Question

Dr.Bob you had told me to ignore 2x so I wondering if I wrote my did these steps correctly.

Calculate the molar solubility of nickel (II) hydroxide in 0.100 M sodium hydroxide
The solubility product of nickel (II) hydroxide at 25°C is 1.6 x 10-16.

Ni(OH)2 <=> Ni2+ + 2OH–

0.10/Ksp= 0.10/1.6 E-16
= 6.25 E14 > 500

Ksp = [Ni2+][OH-]2 = 1.6 x 10-16
Listen fellas,,, a gr8 chemist once said: "whats in the water?"

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