Answer: The molar solubility of Ni(OH)2 when buffered at pH=8.0 is 1.0×10^-7 M.
The molar solubility of Ni(OH)2 when buffered at pH=10.3 is 1.6×10^-6 M.
The molar solubility of Ni(OH)2 when buffered at pH=11.9 is 2.5×10^-5 M.
Explanation: The solubility of a sparingly soluble salt is affected by pH of the solution. At a given pH, the solubility of the salt is related to the solubility product constant, Ksp, of the salt. The expression for the solubility product constant is:
Ksp = [Ni2+][OH-]^2
where [Ni2+] and [OH-] are the concentrations of Ni2+ and OH- ions, respectively.
The solubility of Ni(OH)2 in water is 1.4 × 10^-15 M.
At pH 8.0, Ni(OH)2 is insoluble as the pH is below the pKa of the hydroxide ion. Therefore, the molar solubility of Ni(OH)2 is 0.
At pH 10.3, the concentration of OH- ions is 5.0 × 10^-4 M. Thus, the concentration of Ni2+ ions is given by:
Ksp = [Ni2+][OH-]^2
1.4 × 10^-15 = [Ni2+][5.0 × 10^-4]^2
[Ni2+] = 5.6 × 10^-8 M
Therefore, the molar solubility of Ni(OH)2 is 5.6 × 10^-8 M.
At pH 11.9, the concentration of OH- ions is 5.0 × 10^-3 M. Thus, the concentration of Ni2+ ions is given by:
Ksp = [Ni2+][OH-]^2
1.4 × 10^-15 = [Ni2+][5.0 × 10^-3]^2
[Ni2+] = 5.6 × 10^-12 M
Therefore, the molar solubility of Ni(OH)2 is 5.6 × 10^-12 M.