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The ionic product of Ni(OH)2 is 2.0 × 10-15. The molar solubility of Ni(OH)2 in 0.10 M NaOH is-
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Verified Answer
The correct answer is:
2.0 × 10-13
Ni(OH)2 Ni2+ + 2OH-
s 2s
Total [OH-] = 2s + 0.1
Ionic product = [s] [2s + 0.1]2
0.01 s = 2 × 10-15
s = 2 × 10-13
s 2s
Total [OH-] = 2s + 0.1
Ionic product = [s] [2s + 0.1]2
0.01 s = 2 × 10-15
s = 2 × 10-13
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