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

Course: Chemistry library > Unit 11

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_{a}

and base ionization constant

K_{b}

are related by the following equations:

$K_{a} \cdot K_{b} = K_{w}$ ‍
where $K_{w}$ ‍ is the autoionization constant

$p K_{a} + p K_{b} = 14 at 25^{\circ} C$ ‍

Introduction: Weak acid and bases ionize reversibly

Weak acids, generically abbreviated as

HA

, donate

H^{+}

(or proton) to water to form the conjugate base

A^{-}

and

H_{3} O^{+}

HA (a q) + H_{2} O (l) ⇌ H_{3} O^{+} (a q) + A^{-} (a q)

acid base acid base

Similarly, a base (abbreviated as

B

) will accept a proton in water to form the conjugate acid,

{HB}^{+}

, and

{OH}^{-}

B (a q) + H_{2} O (l) ⇌ {HB}^{+} (a q) + {OH}^{-} (a q)

base acid acid base

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_{a}

and

K_{b}

for a conjugate acid-base pair.

Note: For this article, all solutions will be assumed to be aqueous solutions.

Finding $K_{a}$ ‍ for $HA$ ‍ reacting as an acid

Let's look more closely at the dissociation reaction for a monoprotic weak acid

HA

HA (a q) + H_{2} O (l) ⇌ H_{3} O^{+} (a q) + A^{-} (a q)

The products of this reversible reaction are

A^{-}

, the conjugate base of

HA

, and

H_{3} O^{+}

. We can write the following expression for the equilibrium constant

K_{a}

K_{a} = \frac{[H_{3} O^{+}] [A^{-}]}{[HA]}

Finding $K_{b}$ ‍ for $A^{-}$ ‍ reacting as a base

Since

A^{-}

is a base, we can also write the reversible reaction for

A^{-}

acting as a base by accepting a proton from water :

A^{-} (a q) + H_{2} O (l) ⇌ HA (a q) + {OH}^{-} (a q)

The products of this reaction are

HA

and

{OH}^{-}

. We can write out the equilibrium constant

K_{b}

for the reaction where

A^{-}

acts as a base:

K_{b} = \frac{[HA] [{OH}^{-}]}{[A^{-}]}

Even though this almost looks like the reverse of

HA

acting as an acid, they are actually very different reactions. When

HA

acts as an acid, one of the products is

H_{3} O^{+}

. When the conjugate base

A^{-}

acts as a base, one of the products is

{OH}^{-}

Relationship between $K_{a}$ ‍ and $K_{b}$ ‍ for conjugate acid-base pair

If we multiply

K_{a}

for

HA

with the

K_{b}

of its conjugate base

A^{-}

, that gives:

\begin{aligned} K_{a} \cdot K_{b} & = (\frac{[H_{3} O^{+}] [A^{-}]}{[HA]}) (\frac{[HA] [{OH}^{-}]}{[A^{-}]}) \\ = [H_{3} O^{+}] [{OH}^{-}] \\ = K_{w} \end{aligned}

where

K_{w}

is the water dissociation constant. This relationship is very useful for relating

K_{a}

and

K_{b}

for a conjugate acid-base pair!! We can also use the value of

K_{w}

25^{\circ} C

to derive other handy equations:

\begin{aligned} K_{a} \cdot K_{b} & = K_{w} \\ = 1.0 \times 10^{- 14} at 25^{\circ} C (Eq. 1) \end{aligned}

If we take the negative

\log_{10}

of both sides of the Eq. 1, we get:

p K_{a} + p K_{b} = 14 at 25^{\circ} C (Eq. 2)

We can use these equations to determine

K_{b}

(or

p K_{b}

) of a weak base given

K_{a}

of the conjugate acid. We can also calculate the

K_{a}

(or

p K_{a}

) of a weak acid given

K_{b}

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 $K_{b}$ ‍ of ${NH}_{3}$ ‍ at $25^{\circ} C$ ‍?

Example: Finding $K_{b}$ ‍ of a weak base

The

p K_{a}

of hydrofluoric acid

(HF)

3.36

25^{\circ} C

What is $K_{b}$ ‍ for fluoride, $F^{-} (a q)$ ‍?

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

HF

HF (a q) + H_{2} O (l) ⇌ F^{-} (a q) + H_{3} O^{+} (a q)

We can see that

HF

donates its proton to water to form

H_{3} O^{+}

and

F^{-}

. Therefore,

F^{-}

is the conjugate base of

HF

. That means we can use the

p K_{a}

HF

to find the

p K_{b}

F^{-}

. Hooray!

Step 2: Use Eq. 2 to find $p K_{b}$ ‍ from $p K_{a}$ ‍

Rearranging Eq. 2 to solve for

p K_{b}

, we have:

p K_{b} = 14.00 - p K_{a}

Plugging in our known

p K_{a}

for

HF

, we get:

p K_{b} = 14.00 - (3.36) = 10.64

Therefore, the

p K_{b}

for

F^{-}

10.64

Step 3: Calculate $K_{b}$ ‍ from $p K_{b}$ ‍

Finally, we can convert

p K_{b}

K_{b}

using the following equation:

p K_{b} = - \log (K_{b})

Solving this equation for

K_{b}

, we get:

K_{b} = 10^{- p K_{b}}

Substituting our known value of

p K_{b}

and solving, we get:

K_{b} = 10^{- (10.64)} = 2.3 \times 10^{- 11}

Therefore, the $K_{b}$ ‍ of $F^{-}$ ‍ is $2.3 \times 10^{- 11}$ ‍.

Summary

For conjugate-acid base pairs, the acid dissociation constant

K_{a}

and base ionization constant

K_{b}

are related by the following equations:

$K_{w} = K_{a} \cdot K_{b}$ ‍

$p K_{a} + p K_{b} = 14 at 25^{\circ} C$ ‍

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Ayudh Saxena
Posted 8 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?
Button navigates to signup pageButton navigates to signup page
(13 votes)
Answer
- Joseph Noh
  Posted 8 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...”
  (24 votes)
Gray
Posted 7 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 6 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
  Button navigates to signup page
  (6 votes)
Alfonsus Ardani
Posted 6 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 6 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.
  Button navigates to signup page
  (4 votes)
ajbaiden
Posted 6 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 6 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+
  Button navigates to signup page
  (5 votes)
Lily Martin
Posted 7 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
Ramsa Nadeem
Posted 6 years ago. Direct link to Ramsa Nadeem's post “what is the relationship ...”
what is the relationship between the Ka/Kb for conjugate acids/bases?
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(2 votes)
Answer
earl kraft
Posted 6 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 6 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 6 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 5 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.
  Button navigates to signup page
  (2 votes)
quinterius.thurman11
Posted 6 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 6 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)
Perseus
Posted 10 months ago. Direct link to Perseus's post “When Ka = 1.0*10^-14, doe...”
When Ka = 1.0*10^-14, does Kb = 1? This would be under the equilibrium of water.

#MCATprep
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(1 vote)
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- aphranm
  Posted 8 months ago. Direct link to aphranm's post “yeah it does”
  yeah it does
  Comment on aphranm's post “yeah it does”
  (2 votes)

Course: Chemistry library > Unit 11

Relationship between Ka and Kb

Key points

Introduction: Weak acid and bases ionize reversibly

Finding $K_{a}$ ‍ for $HA$ ‍ reacting as an acid

Finding $K_{b}$ ‍ for $A^{-}$ ‍ reacting as a base

Relationship between $K_{a}$ ‍ and $K_{b}$ ‍ for conjugate acid-base pair

Example: Finding $K_{b}$ ‍ of a weak base

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

Step 2: Use Eq. 2 to find $p K_{b}$ ‍ from $p K_{a}$ ‍

Step 3: Calculate $K_{b}$ ‍ from $p K_{b}$ ‍

Summary

Attributions

Additional References

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Chemistry library

Course: Chemistry library > Unit 11

Key points

Introduction: Weak acid and bases ionize reversibly

Finding Ka‍ for HA‍ reacting as an acid

Finding Kb‍ for A−‍ reacting as a base

Relationship between Ka‍ and Kb‍ for conjugate acid-base pair

Example: Finding Kb‍ of a weak base

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

Step 2: Use Eq. 2 to find pKb‍ from pKa‍

Step 3: Calculate Kb‍ from pKb‍

Summary

Want to join the conversation?

Finding $K_{a}$ ‍ for $HA$ ‍ reacting as an acid

Finding $K_{b}$ ‍ for $A^{-}$ ‍ reacting as a base

Relationship between $K_{a}$ ‍ and $K_{b}$ ‍ for conjugate acid-base pair

Example: Finding $K_{b}$ ‍ of a weak base

Step 2: Use Eq. 2 to find $p K_{b}$ ‍ from $p K_{a}$ ‍

Step 3: Calculate $K_{b}$ ‍ from $p K_{b}$ ‍