When 12.0 mL of a 7.13 * 10-4 M manganese (II) iodide, MnI2, solution is combined with 22.0 mL of 5.17 * 10-4 M potassium hydroxide, does a precipitate form? (Ksp for manganese (II) hydroxide is 4.6 x 10-14) Group of answer choices The solution is unsaturated - no precipitate forms The concentration of iodide ions is reduced by the addition of OH - ions The presence of KOH will raise the solubility of MnI2 One must know Ksp for MnI2 to make a prediction Mn (OH) 2 will precipitate until the solution becomes saturated

Mn (OH) ₂ will precipitate until the solution becomes saturated

Explanation:

The Ksp of manganese (II) hydroxide is:

Mn (OH) ₂ (s) ⇄ Mn²⁺ (aq) + 2 OH⁻ (aq)

Ksp is defined as:

Ksp = 4.6x10⁻¹⁴ = [Mn²⁺][OH⁻]²

12.0 mL of a 7.13 * 10⁻⁴M manganese (II) iodide are:

0.0120L * (7.13 * 10⁻⁴mol / L) = 8.556x10⁻⁶moles of Mn²⁺

And 22.0 mL of 5.17 * 10⁻⁴ M potassium hydroxide are:

0.0220L * (5.17 * 10⁻⁴mol / L) = 1.1374x10⁻⁵moles of OH⁻

After solutions are combined, total volume is 34.0mL. Thus, concentrations are:

[Mn²⁺] = 8.556x10⁻⁶moles of Mn²⁺ / 0.034L = 2.52x10⁻⁴M

[OH⁻] = 1.1374x10⁻⁵moles of OH⁻ / 0.034L = 3.35x10⁻⁴M

Reaction quotient, Q, is:

Q = [2.52x10⁻⁴M] [3.35x10⁻⁴M]² = 2.81x10⁻¹¹

As Q > Ksp, the reaction will decrease concentration of both [Mn²⁺] [OH⁻] until Q = Ksp, that means insoluble Mn (OH) ₂ (s) is produced. Thus, a precipitate is formed

Right answer is:

Mn (OH) ₂ will precipitate until the solution becomes saturated