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Chemistry · EQUILIBRIUM
हिन्दीThe solubility product of \( \text{Ni(OH)}_2 \) is \( 2.0 \times 10^{-15} \) at 298 K. What is the molar solubility of \( \text{Ni(OH)}_2 \) in a 0.05 M aqueous \( \text{NaOH} \) solution?
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Correct answer · Option B
\( 8.0 \times 10^{-13}\text{ M} \)
Explanation
In 0.05 M NaOH, \( [\text{OH}^-] = 0.05 + 2S \approx 0.05\text{ M} \). Then \( K_{sp} = [\text{Ni}^{2+}][\text{OH}^-]^2 = S(0.05)^2 = 2.0 \times 10^{-15} \). Hence, \( S = \frac{2.0 \times 10^{-15}}{2.5 \times 10^{-3}} = 8.0 \times 10^{-13}\text{ M} \).
