Henderson Hasselbalch Equation - Video Tutorials & Practice Problems
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concept
The Henderson-Hasselbalch Equation
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Video transcript
In this video, we're gonna brush up on the Henderson Hasselbalch equation. So in our previous lesson videos, we covered the Ka and the pKa of acids, and so we already know that the smaller the pKa is, the stronger the acid will be. And so strong acids have really small pKa's and they have a tendency to completely dissociate. Whereas, weak acids, on the other hand, they have really high pKa's, and they do not completely dissociate. And by completely dissociate 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 dissociate, 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 ion the hydrogen ion concentration, to determine the pH of the solution. And so calculating the pH of a strong acid solution is an easy process, but calculating the pH of a weak acid solution is not as easy. And that's because weak acids do not completely dissociate. And this is where the Henderson Hasselbalch 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 Hasselbalch equation is important for biochemists. Now recall from your previous chemistry courses that the Henderson Hasselbalch equation expresses the relationship between the solution pH and the pKa of the acid. And so typically, the Henderson Hasselbalch equation is used to determine one of 2 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 Hasselbalch equation, which is expressed as the pH equaling to the pKa 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 equation. So I'll see you guys in that example video.
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example
Henderson-Hasselbalch Equation
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Video transcript
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 pKa is 3.4 and the pH of blood is 7.4. And so we know that we can use the the Henderson Hasselbeck equation to determine the ratio of the concentration of conjugate base to the concentration of conjugate acid. And we covered the Henderson Hasselbeck equation in our previous lesson video. So we know that it's equal to the pH, which is equal to the pka of the acid plus the log of the final concentration of conjugate base over the final concentration of conjugate acid. And so all we need to do is plug our variable straight into this Henderson Hasselbalch 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 pKa, 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 CB, over the concentration of conjugate acid, which I'll abbreviate with CA. And so we're solving for this ratio here, that's really what it's asking us to determine. So we want to isolate for this ratio. So what we can do is subtract 3.4 from both sides of the equation, and so 7.4 minuteus 3 point 4 is equal to 4. So we have 4 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 antilog, but if we take the antilog of one side of the equation, that means that we need to take the antilog of the other side of the equation in order to make sure that it stays equal. And so the antilog of a number is literally just 10 raised to the power of the same number, So it's 10 raised to the power of 4. And so, remember taking the antilog 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 4th, if you type that into your calculators, it's equal to 10,000, which is essentially 10,000 over 1. So this is the ratio that we were looking for all along for the concentration of conjugate base to the concentration of conjugate acid. And so what this ratio means is that for every 10,000, conjugate base molecules, we will have 1 conjugate acid molecule. And so notice that this ratio here of 10,000 is equal to answer option b, so we can go ahead and indicate that b here is correct. And that concludes this example problem, so we'll be able to get some practice in 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
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Problem
Problem
What is the ratio of of [CH 3COO-] / [CH3COOH] 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.