Calculate the minimum pH needed to precipitate Mn (OH)2 so completely that the remaining concentration of Mn2+ in solution is less than 1.82 x 10^ -8 M (which is the equivalence of 1 ppb). The Ksp for Mn(OH)2 = 1.6 x 10^ -13Mn(OH)2 ⇌ Mn2+ (aq) + 2OH- (aq)

To calculate the minimum pH needed to precipitate Mn(OH)2 completely, we need to consider the equilibrium expression for the dissociation of Mn(OH)2:

Mn(OH)2 ⇌ Mn2+ (aq) + 2OH- (aq)

The equilibrium constant for this reaction is given as Ksp = [Mn2+][OH-]^2 = 1.6 x 10^-13.

We are given that the remaining concentration of Mn2+ in solution should be less than 1.82 x 10^-8 M, which is equivalent to 1 ppb.

To find the minimum pH, we need to determine the concentration of OH- ions at which the concentration of Mn2+ is less than 1.82 x 10^-8 M.

Let's assume x mol/L is the concentration of OH- ions at equilibrium. Since 2 moles of OH- ions are produced for every mole of Mn(OH)2 that dissociates, the concentration of Mn2+ at equilibrium will be 2x mol/L.

Using the equilibrium constant expression, we can write:

Ksp = [Mn2+][OH-]^2 1.6 x 10^-13 = (2x)(x)^2 1.6 x 10^-13 = 2x^3

Solving for x, we find:

x = (1.6 x 10^-13 / 2)^(1/3) x ≈ 1.26 x 10^-5 M

Since the concentration of OH- ions is equal to x, we can calculate the pOH:

pOH = -log10(x) pOH = -log10(1.26 x 10^-5) pOH ≈ 4.9

Finally, we can calculate the minimum pH by subtracting the pOH from 14:

pH = 14 - pOH pH ≈ 14 - 4.9 pH ≈ 9.1

Therefore, the minimum pH needed to precipitate Mn(OH)2 completely, such that the remaining concentration of Mn2+ in solution is less than 1.82 x 10^-8 M (1 ppb), is approximately 9.1.