Relationship between pka, ka, and acidity strength - CHEMISTRY COMMUNITY
Sorry, you have misunderstood the concept of acidity. Low p K a means Surely, hydrochloric acid is a much stronger acid than acetic acid. Therefore pKa is also a measure of how stable the conjugate base is. Put another way, strong acids have weak conjugate bases, and vice. Acids and Bases. Your organic teachers are quite likely to ask you questions like identify the most acidic protons or the most basic site in a molecule. These facts.
Factor 2 — The Role of the Atom This point causes a lot of confusion due to the presence of two seemingly conflicting trends. This makes sense, right? It makes sense that HF is more electronegative than H2O, NH3, and CH4 due to the greater electronegativity of fluorine versus oxygen, nitrogen, and carbon. A fluorine bearing a negative charge is a happy fluorine.
You could make two arguments for why this is. The first reason has to do with the shorter and stronger H-F bond as compared to the larger hydrogen halides. The second has to do with the stability of the conjugate base. The fluoride anion, F — is a tiny and vicious little beast, with the smallest ionic radius of any other ion bearing a single negative charge. Its charge is therefore spread over a smaller volume than those of the larger halides, which is energetically unfavorable: Factor 3 — Resonance.
A huge stabilizing factor for a conjugate base is if the negative charge can be delocalized through resonance. Factor 4 — Inductive effects. Electronegative atoms can draw negative charge toward themselves, which can lead to considerable stabilization of conjugate bases. Check out these examples: Predictably, this effect is going to be related to two major factors: Factor 5 — Orbitals.
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. 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. Let me do it right down here. So I'll call that my K sub b. This is a base equilibrium. 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. 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. 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.
pKa and pKb relationship (video) | Khan Academy
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.
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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?
What do we know about our hydrogen times our hydroxide concentration in an aqueous solution? For example, let me review just to make sure I'm jogging your memory properly. We could have H2O. It can autoionize into H plus. And this has an equilibrium. You just put the products. So the concentration of the hydrogen protons times the concentration of the hydroxide ions. And you don't divide by this because it's the solvent. And we already figured out what this was. If we have just completely neutral water, this is 10 to the minus 7.
And this is 10 to the minus 7. So this is equal to 10 to the minus Now, these two things could change. I can add more hydrogen, I could add more hydroxide. And everything we've talked about so far, that's what we've been doing. That's what acids and bases do.