Lactic acid, CH3 CH(OH) COOH, is found in sour milk. A...

Key Concepts

Buffer Solutions

A buffer solution is a mixture of a weak acid and its conjugate base that resists changes in pH when small amounts of an acid or base are added. This concept is crucial because the combination allows the solution to maintain a relatively constant pH, which is central to understanding how mixtures like a weak acid with its salt function in acid-base equilibrium.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation provides a relationship between the pH of a buffer solution, the acid dissociation constant (pKa) of the weak acid, and the ratio of the concentrations of the conjugate base to the weak acid. This equation is fundamental in determining the pKa of an acid or in formulating buffers with desired pH values, making it a key tool in acid-base chemistry.

Acid Dissociation Constant (Ka)

The acid dissociation constant, Ka, quantifies the extent to which a weak acid dissociates into its ions in solution. This concept is central to understanding the strength of an acid; a lower Ka indicates a weaker acid. Determining the Ka of an acid, such as lactic acid, is important for predicting its behavior in chemical reactions and buffer solutions.

Weak Acid Equilibrium

Weak acid equilibrium governs the balance between the undissociated acid and its ions in solution. Understanding this equilibrium is critical in systems where the acid does not fully dissociate, influencing the pH and overall behavior of the solution. The equilibrium concept underlies the calculation of parameters like Ka and the application of the Henderson-Hasselbalch equation in buffer systems.

Transcript

00:01 This problem, we are given 1 .00 grams of the salt compound.

00:06 I'm just going to write it out as an a dot, dot, dot, instead of writing it all out.

00:09 Just know that's how i'm going to write some things out, especially at the beginning so that i don't take up too much space.

00:15 And we also have 100 .0 .0 milliliters of 0 .0 .0 .000 molar.

00:25 That dot dot dot, c -o -o -h, which that is the entire lactic acid compound.

00:33 I just, as you know, i'm not writing it all out, just the important parts.

00:37 So it's going to be dot -d -d -c -o -h for now.

00:41 And we also know the ph value is 4 .11.

00:46 What do we need to find? we need to find the k -a value.

00:53 So let's look at units.

00:55 Units, that's the first thing i always think of before i even start a problem.

01:00 Like, are the units right for us to continue forward? so let's look at this.

01:05 We have molar.

01:06 What is molar? it equals moles over a liter.

01:10 That's going to be important.

01:11 That's going to make it so that we can move forward in every part of every equation.

01:17 Okay.

01:18 So over here, if we look at the salt, we have grams.

01:24 Well, we're going to need to get grams to moles.

01:27 What does that make you think of stoichiometry? and over here, you see we have milliliters.

01:35 Well, we need to get those milliliters over to liters.

01:39 So let's just do that real quick.

01:40 We know that's going to give us 0 .1 -000 -0 -0 liters.

01:45 As you can see, i kept all of the significant figures because that's what we're going to need at the end to make sure that we've got the right answer.

01:55 Also, what do we know about ka? it's not going to have any units, actually.

02:00 It's a constant.

02:02 So what else do we know? well, when we see the acid ending in c -e -o -o -o -h, it's actually a carboxal group, those four elements right there.

02:18 And it's typical in weak acids.

02:25 So that tells us k -a is less than one.

02:29 If you didn't know this, it's actually in chapter 16, so you can go back and look for it.

02:34 But that's just something if you didn't know it, i guess.

02:40 We also know that the salt in a dot dot dot that we began with is going to disassociate entirely.

02:49 And it will give us the initial concentration of our dot dot dot dot c -o -0 minus, which is our conjugate base.

02:59 It's going to give us that initial concentration.

03:05 And another thing that we know is that the log of our concentration of h3o plus is going to give us a negative ph value.

03:15 Well, how about we go ahead and use that in our first step? let's plug in what we know.

03:25 So i'm just copying it for now equals negative 4 .11.

03:35 So therefore, let's rearrange this out.

03:38 Remember it's log base 10 so that gives us a concentration of the h3o and i should have written a plus in there sorry about that and that equals 10 to the negative 4 .11 so what you might not realize is this is actually going to be our x value later but it's okay if that doesn't make sense now we'll go over it later all right we've used that equation next let's do step number two let's go through the salt dissociation to create that initial conjugate base concentration.

04:21 So let's plug in what we know.

04:23 We know that we have 1 .00 grams of the salt.

04:30 All right? remember, we're going to do stoichiometry now, right? we're trying to get from grams to moles.

04:38 So that means we're going to go find the molar mass of that salt, which if you don't remember, you're just adding up the molar mass of all of the elements.

04:50 So that gives us 112 .06 grams of that salt.

04:56 And of course, molar mass is over one mole.

05:01 Okay.

05:02 Next, since the salt disassociates entirely to give us that conjugate base, it's just going to be a one to one mole ratio.

05:19 I'm just putting that in the bottom.

05:21 So that it cancels out later.

05:23 The one mole of the conjugate base, which is what we're trying to get to.

05:27 Dot, dot, dot, c -o -o -m -a -0 -minesis.

05:30 All right.

05:32 So the grams of the salt cancel out, the moles of the salt cancel out, and that gives us 8 .92 times 10 to the negative 3 moles of the dot -da -c -0 -minus conjugate base.

05:51 Well, we're at moles, which is great, but we actually still need to get it over to molar, moles per liter.

06:04 What we're going to do is we're going to assume that the volume remains the same.

06:15 So we're going to take this mole volume that we just solved for, 8 .92 times 10 to the negative 3 moles of the conjugate base, and we're going to divide it by the initial.

06:33 The given volume...

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