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Acid-Base Equilibrium: Calculating the Ka or Kb of a Solution

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  • 0:01 Review: Acid & Base Strength
  • 2:49 Intro to Ka & Kb
  • 5:14 Ka & Kb in Action
  • 8:40 Finding pH Given Ka
  • 10:40 Lesson Summary
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Lesson Transcript
Instructor: Elizabeth (Nikki) Wyman

Nikki has a master's degree in teaching chemistry and has taught high school chemistry, biology and astronomy.

In this lesson, you will review acid and base strength and acid and base dissociation. You will then learn what acid and base dissociation constants (Ka and Kb) are, what they mean, and how to perform calculations involving them.

Review: Acid and Base Strength

Why is it that some acids can eat through glass, but we can safely consume others? Why can you cook with a base like baking soda, but you should be extremely cautious when handling a base like drain cleaner?

The answer lies in the ability of each acid or base to break apart, or dissociate: strong acids and bases dissociate well (approximately 100% dissociation occurs); weak acids and bases don't dissociate well (dissociation is much, much less than 100%).

Let's go to the lab and zoom into a sample of hydrochloric acid to see what's happening on the molecular level. Given that hydrochloric acid is a strong acid, can you guess what it's going to look like inside?

Hydrochloric Acid
hydrochloric acid at molecular level

There are no HCl molecules to be found because 100% of the HCl molecules have broken apart into hydrogen ions and chloride ions. In fact, the hydrogen ions have attached themselves to water to form hydronium ions (H3O+). We would write out the dissociation of hydrochloric acid as HCl + H2O --> H3O+ + Cl-.

HCl is the parent acid, H3O+ is the conjugate acid, and Cl- is the conjugate base. The conjugate acid and conjugate base occur in a 1:1 ratio. In case it's not fresh in your mind, a conjugate acid is the protonated product in an acid base reaction or dissociation. A conjugate base is the negatively charged particle that remains after a proton has dissociated from an acid.

If we were to zoom into our sample of hydrofluoric acid, a weak acid, we would find that very few of our HF molecules have dissociated. However, we would still write the dissociation the same: HF + H2O --> H3O+ + F-. In fact, for all acids we can use a general expression for dissociation using the generic acid HA: HA + H2O --> H3O+ + A-.

For all bases, we can use a general equation using the generic base B: B + H2O --> BH+ + OH-. B is the parent base, BH+ is the conjugate acid, and OH- is the conjugate base. Great! We know what is going on chemically, but what if we can't zoom into the molecular level to see dissociation? How is acid or base dissociation measured then?

Intro to Ka and Kb

We use dissociation constants to measure how well an acid or base dissociates. For acids, these values are represented by Ka; for bases, Kb. These constants have no units.

All chemical reactions proceed until they reach chemical equilibrium, the point at which the rates of the forward reaction and the reverse reaction are equal. We use the equilibrium constant, Kc, for a reaction to demonstrate whether or not the reaction favors products (the forward reaction is dominant) or reactants (the reverse reaction is dominant). High values of Kc mean that the reaction is product-favored, while low values of Kc mean that the reaction is reactant-favored.

For acid and base dissociation, the same concepts apply, except that we use Ka or Kb instead of Kc. High values of Ka mean that the acid dissociates well and that it is a strong acid. Low values of Ka mean that the acid does not dissociate well and that it is a weak acid. The same logic applies to bases.

There is a relationship between the concentration of products and reactants and the dissociation constant (Ka or Kb). For acids, this relationship is shown by the expression: Ka = [H3O+][A-] / [HA].

The products (conjugate acid H3O+ and conjugate base A-) of the dissociation are on top, while the parent acid HA is on bottom. Notice that water isn't present in this expression. We get to ignore water because it is a liquid, and we have no means of expressing its concentration.

For bases, this relationship is shown by the equation Kb = [BH+][OH-] / [B]. The products (conjugate acid and conjugate base) are on top, while the parent base is on the bottom. Once again, water is not present.

Both the Ka and Kb expressions for dissociation can be used to determine an unknown, whether it's Ka or Kb itself, the concentration of a substance, or even the pH.

Ka and Kb in Action

Let's go into our cartoon lab and do some science with acids!

We need a weak acid for a chemical reaction. We have an acetic acid (HC2H3O2) solution that is 0.9 M. Its hydronium ion concentration is 4 * 10^-3 M. What is the Ka for acetic acid? Is this a strong or a weak acid?

To solve this problem, we will need a few things: the equation for acid dissociation, the Ka expression, and our algebra skills.

The equation is for the acid dissociation is HC2H3O2 + H2O <==> H3O+ + C2H3O2-.

The Ka expression is Ka = [H3O+][C2H3O2-] / [HC2H3O2].

The problem provided us with a few bits of information: that the acetic acid concentration is 0.9 M, and its hydronium ion concentration is 4 * 10^-3 M.

Since the equation is in equilibrium, the H3O+ concentration is equal to the C2H3O2- concentration. We plug the information we do know into the Ka expression and solve for Ka.

Ka = (4.0 * 10^-3 M) (4.0 * 10^-3 M) / 0.90 M

Ka = 1.8 * 10^-5

This Ka value is very small, so this is a weak acid.

In another laboratory scenario, our chemical needs have changed. We absolutely need to know the concentration of the conjugate acid for a super concentrated 15 M solution of NH3. We know that the Kb of NH3 is 1.8 * 10^-5. Like with the previous problem, let's start by writing out the dissociation equation and Kb expression for the base.

The equation is NH3 + H2O <==> NH4+ + OH-. NH4+ is our conjugate acid.

Our Kb expression is Kb = [NH4+][OH-] / [NH3].

We know that Kb = 1.8 * 10^-5 and [NH3] is 15 M. We can make the assumption that [NH4+] = [OH-] and let these both equal x.

The equation then becomes Kb = (x)(x) / [NH3].

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