α=1 for NaOH
NaOH(aq)→0.1MNa+(aq)+0.1MOH−(aq)
Ni(OH)2(s)⇌s′Ni+2(aq)+0.1+2s′2OH−(aq)
Ionic product =(s′)(0.1+2s′)2
2×10−15=s′(0.1)2
s′=2×10−13M
Find out the solubility of Ni(OH)2 in 0.1MNaOH. Given that the ionic product of Ni(OH)2 is 2×10–15.
Held on 30 Apr 2020 · Verified 9 Jul 2026.
2×10–8M
1×10–13 M
1×108 M
2×10–13M
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