Relationship Between Ka and Kb: Values

A conjugate pair refers to a set of two species that are connected to each other by the loss or gain of a proton (H⁺). The idea of conjugate acid-base pairs was developed as part of the wider development of the acid-base theory of Johannes Nicolaus Bronsted and Thomas Martin Lowry they independently formed the Brønsted-Lowry acid-base theory in the early 20th century. Their work gives the idea that acids are proton donors and bases are proton acceptors, leading to the concept of conjugate acid-base pairs. The conjugate base of a strong acid is weak. The conjugate acid of a strong base is a weak acid.

Conjugate acid-base pairs are closely related to the acid dissociation constant Ka, base dissociation constant Kb, and the ion product of water Kw. The acid dissociation constant Ka expresses the strength of an acid in solution. For a weak acid (HA) dissociating into its conjugate base and a proton. The base dissociation constant Kb measures the strength of a base. For a weak base (B) that accepts a proton from water to form its conjugate acid and hydroxide ion. For an acid (HA): (Ka) is a measure of how it is dissociates in water to form its conjugate base. For a base (B) (Kb) tells its ability to accept a proton to form its conjugate acid. The product of (Ka) of an acid and (Kb) of its conjugate base equals (Kw), showing the inverse relationship between the strength of an acid and the strength of its conjugate base. The dissociation constant for acids and bases is commonly measured in moles per liter. The acid dissociation constant Ka is the -log of this constant is pKa. and the base dissociation constant is Kb, while the -log of the constant is pKb.

Relation Between Ka, Kb, And Kw For Conjugate Pair

The relation between $\mathrm{K}_{\mathrm{a}}, \mathrm{K}_{\mathrm{b}}$ and $\mathrm{K}_{\mathrm{w}}$ can be understand by the following reaction.
$
\begin{array}{ll}
\mathrm{NH}_3+\mathrm{H}_2 \mathrm{O} & \rightleftharpoons \mathrm{NH}_4^{+}+\mathrm{OH}^{-} \\
\text {Base } & \text { Conjugate acid }
\end{array}
$

The equilibrium constant $\mathrm{K}_{\mathrm{b}}$ for base $\mathrm{NH}_3$ is given as:
$
\mathrm{K}_{\mathrm{b}}=\frac{\left[\mathrm{NH}_4^{+}\right]\left[\mathrm{OH}^{-}\right]}{\mathrm{NH}_3}
$

Again, the reaction is:
$
\mathrm{NH}_4^{+}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{NH}_3+\mathrm{H}_3 \mathrm{O}^{+}
$

The equilibrium constant $\mathrm{K}_{\mathrm{a}}$ for conjugate acid $\mathrm{NH}_4{ }^{+}$is given as:
$
\mathrm{K}_{\mathrm{a}}=\frac{\left[\mathrm{NH}_3\right]\left[\mathrm{H}_3 \mathrm{O}^{+}\right]}{\mathrm{NH}_4^{+}}
$

Thus,
$
\mathrm{K}_{\mathrm{a}} \times \mathrm{K}_{\mathrm{b}}=\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]=\mathrm{K}_{\mathrm{w}}
$

Thus, $\mathrm{K}_{\mathrm{a}} \propto 1 / \mathrm{K}_{\mathrm{b}}$ at a given temperature

Therefore, if K_a of acid increases, then K_b of the conjugate base decreases. In other words, the conjugate base of a strong acid is a weak base, and vice-versa.

Buffer Solution

A buffer solution is a special type of solution that resists changes in pH when small amounts of an acid or base are added. It typically consists of a weak acid and its conjugate base, or a weak base and its conjugate acid.

Basic buffer solution

A basic buffer solution contains equimolar quantities of a weak base and its salt with strong acid. For example ammonium hydroxide i.e. NH₄OH and ammonium chloride i.e. NH₄Cl.On Adding Acid: H⁺ release and combines with OH^- of the base. On Adding Base: OH^- releases and combines with NH₄⁺ of salt. On adding acid to the basic buffer, its H⁺ ions react with OH^- ions of the base and form H₂O. Thus, in this case, the solution feels that its [OH^-] has decreased, thus to neutralize this effect, NH₄OH dissociates in small amounts and gives [OH^-] to restore the concentration of [OH^-]On adding base to the basic buffer, its [OH^-] ions react with NH₄⁺ ions and forms NH₄OH. In this case, the solution feels that its NH₄OH concentration is increased. Thus, in this case, the reaction will not proceed forward because of the common ion effect.

Recommended video (Ka and Kb Relationship)

Some Solved Example

Example.1

1. Which one of the following is true regarding the relation between $\mathrm{k}_{\mathrm{a}}$ and $\mathrm{k}_{\mathrm{b}}$

1) $\mathrm{Pk}_{\mathrm{a}}+\mathrm{pk}_{\mathrm{b}}=\mathrm{pk}_{\mathrm{w}}$
2) $\mathrm{k}_{\mathrm{a}} \times \mathrm{k}_{\mathrm{b}}=\mathrm{k}_{\mathrm{w}}$
3) $\mathrm{Pk}_{\mathrm{a}}+\mathrm{pk}_{\mathrm{b}}=14$

Solution

We know,
$
\begin{aligned}
& \mathrm{k}_{\mathrm{a}} \times \mathrm{k}_{\mathrm{b}}=\mathrm{k}_{\mathrm{w}} \\
& \therefore \mathrm{pk}_{\mathrm{a}}+\mathrm{pk}_{\mathrm{b}}=\mathrm{pk}_{\mathrm{w}}=14
\end{aligned}
$

So,

All of the given equations are true regarding the relation between $\mathrm{k}_{\mathrm{a}}$ and $\mathrm{k}_{\mathrm{b}}$

Hence, the answer is the option (4).

Example.2

2. Which one of the following is equal to the $\frac{p+1}{p K_1}$ of a weak acid?

1)Its relative molecular mass

2)The $p k_b$ its conjugate base

3) (correct)The pH of a solution containing equal amount of acids and its conjugate base

4)The equilibrium concentration of its conjugate

Solution

Pk_a represents the pH of the solution when the acid is half dissociated, that is when the solution contains an equal amount of the acid and its conjugate base.

Hence, the answer is the option (3).

Example.3

3. Which statement is correct regarding the relationship between K_a and K_b of the conjugate acid-base pair?

${ }_{1)} K_w=\frac{K_b}{K_a}$
2) $K_w=\frac{K_a}{K_b}$
3) $K_w=K_a+K_b$
4) (correct) $K_w=K_a \times K_b$

Solution

Let's consider a conjugate acid-base pair HA and A^-.

The acid reacts with water as :
$
\begin{aligned}
& \mathrm{HA}(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}(a q)+A^{-}(a q) \\
& K_a=\frac{\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[A^{-}\right]}{[\mathrm{H} A]}
\end{aligned}
$

Conjugate base reacts with water as :
$
\begin{aligned}
& A^{-}(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{HA}+\mathrm{OH}^{-} . \\
& K_b=\frac{[\mathrm{H} A]\left[\mathrm{OH}^{-}\right]}{\left[A^{-}\right]}
\end{aligned}
$

Adding equations (A) and (B)
$
\begin{gathered}
2 \mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q) \\
K_w=\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]
\end{gathered}
$

By equations (1), (2), and (3)
$
K_w=K_a \times K_b
$

Hence, the answer is the option(4).

Example.4

4. The ionization constant of $N H_4^{+}$water is $6^{\star} 10^{-10}$ at 25 degrees Celcius. The rate constant for reaction of $N H_4^{+}$and $O H^{-}$to form $N H_3$ and $H_2 O$ at 25 degrees Celcius is $3.5^{\star} 10^{10}$ I/mol.sec. Calculate the rate constant for proton transfer from water to $\mathrm{NH}_3$.

1) $6.66 \times 10^3$
2) (correct) $5.83 \times 10^5$
3) $6.66 \times 10^6$
4) $7.64 \times 10^5$

Solution

The reactions according to the question will be:
$
\begin{aligned}
& \mathrm{NH}_3+\mathrm{H}_2 \mathrm{O} \stackrel{K_f}{K_b} \mathrm{NH}_4^{+}+\mathrm{OH}^{-} \quad K_{\mathrm{b}}=3.5 \times 10^{10} \\
& \mathrm{NH}_4^{+}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{NH}_4 \mathrm{OH}+\mathrm{H}^{+} ; \quad K_{\text {acid }}=6 \times 10^{-10} \\
& K_{\text {base }}=\frac{K_f}{K_b} \\
&
\end{aligned}
$

Also, $K_{\text {base }}=\frac{K_w}{K_{\text {acid }}}$
$
\begin{aligned}
& \frac{K_f}{3.5 \times 10^{10}}=\frac{10^{-14}}{6 \times 10^{-10}} \\
& K_f=5.83 \times 10^5
\end{aligned}
$

Hence, the answer is the option (2).

Example.5

5. Calculate the ionization constant of the conjugate acid of NH₃.

Given : $\mathrm{K}_{\mathrm{b}}=1.77 \times 10^{-5}$

1) $\left(\right.$ correct) $5.64 \times 10^{-10}$
2) $5.64 \times 10^{-19}$
3) $1.77 \times 10^{-10}$
4). $1.77 \times 10^{-5}$

Solution

We know that, for a conjugate acid-base pair
$
\mathrm{k}_{\mathrm{a}} \times \mathrm{k}_{\mathrm{b}}=\mathrm{k}_{\mathrm{w}}
$

Thus,
$
\mathrm{k}_{\mathrm{a}}=\frac{\mathrm{k}_{\mathrm{w}}}{\mathrm{k}_{\mathrm{b}}}=\frac{10^{-14}}{1.77 \times 10^{-5}}=5.64 \times 10^{-10}
$

Hence, the answer is the option (1).

6. HF and F^- represent a conjugate acid-base pair.

Which of the following is the correct relation between $\mathrm{k}_{\mathrm{a}}$ of $\mathrm{HF}_{\text {and }} \mathrm{k}_{\mathrm{b}}$ of $\mathrm{F}^{-}$at temperature T ?

(Given, at Temperature T, the self-ionization constant of Water is 10^-12)

1) $\left(\right.$ correct) $\mathrm{pk}_{\mathrm{a}}+\mathrm{pk}_{\mathrm{b}}=12$
2) $\mathrm{pk}_{\mathrm{a}}+\mathrm{pk}_{\mathrm{b}}=14$
3) $\mathrm{pk}_{\mathrm{a}}+\mathrm{pk}_{\mathrm{b}}=6$
4) $\mathrm{Pk}_{\mathrm{a}}+\mathrm{pk}_{\mathrm{b}}=7$Solution

As we learned,
For a conjugate acid-base pair,
$
\begin{aligned}
& \mathrm{k}_{\mathrm{a}} \times \mathrm{k}_{\mathrm{b}}=\mathrm{k}_{\mathrm{w}} \\
& \Rightarrow \mathrm{pk}_{\mathrm{a}}+\mathrm{pk}_{\mathrm{b}}=\mathrm{pk}_{\mathrm{w}}=12 \\
& {\left[\because \text { At Temperature T K, } \mathrm{k}_{\mathrm{w}}=10^{-12}\right]}
\end{aligned}
$

Hence, the answer is the option (1).

Summary

Acid-base conjugate pairs are important in various chemical and biological processes. Some of the uses of conjugate acid-base pairs are Buffer Solutions in which acid-base conjugate pairs are fundamental in buffer solutions, which stops changes in pH when small amounts of acid or base are added. Ph is an important component in maintaining G the stability between the biological and the industrial process. It has several applications and one of its applications includes the regulation of pH in the components either It is biological or chemical. And the biological systems (like blood) to chemical manufacturing. For example, the bicarbonate (HCO₃⁻)/carbonic acid (H₂CO₃) pair helps maintain blood pH. Conjugate acid-base pairs are also used for chemical reactions in many reactions, conjugate pairs take part in acid-base equilibrium, affecting reaction direction and equilibrium positions. This is important in controlling the progress and outcome of chemical reactions It also has its biological functions in biology, conjugate pairs are involved in enzyme activity, metabolism, and cellular processes. They play a role in enzyme-catalyzed reactions where pH changes can affect enzyme activity and stability. In a titration, conjugate pairs help in identifying the last point of the reaction by providing a visual change in the shift in pH. The relationship between ( Ka ) (acid dissociation constant), ( Kb ) (base dissociation constant), and ( Kw ) (ion-product constant for water) is important for understanding the behavior of acids and bases in aqueous solutions. It helps in measuring pH and pOH by Knowing ( Ka ) and ( Kb ) allows you to find the pH of acidic solutions and the pOH of basic solution relation also used in the equilibrium calculations which interact with equilibrium questions in acid-base chemistry, these constants are used to set up and solve equilibrium problem, helping in telling the direction of the reaction and the concentrations of various species at equilibrium.