 ## GOB Chemistry

Learn the toughest concepts covered in your GOB - General, Organic, and Biological Chemistry class with step-by-step video tutorials and practice problems.

Acids and Bases

# Henderson-Hasselbalch Equation

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concept

## Henderson-Hasselbalch Equation Concept 1 1m
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Hey, guys in this new video will finally get to see how do we calculate the Ph of a buffer? So we're gonna say we learned that whenever we had a weak acid or a base, we had to use our favorite friend the ice chart in order to find ph or P O. H. Now, the great thing is, now that we're dealing with buffers, we no longer necessarily need to do a nice chart. We're gonna say anytime we know we have a buffer for shore. We can skip the ice chart altogether and just use our Henderson Hasselbach equation. So Henderson hassle back, and it's also known as the buffer equation. Now it says that pH equals P K A plus the log of conjugate base over weak acid. Now, when I say piquet, remember, P just means negative log. So P k equals the negative log of K of K And here what I'm talking about conjugate based or weak acid. The units here could be either in polarity or in moles. It all depends if they give you only polarity used polarity. If they give you volumes off a particular polarity, remember, the word off means multiply. So we multiply leaders times, molar, iti to give ourselves moles 2
example

## Henderson-Hasselbalch Equation Example 1 2m
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So let's look at this first question and see how exactly do we use it? So here it says, What is the Ph of a solution? Consisting off 2.75 Moeller Sodium Finale, which is this compound and 3.0 Moeller phenol, which is this compound? The K of female is 1.0 times 10 to the negative 10. Now what you should realize here is phenol has what it has. H is non metals and oxygen, so phenol is an oxy acid. Use the math that we've learned. You take the number of oxygen's minus by the number of hydrogen. We don't get to oxygen's left, so this is a week oxy acid. We definitely know it's weak because, look, it's it's K a value. It's K value is extremely small. If you're K value is less than one. You're very weak acid. So here we have a weak acid sodium final. It has one less h available. So this is the conjugate base. So we have a weak acid. We have a conjugate base, and therefore we have a buffer. And because we have a buffer, we need to use the buffer equation. Ph equals negative log of K plus the log off. Now the only give us more clarity here. So that's what we're gonna plug in 2.75 Moeller the conjugate base over 3.0 moller are weak acid. When we plug all that in, we'll get back the answer of 9. So that will be our Ph four solution. So remember, if you know for sure you have a weak acid and conjugate base, then you know you have a buffer. And if you have a buffer, just use the Henderson Hasselbach equation. It was designed to find the ph of buffers. Now that we've seen this example, I want you guys to attempt to do practice question one in this one, I'm telling you, calculate the pH of a solution formed by mixing 200 MLS off 2000.400 Mueller Ethel Amine solution with 250 MLS off a point 450 Mueller solution off Ethel. Ammonium solution K B is 5.6 times 10 to the negative four. So figure out first if you have a buffer. If you do manipulate things so that you get the correct units and plug it into the Henderson Hasselbach equation. Doing that, we'll give you the correct pH. Good luck, guys.
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Problem

Calculate the pH of a solution formed by mixing 200 mL of a 0.400 M C2H5NH2 solution with 350 mL of a 0.450 M C2H5NH3solution. (Kb of C2H5NHis 5.6 x 10 -4).

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example

## Henderson-Hasselbalch Equation Example 2 3m
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Hey, guys, In this new video, we're gonna continue with our discussion and calculating Ph for a buffer. So here in this first example, it says, What is the buffet component concentration ratio, which really is just conjugate base over weak acid off a buffer that has a pH off 5.11 here, I tell you, the K of the acid is 1.30 times 10 to the negative five. Now, since we're dealing with a buffer, let's write out the Henderson Household back equation. The buffer equation so ph equals P K plus log of conjugate base over weak acid. Here, we're gonna say that the pH I tell you is 5.11 PK. Remember, PK is just simply the negative log of K. So we're gonna take the negative log of 1.30 times 10 to the negative five. When we do that, we get 4.89 Take that 4.89 Plug it in here, plus log of conjugate base over weak acid. Now, we're trying to find just this portion right here, so we just have to move everything away from it. Toe isolated. So here we're gonna say so. Track 4.89 from here. So track 4.89 from here. So we're gonna get 0.22 equals Log off. Conjugate base over weak acid. Now, remember, we need toe isolate conjugate base over weak acid. So we have to get rid of that log. How do we do that? Divide both sides by log, which is similar to doing the anti log. So all you say to yourself here is I'm doing the anti log of something. I'm dividing it by log. When I divide anything by log, it becomes 10 to that number. Okay, Now, just realize here that the sign does not change. It was positive here. It's still positive here. Taking the anti log of something does not change the sign. Now, when we take that number and plug it into our calculator, we're gonna get 166 And since this is a ratio, the answer you get is always gonna be over one. So we have 1.6 6/1. What does that mean? That means for every 1.66 conjugate base, I have one weak acid involved. So here we're basically saying that conjugate base is greater in amount than weak acid. These are all the little things you need to know. Professor could just simply say, What is the ratio of conjugate based a weak acid, you would say 1.66 to 1. Or they could ask which one is greater in amount, conjugate base or weak acid? Because the conjugate base is 1.66 while the weak acid is only one. You would say that conjugate base is greater than weak acid, and it all starts from knowing. First we're dealing with a buffer. And second, we have to use the Henderson Hasselbach equation, so we just simply have to manipulate it through algebra. Now that we've done this one, let's go to example, too.
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example

## Henderson-Hasselbalch Equation Example 3 2m
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okay, An example to I'm saying over what ph range will in oxalic acid, H two, C 204 and a sodium oxalate. NHC 204 Solution work most effectively the acid association constant of oxalic acid is 6.0 times 10 to the negative, too. So when I say acid association constant, I'm just saying, K, now we're talking about the ph Range or, in this case, the buffer range, because here we have a weak acid and its conjugate base. Now you need to realize that buffers can't work at any pH. There is a range to their effectiveness, and you're just gonna simply say that a buffer works best when pH equals P K plus or minus one. That is the range of a buffer, plus or minus one. The P. K that is called our buffer range. So we're gonna do here is we're gonna say p k equals negative log of K A so equals negative log of 6.0 times 10 to the negative, too. And that is 1.22 So you're Ph is plus or minus that number, so you're gonna say 1.2 to minus one is 10.22 1.22 plus one is 2.22 So the range would be 0.22 to 2. This would be our buffer range. Within this range, this buffer is most effective. Okay, so the pH would have to be between 0.22 or 2.22 If it's anywhere outside that range, then this buffer will not be able to work. So that means a would be our answer. So just remember, a buffer works effectively within a buffer range, which is a simply p k plus or minus one. That's the pH range it'll work in. Now that we've attempted these two questions, guys, I want you to attempt to do the practice question I left on the bottom again. It involves. Are you able to manipulate the Henderson Hasselbach equation? And again, if you get stuck, don't worry about it. Just come back, take a look at look at my video to see how best I approach and explain the best way to get the answer. Good luck, guys.
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Problem

Determine how many grams of sodium acetate, NaCH3CO2 (MW:82.05 g/mol), you would mix into enough 0.065 M acetic acid CH3CO2H (MW:60.05 g/mol) to prepare 3.2 L of a buffer with a pH of 4.58. The Ka is 1.8 x 10-5.

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example

## Henderson-Hasselbalch Equation Example 4 2m
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