OnlineCBT
Chemistry · EQUILIBRIUM
हिन्दीThe solubility product of \( \text{Ni(OH)}_2 \) is \( 2.0 \times 10^{-15} \). What is its molar solubility in a \( 0.10\text{ M NaOH} \) solution?
Correct answer · Option 2
\( 2.0 \times 10^{-13} \text{ M} \)
Explanation
In \( 0.10\text{ M NaOH} \), \( [\text{OH}^-] \approx 0.10\text{ M} \). Thus, \( K_{\text{sp}} = [\text{Ni}^{2+}][\text{OH}^-]^2 = S'(0.10)^2 \implies S' = \frac{2.0 \times 10^{-15}}{0.01} = 2.0 \times 10^{-13}\text{ M} \).
