What products are obtained when \(\mathrm{Mg}^{2+}(\mathrm{aq})\) and
\(\mathrm{Cr}^{3+}(\mathrm{aq})\) are each treated with a limited amount of
NaOH(aq)? With an excess of \(\mathrm{NaOH}(\) aq)? Why are the results
different in these two cases?A certain steel is to be analyzed for
\(\mathrm{Cr}\) and \(\mathrm{Mn}\). By suitable treatment, the Cr in the steel is
oxidized to \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(\mathrm{aq})\) and the
\(\mathrm{Mn}\) to \(\mathrm{MnO}_{4}(\mathrm{aq}) . \mathrm{A} 10.000
\mathrm{g}\)
sample of steel is used to produce \(250.0 \mathrm{mL}\) of a solution
containing \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(\mathrm{aq})\) and
\(\mathrm{MnO}_{4}^{-}(\mathrm{aq}) . \mathrm{A} 10.00 \mathrm{mL}\)
portion of this solution is added to \(\mathrm{BaCl}_{2}(\mathrm{aq}),\) and by
proper adjustment of the \(\mathrm{pH}\), the chromium is completely
precipitated as \(\mathrm{BaCrO}_{4}(\mathrm{s}) ; 0.549 \mathrm{g}\) is
obtained. A second \(10.00 \mathrm{mL}\) portion of the solution requires
exactly \(15.95 \mathrm{mL}\) of \(0.0750 \mathrm{M}
\mathrm{Fe}^{2+}(\mathrm{aq})\) for its titration in acidic solution. Calculate
the \(\%\) Cr and \% \(\mathrm{Mn}\) in the steel sample. [Hint: In the titration
\(\mathrm{MnO}_{4}^{-}(\mathrm{aq})\) is reduced to
\(\mathrm{Mn}^{2+}(\mathrm{aq})\) and \(\mathrm{Cr}_{2}
\mathrm{O}_{7}^{2-}(\mathrm{aq})\) is reduced to
\(\left.\mathrm{Cr}^{3+}(\mathrm{aq}) ; \text { the }
\mathrm{Fe}^{2+}(\mathrm{aq}) \text { is oxidized to }
\mathrm{Fe}^{3+}(\mathrm{aq}) \cdot\right]\)