# Calculating keq from ka and kb relationship

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

A pH scale is a measure of how acidic or basic a substance is. Ksp is really just an equilibrium constant (Keq), but it's for a solid dissolving in water. . If this results in a quadratic equation, assume that x is much smaller than [HA] and omit it. I have a practice exam question where we have Ka but no Keq, but the formula requires Keq. I am just missing the Keq and am confused of how. Conjugate Acids of Bases - Ka Kb and Kw. Last updated Treat the conjugate acid of a base as an acid in numerical calculations. Reverse the If so, what is the relationship between Ka of the conjugate acid and Kb of the base? We are Then the expression for the equilibrium constant for the reaction.

If NH4 plus donates a proton you're left with NH3.

The Ka for this reaction is 5. Now let's look at NH3 which we know is a weak base, and it's going to take a proton from water, therefore forming NH4 plus. If we take a proton from water we're left with OH minus.

Since we talked about a base here we're gonna use Kb, and Kb for this reaction is 1. What would happen if we add these two reactions together?

## Acids and Bases: Ionization Constants

We have two water molecules for our reactants, so let me go ahead and write H2O plus H2O here. We have ammonium on the left side for reactant, we also have ammonium over here for our product. Same thing happens with ammonia, NH3. We have NH3 on the left. We have NH3 on the right. We have NH3 as a reactant, NH3 as a product.

We can cancel those out too.

## Calculating the equilibrium constant from the standard cell potential

Our only reactants would be two H2O. This reaction should sound familiar to you. This net reaction is the auto-ionization of water where one water molecule acts as an acid, one water molecule acts as a base. We get H3O plus and OH minus. The equilibrium constant for the auto-ionization of water you've already seen that Kw is equal to 1. We added these two reactions together and we got this for our net reaction. What would we do with Ka and Kb to get Kw?

It turns out that you multiply them, Ka times Kb for a conjugate acid-base pair is equal to Kw. Let's do that math. Kb is equal to 1. We're going to multiply that by 1. This is equal to 1. When you add reactions together to get a net reaction you multiply the equilibrium constants to get the equilibrium constant for the net reaction which in this case is Kw for the auto-ionization of water.

Let's go ahead and go in even more detail here. Ka, that's your products over your reactants. That'd be H3O plus, the concentration of H3O plus times the concentration of ammonia. Let's go ahead and do that. The concentration of H3O plus times the concentration of ammonia, and that's all over the concentration of ammonium. This is all over the concentration of ammonium.

This represents, let me go ahead and highlight this here. Next let's think about Kb. Let's put this in parenthesis here. We have the concentration of ammonium, NH4 plus, times the concentration of OH minus.

That's over the concentration of NH3. That's over the concentration of NH3 here. What do we get? The ammonium would cancel out. Then the NH3 cancels out. We're left with H3O plus times OH minus which we know is equal to 1. Just another way to think about this. This can be important, relating Ka and Kb to Kw. If you know one you can figure out the other.

### Ionization of Acids and Bases

Let's think about a strong acid for a second here. Let's think about HCl.

The conjugate base to HCl would be Cl minus, the chloride anion here. Let's think about what this equation means. HCl is a strong acid which means a very high value for Ka. An extremely, extremely high value for Ka.

What does that say about Kb for the conjugate base? So, for the reduction half reaction, the standard reduction potential is negative 1. We reversed the reaction, so we need to change the signs. So the standard oxidation potential is positive 1. For our reduction half reaction, we left it how it was written, all right.

So we're just gonna write our standard reduction potential as positive. Next, we need to look at our balanced equation, all right. We need to make the number of electrons equal for our half reactions. So for this first half reaction, I'm just gonna draw a line through, I'm just gonna draw a line through this half reaction so we don't get ourselves confused. For our first half reaction here, we have two electrons. Then over here, right, for our oxidation half reaction, we have three electrons.

We need to have the same number of electrons. So we need to have six electrons for both half reactions, because remember the electrons that are lost are the same electrons that are gained.

So we need to multiply our first half reaction by three. All right, if you multiply our first half reaction by three, we'll end up with six electrons. And our second half reaction, we would need to multiply the oxidation half reaction by two, in order to end up with six electrons. So let's rewrite our half reactions.

So, first we'll do the reduction half reaction, so we have, let me change colors again here. And let's do this color. So we have three times, since we have three I two now, We have three I two, and three times two gives us six electrons.

So three I two plus six electrons, and then three times two, all right, gives us, three times two gives us six I minus. All right, so we multiplied our half reaction by three, but remember, we don't multiply the voltage by three, 'cause voltage is an intense of property.

So the standard reduction potential is still positive. So we have positive. Next, we need to multiply our oxidation half reaction by two. So we have two Al, so this is our oxidation half reaction, so two Al, so two aluminum, and then we have two Al three plus, so two Al three plus, and then two times three gives us six electrons. So now we have our six electrons and once again, we do not multiply our standard oxidation potential by two, so we leave that, so the standard oxidation potential is still positive 1.

Next we add our two half reactions together. And if we did everything right, we should get back our overall equation. So our overall equation here.

### Calculating the equilibrium constant from the standard cell potential (video) | Khan Academy

We have six electrons on the reactant side, six electrons on the product side, so the electrons cancel out. And so we have for our reactants three I two, so we have three I two, plus two Al, and for our products right here, we have six I minus, so six I minus, plus two Al three plus, plus two Al three plus.

So this should be our overall reaction, all right, this should be the overall reaction that we were given in our problem. Let's double check that real fast. So three I two plus two Al, all right, so right up here, so three I two plus two Al, should give us six I minus plus two Al three plus. So six I minus plus two Al three plus.

So we got back our original reaction. Remember our goal was, our goal was to find the standard cell potential E zero, because from E zero we can calculate the equilibrium constant K. So we know how to do that, again, from an earlier video. To find the standard cell potential, all right, so to find the standard cell potential, all we have to do is add our standard reduction potential and our standard oxidation potential.

So if we add our standard reduction potential and our standard oxidation potential, we'll get the standard cell potential. So that would be positive. So the standard potential for the cell, so E zero cell is equal to. All right, now that we've found the standard cell potential, we can calculate the equilibrium constant.