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Chemistry · EQUILIBRIUM
हिन्दीA 0.02 M aqueous solution of hydrazine (\( \text{N}_2\text{H}_4 \)) has a pH of 10.30 at 298 K. Given \( \log 2 = 0.30 \) and \( K_w = 1.0 \times 10^{-14} \), the ionization constant \( K_b \) and \( pK_b \) of hydrazine are respectively:
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Correct answer · Option B
\( 2.02 \times 10^{-6} \) and \( 5.69 \)
Explanation
\( pOH = 14 - 10.30 = 3.70 \), so \( [\text{OH}^-] = 10^{-3.70} = 2.0 \times 10^{-4}\text{ M} \). Then \( K_b = \frac{[\text{OH}^-]^2}{C - [\text{OH}^-]} = \frac{(2.0 \times 10^{-4})^2}{0.02 - 0.0002} = \frac{4.0 \times 10^{-8}}{0.0198} \approx 2.02 \times 10^{-6} \). Thus \( pK_b = -\log(2.02 \times 10^{-6}) \approx 5.69 \).
