Henderson Hasselbalch Equation - Video Tutorials & Practice Problems

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concept

The Henderson-Hasselbalch Equation

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in this video, we're gonna brush up on the Henderson Hustle back equation. So in our previous lesson videos, we covered the K and the PK of acids. And so we already know that the smaller the PK is, the stronger the acid will be, and so strong acids have really small P. K s and they have a tendency to completely disassociate, whereas weak acids. On the other hand, they have really high P. K s and they do not completely associate. And by completely disassociate here, what we really mean is that these strong acids will completely break down into their conjugate base and the hydrogen ion. But again, weak acids do not completely break down like that. And so because strong acids completely disassociate, calculating the pH of strong acid solutions is a relatively easy process, since the initial acid concentration is going to equal the final concentration of hydrogen ions. And so, for example, we know that hydrochloric acid, or HCL, is a strong acid, and so the initial concentration of hydrochloric acid is going to equal the final concentration of hydrogen ions, and then we can use our equation for pH, which is the negative log of the hydrogen ion concentration to determine the pH of the solution. And so calculating the pH of a strong acid solution isn't easy process. But calculating the pH of a weak acid solution is not as easy, and that's because weak acids do not completely dissociated. And this is where the Henderson household back equation comes into play because we'll need to use it to calculate the pH of weak acid solutions. And this is an especially important point for biochemistry because most biological acids are actually weak acids. And so that shows why the Henderson Hasselbach equation is important for biochemists. Now. Recall from your previous chemistry courses that the Henderson Hasselbach equation expresses the relationship between the solution pH and the peak A of the acid and so typically, the Henderson Hasselbach equation is used to determine one of two different things. The first is the final pH of a weak acid solution after it reaches equilibrium, and the second is the ratio of the concentration of conjugate base to the concentration of conjugate acid. When the pH is given to us and so down below, you can see we have an image of the Henderson hustle back equation, which is expressed as the pH Equalling to the P K of the acid, plus the log of the final concentration of conjugate base over the final concentration of conjugate acid. And so, in our next video, we'll be able to get an example of how to apply the Henderson Hustle back equation. So I'll see you guys in that example video.

2

example

Henderson-Hasselbalch Equation

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Alright. So here's an example problem that wants us to determine the ratio of the concentration of conjugate base to the concentration of conjugate acid for aspirin in the blood. And it tells us that aspirin's Pekka is 34 and the pH of blood is 7.4. And so we know that we can use the Henderson Hustle back equation to determine the ratio of the concentration of conjugate base to the concentration of conjugate acid. And we covered the Henderson Hasselbach equation in our previous lesson video. So we know that it's equal to the pH, which is equal to the peak a of the acid, plus the log of the final concentration of conjugate base over the final concentration of kinda get acid. And so all we need to do is plug are variable straight into this Henderson Hasselbach equation. And so we're told that we have a pH of 7.4 so we can plug that in right down here 7.4, which is equal to the peak A, which is given to us as 3.4 so we can plug that in here, plus the log of the concentration of conjugate base, which I'll abbreviate with C B over the concentration of conjugate acid, which I'll abbreviate with C A. And so we're solving for this ratio here. That's really what it's asking us to determine, so we want toe isolate for this ratio. So what we can do is subtract 3.4 from both sides of the equation, and so 7.4 minus 3.4 is equal to four. So we have four is equal to the log of the concentration of conjugate base over the concentration of conjugate acid. And so now we want to get rid of this log here so that we can continue to isolate for this ratio. And the way that we get rid of a log is by taking the anti log. But if we take the anti log of one side of the equation, that means that we need to take the anti log of the other side of the equation in order to make sure that it stays equal. And so the anti log of a number is literally just 10 raised to the power of the same number, so it's 10 raised to the power of four and So remember, taking the anti log of this side gets rid of our log here. So we're just left with the concentration of conjugate base over the concentration of conjugate acid. And so, 10 to the fourth of you type that into your calculators. It's equal to 10,000, which is essentially 10,000 over one. So this is the ratio that we were looking for all along for the concentration of congregate based to the concentration to conjugate acid. And so what this ratio means is that for every 10,000, uh, conjugate base molecules, we will have one conjugate acid molecule. And so notice that this ratio here of 10,000 is equal to answer option B so we could go ahead and indicate that be here is correct. And that concludes this example problems. So we'll be able to get some practice and our next practice video. So I'll see you guys there

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Problem

Problem

What is the pH of a mixture of 0.02 M sodium formate & 0.0025 M formic acid (pK _{a} = 3.75)

A

pH = 4.21

B

pH = 1.27

C

pH = 4.65

D

pH = 9.34

4

Problem

Problem

What is the ratio of of [CH _{3}COO^{-}] / [CH_{3}COOH] in an acetate buffer at pH = 7? pK _{a} = 4.76.

A

122.43

B

173.78

C

39.84

D

96.31

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Problem

Problem

Consider 100 mL of a 1M acid solution (pK _{a} = 7.4) at pH = 8. Calculate final pH if 30 mL of 1M HCl is added.