 # Video Transcript : Henderson-Hasselbalch MCAT Trick for Buffer pH Without a Calculator Below is the written transcript of my YouTube tutorial video Henderson-Hasselbalch MCAT Trick for Finding pH Without a Calculator.

(click here to watch the video on YouTube)

[Start Transcript]

Leah here from Leah4sci.com and in this video I’ll show you how to easily calculate the pH of a buffer without a calculator when givne different acid and conjugate base concentrations. I’m teaching this video to go with my Zwitterion tutorial which you can find on my website at leah4sci.com/AminoAcids.

On the MCAT you can’t use a calculator which means you have to know how to estimate values. More importantly you have to understand what is going on in the solution because if you find yourself faced with a non math question where instead you have to figure out what form of molecule will look like or which type of molecule would dominate at a given PH you don’t have the time to waste on complicated Math.

The first thing you have to understand is “what is a  weak acid?” I highly recommend that you memorize the list of strong acids for the MCAT and these are acids that when dissolved in water will dissociate 100%. For example, HCl, if I put 10 moles of HCl solution, I will get 10 moles of H+ ions and Cl- ions. If I put a thousand moles of HCl into a liter solution, I will have 8 thousand molar H+ and a thousand molar Cl- because there is a 100% dissociation. Notice we have a one way arrow? Because this is a reaction that goes to completion.

But now, if we look at a weak acid, we don’t have a reaction that goes to completion, instead we have something that will have an equilibrium somewhere between the protonated and deprotonated form. For this video we’ll use acetic acid as an example but you have to understand what’s going on so that you can apply it to everything else. Acetic acid or Ethanoic Acid is CH3CO2H without last hydrogen is acidic. When dissolved in water, you’re going to get equilibrium between acetate which is CH3CO2- and H+. That H+ will dissolve in water to give you a proton concentration which contributes to the pH value of the solution.

In Gen Chem you learned how to calculate the pH and calculate the concentration by using the Ka value where Ka stands for acid dissociation constant and gives you the concentration of H+ in solution A- which is the conjugate base divided by the HA your un-dissociated acid. In this case the Ka would equal to H+, acetate, divided by acetic acid. But that’s not our focus for today. What you want to look at is the fact that Ka is proportional to the H+ concentration and so the higher the Ka, the higher your H+ concentration, the  stronger your acid. Because the more of the acid that dissociates, the more H+  you get in the solution, the more the pH will be lowered which is your strong acid.

To find the p of anything, we do negative log. So to find the pKa, we’re going to use negative log of the Ka and this is important because pKa’s are easier to visualize. We’re not dealing with exponents, we’re dealing with whole numbers and when we’re looking at buffers, this is the form that we’re looking at. So pKa is simply negative log of the Ka.

Negative log, that’s a base 10 and so for every one unit of ten we get one pka value. 10 to the second is a hundred or 2 pka values and so on. The negative tells us that we flip the concept. So if we have a high ka for a strong acid, a low pka is how we know our strong acid. The pKa for acetic acid is 4.8 but you don’t have to memorize this. So now let’s take a look at what form acetic acid will take when dissolved in different types of acetic or basic solutions. We’ll start with a pure water solution; if we take acetic acid and dissolve them in water, to find the pH, all you have to do is set up ka equals H plus A minus over HA.

In Gen Chem we learned how to do an Ice Chart for the MCAT you wanna simplify H squared HA, again not a topic for today, but what I wanna do is take you to the next step.

So now what happens, if I  take a solution and I dissolve 100 moles of acetic acid and then I start slowly adding molecules of NaOH which will dissolve to give me OH minus in solution.  Every time I have an OH minus enter the solution, it will grab an H+ take it off of acetic acid and give me acetate and water. So again, acetate which is the de-protonated form  acetic acid and then that hydrogen that came off is now bound to the OH to give me water which is neutral. When we look at the K equilibrium we’re only looking at what acetic acid will on its own. But if we go back to that equation, and then look at Le Chatelier’s principle that says “If we disturb the system the system will compensate”. Well, every time I have an OH minus still an H+, it doesn’t matter which side of the equation, the fact is, every time OH- steals an H+ another acetic acid will dissociate to compensate for that in solution.

Now if we have one or two, that’s fine. We still have about a hundred acetic acid. But as that number gets more and more significant, the pH is going to start changing. For example if I had 50 moles of OH- then half of the acetic acid will react with OH- to form acetate and the other half will remain untouched. Let’s break up the hundred moles of acetic acid as follows. We have 50 moles of acetic acid that will react with 50 moles of OH- then we’ll react 100 percent because OH- is a strong base and will definitely attack it that gives me 50 moles of acetate. And then we have 50 moles of acetic acid that wasn’t reacted because we don’t have any more OH-. That means my final solution after the OH- reacted is 50 acetate and 50 acetic acid or acid and conjugate base on a one is to one ratio.

Whenever you see something like that, you have to recognize the buffer. A buffer as most students memorize is a solution made up of an acid and its conjugate base for the purpose of resisting 6:36 change the pH. But I want you to understand how this Buffer works.

In this acetic acid buffer, if I decide to add in some HCl which is a very strong acid, instead of lowering the pH because of the H+ dissolving in the solution, that H+ will get consumed by the acetate. The acetate will attack it and if you have no more free H plusses, the pH is not gonna drop. If I then decide to add a strong base like OH minus, instead of raising the pH because of the OH minus concentration going up, I will have that OH minus grab that H off of the acetic acid giving me the conjugate acetate but again, no drastic change to pH. And the keyword here is NO DRASTIC CHANGE, but you might see a change in pH depending on how much acid or base you add.

And the way you calculate that is by using the Henderson-Hasselbalch or simply Buffer Equation.  This is an equation that you definitely have to memorize. The Henderson-Hasselbalch equation tells you that the pH is equal to the pKa plus the log of the conjugate base over acid (pH=pka + log [CB/A]). You’ll also see this written as pH is equal to pKa plus the log of A minus which is your conjugate base over A which is your acid. They both tell the exact same thing.

What you need to take away from here is when the pH  is equal to somewhere near the pKa, that difference, the amount of the pH goes up or down is directly influenced by the ratio of the conjugate base of the acid. Now think about it logically, if your solution has more base than acid you would expect your pH to be slightly more basic which means an increase in pH and if your buffer has slightly more acid than base you would expect your pH to be more acidic or slightly lower than the pKa value.

Let’s first use the example we saw above, we had 50 moles of acid and 50 moles of conjugate base.  You can use concentration for CB over A but if they’re dissolved in the same solution and the volume is the same, you can use moles just as easily and the numbers will turn out the same.

So we’ll set it out so that pH is equal to pKa plus the log, conjugate base which was 50 divided by the acid which was also 50. 50 over 50 is one and the log of one is something that you have  to know as equal to zero. In an ideal buffer, when your conjugate base is exactly equal to the number of moles or the concentration of your acid, the pH will be the exact pKa value. And knowing that our pKa was 4.8 for acetic acid we get a pH of 4.8 when the solution has 50 percent acetic acid and 50 percent acetate.

In an ideal buffer, the acid and conjugate base will be so much close so we’re looking at a difference of maybe 1 to a thousand or a thousand to one as your extremes. But you wanna see 1 to 10 or 1 to a hundred. On the MCAT, if your number is not that clean, meaning a ration of 1 to 10, 1 to a hundred, round it, take the two factors of ten that are closest, round up, round down and then find something in the middle.

Let’s look at an example where the conjugate base to acid ratio is 10:1. That means that for every one acetic acid, I have ten acetates in solution. So I plugged in into my equation and I say that pH is equal to 4.8 plus the log of ten over one, just simply clause the log of ten. Great! But we don’t have a calculator, how do you figure this out? I show you how to do logs without a calculator on the MCAT Math series which you can find on my website at leah4sci.com/MCATMath. If we look here, we have log base ten equals ten. Ten to the power of one gives me ten and the answer is one. Or you can simply re-write the number in scientific notation. This is really ten to the first and that means log ten to the first is one.  So our pH is equals to 4.8 plus one which is 5.8. But this video isn’t about Math. On the MCAT it’s about the logic so what happened?

Because that conjugate base, that basic higher pH molecule is found in solution at a concentration ten times that of the acid the pH is going to go up by that factor of 10. Since every pH unit represents a  10 full change, if we have 10 times more base we expect a difference of one pH unit.  So the question is, does it go up or does it go down? Don’t memorize it, ask yourself what’s happening?  If I have a base, base tends to be higher in pH, if I have a change of one, I expect it to go up because up is more basic and down is more acidic.

So without the Math, if I have a 10 to 1 concentration where 10 is the base, the pH is also change by unit of 1 in the direction of the base.  Let’s look at the same thing but this time, the concentration of base to acid will be a hundred to one. A hundred times stronger. Ones again, pH is equal to the pKa which is 4.8 plus the log of one hundred over one where one hundred can be written in scientific notation as ten to the second. So get rid of that and just use the 2. Our pH is going to change by a factor of two and do we go up or down,  well yes, we see that it’s plus but you can also recognize that it’s a factor of two in the basic direction and since a higher pH is basic it’ll be 6.8 which is 4.8 plus basic 2.

Now what happens if the ratio between the conjugate base and acid is favoring the acidic direction so we’ll have one to ten. We’ll setup the equation the same way so that pH is equal to 4.8 plus the log of one over ten but now where the only decimal and it gets a lot trickier. One over ten is equal to zero point one and how do you do that without the calculator? Well, what did I tell you? Turn it into scientific notation. Zero point one is really one times ten to the minus one and that means the change in pH will be negative one, it’ll go down.  How does that make sense logically? If we have more acid compared to base, we expect the pH to change in favor of the acid and the lower pH is more acidic.  And so we get that out pH is equal to 4.8 minus one and that is 3.8 which is definitely more acidic.

Now ones again if I look at the ratio of one to one hundred I will find the same logic.  pH is equal to 4.8 plus the log of one over one hundred. One over one hundred is one times ten to the minus two which means this entire expression comes to negative two. And so pH is equal to 4.8 minus 2 which is 2.8. but again, keep in mind, you don’t need to do the Math.  If you see that your solution has one hundred times more acid than base, change your pH in favor of two acidic units which means that it has to go down from 4.8 to  2.8.

I mentioned that this tutorial is to help you understand the Zwitterions tutorial but when we look at Amino Acids we do have that carboxylic acid but then we also have the amino group which is a base.  So how do we calculate the base using the Henderson-Hasselbalch equation? There are two ways to approach it. One you can turn the pH equals pKa into a basic version where pOH is pKb or if you’re looking at the entire amino acidic where pH system, just turn your pKb into a pKa. Remember that Ka times Kb is equal to Kw where Kw is one times ten to the minus fourteen.

But if we take the negative log of every term of this equation, we get pKa plus pKa is equal to pKw which is fourteen. So if you’re given a pKb for your Amino group just do fourteen minus the pKb, that’ll give a pKa of the conjugate acid and you can apply the same equation. What do we mean by the conjugate acid? If your amino acid zwitterions form has a protonated NH2 or NH3+, treat that as your acid and then acid is going to have a pKa value and that acid will dissociate in solution to give you the NH2 neutral as your conjugate base.

Remember you can find my Amino Acid tutorials including cheat sheet, zwitterions and more by visiting my website leah4sci.com/AminoAcids.

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