# Relationship between ka and kb values

### Relationship between Ka and Kb (article) | Khan Academy

First of all, let me put a statement to clarify the proper meanings of K a and K b for this case. The K a is the acid dissociation constant of a weak acid, and the K b. Ka Values of Conjugate Acids of Bases If so, what is the relationship between Ka of the conjugate acid. Relationship between Ka of a weak acid and Kb for its conjugate base. .. We can also use the value of KwK_\text{w}K​w​​K, start subscript, w, end subscript .

The ammonium ion would function as an acid and donate a proton to water to form H3O plus. 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.

### Conjugate Acids of Bases - Ka Kb and Kw - Chemistry LibreTexts

What would happen if we add these two reactions together? 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. 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. Let's think about a strong acid for a second here.

### Relation between Ka and Kb - Chemistry!!! Not Mystery

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. For more background on weak acids and bases, check out the article on weak acid-base equilibria. Diagram of beaker containing aqueous solution of hydrofluoric acid. Ions of water, neutral HF molecules, and fluoride and hydrogen ions are all shown in solution.

For this article, all solutions will be assumed to be aqueous solutions. For a refresher on how to write equilibrium constants for reversible reactions, see the article on the equilibrium constant K.

An important thing to remember is that these equations only work for conjugate acid-base pairs!! For a quick review on how to identify conjugate acid-base pairs, check out the video on conjugate acid-base pairs.

Please choose from one of the following options. Let's work through this problem step-by-step.

Let's look at NH4 plus. The ammonium ion would function as an acid and donate a proton to water to form H3O plus. 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? 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.

## Conjugate Acids of Bases - Ka Kb and Kw

We have NH3 on the right. We have NH3 as a reactant, NH3 as a product. We can cancel those out too. 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. Let's think about a strong acid for a second here.