Final answer:
The K_sp for a NaF solution with equal concentrations of Na^+ and F^- ions, both at 4.0 x 10^-4 M, is 1.6 x 10^-7, corresponding to option B.
Step-by-step explanation:
The question pertains to the solubility product constant, commonly referred to as Ksp, which is a measure of the solubility of an ionic compound in a solution. In a solution of sodium fluoride (NaF), if the concentration of Na+ ions is 4.0 x 10-4 M and the concentration of F- ions is also 4.0 x 10-4 M, the Ksp can be calculated using the formula Ksp = [Na+][F-].
By plugging the values into this formula, we find that:
Ksp = (4.0 x 10-4 M) * (4.0 x 10-4 M) = 1.6 x 10-7
Therefore, the value of Ksp for NaF is 1.6 x 10-7. This corresponds to option B. Understanding Ksp values is important for predicting whether a precipitate will form in a solution under certain conditions.