Video Transcript : pH pOH for Strong Bases in MCAT Chemistry

pH and pOH Calculations for Strong Bases in MCAT Chemistry Video 3Below is the written transcript of my YouTube tutorial video pH pOH for Strong Bases in MCAT Chemistry.

If you prefer to watch it, see Video HERE, or catch the entire MCAT Acid Base Series.




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Leah here from and in this video we’ll take a look at pOH and pH calculations for weak bases. You can find this entire video series along with practice quiz and cheat sheet by visiting my website

Recall that a weak base is something that does not dissociate or does not react a hundred percent in solution which means that there will be an equilibrium, there will be a back and forth reactants going to products and products back to reactants. A simple reaction to look at is ammonia which will react with water to form ammonium and hydroxide. We have equilibrium arrows to show that the reaction is constantly going back and forth. So if we want to find something like a concentration or a pH we have to take into account this equilibrium. To understand how you set this up for a base, let’s look back at how you set up a keq or an equilibrium expression. You put the concentration of the products divided by the concentration of reactants. For this reaction, the products are ammonium and hydroxide and the reactants are ammonia and water. But don’t forget, for a keq expression you can only have concentrations. Water is a liquid and we don’t include liquids or solids on a keq expression. That gives us ammonium which is the protonated base, you can look at that sa BH+. Hydroxide divided by ammonia which is the starting base. The keq for a base dissociation can be thought of as a keq of base or simply kb, where the kb expression is written out as protonated base hydroxide divided by your starting base.

Let’s take a look at this example. Find the ph of a 0.0008M solution of NH3 given that the kb for NH3 is 1.8 x 10 to the minus 5. For us to find the pH which is an acidic property, but we’re given the kb of a base. So the first equation we have to keep in mind is that pH plus pOH is equal to pkw which is equal to 14. Use the information given to find the pOH and then use that to find the pH. When you cover this in GenChem, you were taught to write out the reaction and draw detailed ICE chart. Now it gives us NH3 plus H2O in equilibrium with NH4 plus OH minus. We’ll look for the initial change and equilibrium and cross out water because we don’t include concentrations for a liquid. The start of concentration of NH3 is 0.008, we’ll assume that the concentrations hydroxide are negligible write zeros. Some quantity x will dissociate so that’s some quantity x will form for ammonium and hydroxide so that at equilibrium my ammonia has a concentration of 0.008 minus x. Ammonium and hydroxide each has a concentration of x. And then we set it up so that kb is equal to ammonium and hydroxide divided by the molarity of ammonia which is 0.008 minus x. All of that is equal to 1.8 times 10 the minus 5 which is the kb. And then you turn this into a quadratic equation waste a good ten minutes and hopefully got the correct answer. Luckily you have no calculators on the MCAT and you can’t solve it this way on the MCAT. Instead you’ll have to do some estimations.

The first one is that given such a small dissociation value, we’re talking about an average of one dissociation per one times ten to the fifth on dissociate molecules so that if we subtract the tiny tiny amount of x from 0.008, that’s negligible. We’ll gonna cancel that out and when we cancel that out we get a much much more simplified equation. This is what you want to use instead, kb is equal to x squared over the molarity initial and that you can solve without a calculator.So let’s set this up. If kb is equal to x squared over the molarity initial, and we’re looking for x because that’ll give us the OH minus concentration then we multiply both sides by the molarity initial to bring it to the top. And then we take the square root on both sides to isolate x. This gives me a new expression. X is equal to the square root of the molarity initial times the kb which we’ll plug in. The molarity initial is 0.008 and the kb is 1.8 times 10 to the minus 5. And how do we solve this quickly without a calculator? Let’s run some estimations. 1.8 is approximately 2, so we have 2 times 8 which is 16. And then we have to account for the decimals. We have 10 to the minus 5 and this right here to the minus 1 to 3 so we combined the two, negative five and negative three is negative eight which give me 60 times 10 to the minus 8. Now this is not improper scientific notation so we’d have to fix it. 1.6 times 10 to the minus 7 and the reason we did it that way is because we’re doing the times ten divided by ten trick. 16 divided by 10 is 1.6. 10 to the minus 8 times 10 makes it one order of 10 bigger which is minus 7 and that’s the value we’re looking at. The problem is, if we now wanna take the square root we do not have an even exponent. The shortcut for square root of exponents is to divide it by two if it’s an even number. So in situations like this look for a pattern. Right here we have an even exponent. Forget the proper scientific notation, this is so much easier to work with because we can break it down to say the square of 16 is 4, the square of 10 to the minus 8 is just half, that’s ten to the minus 4 and that means our OH minus concentration which is equal to X is equal to 4 times ten to the minus 4. Now we want to use this to find the pOH which is equal to negative log the OH minus concentration where negative log of 4 times ten to the minus four. If we want to find an estimate, we can say it’s approximately 4 discounting that initial number. If we want to get more a little specific, the first thing we want to do is find if that pOH is slightly greater than 4 or slightly less than 4 and that means we have to find the closest unit of 10 next to it. 4 times ten to the minus 4 doesn’t let us do our perfect trick of 1 times 10 to the minus the power. So if we round it to the nearest ones meaning we round it down to 1 times 10 to the minus 4 and we round it up to 10 times 10 to the minus 4, now we know which pOH values are slightly greater than or slightly less than our answer. 1 times 10 to the minus 4 gives me 4, 10 times 10 to the minus 4 is not in proper scientific notation, this is really equal to 1 times 10 to the minus 3 which gives me a pOH of 3 and that tells me that the pOH is somewhere between 3 and 4. If you want to take this to a step further, you can think of the numbers you memorized for logs. If your number starts with a 3, you know that the negative log of that is gonna be .5. If the number starts with a 5, you know the negative log is going to be point 3. Since we have a 4, the answer is going to be somewhere between 3 and 5, 4 is a perfect in-between and we need something between 3 and 4. So 3.4 is going to be equal to our pOH. The 3 because that’s between 3and 4 and the .4 because 3 and 5 gives me .5 and .3 and the fourth is right in the middle.

So we know our pOH is 3.4, 14 minus 3.4 equals our pH and that is equal to 10.6. Let’s try a slightly trickier example, this time let’s say we’re given a 0.01molar solution of ammonium fluoride and we’re asked to find the pH. Once again we’re given that kb of ammonia is 1.8 times 10 to the minus 5. We’re asked for a pH, we’re given an acid but the only value is a kb which is a base. How do we calculate this from a base? The first thing you want to recognize is that a weak acid will be in equilibrium with it’s weak base. And the equilibrium here is NH4+. Cl minus is a spectator we ignore it. It will react with OH minus to give me NH3 plus H2O. Even though this is a reaction at equilibrium, we can just as easily write the reverse reaction, NH3 plus H2O will be in equilibrium to give me NH4+ plus OH-. If we set it up this way, we can use the kb but the problem is our starting concentration is given for ammonia. So we have to find a way to calculate the ka so that we can look at the acidic version of this reaction rather than the reverse basic version. And this is the equation to know: ka times kb is equal to kw which is equal to 1 times 10 to the minus 14 at 1.5 degrees Celsius. I want to prove this to you ones so that you understand this and then you have a much easier time remembering than memorize the expression. The equation to know for ka is that H+ times A- is equal to HA. We’re multiplying that by kb which is BH+ times OH- divided by B for base. But this just looks like an entire mess, nothing seems to relate to each other. And the problem with just memorizing is that you’re not seeing the connection. BH+ is simply a protonated base, HA is a protonated conjugate base so it’s actually the same thing. And if one’s in the numerator and one’s in the denominator they cancel. A minus is the deprotonated base, B is a base meaning it’s deprotonated and once again they cancel leaving me with just H+ and OH-, now isn’t that the expression for kw?

Going back to our example, the first thing we want to do is finding out our ka. So we have ka times kb is equal to kw divide both sides by kb which gives me ka is equal to 1 times 10 to the minus 14 divided by 1.8 times 10 to the minus 5. Round the 1.8 to a 2 so that 1 divided by 2 is equal to 0.5. Negative 14 divided by negative 5 is simply negative 14 minus negative 5 or negative 14 plus 5 which is equal to negative 9. And if we put this into proper scientific notation we get point 5 times 10 which is 5. 10 to the minus 9 divided by 10 which is minus 10. 5 times 10 to the minus 10. Now we’re skipping the ICE chart and jumping straight to the expression of ka is equal to x squared over molarity initial just like we did in the last video. Solve for x so that x is equal to the square root of ka times molarity initial and then we plug it in. X is equal to 5 times 10 to the minus 10, times the initial molarity of 0.01. Point zero one, move the decimal to the right, that’s multiplying by a hundred. 10 to the minus 10 has to be divided by a hundred so we drop the exponent by two and that means x is equal to the square root of 5 times 10 to the minus 12. I don’t know the square root of 5 but since 5 is very close to 4 we’ll estimate and we’ll say that it’s approximately 2. 10 to the minus 12 divide that by 2, it gives me times 10 to the minus 6, that means our pH is equal to negative log of 2 times 10 to the minus 6. If you want to round it to 1 times 10 to the minus 6, your pH is going to equal to 6, but the MCAT is likely gonna give you something slightly greater than 6 and slightly less than 6 so you have to know which direction to go. Since our proton concentration is slightly more acidic than 1 times 10 to the minus 6, we want our pH to be slightly more acidic and therefore we’re looking for an answer slightly less than 6. If you have you log values memorized you’ll know that if it starts with a 2, you actually get a point 7. What is a point 7 slightly less than 6? 5.7!

Just to show you how close we are, I punched this entire example into the calculator, from the kb to ka conversion and I got a final answer of 5.63 for our pH. Now how close is that? Be sure to join me in the next video where we look at calculations related to the Autoionization of water.

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