JEE MAIN - Chemistry (2025 - 23rd January Morning Shift - No. 23)
If 1 mM solution of ethylamine produces $\mathrm{pH}=9$, then the ionization constant $\left(\mathrm{K}_{\mathrm{b}}\right)$ of ethylamine is $10^{-x}$. The value of $x$ is _________ (nearest integer).
[The degree of ionization of ethylamine can be neglected with respect to unity.]
Explanation
$\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2(\mathrm{aq})+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{C}_2 \mathrm{H}_2 \mathrm{NH}_3^{+}+\stackrel{\ominus}{\mathrm{O}} \mathrm{H}$
$$\begin{array}{ccc} \mathrm{C}=10^{-3} \mathrm{M} & - & - \\ \mathrm{C}(1-\alpha) & \mathrm{C} \alpha & \mathrm{C} \alpha \\ \Rightarrow \mathrm{C}=10^{-3} & =10^{-5} & =10^{-5} \end{array}$$
$1-\alpha \simeq 1$
Given, $\mathrm{P}^{\mathrm{H}}=9 \Rightarrow \mathrm{P}^{\mathrm{OH}}=5 \Rightarrow[\stackrel{\ominus}{\mathrm{O}} \mathrm{H}]=10^{-5} \mathrm{M}$
Now, $K_b=\frac{\left[\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_3^{+}\right][\stackrel{\ominus}{\mathrm{O}} \mathrm{H}]}{\left[\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2\right]}$
$$\Rightarrow \mathrm{K}_{\mathrm{b}}=\frac{10^{-5} \times 10^{-5}}{10^{-3}}=10^{-7}$$
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