Ka and Kb calculations

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  1. Applying the Ka expression
  2. Weak acids/bases at equilibrium.
  3. Assumptions made in calculation.
  4. Calculating pH and percentage dissociation.
  5. Calculations using Kb.
  1. Find the pH of the weak acid solution and the percentage dissociation of the weak acid/base.
    1. What is the pH of a solution of 0.5 M ethanoic acid, CH3COOH? Find the percentage dissociation of this ethanoic acid solution.
    2. What is the pH of a solution of 0.1 M ammonia, NH3? Find the percentage dissociation of this ammonia solution.
  2. Find the Ka of an unknown weak acid, given pH and concentration.
    1. A solution of carbonic acid had a pH of 3.72. What was the initial concentration of this acid solution?
Topic Notes

In this lesson, we will learn:

  • How to apply the Ka expression to find the pH of weak acid solutions.
  • The assumptions made in Ka calculations at equilibrium and how to justify them.
  • How to solve for Kb using Kw and find the pH of weak base solutions.


  • We know that weak acids and bases are any acid/base species that does not completely dissociate in water. Dissolving a weak acid in water then has two chemical effects:
    • Some of the weak acid, HX, will interact with water and dissociate into H3O+ and X- ions.
    • The rest of the HX will stay un-dissociated.
    We can write an equilibrium for the dissociation of weak acid HX (with A concentration) in water, and show amounts in a table, in a β€˜before and after’ format like below:

    HX + H2O β‡Œ\rightleftharpoons H3O+ + Cl-

    Start concentration (M)

    Equilibrium conc. (M)



    A - B


    β‰ˆ\approx 0





    The acid dissociation constant, Ka can be expressed in these terms:

    Ka = [Xβˆ’][H3O+][HX]\frac{[X^{-}][H_3O^+]}{[HX]} = [X2][A]βˆ’B\frac{[X^2]}{[A] -B}

  • Taking an example with 0.1M methanoic acid HCOOH, we can write the following:

    Start concentration (M)

    Equilibrium conc. (M)



    0.1 - B







    We can apply the acid dissociation constant, Ka to this equilibrium. Methanoic acid1 has a Ka value of 1.8βˆ—*10-4 so the Ka equation can be written fully:

    Ka = [HCOOβˆ’][H3O+][HCOOH]\frac{[HCOO^-][H_3O^+]}{[HCOOH]} = [H3O+]2[0.1]βˆ’[H3O+]\frac{[H_3O^+]^2}{[0.1] -[H_3O^+]} = 1.8 βˆ—* 10-4

    Some assumptions are made to complete this calculation:
    • The starting concentration of H3O+ ions, in neutral water is only 1*10-7 M (see Autoionization of water) and an equally tiny amount of hydroxide ions are also present. This is an incredibly small amount, so this H3O+ is not taken into the calculation; only H3O+ due to the weak acid is used in the calculation.
    • With weak acids, we assume that the acid is weak enough that the amount of dissociation doesn’t affect acid concentration. Using the table above, 0.1 M HCOOH added to neutral water will still have concentration of approximately 0.1M at equilibrium, and we can ignore the β€˜β€“ B’ in β€˜A-B’. Where [HX]eq = equilibrium concentration, [HX]i = start concentration of HX:

      [HX]eq - B β‰…\cong [HX]i


    With the assumptions, we have a final expression:

    Ka = [HCOOβˆ’][H3O+][HCOOH]\frac{[HCOO^-][H_3O^+]}{[HCOOH]} = [H3O+]2[0.1]\frac{[H_3O^+]^2}{[0.1] } = 1.8 βˆ—* 10-4

    0.1 βˆ—* 1.8 βˆ—* 10-4 = [H3O+]2

    1.8βˆ—10βˆ’5\small\sqrt{1.8 * 10^{-5}} = [H3O+] = 4.24 βˆ—* 10-3

    pH = -log [H3O+] = 2.37

    Note that the assumption we made was that [HX] – B is approximately equal to [HX]. We now know that B = 4.24*10-3, so our assumption in this example was to say that 0.1 – 4.24*10-3 = 0.0958.
    The assumption can be justified if percentage dissociation is less than 5% which we can work out:

    % dissociation = [H3O+]eq[HX]i\frac{[H_3O^+]_{eq}}{[HX]_i} βˆ—* 100


    % dissociation = [4.24βˆ—10βˆ’3][0.1]\frac{[4.24 *10^{-3}]}{[0.1]} βˆ—* 100 = 4.24%

    As the calculation shows, the assumption was justified as only 4.24% dissociation occurs.

  • Kb calculations are similar to Ka calculations with some changes:
    • Because acidity strength tables give only Ka, Kb of a weak base will need to be found by the calculation in the autoionization of water expression. You will need to find the Ka of the conjugate acid in the acidity strength table to do this.
    • The equilibrium concentrations you obtain using Kb will give you [OH-], so pH will need to be found by solving: pH = 14 – pOH.

    Taking an example with 0.5M of the weak base ammonia, NH3, we can write the following:

    NH3 + H2O β‡Œ\rightleftharpoons NH4+ + OH-

    Start concentration (M)

    Equilibrium conc. (M)



    0.5 - B







    The conjugate acid of ammonia is the ammonium ion, NH4+ which has a Ka value1 of 5.6 βˆ—* 10-10. Solving the autoionization expression for Kb(NH3) gives:

    Kb = KwKa\frac{K_w}{K_a} = 10βˆ’145.6βˆ—10βˆ’10\frac{10^{-14}}{5.6 * 10^{-10}} = 1.79 βˆ—* 10-5

    Using our value for Kb (now rounding to 1.8 βˆ—* 10-5 or 2 significant figures) we can solve for the equilibrium concentration of hydroxide ions. Again, we make the assumption that the concentration of NH3 isn’t significantly affected by the dissociation into NH4+:

    [NH3]eq - B β‰…\cong [NH3]i

    Now we can find the hydroxide ion concentration:

    Kb = [NH4+][OHβˆ’][NH3]\frac{[NH_{4}^{+}][OH^-]{}} {[NH_3]} = [OHβˆ’]2[0.5]\frac{[OH^-]^2}{[0.5]} = 1.8 βˆ—* 10-5

    [OH-] = ((1.8βˆ—10βˆ’5)βˆ—0.5)\sqrt{((1.8 * 10^{-5}) * 0.5)} = 3 βˆ—* 10-3

    pOH = -log[OH-] = -log (3 βˆ—* 10-3) = 2.52

    pH = 14 - pOH = 14 - 2.52 = 11.48

    Testing the assumption can now be done:

    % dissociation = [OHβˆ’]eq[NH3]i\frac{[OH^-]_{eq}}{[NH_3]_i} βˆ—* 100 = 3βˆ—10βˆ’30.5\frac{3 * 10^{-3}}{0.5} βˆ—* 100 = 0.6 %

    With only 0.6% dissociation, the assumption is justified and the pH has been found.

  • Another calculation that shows the difference between strong and weak acids and bases is the effect of dilution on pH.
    The effect on strong acids is straightforward, because we assume 100% dissociation:
    • If a solution of strong acid, e.g. HCl is 1M, then [H3O+ (aq)] = 1M. Taking the negative log of this:
      pH = -log[1] = 0
      Diluting this by a factor of 10 will give a concentration of 0.1M
      pH = -log[0.1] = 1
      A further 10, or 100 fold from the original:
      pH = -log[0.01] = 2
      In short, diluting a strong acid or base has a direct logarithmic effect on pH.
    The effect of dilution on weak acids and bases is different:
    • If a solution of weak acid, e.g. CH3COOH is 1M, then pH and [H3O+ (aq)] is worked out using the Ka expression:
      Ka (CH3COOH) = 1.4*10-5

      Ka = [H3O+][CH3COOβˆ’][CH3COOH]\frac{[H_{3}O^{+}][CH_{3}COO^{-}]}{[CH_{3}COOH]}

      The calculation to find pH using this expression has been explained above, so moving forward to an answer (using the assumptions needed)

      1.4 * 10-5 = [H3O+][CH3COOβˆ’][1]\frac{[H_{3}O^{+}][CH_{3}COO^{-}]}{[1]} where [H3O+] = [CH3COOH-]

      1.4βˆ—10βˆ’5\sqrt{1.4 * 10^{-5}} = [H3O+] = 3.74 * 10-3

      pH = -log[ 3.74 * 10-3] = 2.42

      A dilution of this weak acid solution to make it 0.1M would have the following effect on the calculation:

      1.4 * 10-5 = [H3O+][CH3COOβˆ’][0.1]\frac{[H_{3}O^{+}][CH_{3}COO^{-}]}{[0.1]} where [H3O+] = [CH3COOH-]

      1.4βˆ—10βˆ’6\sqrt{1.4 * 10^{-6}} = [H3O+] = 1.18 * 10-3

      pH = -log[ 1.18 * 10-3] = 2.92

      Another dilution by a factor of ten:

      1.4 * 10-5 = [H3O+][CH3COOβˆ’][0.01]\frac{[H_{3}O^{+}][CH_{3}COO^{-}]}{[0.01]} where [H3O+] = [CH3COOH-]

      1.4βˆ—10βˆ’7\sqrt{1.4 * 10^{-7}} = [H3O+] = 3.74 * 10-4

      pH = -log[ 3.74 * 10-4] = 3.42

      In short, diluting a weak acid has a lesser effect on pH than in strong acids.