Show that under the following conditions, \(\mathrm{Ba}^{2+}(\mathrm{aq})\) can
be separated from \(\mathrm{Sr}^{2+}(\mathrm{aq})\) and \(\mathrm{Ca}^{2+}(\) aq)
by precipitating \(\mathrm{BaCrO}_{4}(\mathrm{s})\) with the other ions
remaining in solution: $$\begin{aligned}
&\left[\mathrm{Ba}^{2+}\right]=\left[\mathrm{Sr}^{2+}\right]=\left[\mathrm{Ca}^{2+}\right]=0.10
\mathrm{M}\\\
&\left[\mathrm{HC}_{2} \mathrm{H}_{3}
\mathrm{O}_{2}\right]=\left[\mathrm{C}_{2} \mathrm{H}_{3}
\mathrm{O}_{2}\right]=1.0 \mathrm{M}\\\
&\left[\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\right]=0.0010 \mathrm{M}\\\
&\mathrm{K}_{\mathrm{sp}}\left(\mathrm{BaCrO}_{4}\right)=1.2 \times
10^{-10}\\\
&K_{\mathrm{sp}}\left(\mathrm{SrCrO}_{4}\right)=2.2 \times 10^{-5}
\end{aligned}$$ Use data from this and previous chapters, as necessary.