Calculate the number of mg of Mn2+ left
1 answer:
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:
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.
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