Final answer:
When the carbonate level in a contaminated solution with 1.0x10⁻⁶ M Pb²⁺ and Ag⁺ ions is increased to 0.00013 M, only silver ions will precipitate as the ion product exceeds the solubility product constant for Ag₂CO₃, while lead ions will not precipitate.
Step-by-step explanation:
If a contaminated solution contains 1.0x10⁻⁶ M each for both Pb²⁺ and Ag⁺ ions and the carbonate level is raised to 0.00013 M, we must consider which salts will precipitate by comparing the ion product (Q) with the solubility product constant (Ksp) for both potential precipitates, PbCO₃ and Ag₂CO₃.
To determine if the salts will precipitate, we calculate Q for each:
- For PbCO₃, Q = [Pb²⁺][CO₃²⁺] = (1.0x10⁻⁶)(0.00013) = 1.3x10⁻⁺
- For Ag₂CO₃, since Ag⁺ is 1.0x10⁻⁶ M and the formula of the salt is Ag₂CO₃, two Ag⁺ ions are involved in its formation, so Q = [Ag⁺]²[CO₃²⁺] = (1.0x10⁻⁶)²(0.00013) = 1.3x10⁻⁻²
Now we compare Q to the Ksp for each compound:
- For PbCO₃, since Q (1.3x10⁻⁺) < Ksp (7.4x10⁻ⁱ⁴), no precipitation of Pb²⁺ will occur.
- For Ag₂CO₃, since Q (1.3x10⁻⁻²) > Ksp (8.1x10⁻ⁱ), Ag⁺ ions will precipitate as Ag₂CO₃.
Therefore, the correct answer is A. only silver ions will precipitate.