Question

Calculate the number of mg of Mn2+ left

unprecipitated in 100 mL of a 0.1000M solution of MnSO4
to whichenough Na2S has been added to makethe final
sulfide ion (S2-)concentration equal to 0.0900 M. Assume
no change in volume due tothe addition of Na2S.
ThepKsp of MnS is 13.500.

Answer 1

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

Step-by-step 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.