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  • Find out the solubility of   in <img alt="0.1 M NaOH" src="https://entrancecorner.onc

Find out the solubility of Ni(OH)_{2}  in 0.1 M NaOH. Given that the ionic product of Ni(OH)_{2} is 2 \times 10^{-15}

Option: 1

1 \times 10^{8}M
 


Option: 2

2 \times 10^{-13}M


Option: 3

2 \times 10^{-8}M

 


Option: 4

1 \times 10^{-13}M


Answers (1)

best_answer

The reaction occurs as follows:

\mathrm{Ni(OH)_{2}\: \rightleftharpoons \: Ni^{2+}\: +\: 2OH^{-}}

      (s)                     (s)                0.1+2s \sim 0.1

Thus, s(0.1)^{2}=2 \times 10^{-15}

s=2 \times 10^{3}

Therefore, Option(2) is correct.

Posted by

vinayak

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