Find out the solubility of Ni(OH)_2 in 0.1 M NaOH . Given that the ionic product of Ni(OH)_2 is 2 × 10^-15 . - Chemistry MCQ NEET 2020 E1 2026 | ExamTest.live

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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 1 0^{- 15} .

Solution & Step-by-step Explanation

Ni (OH)_{2} ⇌ Ni^{2 +} + 2 OH^{-} Let solubility of Ni (OH)_{2} be s . [Ni^{2 +}] = s [OH^{-}] = 0.1 (from NaOH, neglecting 2 s from Ni (OH)_{2} as it is very small). K_{s p} = [Ni^{2 +}] [OH^{-}]^{2} 2 \times 1 0^{- 15} = s \times (0.1)^{2} 2 \times 1 0^{- 15} = s \times 1 0^{- 2} s = \frac{2 \times 1 0 ^{- 15}}{1 0 ^{- 2}} = 2 \times 1 0^{- 13} M .

Try it yourself before checking the explanation above.

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 1 0^{- 15} .

2 \times 1 0^{- 13} M

2 \times 1 0^{- 8} M

1 \times 1 0^{- 13} M

1 \times 1 0^{8} M

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