Question

Calculate the number of mg of Mn2+ left

Answer 1

Answer:

1.930 * 10⁻⁹ mg of Mn⁺² are left unprecipitated.

Explanation:

The reaction that takes place is:

Mn⁺² + S⁻² ⇄ MnS(s)  

ksp = [Mn⁺²] [S⁻²]

If the pksp of MnS is 13.500, then the ksp is:

$ksp=10^{-13.500}=3.1623*10^{-14}$

From the problem we know that [S⁻²] = 0.0900 M

We use the ksp to calculate [Mn⁺²]:

3.1623*10⁻¹⁴= [Mn⁺²] * 0.0900 M

[Mn⁺²] = 3.514 * 10⁻¹³ M.

Now we can calculate the mass of Mn⁺², using the volume, concentration and atomic weight. Thus the mass of Mn⁺² left unprecipitated is:

3.514 * 10⁻¹³ M * 0.1 L * 54.94 g/mol = 1.930 * 10⁻¹² g = 1.930 * 10⁻⁹ mg.