Final answer:
The pH at which MnS begins to precipitate in a 0.125-M solution of Mn(NO₃)₂ saturated with H₂S can be determined using the stoichiometry of the reaction and the dissociation of water.
Step-by-step explanation:
The pH at which MnS begins to precipitate can be determined by considering the equilibrium between Mn(NO₃)₂ and H₂S.
First, we need to write the balanced equation for the reaction between Mn(NO₃)₂ and H₂S:
Mn(NO₃)₂(aq) + H₂S(aq) → MnS(s) + 2HNO₃(aq)
This reaction shows that one mole of Mn(NO₃)₂ reacts with two moles of H₂S to form one mole of MnS and two moles of HNO₃. Since Mn(NO₃)₂ is in excess, we can assume that all of the H₂S will react. Therefore, we can find the concentration of H⁺ ions (which determine the pH) using the stoichiometry of the reaction:
0.125 M Mn(NO₃)₂ × (2 mol H₂S / 1 mol Mn(NO₃)₂) × (2 mol H⁺ / 1 mol H₂S) = 0.25 M H⁺
Using the equation for the dissociation of H₂O, we can find the pOH and pH:
pOH = -log(0.25 M) ≈ 0.60
pH = 14 - pOH ≈ 13.40
Therefore, the pH at which MnS begins to precipitate is approximately 13.40.