Answer
a) $[F^-]=1.36\times10^{-3}\ M$
b) $[Ca^{2+}]=1.875\times10^{-6}\ M$
Work Step by Step
a) To accomplish this, the solution must be saturated with BaF2(aq).
$Ksp=[Ba^{2+}][F^-]^2$
$1.84\times10^{-7}=0.10\times[F^-]^2$
$[F^-]=1.36\times10^{-3}\ M$
b)
$Ksp=[Ca^{2+}][F^-]^2$
$3.45\times10^{-11}=[Ca^{2+}]\times (1.36\times10^{-3})^2$
$[Ca^{2+}]=1.875\times10^{-6}\ M$
