# Lesson Video: Acid Dissociation Constants Chemistry

In this video, we will learn how to write equations for the dissociation constants of acids and bases and calculate their values.

16:14

### Video Transcript

In this video, we will learn about K𝑎 and K𝑏, the acid and base dissociation constants. We will learn how to write their expressions and practice calculating their values for some acids and bases.

What are K𝑎 and K𝑏? K𝑎 and K𝑏 are equilibrium constants for the dissociation of weak acids and bases in solution, usually aqueous solution. K𝑎 is the acid dissociation constant, sometimes known as the acid ionization constant, and K𝑏 is the base dissociation constant or base ionization constant. The value of K𝑎 or K𝑏 indicates the strength of an acid or base. K𝑎 is a measure of the strength of an acid in solution, and K𝑏 is a measure of the strength of a base in solution.

Let’s imagine this diagram represents an acid or a base, which breaks apart or dissociates in solution into ions. When all of the reactant dissociates into ions, we say that that acid or base is strong. When some of the reactant dissociates into ions and some of the ions reassociate to form the reactant, we say the acid or base is weak. An equilibrium is formed when a weak acid or base dissociates in solution. And therefore we can determine the equilibrium constant for that acid or base in much the same way as other equilibrium constants are determined.

Let’s derive a general expression for K𝑎, the acid dissociation constant. To derive a general expression for K𝑎, we need a general reaction equation for a weak acid HA dissociating or ionizing in solution. When a weak acid HA dissociates in aqueous medium, the reaction equation is HA plus H2O reacting reversibly to give A− plus H3O+. The hydronium ion produced comes from the reaction of water with a hydrogen ion from the weak acid HA. Water plus a hydrogen ion forms the hydronium ion.

We can derive the equilibrium constant for this reversible reaction in much the same way as other equilibrium constants are determined. K is equal to the concentration of all the products multiplied by each other divided by the concentration of all the reactants multiplied by each other. Substituting the products and reactants from our general reaction equation, we get the concentration of the anions A− multiplied by the concentration of the cations, in this case the hydronium ions, H3O+, divided by the multiplication product of the reactant concentrations. Square brackets mean concentration in terms of molar units, or molarity, which is moles per liter or moles per decimeter cubed.

When the solution is dilute, the concentration of water is very large relative to the concentration of the other species and is essentially unchanging. Or at least we assume that the concentration of water doesn’t change. For this reason, we do not include the concentration of water in the denominator. Instead, we make its concentration a constant, lowercase k. If we then take water’s constant to the left-hand side, we get capital K multiplied by small k equals the concentration of the products multiplied with each other divided by the concentration of the weak acid. And the left-hand side value we call K𝑎. This is the general expression for the acid dissociation constant K𝑎.

Because we don’t need to consider water’s concentration, we can rewrite this reaction equation more simply: HA in equilibrium with its ions A− and H+. Since we have removed water, we can change the hydronium ion to a simple hydrogen ion. For this reason, we might sometimes find the K𝑎 expression without a hydronium ion. And instead a hydrogen ion will be in its place. The equilibrium constant for a weak base dissociates in water K𝑏 is derived in much the same way.

Let’s simplify by removing water right from the start so we don’t need to do a long derivation. And we get the general reaction equation for a weak base BOH, which is BOH dissociating reversibly into its ions OH−, the hydroxide ion, and B+. Then, the base dissociation constant K𝑏 is equal to the molar concentrations of the ion products multiplied with each other divided by the molar concentration of the weak base. This is the general equation for the base dissociation constant K𝑏 when a weak base dissociates in aqueous medium.

Let’s put the acid dissociation expression back and have a look at these two expressions. Because the base and acid are weak, they only disassociate to a small extent. And sometimes the denominators are referred to as the original or starting concentrations. But we must remember that these expressions are actually for equilibrium situations. K𝑎 and K𝑏, like other equilibrium constants, depend on temperature. Temperature influences or affects how far an equilibrium sits to the right or to the left. In other words, the equilibrium will sit more to the right or more to the left depending on whether the forward reaction is endothermic or exothermic and depending on the temperature under which the equilibrium exists. For this reason, 25 degrees Celsius is a common temperature at which K𝑎 or K𝑏 is determined. In this way, these values have meaning when they are compared with each other or have meaning when the temperature changes.

Now, let’s practice writing K𝑎 and K𝑏 for a weak acid and a weak base.

HCN, or hydrocyanic acid, is a weak acid. This is the equation for the dissociation of this weak acid in solution. We can write the acid dissociation constant for this acid K𝑎 is equal to the molar concentrations of the ion products multiplied with each other divided by the molar concentration of the acid HCN. Note that because the mole ratio of H+ to CN− is one as to one, these two concentration values will be the same. If the stoichiometry was not one as to one, the concentrations would not be the same. However, the stoichiometry of the products is usually one as to one, even in the case of polyprotic acids, which lose their protons one at a time, forming different equilibrium reactions. In other words, they undergo a stepwise dissociation. But this is a discussion for another video.

When ammonia gas dissolves in water, ammonium hydroxide is formed. This is a weak base. Ammonium hydroxide dissociates to form an equilibrium with its ions, NH4+ and OH−, the ammonium and hydroxide ions. We can write the K𝑏 expression for this base dissociation as follows. K𝑏 is equal to the molar concentrations of the ion products multiplied with each other divided by the molar concentration of ammonium hydroxide. In this example, we again have a molar ratio of the product ions one as to one. And so we know these values are the same.

In a moment, we will practice calculating the actual value of K𝑎 for a weak acid. But first let’s investigate what the size of a K𝑎 or K𝑏 value actually means. What information does it give us? What does the size of K𝑎 or K𝑏 tell us?

We know that K𝑎 or K𝑏 is equal to the molar concentrations of the products multiplied with each other divided by the molar concentration of the reactant. If K𝑎 or K𝑏 is a large value, this tells us that the numerator value is large relative to the denominator value. In other words, the products of the dissociation reaction are favored, meaning the acid or base is relatively strong in that it dissociates to a large degree. Now, I say relatively because K𝑎 and K𝑏 are only for weak acids and bases. But even weak acids or bases have relative strengths.

When K𝑎 or K𝑏 is a small value, this tells us that the numerator value is small relative to the denominator value, meaning the acid or base in its undissociated form is favored. That acid or base is relatively weak, meaning only a little bit of it dissociates. So, again, K𝑎 or K𝑏’s size indicates or is related to the acid or base strength.

Now it’s time to practice.

Calculate the K𝑎 value of a 0.2-molar aqueous solution of propanoic acid with a concentration of H+ ions of 1.62 times 10 to the negative three molar. Give your answer to one decimal place. (A) 4.8 times 10 to the negative four moles per liter. (B) 6.6 times 10 to the negative five moles per liter. (C) 1.6 times 10 to the negative two moles per liter. (D) 5.2 times 10 to the negative seven moles per liter. Or (E) 1.3 times 10 to the negative five moles per liter.

Propanoic acid is CH3CH2COOH. We know this from the name stem “prop,” meaning three carbons in the chain, “an” telling us there are single bonds between carbon atoms, and the suffix -oic acid telling us this is a carboxylic acid with a carboxyl group. Propanoic acid is a weak acid, meaning in solution it dissociates to a small degree, forming an equilibrium according to this equation. The dissociation products are the ions CH3CH2COO− and the hydrogen ion. We are asked to find K𝑎, the acid dissociation constant for this weak acid.

We are told the starting concentration of the acid, 0.2 molar. And we are told the hydrogen ion concentration in solution, 1.62 times 10 to the negative three molar. Let’s start by writing the expression for K𝑎. K𝑎 is equal to the molar concentrations of the ion products, which are the propanoate ion and the hydrogen ion, multiplied with each other divided by the molar concentration of propanoic acid, the reactant. Square brackets refer to molar concentration, or molarity, which is moles per liter or moles per decimeter cubed.

We are given the denominator value in terms of molarity, 0.2 molar. And we are given the hydrogen ion concentration in terms of molarity too, 1.62 times 10 to the negative three molar. But we don’t know the concentration of the propanoate ion. However, from the balanced equation, we know that the molar ratio of the anion to the hydrogen cation is one as to one. And so the concentration of the propanoate anion must be the same as that of the hydrogen ion.

Let’s now put in our values. We get 1.62 times 10 to the negative three molar for the concentration of the anions multiplied by 1.62 times 10 to the negative three molar for the concentration of the hydrogen ions divided by 0.2 molar, which is the concentration of propanoic acid. And the value of the answer is 1.31 times 10 to the negative five. These two units can cancel, and so the unit in the answer is molar, which is moles per liter. Note that sometimes K𝑎 and its relative K𝑏, the base dissociation constant, are expressed without units.

We were asked to give our answer to one decimal place. So let’s round off, and we get 1.3 times 10 to the negative five moles per liter, which corresponds with answer option (E). The K𝑎 value for this propanoic acid equilibrium is 1.3 times 10 to the negative five moles per liter.

Let’s sum up what we’ve learnt in this video. We learnt about K𝑎 and K𝑏, the acid and base dissociation or ionization constants. We saw that for a weak acid in solution HA dissociating to form an equilibrium with its ions, A− and H+, the K𝑎 expression will be the molar concentrations of the ion products multiplied with each other divided by the molar concentration of the reactant, which is the weak acid. And for a weak base dissociating into its ions in solution and forming an equilibrium, the K𝑏 expression is K𝑏 is equal to the molar concentrations of the ion products multiplied with each other divided by the molar concentration of the reactant, which is the weak base.

We learnt that K𝑎 and K𝑏, like other equilibrium constants, are temperature dependent and change according to the temperature. We saw that this is because the values in the numerator and denominator of a K𝑎 or K𝑏 expression will change as an equilibrium shifts due to temperature change. We also learnt that the value or size of K𝑎 or K𝑏 indicates or is a measure of the acid or base strength.