Final answer:
The molar solubility of nickel(II) hydroxide, given its Ksp of 5.84×10−16, is calculated as 1.13×10−6 M by setting up the equilibrium expression and solving for the solubility.
Step-by-step explanation:
Calculate the Molar Solubility of Nickel(II) Hydroxide
The question asks to calculate the molar solubility of nickel(II) hydroxide, Ni(OH)2, given its Ksp is 5.84×10−16. Molar solubility is the number of moles of a substance that can dissolve in a liter of solution before the solution becomes saturated.
To find the molar solubility of Ni(OH)2, we set up the equilibrium expression based on its dissolution equation:
Ni(OH)2 (s) → Ni2+ (aq) + 2OH− (aq)
Letting 's' be the molar solubility, the ion concentration at equilibrium are [Ni2+] = s and [OH−] = 2s. Inserting these into the Ksp expression gives: Ksp = [Ni2+]×[OH−]2 = s×(2s)2 = 4s3.
Solving for 's' with the given Ksp value:
4s3 = 5.84×10−16
s3 = 1.46×10−16
s = (1.46×10−16)1/3 = 1.13×10−6 M.
Therefore, the molar solubility of nickel(II) hydroxide in water is 1.13×10−6 M.