curve. So we put the pH on the
y-axis, and on the x-axis we put the volume of
base that we are adding. So in part A, our goal is to find the pH before we've added any
of our sodium hydroxide, so the addition of 0.0 mL of our base. So if we haven't added any base, the only thing we have present is acid. We know that HCl is a strong acid, so it is going to ionize 100%. So if HCl donates a proton to H2O, then we get H3O plus, and
we would get the conjugate base to HCl, which is Cl minus. If we're starting with a
concentration of .500 molar HCl, let's go ahead and write
that, .500 molar HCl, and since we know this is 100% ionization because we have a strong acid,
that's the same concentration of hydronium ions that we'll
have in solution, so we have .500 molar for the
concentration of hydronium ions. Now it's easy to find
the pH because we know that the pH is equal to the negative log of the concentration of hydronium ions. So let's go ahead and do the calculation. We take the negative log of
the concentration of hydronium which is .500, so that gives us .301. So we plug this into here, and
we get a pH equal to 0.301. So we can find that point
on our titration curve. It's right here at the beginning. We've added 0.0 mL of
base, and our pH looks like it's just above zero here
on our titration curve, and we calculated it to be .301. Let's find another point
on our titration curve, this time after we add 10 mL of our base. So we find 10 mL of our base right here, and we're trying to find this point. So what is the pH after we add 10 mL? Well it looks like it's
pretty close to one. Let's see if we can
calculate what the pH is. So if we're adding base,
we know that the base that we're adding, the hydroxide ions that we're adding, are going to neutralize the hydronium ions that
are already present. So first, let's calculate how many moles of hydronium ions that
we had present here. So the concentration of
hydronium ions is .500. Let's get some more room down here. We know the concentration
of hydronium ions is equal to .500 molar, and
we know that concentration, or molarity, is equal
to moles over liters. So if this is the molarity,
that's equal to the moles of H3O plus over the liters,
so what was the original volume of acid that we started with? Let's go back up here to our problem. We started with 20 milliliters, so let's write that right here, 20.00 mL. If I want to convert that into liters, I move my decimal place three to the left. One, two, three. So that's .02 liters. Let me go ahead and write
that in here, so we have... We have .02 liters, and they're
going to be a lot of zeros in this video and it would be
annoying if I said them all, so I'm just going to say .02 and not list all of those zeros that we have. So to solve for moles, we
just multiply .5 by .02. So we would come out with
the moles of H3O plus, and this is equal to
.01 moles of H3O plus. So that's the moles of acid
that we're starting with and we're going to add some base. We're adding some sodium hydroxide, and we know sodium
hydroxide is a strong base, so the concentration of sodium hydroxide is the same as the
concentration of hydroxide ions. Na plus and OH minus. So .500 molar solution of sodium hydroxide is .500 molar for hydroxide
ions, so we write down here the concentration of
hydroxide ions is equal to .500 molar, and once again
we need to figure out how many moles of hydroxide ions that we have, and what's the volume. We have 10 milliliters, so
in liters that would be, move our decimal place
three, so that's .01. So this is equal to .01 for our liters. Solve for moles. So we would just multiply .5 by .01 and we would get our moles equal to .005. So that's how many moles
of hydroxide ions we have. Now that we've found moles of
both our acid and our base, we can think about the
neutralization reaction that occurs. The base that we add
is going to neutralize the acid that we had present. So we had hydronium ions
present, and we added some base. So the acid donates a proton to the base. If OH minus picks up an
H plus, then we get H2O. If H3O plus donates a
proton, we get another molecule of H2O, so we end
up with two H2O over here. Or, if you prefer, instead
of writing H3O plus, you could have just written
H plus as your proton, and your proton reacts with
hydroxide to give you water. So either way of representing the neutralization reaction is fine. So let's plug in our moles here. We know that we started
with .01 moles of H3O plus, so let me write that down here. .01 moles of H3O plus,
and we started with, and we added I should say,
.005 moles of hydroxide. We added .005 moles of hydroxide ions. All of the hydroxide is going to react. It's going to neutralize the
same amount of hydronium ions. So we're going to lose
all of our hydroxide ions because the base is going to
completely react with our acid, so we're left with zero moles of our base. If we're losing that much hydroxide, we're also going to lose
that much hydronium, so that much is reacting
with the hydroxide ions. We're losing the same
amount of hydronium ions. So .01 minus .005 is of
course equal to .005. That's how many moles of
hydronium are left over. We've neutralized half of
the hydronium ions present, and so we have another half left over. So one half of the acid
has been neutralized, so one half of the acid is left. Let's think about the new concentration of hydronium ions in
solution, so the concentration of hydronium is equal
to moles over liters. That would be .005 moles of H3O plus. Well, what is the new volume? We started with 20 mL
of our acid, and to that we added 10 mL of our base. So right up here, we
added 10 mL of our base. The new volume is 30 mL. We added 10 mL, so our
new volume is 30 mL. So to find the new
concentration of hydronium ions, we need to use 30 mL, or .03 liters. One, two, three. So this would be .03 liters, and we can get our concentration. So let's get out the calculator here and let's take .005, we're
going to divide that by .03, and we get our concentration
of hydronium ions to be .17. So our concentration
is equal to .17 molar. Now that we know our
concentration of hydronium ions, so our concentration is
.17, we can find the pH, because of course the pH is equal to the negative log of the
concentration of hydronium ions, so the negative log of .17. So we get out our
calculator one more time, and we find the negative log of .17. That gives us a pH of .77. So our pH is equal to,
our pH is equal to .77. So now, now we know, we add 10 mL of base and the pH is .77, so let's go back to our titration curve up here and find that point we talked about earlier. So we found this point right
here after 10 mL of base. This pH right here is .77. We just did the calculation to show it. On the next video, we'll
analyze this titration curve. We'll looks at some more points on it.