JEE Chemistry 2025 Question with Solution

Given: A $1 \, \text{mM}$ solution of ethylamine has $\text{pH} = 9$.

Find: The value of $x$ if $K_b = 10^{-x}$.

Ethylamine ionizes in water as:

$$\mathrm{CH_3CH_2NH_2(aq) + H_2O(l) \rightleftharpoons CH_3CH_2NH_3^+(aq) + OH^-(aq)}$$

From $\text{pH} = 9$, we get:

$$\text{pOH} = 14 - 9 = 5$$

Hence,

$$[OH^-] = 10^{-5} \, \text{M}$$

The initial concentration of ethylamine is:

$$1 \, \text{mM} = 10^{-3} \, \text{M}$$

Using the base ionization expression,

$$K_b = \frac{[\mathrm{CH_3CH_2NH_3^+}][OH^-]}{[\mathrm{CH_3CH_2NH_2}]}$$

Since the degree of ionization is small, we take:

$$[\mathrm{CH_3CH_2NH_2}] \approx 10^{-3} \, \text{M}$$

and

$$[\mathrm{CH_3CH_2NH_3^+}] = [OH^-] = 10^{-5} \, \text{M}$$

Therefore,

$$K_b = \frac{(10^{-5})(10^{-5})}{10^{-3}} = \frac{10^{-10}}{10^{-3}} = 10^{-7}$$

So, if $K_b = 10^{-x}$, then:

$$x = 7$$

Therefore, the value of $x$ is $7$.

Given: $C = 1 \times 10^{-3} \, \text{M}$ and $\text{pH} = 9$.

Find: The nearest integer value of $x$ in $K_b = 10^{-x}$.

For the reaction

$$\mathrm{CH_3CH_2NH_2 + H_2O \rightleftharpoons CH_3CH_2NH_3^+ + OH^-}$$

first convert pH to pOH:

$$\text{pOH} = 14 - 9 = 5$$

Then,

$$[OH^-] = 10^{-\text{pOH}} = 10^{-5} \, \text{M}$$

For a weak base,

$$K_b = \frac{[BH^+][OH^-]}{[B]}$$

Here, because ionization is negligible compared to unity, the base concentration at equilibrium is approximated by its initial concentration:

$$[B] \approx 10^{-3} \, \text{M}$$

Also, from stoichiometry,

$$[BH^+] = [OH^-] = 10^{-5} \, \text{M}$$

Substituting,

$$K_b = \frac{(10^{-5})^2}{10^{-3}} = \frac{10^{-10}}{10^{-3}} = 10^{-7}$$

Thus,

$$K_b = 10^{-7}$$

So the required value is $x = 7$.