Question

If 200 mL of a 2.5 x 10-5 mol/L solution of Ni(NO3), was mixed with 200 mL of a 4.86 X

102 mol/L solution of KOH, determine if a precipitate will form. The Kg, of Ni(OH)21, is 6
X 10-16 SHOW ALL WORK.

Answer 1

No precipitate will form since the ion product is smaller than the solubility product constant.

Step-by-step explanation:

To determine if a precipitate will form, we need to compare the ion product, [Ni²+][OH-]², with the solubility product constant, Ksp. First, calculate the moles of Ni²+ and OH- in the solutions.

For the Ni(NO3)2 solution:

• Moles of Ni²+ = volume (L) x concentration (mol/L) = 0.200 L x 2.5 x 10^(-5) mol/L = 5.0 x 10^(-6) mol

For the KOH solution:

• Moles of OH- = volume (L) x concentration (mol/L) = 0.200 L x 4.86 x 10^(2) mol/L = 9.72 x 10^(-2) mol

Now multiply the moles of Ni²+ and OH- to get the ion product:

• Ion product = (5.0 x 10^(-6) mol) x (9.72 x 10^(-2) mol)^2 = 4.85 x 10^(-8)

Since the ion product is smaller than the solubility product constant (6 x 10^(-16)), no precipitate will form.