Calcium fluoride, CaF₂ (M = 78.08 g mol⁻¹), dissolves only slightly in water: Ksp = 3.9 × 10⁻¹¹ at…

(a) Pure water. $K_{sp} = s(2s)^2 = 4s^3$, so $s = \sqrt[3]{K_{sp}/4} = \sqrt[3]{9.750 \times 10^{-12}} = 2.14 \times 10^{-4}\ \text{mol L}^{-1}$.

(b) In mass. $2.136 \times 10^{-4} \times 78.08\ \text{g L}^{-1} = 16.7\ \text{mg L}^{-1}$.

(c) Common ion. $K_{sp} = s(0.020 + 2s)^2 \approx s \times 4.0 \times 10^{-4}$, so $s = \dfrac{3.9 \times 10^{-11}}{4.0 \times 10^{-4}} = 9.75 \times 10^{-8}\ \text{mol L}^{-1}$. That is more than two thousand times smaller than in pure water, and $2s$ is indeed tiny next to $0.020$.

(d) In acid. $[\text{H}^+] = 1.0 \times 10^{-2}\ \text{mol L}^{-1}$. From $K_a = \dfrac{[\text{H}^+][\text{F}^-]}{[\text{HF}]}$, $\dfrac{[\text{HF}]}{[\text{F}^-]} = \dfrac{[\text{H}^+]}{K_a} = 14.71$, so only a fraction $f = \dfrac{K_a}{K_a + [\text{H}^+]} = 0.0637$ of the fluoride stays as $\text{F}^-$. Then

(e) Ratio. $\dfrac{1.340 \times 10^{-3}}{2.136 \times 10^{-4}} = 6.27$. Acid takes the fluoride ions away as HF, so more of the solid dissolves. This is why acidic drinks attack the fluoride in tooth enamel.