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 \(\left.\mathrm{Cr}^{3+}(\mathrm{aq}) ; \text { the } \mathrm{Fe}^{2+}(\mathrm{aq}) \text { is oxidized to } \mathrm{Fe}^{3+}(\mathrm{aq}) \cdot\right]\)

Step by step solution

01

- Reactions with NaOH

When treated with a limited amount of NaOH, Mg2+ ions form a white precipitate of magnesium hydroxide (Mg(OH)2) and Cr3+ ions form a green precipitate of chromium(III) hydroxide [Cr(OH)3]. With excess NaOH, Mg2+ continue to precipitate as Mg(OH)2, while Cr3+ transform into a dark green complex ion [Cr(OH)4]- . The reason for the different results in these two cases is due to the excess OH- ions reacting with Cr3+ to form a negatively charged complex ion.

02

- Calculate Molarity of Cr2O72-

In this case, the chromium is completely precipitated as BaCrO4. The balanced equation for this reaction is: Cr2O72- + 2Ba2+ + H2O → 2BaCrO4 + 4H+. This equation tells that 0.549 g of BaCrO4 is formed by 1 mole of Cr2O72-. We calculate the molarity of Cr2O72- based on this reaction and result, using the formula Moles = Mass/Molar Mass

03

- Calculate Molarity of MnO4-

The balanced equation of the reaction is MnO4- + 5Fe2+ + 8H+ → Mn2+ + 5Fe3+ + 4H2O. So 1 mole of MnO4- reacts with 5 moles of Fe2+. By knowing the molarity and volume of Fe2+ used in the reaction, the molarity of MnO4- can be calculated.

04

- Calculate Percentage of Cr and Mn in Steel Sample

From the molarity and volume of the solution of Cr2O72- and MnO4-, calculate the amount (moles) of Cr and Mn. Convert these mole values to mass using their molar mass and express these mass values as a percentage of the total mass of the steel sample.

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