If 1 mM solution of ethylamine produces $\mathrm{pH}=9$, then the ionization constant ( $K_{b}$ ) of

Question

If 1 mM solution of ethylamine produces $\mathrm{pH}=9$, then the ionization constant ( $K_{b}$ ) 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.]\

Professor PC Thomas Classes · Answer

(7) : Given, Initial concentration of ethylamine $$ \begin{aligned} & =1 \mathrm{mM}=0.001 \mathrm{M} \ & \quad \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}+\mathrm{H}_{2} \mathrm{O} ightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}+\mathrm{OH}^{-} \end{aligned} $$  $\mathrm{pH}=9 \Rightarrow \mathrm{pH}+\mathrm{pOH}=14$\ $	herefore \mathrm{pOH}=5 \Rightarrow\left[\mathrm{OH}^{-}ight]=10^{-\mathrm{pOH}}$\ $=10^{-5} \mathrm{M}$ $$ \begin{aligned} & 	ext { Now, } K_{b}=\frac{\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}ight]\left[\mathrm{OH}^{-}ight]}{\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}ight]} \ & \quad=\frac{x \cdot x}{0.001-x} \ & =\frac{x^{2}}{0.001} \quad(\because x \ll 0.001) \ & \begin{array}{l} K_{b}=\frac{\left(10^{-5}ight)^{2}}{0.001}=10^{-7} \mathrm{M} \quad\left(\because x=\left[\mathrm{OH}^{-}ight]ight. \ \left.=10^{-5} \mathrm{M}ight) \end{array} \end{aligned} $$