Ph pka and pkb relationship questions

pH, pKa, Ka, pKb, Kb - Organic Chemistry | Socratic

ph pka and pkb relationship questions

This can be seen through the equation pKa + Pkb = pKw. I hope this could answer your questions! I think Dr. Lavelle mentioned this way to remember that the low pKa = strong acid relationship is similar to pH during lecture. pH describes the acidity of a solution. pKa and pKb are the logarithmic acid and base dissociation constants, respectively. Ka and Kb Key Questions. What are . Sample Problems and Answers - Lecture 1. What is the pH of a pKa + pKb = so. pKb = – = equation, the pH is: pH = pKa +. [base].

A minus, and that's also going to be in an aqueous solution. And that's the general pattern. We've seen the case where A could be an NH3, right? And this is just NH3. A could be a fluorine molecule right there, because then this would be hydrogen fluoride or hydrofluoric acid. And this would just be the negative ion of fluorine. Or a fluorine with an extra electron. So it could be a bunch of stuff. You can just throw in anything there, and it'll work.

Especially for the weak acids. So we learned a last video that if this is the acid, then this is the conjugate base. And we could write the same reactions, essentially, as kind of more of a basic reaction.

So we could say, if I start with A minus-- it's in an aqueous solution-- that's in equilibrium with-- This thing could grab a hydrogen from the surrounding water and become neutral then. It's still in an aqueous solution. And then one of those water molecules that it plucked that hydrogen off of is now going to be a hydroxide molecule.

ph pka and pkb relationship questions

Remember, whenever I say pluck the hydrogen, just the proton, not the electron for the hydrogen. So the electron stays on that water molecule, so it has a negative charge.

It's in an aqueous solution. So we could write the same reaction both ways. And we can write equilibrium constants for both of these reactions.

pKa and pKb relationship

So let's do that. Let me erase this, just because I can erase this stuff right there, and then use that space. So an equilibrium reaction for this first one. I could call this the K sub a, because the equilibrium reaction for an acid.

And so this is going to be equal to its products. So the concentration of my hydrogen times my concentration of whatever my conjugate base was, divided by my concentration of my original acid. So this would be the concentration of HA. I could also write an equilibrium constant for this basic reaction.

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Let me do it right down here. So I'll call that my K sub b. This is a base equilibrium.

Relationship between pka, ka, and acidity strength - CHEMISTRY COMMUNITY

And so this is equal to the concentration of the products. It's becoming tedious to keep switching colors. Actually, I'll do it. Because it makes it easier look at, at least for me.

ph pka and pkb relationship questions

HA times the concentration of my hydroxide ions divided by my concentration of my weak base. Remember, this can only be true of a weak base or a weak acid.

If we were dealing with a strong acid or is or a strong base, this would not be an equilibrium reaction.

Ph vs PKA

It would only go in one direction. And when it only goes in one direction, writing this type of equilibrium reaction makes no sense-- or equilibrium constant-- because it's not in equilibrium. It only goes in one direction. If A was chlorine, if this was hydrochloric acid, you couldn't do this. You would just say look, if you have a mole of this, you're just dumping a mole of hydrogen protons in that solution and then a bunch of chlorine anions who are not going to do anything.

Even though they are the conjugate base, they wouldn't do anything. So you can only do this, remember, for weak acids and bases. So with that said, let's see if we can find a relationship between Ka and Kb.

What do we have here? We have an A minus on both sides of this. We have H over-- OH over A minus. Let's solve for A minus. If we multiply both sides of this equation by HA over H plus, on the left-hand side we get Ka times the inverse of this. So you have your HA over H plus is equal to your concentration of your conjugate base.

And let's do the same thing here. Solve for A minus. So to solve for A minus here, we might have to do 2 steps. So if we take the inverse of both sides, you get 1 over Kb is equal to A minus over H, the concentration of my conjugate acid times the concentration of hydroxide.

Multiply both sides by this. And I get A minus is equal to my concentration of my conjugate acid times concentration of hydroxide. All of that over my base equilibrium constant. Now, these are the same reactions. In either reaction for given concentrations, I'm going to end up with the same concentration. This is going to equal that. These are two different ways of writing the exact same reaction.

So let's set them equal to each other. So let me copy and paste it, actually. So I'm saying that this thing, copy, is equal to this thing right here. So this is equal to-- let me copy and paste this-- that. That's equal to that. So let's see if we can find a relationship between Ka and Kb. Well, one thing we can do is we can divide both sides by HA. So if we divide both sides by HA. Actually, I could probably have that earlier on to the whole thing.

If we ignore this part right here, this is equal to that. Let me erase all of this. I'm using the wrong tool. So we could say that they both equal the concentration of A minus. So that's equal to that. We can divide both sides by HA. This over here will cancel with this over here. And we're getting pretty close to a neat relationship. And so we get Ka over our hydrogen proton concentration is equal to our hydroxide concentration divided by Kb.

You can just cross-multiply this. So we get Ka, our acidic equilibrium concentration, times Kb is equal to our hydrogen concentration times our hydroxide concentration. Remember, this is all in an aqueous solution. What do we know about this?

A lot of times, you just wanna know, you know, what's in your solution, depending on what you wanna do to your solution, if you wanna add things to it, maybe you wanna add some acid, you wanna add some base. You wanna know what's going on. The Henderson-Hasselbalch equation gives you a really quick and easy way of doing that. So what we're gonna do, is we're gonna rearrange this equation to solve for this ratio that we might be interested in.


And I don't know about you, but I actually find, well, laughs I find logs not super-intuitive sometimes. So I'm actually going to get rid of the log by raising both sides to the 10th power. So what does this tell us? It may not look like it tells us a whole lot more, but actually, it tells us a lot. It tells us about the relative relationship and size between A minus and HA concentration.

So if we look at this, we can derive a couple relationships. So let's go ahead and look at all the possible scenarios for these three things. So anything to the zeroth power is equal to one. Which tells us that this ratio is equal to one. And if A minus concentration over HA concentration is equal to one, that means that they have the same concentration.

I forgot a minus sign there. This is a really helpful thing to remember. And this comes up a lot not just when you're talking about buffers by themselves, but also when you're doing titrations. And the point in your titration where the HA is equal to A minus is called the half-equivalence point.