નીચે આપેલ મુજબ નિર્બળ એસીડ $HX$ નું વિયોજન ધ્યાનમાં લો.

$\mathrm{HX}(\mathrm{aq}) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq})+\mathrm{X}(\mathrm{aq}), \mathrm{Ka}=1.2 \times 10^{-5}$

$\left[\mathrm{K}_{\mathrm{n}}:\right.$ વિયોજન અચળાંક]

$300 \mathrm{~K}$ પર $HX$ ના $0.03 \mathrm{M}$ જલીય દ્રાવણ નું અભિસરણ (પરાસરણ) દબાણ .......... $\times 10^{-2}$ bar છે.

$\left[\right.$ આપેલ : $\mathrm{R}=0.083 \mathrm{~L} \mathrm{bar} \mathrm{Mol}^{-1} \mathrm{~K}^{-1}$ ]

Question

નીચે આપેલ મુજબ નિર્બળ એસીડ $HX$ નું વિયોજન ધ્યાનમાં લો.

$\mathrm{HX}(\mathrm{aq}) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq})+\mathrm{X}(\mathrm{aq}), \mathrm{Ka}=1.2 \times 10^{-5}$

$\left[\mathrm{K}_{\mathrm{n}}:\right.$ વિયોજન અચળાંક]

$300 \mathrm{~K}$ પર $HX$ ના $0.03 \mathrm{M}$ જલીય દ્રાવણ નું અભિસરણ (પરાસરણ) દબાણ .......... $\times 10^{-2}$ bar છે.

$\left[\right.$ આપેલ : $\mathrm{R}=0.083 \mathrm{~L} \mathrm{bar} \mathrm{Mol}^{-1} \mathrm{~K}^{-1}$ ]

JEE MAIN 2024, Medium

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Solution

a
$\mathrm{HX} \rightleftharpoons \mathrm{H}^{+}+\mathrm{X}^{-} \quad \mathrm{K}_{\mathrm{a}}=1.2 \times 10^{-5}$

$0.03 \mathrm{M}$

$0.03-\mathrm{x} \quad \mathrm{x} \quad \mathrm{x}$

$\mathrm{K}_{\mathrm{a}}=1.2 \times 10^{-5}=\frac{\mathrm{x}^2}{0.03-\mathrm{x}}$

$0.03-\mathrm{x} \approx 0.03\left(\mathrm{~K}_{\mathrm{a}}\right. \text { is very small) }$

$\frac{\mathrm{x}^2}{0.03}=1.2 \times 10^{-5}$

$\mathrm{x}=6 \times 10^{-4}$

$\text { Final solution }: 0.03-\mathrm{x}+\mathrm{x}+\mathrm{x}$

$=0.03+\mathrm{x}=0.03+6 \times 10^{-4}$

$\Pi=\left(0.03+\left(6 \times 10^{-4}\right)\right) \times 0.083 \times 300$

$=76.19 \times 10^{-2} \approx 76 \times 10^{-2}$

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