Question

Calculate the minimum pH needed to precipitate Mn(OH)2 so completely that the concentration of Mn2+ is less than 1μg per liter [1 part per billion (ppb)]

Answer 1

Solution : Here we calculate [OH]- by using ksp expression and ksp value.

Ppb = 1 microgram per L

We calculate concentration of Mn2+ by for 1 ppb

For this we assume solution is 1.0L

And in that 1 micro gram Mn2+ is present

Mass in g = 1 micro gram * 1 E-6 g / 1 micro gram = 1 E-6 g

Mol of Mn2+ = 1.0E-6 g / molar mass of Mn2+

=1.0E-6 g * 1 mol Mn/ 54.94 g Mn2+

= 1.82E-8 mol Mn2+

[Mn] = 1.82E-8 mol / 1.0 L

= 1.82E-8 M Mn2+

We got concentration we use it in ksp expression.

Mn(OH)2    ----- > Mn²⁺ + 2 OH⁻

Ksp = [ Mn²⁺ ][OH⁻]²

Ksp value of Mn(OH)2 = 2.0E-13

[ Mn²⁺ ][OH⁻]² = 2.0E-13

1.82E-8 M * [OH⁻]²= 2.0E-13

[OH⁻]² = 2.0E-13/ 1.82 E-8 M

= 1.1E-5

[OH-]= 0.003314 M

pOH = -log [OH-]= -log 0.003314 = 2.50

pH = 14 – 2.50 = 11.52

The pH 11.52 is the answer.