What is the ion product (Q) in the given scenario, and will a precipitate form?
The ion product (Q) is calculated using the concentrations of the reactants. Calculating Q and comparing it to the solubility product constant (Ksp) determines whether a precipitate will form. In this case, Q is less than Ksp, so no precipitate will form.
Understanding Ion Product (Q) and Precipitation
Precipitation in Chemistry: Precipitation is the formation of a solid, insoluble substance from a solution due to a chemical reaction.
Calculation of Ion Product (Q):
The ion product Q is calculated by multiplying the concentrations of the ions involved in the reaction. In this case, we are dealing with the dissociation of Pb(NO₃)₂ and NaI to form Pb²⁺ and I⁻ ions.
Given Concentrations:
- Volume of Pb(NO₃)₂ solution = 150.0 mL, concentration = 0.0050 M
- Volume of NaI solution = 350 mL, concentration = 0.0015 M
Calculating Q:
The concentration of Pb²⁺ is 0.0050 M since its concentration is the same as Pb(NO₃)₂. The concentration of I⁻ can be calculated using the molarity and volume of NaI solution.
Q = [Pb²⁺] * [I⁻]²
Q = (0.0050 M) * (0.0015 M)²
Q = 1.125 x 10⁻⁸
Determining Precipitation:
To determine if a precipitate will form, we compare Q to the Ksp of PbI₂, which is 3.2 x 10⁻⁸ in this case. If Q is greater than Ksp, a precipitate will form. If Q is less than Ksp, no precipitate will form.
Final Answer: Based on the calculations, the ion product (Q) is 1.125 x 10⁻⁸, which is less than the Ksp of PbI₂. Therefore, no precipitate will form in this scenario.