Will a precipitate of magnesium fluoride form when 200. mL of 1.9 * 10^-3 M MgCl₂ are added to 300. mL of 1.4 * 10^-2 M NaF? [K_sp (MgF₂) = 6.9 * 10^-9]

Question

Quizplus · Accepted Answer

To determine if a precipitate will form, we calculate the reaction quotient (Q) and compare it to the solubility product constant (Ksp). The concentration of Mg2+ is $1.9 \times 10^{-3} \, \text{M}$ in 200 mL, and the concentration of F- is $1.4 \times 10^{-2} \, \text{M}$ in 300 mL. When mixed, the total volume is 500 mL. The final concentrations are $ \frac{1.9 \times 10^{-3} \, \text{M} \times 200 \, \text{mL}}{500 \, \text{mL}} = 7.6 \times 10^{-4} \, \text{M}$ for Mg2+ and $ \frac{1.4 \times 10^{-2} \, \text{M} \times 300 \, \text{mL}}{500 \, \text{mL}} = 8.4 \times 10^{-3} \, \text{M}$ for F-. The reaction quotient, Q, is $Q = [Mg^{2+}][F^-]^2 = (7.6 \times 10^{-4}) \times (8.4 \times 10^{-3})^2 = 5.36 \times 10^{-8}$. Since $Q = 5.36 \times 10^{-8}$ is less than $K_{sp} = 6.9 \times 10^{-9}$, the correct answer is A, indicating that a precipitate will form because Q is greater than Ksp.

Will a Precipitate of Magnesium Fluoride Form When 200