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Course: Chemistry library > Unit 13
Lesson 2: Acid-base equilibriaRelationship 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:
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:
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.
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:
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:
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 :
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:
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:
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:
If we take the negative log, start base, 10, end base of both sides of the Eq. 1, we get:
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?
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:
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:
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:
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:
Solving this equation for K, start subscript, start text, b, end text, end subscript, we get:
Substituting our known value of start text, p, end text, K, start subscript, start text, b, end text, end subscript and solving, we get:
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?
- 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)
- 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.(23 votes)
- What's the difference between Kb and pKb?(3 votes)
- pKb is the negative logarithm of Kb.
If you're not introduced to logs, go to this link: https://www.khanacademy.org/math/algebra2/exponential-and-logarithmic-functions/introduction-to-logarithms/v/logarithms(6 votes)
- At what state/situation when [H30+] is equal to [A-]?(2 votes)
- 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)
- What does Ka1 and Ka2 mean?(1 vote)
- 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)
- so basically we are calculating the percentage of acid and base caused by autoionization?(3 votes)
- are there not "medium" acids, and is there not variation in what equilibrium constant value would separate strong from weak acids?(1 vote)
- No there’s no such classification as medium acids.
Strong acids have a Ka > 1, weak acids have a Ka < 1.(3 votes)
- Why can you not find the Ka of NH3? Couldn't you just divide the Kb from 10^-14?(1 vote)
- NH3 is a base. Ka is the ACID dissociation constant, which would be calculated from the concentration of NH4. Kb is the BASE ionization constant, which would be calculated from the concentration of NH3.(2 votes)
- In step 3, where did the 10 come from? Can some one explain the process of rearranging this equation in step 3?(1 vote)
- you use 10 to the negative pKb because that is the inverse of the negative log/how you undo the negative log, which leads you to get Kb instead of pKb(2 votes)
- 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)
- 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)
- 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)