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

An error occurred trying to load this video.

Try refreshing the page, or contact customer support.

Coming up next: Acid-Base Buffers: Calculating the pH of a Buffered Solution

You're on a roll. Keep up the good work!

Take Quiz Watch Next Lesson
Your next lesson will play in 10 seconds
  • 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
Save Save Save

Want to watch this again later?

Log in or sign up to add this lesson to a Custom Course.

Log in or Sign up

Speed Speed Audio mode

Recommended Lessons and Courses for You

Lesson Transcript
Elizabeth (Nikki) Wyman

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

Expert Contributor
Dawn Mills

Dawn has taught chemistry and forensic courses at the college level for 9 years. She has a PhD in Chemistry and is an author of peer reviewed publications in chemistry.

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 the 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.

To unlock this lesson you must be a Member.
Create your account

Additional Activities

Acid-Base Equilibrium

The acid and base strength affects the ability of each compound to dissociate. The following questions will provide additional practice in calculating the acid (Ka) and base (Kb) dissociation constants. Step by step solutions are provided to assist in the calculations.


  1. Vinegar, also known as acetic acid, is routinely used for cooking or cleaning applications in the common household. Calculate the acid dissociation constant for acetic acid of a solution purchased from the store that is 1 M and has a pH of 2.5.
  2. The Ka value of HCO_3^- is determined to be 5.0E-10. Determine the value for the Kb and identify the conjugate base by writing the balanced chemical equation. Based on the Kb value, is the anion a weak or strong base?


1. First, write the balanced chemical equation.

  • CH_3CO_2H + H_2O ---> CH_3CO2^- + H_3O^+

Write the acid dissociation formula for the equation: Ka = [H_3O^+] [CH_3CO2^-] / [CH_3CO_2H]

  • Initial concentrations: [H_3O^+] = 0, [CH_3CO2^-] = 0, [CH_3CO_2H] = 1.0 M
  • Change in concentration: [H_3O^+] = +x, [CH_3CO2^-] = +x, [CH_3CO_2H] = -x
  • Equilibrium concentration: [H_3O^+] = x, [CH_3CO2^-] = x, [CH_3CO_2H] = 1.0 - x

Plug in the equilibrium values into the Ka equation.

  • Ka = x^2 / (1.0 - x)

Determine [H_3O^+] using the pH where [H_3O^+] = 10^-pH.

  • [H_3O^+] = 10^-2.5 = 0.00316 = x

Plug this value into the Ka equation to solve for Ka.

  • Ka = 0.00316 ^2 / (1.0 - 0.00316) = 0.000009986 / 0.99684 = 1.002E-5

2. First, write the balanced chemical equation.

  • CO_3^2- + H_2O --> HCO_3^- + OH^-

The conjugate base is CO_3^2-.

  • Kb = [HCO3^-] [OH^-] / [CO_3^2-]

Recall that Kb = Kw / Ka.

  • Kb = 1E-14 / 5E-10 = 2.0E-5

The Kb value is high, which indicates that CO_3^2- is a strong base.

Register to view this lesson

Are you a student or a teacher?

Unlock Your Education

See for yourself why 30 million people use

Become a member and start learning now.
Become a Member  Back
What teachers are saying about
Try it risk-free for 30 days

Earning College Credit

Did you know… We have over 200 college courses that prepare you to earn credit by exam that is accepted by over 1,500 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level.

To learn more, visit our Earning Credit Page

Transferring credit to the school of your choice

Not sure what college you want to attend yet? has thousands of articles about every imaginable degree, area of study and career path that can help you find the school that's right for you.

Create an account to start this course today
Try it risk-free for 30 days!
Create an account