Consider the dissociation of the weak acid $\mathrm{HX}$ as given below $\mathrm{HX}(\mathrm{aq}) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq})+\mathrm{X}^{-}(\mathrm{aq}), \mathrm{Ka}=1.2 \times 10^{-5}$ $\left[\mathrm{K}_{\mathrm{a}}\right.$ : dissociation constant $]$ The osmotic…

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Consider the dissociation of the weak acid $\mathrm{HX}$ as given below $\mathrm{HX}(\mathrm{aq}) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq})+\mathrm{X}^{-}(\mathrm{aq}), \mathrm{Ka}=1.2 \times 10^{-5}$ $\left[\mathrm{K}_{\mathrm{a}}\right.$ : dissociation constant $]$ The osmotic pressure of $0.03 \mathrm{M}$ aqueous solution of $\mathrm{HX}$ at $300 \mathrm{~K}$ is _______ $\times 10^{-2}$ bar (nearest integer). [Given : $\mathrm{R}=0.083 \mathrm{~L} \mathrm{bar} \mathrm{mol}^{-1} \mathrm{~K}^{-1}$]

Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{HX} ightleftharpoons \mathrm{H}^{+}+\mathrm{X}^{-} \quad \mathrm{K}_{\mathrm{a}}=1.2 	imes 10^{-5} \ & 0.03 \mathrm{M} \ & 0.03-\mathrm{x} \quad \mathrm{x} \quad \mathrm{x} \ & \mathrm{K}_{\mathrm{a}}=1.2 	imes 10^{-5}=\frac{\mathrm{x}^2}{0.03-\mathrm{x}} \ & 0.03-\mathrm{x} \approx 0.03\left(\mathrm{~K}_{\mathrm{a}} 	ext { is very small }ight) \ & \frac{\mathrm{x}^2}{0.03}=1.2 	imes 10^{-5} \ & \mathrm{x}=6 	imes 10^{-4} \end{aligned}$ Final solution : $0.03-x+x+x$ $\begin{aligned} & =0.03+x=0.03+6 	imes 10^{-4} \ & \Pi=\left(0.03+\left(6 	imes 10^{-4}ight)ight) 	imes 0.083 	imes 300 \ & =76.19 	imes 10^{-2} \approx 76 	imes 10^{-2} \end{aligned}$

Consider the dissociation of the weak acid \(\mathrm{HX}\) as given below \(\mathrm{HX}(\mathrm{aq})…

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