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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 : \(\mathrm{R}=0.083 \mathrm{~L} \mathrm{bar} \mathrm{mol}^{-1} \mathrm{~K}^{-1}\)]
\(\mathrm{HX}(\mathrm{aq}) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq})+\mathrm{X}^{-}(\mathrm{aq}), \mathrm{Ka}=1.2 \times 10^{-5}\)
\(\left[\mathrm{K}_{\mathrm{a}}\right.\) : 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 : \(\mathrm{R}=0.083 \mathrm{~L} \mathrm{bar} \mathrm{mol}^{-1} \mathrm{~K}^{-1}\)]
Solution:
1977 Upvotes
Verified Answer
The correct answer is:
76
$\begin{aligned}
& \mathrm{HX} \rightleftharpoons \mathrm{H}^{+}+\mathrm{X}^{-} \quad \mathrm{K}_{\mathrm{a}}=1.2 \times 10^{-5} \\
& 0.03 \mathrm{M} \\
& 0.03-\mathrm{x} \quad \mathrm{x} \quad \mathrm{x} \\
& \mathrm{K}_{\mathrm{a}}=1.2 \times 10^{-5}=\frac{\mathrm{x}^2}{0.03-\mathrm{x}} \\
& 0.03-\mathrm{x} \approx 0.03\left(\mathrm{~K}_{\mathrm{a}} \text { is very small }\right) \\
& \frac{\mathrm{x}^2}{0.03}=1.2 \times 10^{-5} \\
& \mathrm{x}=6 \times 10^{-4}
\end{aligned}$
Final solution : $0.03-x+x+x$
$\begin{aligned}
& =0.03+x=0.03+6 \times 10^{-4} \\
& \Pi=\left(0.03+\left(6 \times 10^{-4}\right)\right) \times 0.083 \times 300 \\
& =76.19 \times 10^{-2} \approx 76 \times 10^{-2}
\end{aligned}$
& \mathrm{HX} \rightleftharpoons \mathrm{H}^{+}+\mathrm{X}^{-} \quad \mathrm{K}_{\mathrm{a}}=1.2 \times 10^{-5} \\
& 0.03 \mathrm{M} \\
& 0.03-\mathrm{x} \quad \mathrm{x} \quad \mathrm{x} \\
& \mathrm{K}_{\mathrm{a}}=1.2 \times 10^{-5}=\frac{\mathrm{x}^2}{0.03-\mathrm{x}} \\
& 0.03-\mathrm{x} \approx 0.03\left(\mathrm{~K}_{\mathrm{a}} \text { is very small }\right) \\
& \frac{\mathrm{x}^2}{0.03}=1.2 \times 10^{-5} \\
& \mathrm{x}=6 \times 10^{-4}
\end{aligned}$
Final solution : $0.03-x+x+x$
$\begin{aligned}
& =0.03+x=0.03+6 \times 10^{-4} \\
& \Pi=\left(0.03+\left(6 \times 10^{-4}\right)\right) \times 0.083 \times 300 \\
& =76.19 \times 10^{-2} \approx 76 \times 10^{-2}
\end{aligned}$
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