Question

The Ksp of nickel hydroxide =6.0×10−16 M.

You may want to reference(Pages 744 - 750) Section 17.5 while completing this problem.
1.) Calculate the molar solubility of Ni(OH)2 when buffered at pH= 8.0.
Express your answer using one significant figure.
2.) Calculate the molar solubility of Ni(OH)2 when buffered at pH= 10.3.
Express your answer using one significant figure.
3.) Calculate the molar solubility of Ni(OH)2 when buffered at pH= 11.9.
Express your answer using one significant figure.

Answer 1

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.