Chemistry - Ionic Equilibrium Question with Solution | TestHub

Find the ${\mathrm{Ba}}^{2+}$ needed to start each precipitate (lowest wins).

${\mathrm{BaF}}_2$: $\frac{1.8\times 10^{-7}}{(10^{-3}{)}^2}=0.18\mathrm{M}$.

${\mathrm{BaCO}}_3$: $\frac{5.1\times 10^{-9}}{10^{-3}}=5.1\times 10^{-6}\mathrm{M}$.

${\mathrm{BaCrO}}_4$: $\frac{1.2\times 10^{-10}}{10^{-3}}=1.2\times 10^{-7}\mathrm{M}$.

${\mathrm{BaSO}}_4$: $\frac{1.1\times 10^{-10}}{10^{-3}}=1.1\times 10^{-7}\mathrm{M}$ (lowest, only just).

Answer: (D).