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.
JEE Main 2025: Chemistry Formula | Study Materials | High Scoring Topics | Preparation Guide
JEE Main 2025: Syllabus | Sample Papers | Mock Tests | PYQs | Study Plan 100 Days
NEET 2025: Syllabus | High Scoring Topics | PYQs
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 Ka of acid increases, then Kb of the conjugate base decreases. In other words, the conjugate base of a strong acid is a weak base, and vice-versa.
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.
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.
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
Pka 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 Ka and Kb 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).
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.
09 Dec'24 11:40 AM
21 Oct'24 12:32 PM
21 Oct'24 11:57 AM
19 Oct'24 03:14 PM
19 Oct'24 03:08 PM
09 Oct'24 10:36 AM
09 Oct'24 10:24 AM
09 Oct'24 10:19 AM
07 Oct'24 05:09 PM
07 Oct'24 04:56 PM