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Relationship between Ka and Kb

Relationship between Ka of a weak acid and Kb for its conjugate base. Equations for converting between Ka and Kb, and converting between pKa and pKb. 

Key points

For conjugate-acid base pairs, the acid dissociation constant K, start subscript, start text, a, end text, end subscript and base ionization constant K, start subscript, start text, b, end text, end subscript are related by the following equations:
  • K, start subscript, start text, a, end text, end subscript, dot, K, start subscript, start text, b, end text, end subscript, equals, K, start subscript, start text, w, end text, end subscript
    where K, start subscript, start text, w, end text, end subscript is the autoionization constant
  • start text, p, end text, K, start subscript, start text, a, end text, end subscript, plus, start text, p, end text, K, start subscript, start text, b, end text, end subscript, equals, 14, space, space, start text, a, t, space, end text, 25, degrees, start text, C, end text

Introduction: Weak acid and bases ionize reversibly

Weak acids, generically abbreviated as start color #1fab54, start text, H, A, end text, end color #1fab54, donate start text, H, end text, start superscript, plus, end superscript (or proton) to water to form the conjugate base start color #1fab54, start text, A, end text, start superscript, minus, end superscript, end color #1fab54 and start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript:
start text, H, A, end text, left parenthesis, a, q, right parenthesis, plus, start text, H, end text, start subscript, 2, end subscript, start text, O, end text, left parenthesis, l, right parenthesis, \rightleftharpoons, start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript, left parenthesis, a, q, right parenthesis, plus, start text, A, end text, start superscript, minus, end superscript, left parenthesis, a, q, right parenthesis
space, space, start color #1fab54, start text, a, c, i, d, end text, end color #1fab54, space, space, space, space, space, space, space, space, space, space, start text, b, a, s, e, end text, space, space, space, space, space, space, space, space, space, space, space, space, start text, a, c, i, d, end text, space, space, space, space, space, space, space, space, space, space, space, space, space, space, start color #1fab54, start text, b, a, s, e, end text, end color #1fab54
Similarly, a base (abbreviated as start color #aa87ff, start text, B, end text, end color #aa87ff) will accept a proton in water to form the conjugate acid, start color #aa87ff, start text, H, B, end text, start superscript, plus, end superscript, end color #aa87ff, and start text, O, H, end text, start superscript, minus, end superscript:
start text, B, end text, left parenthesis, a, q, right parenthesis, plus, start text, H, end text, start subscript, 2, end subscript, start text, O, end text, left parenthesis, l, right parenthesis, \rightleftharpoons, start text, H, B, end text, start superscript, plus, end superscript, left parenthesis, a, q, right parenthesis, plus, start text, O, H, end text, start superscript, minus, end superscript, left parenthesis, a, q, right parenthesis
space, space, start color #aa87ff, start text, b, a, s, e, end text, end color #aa87ff, space, space, space, space, space, space, space, space, space, start text, a, c, i, d, end text, space, space, space, space, space, space, space, space, space, space, start color #aa87ff, start text, a, c, i, d, end text, end color #aa87ff, space, space, space, space, space, space, space, space, space, space, start text, b, a, s, e, end text
For a weak acid or base, the equilibrium constant for the ionization reaction quantifies the relative amounts of each species. In this article, we will discuss the relationship between the equilibrium constants K, start subscript, start text, a, end text, end subscript and K, start subscript, start text, b, end text, end subscript for a conjugate acid-base pair.
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.
An aqueous solution of hydrofluoric acid, a weak acid, contains undissociated start text, H, F, end text molecules which are in equilibrium with protons and fluoride ions.
Note: For this article, all solutions will be assumed to be aqueous solutions.

Finding K, start subscript, start text, a, end text, end subscript for start text, H, A, end text reacting as an acid

Let's look more closely at the dissociation reaction for a monoprotic weak acid start text, H, A, end text:
start text, H, A, end text, left parenthesis, a, q, right parenthesis, plus, start text, H, end text, start subscript, 2, end subscript, start text, O, end text, left parenthesis, l, right parenthesis, \rightleftharpoons, start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript, left parenthesis, a, q, right parenthesis, plus, start text, A, end text, start superscript, minus, end superscript, left parenthesis, a, q, right parenthesis
The products of this reversible reaction are start text, A, end text, start superscript, minus, end superscript, the conjugate base of start text, H, A, end text, and start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript. We can write the following expression for the equilibrium constant K, start subscript, start text, a, end text, end subscript:
K, start subscript, start text, a, end text, end subscript, equals, start fraction, open bracket, start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript, close bracket, open bracket, start text, A, end text, start superscript, minus, end superscript, close bracket, divided by, open bracket, start text, H, A, end text, close bracket, end fraction

Finding K, start subscript, start text, b, end text, end subscript for start text, A, end text, start superscript, minus, end superscript reacting as a base

Since start text, A, end text, start superscript, minus, end superscript is a base, we can also write the reversible reaction for start text, A, end text, start superscript, minus, end superscript acting as a base by accepting a proton from water :
start text, A, end text, start superscript, minus, end superscript, left parenthesis, a, q, right parenthesis, plus, start text, H, end text, start subscript, 2, end subscript, start text, O, end text, left parenthesis, l, right parenthesis, \rightleftharpoons, start text, H, A, end text, left parenthesis, a, q, right parenthesis, plus, start text, O, H, end text, start superscript, minus, end superscript, left parenthesis, a, q, right parenthesis
The products of this reaction are start text, H, A, end text and start text, O, H, end text, start superscript, minus, end superscript. We can write out the equilibrium constant K, start subscript, start text, b, end text, end subscript for the reaction where start text, A, end text, start superscript, minus, end superscript acts as a base:
K, start subscript, start text, b, end text, end subscript, equals, start fraction, open bracket, start text, H, A, end text, close bracket, open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket, divided by, open bracket, start text, A, end text, start superscript, minus, end superscript, close bracket, end fraction
Even though this almost looks like the reverse of start text, H, A, end text acting as an acid, they are actually very different reactions. When start text, H, A, end text acts as an acid, one of the products is start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript. When the conjugate base start text, A, end text, start superscript, minus, end superscript acts as a base, one of the products is start text, O, H, end text, start superscript, minus, end superscript.

Relationship between K, start subscript, start text, a, end text, end subscript and K, start subscript, start text, b, end text, end subscript for conjugate acid-base pair

If we multiply K, start subscript, start text, a, end text, end subscript for start text, H, A, end text with the K, start subscript, start text, b, end text, end subscript of its conjugate base start text, A, end text, start superscript, minus, end superscript, that gives:
KaKb=([H3O+][A][HA])([HA][OH][A])=[H3O+][OH]=Kw\begin{aligned}K_\text{a}\cdot K_\text{b}&=\Bigg(\dfrac{[\text{H}_3\text{O}^+]\cancel{[\text{A}^-]}}{\cancel{[\text{HA}]}}\Bigg)\Bigg(\dfrac{\cancel{[\text{HA}]}[\text{OH}^-]}{\cancel{[\text{A}^-]}}\Bigg)\\ \\ &=[\text{H}_3\text{O}^+][\text{OH}^-]\\ \\ &=K_\text{w}\end{aligned}
where K, start subscript, start text, w, end text, end subscript is the water dissociation constant. This relationship is very useful for relating K, start subscript, start text, a, end text, end subscript and K, start subscript, start text, b, end text, end subscript for a conjugate acid-base pair!! We can also use the value of K, start subscript, start text, w, end text, end subscript at 25, degrees, start text, C, end text to derive other handy equations:
KaKb=Kw=1.0×1014 at 25C(Eq. 1)\begin{aligned}K_\text{a}\cdot K_\text{b}&=K_\text{w}\\ \\ &=1.0\times 10^{-14}~\text{at }25\,^\circ \text C\quad\quad\quad\quad(\text{Eq. 1})\end{aligned}
If we take the negative log, start base, 10, end base of both sides of the Eq. 1, we get:
start text, p, end text, K, start subscript, start text, a, end text, end subscript, plus, start text, p, end text, K, start subscript, start text, b, end text, end subscript, equals, 14, space, space, start text, a, t, space, end text, 25, degrees, start text, C, end text, left parenthesis, start text, E, q, point, space, 2, end text, right parenthesis
We can use these equations to determine K, start subscript, start text, b, end text, end subscript (or start text, p, end text, K, start subscript, start text, b, end text, end subscript) of a weak base given K, start subscript, start text, a, end text, end subscript of the conjugate acid. We can also calculate the K, start subscript, start text, a, end text, end subscript (or start text, p, end text, K, start subscript, start text, a, end text, end subscript) of a weak acid given K, start subscript, start text, b, end text, end subscript of the conjugate base.
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.
Concept check: Which of the following values can we calculate if we know the K, start subscript, start text, b, end text, end subscript of start text, N, H, end text, start subscript, 3, end subscript at 25, degrees, start text, C, end text?
Choose 1 answer:
Choose 1 answer:

Example: Finding K, start subscript, start text, b, end text, end subscript of a weak base

The start text, p, end text, K, start subscript, start text, a, end text, end subscript of hydrofluoric acid left parenthesis, start text, H, F, end text, right parenthesis is 3, point, 36 at 25, degrees, start text, C, end text.
What is K, start subscript, start text, b, end text, end subscript for fluoride, start text, F, end text, start superscript, minus, end superscript, left parenthesis, a, q, right parenthesis?
Let's work through this problem step-by-step.

Step 1: Make sure we have a conjugate acid-base pair

We can check the conjugate acid-base pair relationship by writing out the dissociation reaction for start text, H, F, end text:
start text, H, F, end text, left parenthesis, a, q, right parenthesis, plus, start text, H, end text, start subscript, 2, end subscript, start text, O, end text, left parenthesis, l, right parenthesis, \rightleftharpoons, start text, F, end text, start superscript, minus, end superscript, left parenthesis, a, q, right parenthesis, plus, start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript, left parenthesis, a, q, right parenthesis
We can see that start text, H, F, end text donates its proton to water to form start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript and start text, F, end text, start superscript, minus, end superscript. Therefore, start text, F, end text, start superscript, minus, end superscript is the conjugate base of start text, H, F, end text. That means we can use the start text, p, end text, K, start subscript, start text, a, end text, end subscript of start text, H, F, end text to find the start text, p, end text, K, start subscript, start text, b, end text, end subscript of start text, F, end text, start superscript, minus, end superscript. Hooray!

Step 2: Use Eq. 2 to find start text, p, end text, K, start subscript, start text, b, end text, end subscript from start text, p, end text, K, start subscript, start text, a, end text, end subscript

Rearranging Eq. 2 to solve for start text, p, end text, K, start subscript, start text, b, end text, end subscript, we have:
start text, p, end text, K, start subscript, start text, b, end text, end subscript, equals, 14, point, 00, minus, start text, p, end text, K, start subscript, start text, a, end text, end subscript
Plugging in our known start text, p, end text, K, start subscript, start text, a, end text, end subscript for start text, H, F, end text, we get:
start text, p, end text, K, start subscript, start text, b, end text, end subscript, equals, 14, point, 00, minus, left parenthesis, 3, point, 36, right parenthesis, equals, 10, point, 64
Therefore, the start text, p, end text, K, start subscript, start text, b, end text, end subscript for start text, F, end text, start superscript, minus, end superscript is 10, point, 64.

Step 3: Calculate K, start subscript, start text, b, end text, end subscript from start text, p, end text, K, start subscript, start text, b, end text, end subscript

Finally, we can convert start text, p, end text, K, start subscript, start text, b, end text, end subscript to K, start subscript, start text, b, end text, end subscript using the following equation:
start text, p, end text, K, start subscript, start text, b, end text, end subscript, equals, minus, log, left parenthesis, K, start subscript, start text, b, end text, end subscript, right parenthesis
Solving this equation for K, start subscript, start text, b, end text, end subscript, we get:
K, start subscript, start text, b, end text, end subscript, equals, 10, start superscript, minus, start text, p, end text, K, start subscript, start text, b, end text, end subscript, end superscript
Substituting our known value of start text, p, end text, K, start subscript, start text, b, end text, end subscript and solving, we get:
K, start subscript, start text, b, end text, end subscript, equals, 10, start superscript, minus, left parenthesis, 10, point, 64, right parenthesis, end superscript, equals, 2, point, 3, times, 10, start superscript, minus, 11, end superscript
Therefore, the K, start subscript, start text, b, end text, end subscript of start text, F, end text, start superscript, minus, end superscript is 2, point, 3, times, 10, start superscript, minus, 11, end superscript.

Summary

For conjugate-acid base pairs, the acid dissociation constant K, start subscript, start text, a, end text, end subscript and base ionization constant K, start subscript, start text, b, end text, end subscript are related by the following equations:
  • K, start subscript, start text, w, end text, end subscript, equals, K, start subscript, start text, a, end text, end subscript, dot, K, start subscript, start text, b, end text, end subscript
  • start text, p, end text, K, start subscript, start text, a, end text, end subscript, plus, start text, p, end text, K, start subscript, start text, b, end text, end subscript, equals, 14, space, space, start text, a, t, space, end text, 25, degrees, start text, C, end text

Want to join the conversation?

  • leaf green style avatar for user Ayudh Saxena
    How can we theoritically understand the relation , Kw=Ka.Kb. I mean why does it hold true?What exactly is going on in the solution?
    (13 votes)
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    • blobby green style avatar for user Joseph Noh
      I'm not sure if I need any credentials to answer this... but here I go.

      When writing an equilibrium expression, you MULTIPLY the products and DIVIDE The reactants. In that same sense, Ka * Kb can be conceived as multiplying the products of both Ka and Kb and dividing by the reactants of both Ka and Kb. Reverse the process you use for writing equilibrium expressions: multiplication = add to the products, division = add to the reactants.

      When you follow this process, adding HA + H2O <-> H3O+ + A- (acid) with A- + H2O <-> HA + OH- (base), HA and A- turn out to be "spectators" (not sure if that's the 100% correct term), so you can remove them, resulting in the net equation of H2O <-> H+ + OH-, the equation for the autoionization of water, which is represented by Kw.

      To summarize: Ka * Kb is equivalent to adding the acid and base reactions together, which results in a net equation of the autoionization of water.

      It's not a neutralization/acid-base reaction, but I think the Kw = Ka * Kb is a mathematical relation made to expedite calculations. Which works by the nature of how equilibrium expressions and chemical equations are related.
      (21 votes)
  • winston default style avatar for user Gray
    What's the difference between Kb and pKb?
    (3 votes)
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  • piceratops ultimate style avatar for user Alfonsus Ardani
    At what state/situation when [H30+] is equal to [A-]?
    (2 votes)
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    • piceratops ultimate style avatar for user Sahil
      That could never really happen... for [H30+] to be a conjugate base, it would have to start as [H40]2+, which I have never seen. While it probably does exist, I doubt that it would ever come up.
      (4 votes)
  • blobby green style avatar for user ajbaiden
    What does Ka1 and Ka2 mean?
    (1 vote)
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    • mr pants purple style avatar for user Ryan W
      Ka means the acid dissociation constant, it’s a measure of how much an acid splits up into H+ In solution.

      Acids that have multiple ionisable protons (eg. phosphoric acid H3PO4) have a Ka for each H+ that can be removed.

      Ka1: H3PO4 -> H+ + H2PO4^-
      Ka2: H2PO4^- -> H+ + HPO4^2-
      Ka3: HPO4^2- -> H+ + PO4^3-

      See how it works? Each successive Ka will be smaller in value as it gets harder to remove more H+
      (5 votes)
  • duskpin ultimate style avatar for user Lily Martin
    so basically we are calculating the percentage of acid and base caused by autoionization?
    (3 votes)
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  • blobby green style avatar for user earl kraft
    are there not "medium" acids, and is there not variation in what equilibrium constant value would separate strong from weak acids?
    (1 vote)
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  • starky ultimate style avatar for user Ayush Tarafder
    Why can you not find the Ka of NH3? Couldn't you just divide the Kb from 10^-14?
    (1 vote)
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  • blobby green style avatar for user quinterius.thurman11
    In step 3, where did the 10 come from? Can some one explain the process of rearranging this equation in step 3?
    (1 vote)
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  • blobby green style avatar for user Shelby Lacourse
    When finding the Kb for F- from Ka does that mean you are looking at the strength of F- as a base from the reverse reaction since it is the conjugate base for the forward reaction?
    (1 vote)
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  • blobby green style avatar for user 1chenoli
    H2S has a Ka of 8.9*10^-8 at 25 degree at 0.1mol/L. Why is it still a acid even though the Kb of its conjugate base is higher than the Ka?
    (1 vote)
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    • piceratops tree style avatar for user Daniel Satterfield
      If I'm correct, higher Kb indicates that the base is more willing to give away the H+ attached to it, thus the proton goes back to water to form hydronium, forming more product. The lower Ka for the acid indicates that it's a weak acid that holds tightly onto the donatable proton. The weaker the acid, the stronger the base. The stronger the base, the higher the Kb. The weaker the acid, the lower the Ka. I don't have any examples off the top of my head, but I'd say you'll see this relationship for any conjugate acid-base pair.
      (1 vote)