Find out the solubility of Ni(OH)2Ni(OH)_2Ni(OH)2 in 0.1 M NaOHNaOHNaOH. Given that the ionic product of Ni(OH)2Ni(OH)_2Ni(OH)2 is 2×10−152 \times 10^{-15}2×10−15
2×10−132 \times 10^{-13}2×10−13 M
2×10−82 \times 10^{-8}2×10−8 M
1×10−131 \times 10^{-13}1×10−13 M
1×1081 \times 10^{8}1×108 M
Ni(OH)2⇌Ni2++2OH−Ni(OH)_2 \rightleftharpoons Ni^{2+} + 2OH^-Ni(OH)2⇌Ni2++2OH−. NaOH→Na++OH−NaOH \rightarrow Na^+ + OH^-NaOH→Na++OH−. Total [OH−]=2s+0.1≈0.1[OH^-] = 2s + 0.1 \approx 0.1[OH−]=2s+0.1≈0.1. Ionic product = [Ni2+][OH−]2[Ni^{2+}][OH^-]^2[Ni2+][OH−]2. 2×10−15=s(0.1)22 \times 10^{-15} = s(0.1)^22×10−15=s(0.1)2. s=2×10−13s = 2 \times 10^{-13}s=2×10−13. Solubility of Ni(OH)2=2×10−13Ni(OH)_2 = 2 \times 10^{-13}Ni(OH)2=2×10−13 M.
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