A 25.0 -mL solution of 0.100 M CH3 COOH is titrated with a...

Key Concepts

Excess Strong Base

When more titrant is added than is needed to completely neutralize the weak acid, there is excess strong base in the solution. The pH in this case is determined predominantly by the concentration of the excess hydroxide ions, making the solution strongly basic.

Henderson-Hasselbalch Equation

This equation provides a simplified way to estimate the pH of a buffer solution. By relating the concentration ratio of the conjugate base to the weak acid with the acid's dissociation constant (pKa), one can quickly calculate the pH in the buffer region of a titration.

Equivalence Point

This is the stage in a titration when the amount of titrant added is chemically equivalent to the amount of analyte originally present. In the titration of a weak acid with a strong base, the equivalence point is where all the weak acid has been neutralized, often leading to a solution whose pH is governed by the hydrolysis of the conjugate base.

Reaction Stoichiometry

In titration problems, stoichiometry is used to determine how the moles of acid and base react with each other. It provides the basis for calculating how much of the weak acid is neutralized after each addition of the titrant and how much remains unreacted or is present in excess.

Acid-Base Titration

This concept deals with the gradual addition of a titrant (a solution of known concentration) to an analyte (a solution of unknown concentration) to perform a neutralization reaction. It involves calculating the point at which the moles of acid equal the moles of base, as well as determining the pH at various stages of the titration process.

Buffer Solution

A buffer forms in the titration of a weak acid with a strong base when both the weak acid and its conjugate base are present in appreciable amounts. This system resists changes in pH, and its pH can be estimated well using the Henderson-Hasselbalch equation.

Transcript

00:01 In this video, we're going to be figuring out the ph when we're doing a titration between acetic acid and potassium hydroxide.

00:09 We're going to figure out the ph with different volumes of potassium hydroxide that are being added.

00:16 Now, a couple of things to keep in mind as we begin is an equation that we're going to find quite useful, even before we get into that.

00:23 Now, we know that molarity is mole solute over liters of solution.

00:27 Quite frequently, we're going to find the number of moles by multiplying the molarity times the volume in liters.

00:35 In fact, right away we can do that here.

00:37 So the amount of our acetic acid, the 25 milliliters, and then the 0 .10 molar, of course, we're going to change the volume into liters.

00:47 So it would be 0 .025 liters and multiply that by 0 .1.

00:52 And so we get a value of 0 .025 moles.

00:57 So this is a multiple part question, and we're going to use this multiple times, where molarity times liters is going to equal the number of moles.

01:07 Okay, so that being said, our first part of this question, we're figuring out what is the ph when we haven't added any of the potassium hydroxide? and what that means is that basically we just have a solution of our acetic acid.

01:21 And so we're going to consider how you would, for part, the first part, how you would figure out simply the ph of a solution of a weak acid.

01:31 So the ch3 -c -o -o -h is going to be an equilibrium with the hydrogen ion, and then the acetate ion, the ch3 -c -o -o -m minus.

01:44 We're going to set up an ice table, initial change in equilibrium.

01:48 Initially, we have 0 -1 -0 -0 -n -x, x and x, so we have 0 -10 -0 -x, minus x, we have x, and we have x.

02:02 We have our value of k a, which is going to equal products, in this case x for the hydrogen ion.

02:09 This is our first product, this is our second product, which is also x at equilibrium, over the point one zero minus x.

02:20 Now the value of k a for acetic acid is 1 .8 times 10 to the negative 5.

02:31 That's going to equal x squared as we simplify this term over here.

02:35 And then i'm just going to put in 0 .10.

02:38 We're going to find the value of x relatively small, and so to subtract it from 0 .1 is not going to be of a significant change.

02:45 So we frequently are going to drop that with the weak acids.

02:48 So multiply both sides by 0 .1, take the square root.

02:51 So x, which is also equal to the hydrogen ion concentration, ends up having a value of 1 .34 times 10 to the negative 3 .3.

03:03 Moles per liter.

03:05 The ph is a negative log of 1 .34 times 10 to the negative 3.

03:12 So ph before we've added any potassium hydroxide is 2 .9.

03:18 In the second part, we're figuring out what happens after we add 5 milliliters of our potassium hydroxide and that has a concentration of 0 .200 moles per liter.

03:33 Okay, so let's see what's going to happen.

03:36 Well, first of all, let's figure out the number of moles that's added.

03:38 Here.

03:39 If we multiply the molarity times the liters, that gives us a number of moles, and that's going to be 0 .001 moles of the oh minus.

03:51 Now if we scroll back up to this reaction of what happens when we neutralize it, if we add a certain amount of hydroxide, that's going to then produce some of its conjugate base.

04:01 And now we're all of a sudden producing a buffer, because we're going to have the conjugate base over here, and we're going to also still have some of the weak acids.

04:10 So for these next couple parts, we're going to treat it like a buffer because that's what it is.

04:15 We're also going to notice that the amount of the hydroxide that we add is also going to be equal to the amount of conjugate base that we form in moles.

04:26 And then in addition to that, the amount of base that we add, we're going to subtract that amount of moles from the moles of our acid, and that'll tell us how much of the acid is left.

04:37 And so that will be our strategy as we go through the second part.

04:40 Part.

04:42 Okay, so we know we have 0 .001 moles of the hydroxide.

04:47 And that's also going to give us, as we mentioned, that's going to also equal the conjugate base.

04:55 And so i'm just going to put a minus for the conjugate base.

04:58 That's just going to simplify things.

05:00 Remember, that's because that's producing the acetate ion.

05:05 Now, how much of that acid is left? well, the acid was 0 .0025 moles and we've reacted 0 .001 moles and so that gives us a value of 0 .015 moles of our acid that that's left over that did not get neutralized so i have moles of h .a.

05:29 We're going to use what's called the henderson -heslbeck equation.

05:33 So the ph of a buffer is equal to its p -k -a plus the log and this is why we've bothered with these two substance with these two quantities here plus the log of the concentration of a minus over the concentration of ht -a so hopefully you're familiar with this being the henderson -heselbeck equation so we can fill things in so our ph is going to equal the pk a that's a given quantity found from the k a and for this acid it's going to be 4 .74 take the negative log of the k a so 4 .74 plus the log i am able to leave this in moles.

06:13 So i have 0 .001 moles divided by 0 .0015.

06:19 And the reason for that is that we're going to divide both by the same volume.

06:23 And so we can leave them in moles.

06:26 And so that gives us a ph for the second part of 4 .57.

06:30 You can see the ph has gone up.

06:33 For the next part, it's going to be very similar, but we're adding more potassium hydroxide...

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