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?

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.

What's the difference between Kb and pKb?

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

At what state/situation when [H30+] is equal to [A-]?

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.

What does Ka1 and Ka2 mean?

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+

are there not "medium" acids, and is there not variation in what equilibrium constant value would separate strong from weak acids?

No there’s no such classification as medium acids. Strong acids have a Ka > 1, weak acids have a Ka < 1.

Why can you not find the Ka of NH3? Couldn't you just divide the Kb from 10^-14?

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.

In step 3, where did the 10 come from? Can some one explain the process of rearranging this equation in step 3?

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

Main content

Chemistry library

Course: Chemistry library > Unit 13

Lesson 2: Acid-base equilibria

Relationship between Ka and Kb

Q: Why can you not find the Ka of NH3? Couldn't you just divide the Kb from 10^-14?

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.

Q: In step 3, where did the 10 come from? Can some one explain the process of rearranging this equation in step 3?

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

Google Classroom

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:

[I don't understand the difference between strong vs. weak acids!!]

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.

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:

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

[What about water?]

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:

\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}

[What is Kw? ]

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:

\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?

(Choice A)
K, start subscript, start text, a, end text, end subscript of start text, N, H, end text, start subscript, 3, end subscript
(Choice B)
K, start subscript, start text, b, end text, end subscript of start text, N, H, end text, start subscript, 2, end subscript, start superscript, minus, end superscript
(Choice C)
K, start subscript, start text, a, end text, end subscript of start text, N, H, end text, start subscript, 4, end subscript, start superscript, plus, end superscript
(Choice D)
None of the above

[Show the 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

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Ayudh Saxena
Posted 7 years ago. Direct link to Ayudh Saxena's post “How can we theoritically ...”
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?
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(13 votes)
Answer
- Joseph Noh
  Posted 7 years ago. Direct link to Joseph Noh's post “I'm not sure if I need an...”
  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.
  Comment on Joseph Noh's post “I'm not sure if I need an...”
  (23 votes)
Gray
Posted 6 years ago. Direct link to Gray's post “What's the difference bet...”
What's the difference between Kb and pKb?
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(3 votes)
Answer
- santhosh prabahar
  Posted 5 years ago. Direct link to santhosh prabahar's post “pKb is the *negative loga...”
  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
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  (6 votes)
Alfonsus Ardani
Posted 5 years ago. Direct link to Alfonsus Ardani's post “At what state/situation w...”
At what state/situation when [H30+] is equal to [A-]?
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(2 votes)
Answer
- Sahil
  Posted 5 years ago. Direct link to Sahil's post “That could never really h...”
  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.
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  (4 votes)
ajbaiden
Posted 4 years ago. Direct link to ajbaiden's post “What does Ka1 and Ka2 mea...”
What does Ka1 and Ka2 mean?
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(1 vote)
Answer
- Ryan W
  Posted 4 years ago. Direct link to Ryan W's post “Ka means the acid dissoci...”
  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+
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  (5 votes)
Lily Martin
Posted 6 years ago. Direct link to Lily Martin's post “so basically we are calcu...”
so basically we are calculating the percentage of acid and base caused by autoionization?
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(3 votes)
Answer
earl kraft
Posted 5 years ago. Direct link to earl kraft's post “are there not "medium" ac...”
are there not "medium" acids, and is there not variation in what equilibrium constant value would separate strong from weak acids?
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(1 vote)
Answer
- Ryan W
  Posted 5 years ago. Direct link to Ryan W's post “No there’s no such classi...”
  No there’s no such classification as medium acids.
  Strong acids have a Ka > 1, weak acids have a Ka < 1.
  Comment on Ryan W's post “No there’s no such classi...”
  (3 votes)
Ayush Tarafder
Posted 5 years ago. Direct link to Ayush Tarafder's post “Why can you not find the ...”
Why can you not find the Ka of NH3? Couldn't you just divide the Kb from 10^-14?
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(1 vote)
Answer
- Vikranth
  Posted 4 years ago. Direct link to Vikranth's post “NH3 is a base. Ka is the ...”
  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.
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  (2 votes)
quinterius.thurman11
Posted 5 years ago. Direct link to quinterius.thurman11's post “In step 3, where did the ...”
In step 3, where did the 10 come from? Can some one explain the process of rearranging this equation in step 3?
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(1 vote)
Answer
- maggie
  Posted 5 years ago. Direct link to maggie's post “you use 10 to the negativ...”
  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
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  (2 votes)
Shelby Lacourse
Posted 6 years ago. Direct link to Shelby Lacourse's post “When finding the Kb for F...”
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?
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(1 vote)
Answer
1chenoli
Posted 4 years ago. Direct link to 1chenoli's post “H2S has a Ka of 8.9*10^-8...”
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?
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(1 vote)
Answer
- Daniel Satterfield
  Posted 4 years ago. Direct link to Daniel Satterfield's post “If I'm correct, higher Kb...”
  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.
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  (1 vote)

Chemistry library

Course: Chemistry library > Unit 13

Key points

Introduction: Weak acid and bases ionize reversibly

Finding KaK_\text{a}Ka​K, start subscript, start text, a, end text, end subscript for HA\text{HA}HAstart text, H, A, end text reacting as an acid

Finding KbK_\text{b}Kb​K, start subscript, start text, b, end text, end subscript for A−\text{A}^-A−start text, A, end text, start superscript, minus, end superscript reacting as a base

Relationship between KaK_\text{a}Ka​K, start subscript, start text, a, end text, end subscript and KbK_\text{b}Kb​K, start subscript, start text, b, end text, end subscript for conjugate acid-base pair

Example: Finding KbK_\text{b}Kb​K, start subscript, start text, b, end text, end subscript of a weak base