It is observed that 7.5 mmol of BaF₂ will dissolve in 1.0 L of water. Use these data to calculate the value of K_sp for barium fluoride.

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The correct answer is A.The expression for the solubility product (Ksp) for BaF2 is:Ksp = [Ba2+][F-]^2We know that 7.5 mmol of BaF2 dissolves in 1.0 L of water. This means that the concentration of Ba2+ and F- ions in the solution is:[Ba2+] = 7.5 mmol/L = 7.5 x 10^-3 M[F-] = 2 x 7.5 mmol/L = 15 x 10^-3 MSubstituting these values into the expression for Ksp, we get:Ksp = (7.5 x 10^-3 M)(15 x 10^-3 M)^2 = 1.7 x 10^-6

It Is Observed That 7