Chapter 17 Principles of Chemical Reactivity: Other Aspects of Aqueous Equilibria - Study Questions - Page 677c: 60

See the answer below.

Pure water: $Ksp=[Ba^{2+}]F^-]=s\dot{}(2s)^2$ $s=\sqrt[3]{1.8\dot{}10^{-17}/4}=1.65\dot{}10^{-6}\ M$ $s^*=1.65\dot{}10^{-6}\ mol/L\dot{}175.32\ g/mol=2.89\dot{}10^{-4}\ mg/mL$ With KF: $[KF]=5.0\ g/L\div 58.10\ g/mol=0.086\ M$ $Ksp=s\dot{}(2s+0.086)^2$ $s=2.43\dot{}10^{-15}\ M$ $s^*=2.43\dot{}10^{-15}\ mol/L\dot{}175.32\ g/mol=4.26\dot{}10^{-13}\ mg/mL$