2+ 2 2 + 0.1M 0.1M 0.1M Ni OH Ni + 2OH s 2s s wheresis thesolubilityof Ni(OH) . NaOH Na + OH 2 2+ 2 2 15 15 13 OH = 2s +0.1=0.1 2s <<<0.1 Ionicproduct of Ni OH = Ni OH 2×10 =s 0.1 2×10 s = = 2×10 M 0.1×0.1
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