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### Course: AP®︎/College Chemistry>Unit 12

Lesson 2: Acid-base equilibria

# 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 ${K}_{\text{a}}$ and base ionization constant ${K}_{\text{b}}$ are related by the following equations:
• ${K}_{\text{a}}\cdot {K}_{\text{b}}={K}_{\text{w}}$
where ${K}_{\text{w}}$ is the autoionization constant

## Introduction: Weak acid and bases ionize reversibly

Weak acids, generically abbreviated as $\text{HA}$, donate ${\text{H}}^{+}$ (or proton) to water to form the conjugate base ${\text{A}}^{-}$ and ${\text{H}}_{3}{\text{O}}^{+}$:
$\text{HA}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)⇌{\text{H}}_{3}{\text{O}}^{+}\left(aq\right)+{\text{A}}^{-}\left(aq\right)$
Similarly, a base (abbreviated as $\text{B}$) will accept a proton in water to form the conjugate acid, ${\text{HB}}^{+}$, and ${\text{OH}}^{-}$:
$\text{B}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)⇌{\text{HB}}^{+}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)$
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 ${K}_{\text{a}}$ and ${K}_{\text{b}}$ for a conjugate acid-base pair.
Note: For this article, all solutions will be assumed to be aqueous solutions.

## Finding ${K}_{\text{a}}$‍  for $\text{HA}$‍  reacting as an acid

Let's look more closely at the dissociation reaction for a monoprotic weak acid $\text{HA}$:
$\text{HA}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)⇌{\text{H}}_{3}{\text{O}}^{+}\left(aq\right)+{\text{A}}^{-}\left(aq\right)$
The products of this reversible reaction are ${\text{A}}^{-}$, the conjugate base of $\text{HA}$, and ${\text{H}}_{3}{\text{O}}^{+}$. We can write the following expression for the equilibrium constant ${K}_{\text{a}}$:
${K}_{\text{a}}=\frac{\left[{\text{H}}_{3}{\text{O}}^{+}\right]\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}$

## Finding ${K}_{\text{b}}$‍  for ${\text{A}}^{-}$‍  reacting as a base

Since ${\text{A}}^{-}$ is a base, we can also write the reversible reaction for ${\text{A}}^{-}$ acting as a base by accepting a proton from water :
${\text{A}}^{-}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)⇌\text{HA}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)$
The products of this reaction are $\text{HA}$ and ${\text{OH}}^{-}$. We can write out the equilibrium constant ${K}_{\text{b}}$ for the reaction where ${\text{A}}^{-}$ acts as a base:
${K}_{\text{b}}=\frac{\left[\text{HA}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{A}}^{-}\right]}$
Even though this almost looks like the reverse of $\text{HA}$ acting as an acid, they are actually very different reactions. When $\text{HA}$ acts as an acid, one of the products is ${\text{H}}_{3}{\text{O}}^{+}$. When the conjugate base ${\text{A}}^{-}$ acts as a base, one of the products is ${\text{OH}}^{-}$.

## Relationship between ${K}_{\text{a}}$‍  and ${K}_{\text{b}}$‍  for conjugate acid-base pair

If we multiply ${K}_{\text{a}}$ for $\text{HA}$ with the ${K}_{\text{b}}$ of its conjugate base ${\text{A}}^{-}$, that gives:
$\begin{array}{rl}{K}_{\text{a}}\cdot {K}_{\text{b}}& =\left(\frac{\left[{\text{H}}_{3}{\text{O}}^{+}\right]\overline{)\left[{\text{A}}^{-}\right]}}{\overline{)\left[\text{HA}\right]}}\right)\left(\frac{\overline{)\left[\text{HA}\right]}\left[{\text{OH}}^{-}\right]}{\overline{)\left[{\text{A}}^{-}\right]}}\right)\\ \\ & =\left[{\text{H}}_{3}{\text{O}}^{+}\right]\left[{\text{OH}}^{-}\right]\\ \\ & ={K}_{\text{w}}\end{array}$
where ${K}_{\text{w}}$ is the water dissociation constant. This relationship is very useful for relating ${K}_{\text{a}}$ and ${K}_{\text{b}}$ for a conjugate acid-base pair!! We can also use the value of ${K}_{\text{w}}$ at $25{\phantom{\rule{0.167em}{0ex}}}^{\circ }\text{C}$ to derive other handy equations:
If we take the negative ${\mathrm{log}}_{10}$ of both sides of the Eq. 1, we get:
We can use these equations to determine ${K}_{\text{b}}$ (or $\text{p}{K}_{\text{b}}$) of a weak base given ${K}_{\text{a}}$ of the conjugate acid. We can also calculate the ${K}_{\text{a}}$ (or $\text{p}{K}_{\text{a}}$) of a weak acid given ${K}_{\text{b}}$ 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 ${K}_{\text{b}}$ of ${\text{NH}}_{3}$ at $25{\phantom{\rule{0.167em}{0ex}}}^{\circ }\text{C}$?

## Example: Finding ${K}_{\text{b}}$‍  of a weak base

The $\text{p}{K}_{\text{a}}$ of hydrofluoric acid $\left(\text{HF}\right)$ is $3.36$ at $25{\phantom{\rule{0.167em}{0ex}}}^{\circ }\text{C}$.
What is ${K}_{\text{b}}$ for fluoride, ${\text{F}}^{-}\left(aq\right)$?
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 $\text{HF}$:
$\text{HF}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)⇌{\text{F}}^{-}\left(aq\right)+{\text{H}}_{3}{\text{O}}^{+}\left(aq\right)$
We can see that $\text{HF}$ donates its proton to water to form ${\text{H}}_{3}{\text{O}}^{+}$ and ${\text{F}}^{-}$. Therefore, ${\text{F}}^{-}$ is the conjugate base of $\text{HF}$. That means we can use the $\text{p}{K}_{\text{a}}$ of $\text{HF}$ to find the $\text{p}{K}_{\text{b}}$ of ${\text{F}}^{-}$. Hooray!

### Step 2: Use Eq. 2 to find $\text{p}{K}_{\text{b}}$‍  from $\text{p}{K}_{\text{a}}$‍

Rearranging Eq. 2 to solve for $\text{p}{K}_{\text{b}}$, we have:
$\text{p}{K}_{\text{b}}=14.00-\text{p}{K}_{\text{a}}$
Plugging in our known $\text{p}{K}_{\text{a}}$ for $\text{HF}$, we get:
$\text{p}{K}_{\text{b}}=14.00-\left(3.36\right)=10.64$
Therefore, the $\text{p}{K}_{\text{b}}$ for ${\text{F}}^{-}$ is $10.64$.

### Step 3: Calculate ${K}_{\text{b}}$‍  from $\text{p}{K}_{\text{b}}$‍

Finally, we can convert $\text{p}{K}_{\text{b}}$ to ${K}_{\text{b}}$ using the following equation:
$\text{p}{K}_{\text{b}}=-\mathrm{log}\left({K}_{\text{b}}\right)$
Solving this equation for ${K}_{\text{b}}$, we get:
${K}_{\text{b}}={10}^{-\text{p}{K}_{\text{b}}}$
Substituting our known value of $\text{p}{K}_{\text{b}}$ and solving, we get:
${K}_{\text{b}}={10}^{-\left(10.64\right)}=2.3×{10}^{-11}$
Therefore, the ${K}_{\text{b}}$ of ${\text{F}}^{-}$ is $2.3×{10}^{-11}$.

## Summary

For conjugate-acid base pairs, the acid dissociation constant ${K}_{\text{a}}$ and base ionization constant ${K}_{\text{b}}$ are related by the following equations:
• ${K}_{\text{w}}={K}_{\text{a}}\cdot {K}_{\text{b}}$

## 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?
• 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?
• 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?
(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+
• so basically we are calculating the percentage of acid and base caused by autoionization?
• what is the relationship between the Ka/Kb for conjugate acids/bases?
• 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.
• 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.