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Chapter 5: Problem 67

The titration of \(5.00 \mathrm{mL}\) of a saturated solution of sodium oxalate, \(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4},\) at \(25^{\circ} \mathrm{C}\) requires \(25.8 \mathrm{mL}\) of \(0.02140 \mathrm{M} \mathrm{KMnO}_{4}\) in acidic solution. What mass of \(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) in grams would be present in \(1.00 \mathrm{L}\) of this saturated solution? \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}+\mathrm{MnO}_{4}^{-} \longrightarrow_{\mathrm{Mn}^{2+}}+\mathrm{CO}_{2}(\mathrm{g}) \quad\) (not balanced)

Short Answer

Expert verified
The mass of sodium oxalate in a liter of this saturated solution is 36.992 g.

Step by step solution

01

Balance the Equation

The balanced redox reaction equation is:\[5\, \mathrm{C}_{2} \mathrm{O}_{4}^{2-} + 2\, \mathrm{MnO}_{4}^{-} + 16\, \mathrm{H}^{+} \longrightarrow 10\, \mathrm{CO}_{2} + 2\, \mathrm{Mn}^{2+} + 8\, \mathrm{H}_{2} \mathrm{O}\]
02

Calculate moles of \(\mathrm{KMnO}_{4}\)

From the balanced equation we observe that 5 moles of \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\) react with 2 moles of \(\mathrm{MnO}_{4}^{-}\). Now, using the Molarity (M) of \(\mathrm{KMnO}_{4}\), which is moles/L, we can find the number of moles of \(\mathrm{KMnO}_{4}\) used:\[ n(\mathrm{KMnO}_{4}) = M(\mathrm{KMnO}_{4}) \times V(\mathrm{KMnO}_{4}) = 0.02140\, \mathrm{M} \times 25.8 \, \mathrm{mL} \times (1\, \mathrm{L}/1000\, \mathrm{mL}) = 0.00055242 \, \mathrm{moles}\]
03

Calculate moles of \(\mathrm{Na}_{2}\mathrm{C}_{2} \mathrm{O}_{4}\)

From stoichiometry, we know that the moles of sodium oxalate is given by \( n(\mathrm{Na}_{2}\mathrm{C}_{2} \mathrm{O}_{4}) = n(\mathrm{MnO}_{4}) \times (5/2) = 0.00055242\, \mathrm{moles} \times (5/2) = 0.00138105\, \mathrm{moles}\)
04

Find Concentration of \(\mathrm{Na}_{2}\mathrm{C}_{2} \mathrm{O}_{4}\)

The molarity of sodium oxalate can be determined using the equation \(M = n/V\), thus:\[ M(\mathrm{Na}_{2}\mathrm{C}_{2} \mathrm{O}_{4}) = n(\mathrm{Na}_{2}\mathrm{C}_{2} \mathrm{O}_{4})/V(\mathrm{Na}_{2}\mathrm{C}_{2} \mathrm{O}_{4}) = 0.00138105 \, \mathrm{moles} / 5.00 \, \mathrm{mL} \times (1\, \mathrm{L}/1000\, \mathrm{mL}) = 0.27621\, \mathrm{M}\]
05

Find Mass of \(\mathrm{Na}_{2}\mathrm{C}_{2} \mathrm{O}_{4}\)

From the concentration, the mass of sodium oxalate in 1 L of solution can be found. First, we need to convert the molarity to moles and then to grams using the molar mass (Mm) of sodium oxalate, which is 133.9983 g/mol:\[ Mass = M \times V \times Mm = 0.27621 \, \mathrm{M} \times 1.00 \, \mathrm{L} \times 133.9983 \, \mathrm{g/mol} = 36.992 g\] Note that the volume used is 1L, which is the volume of solution asked in the problem.

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Most popular questions from this chapter

Warfarin, \(\mathrm{C}_{19} \mathrm{H}_{16} \mathrm{O}_{4},\) is the active ingredient used in some anticoagulant medications. The amount of warfarin in a particular sample was determined as follows. A 13.96 g sample was first treated with an alkaline I_ solution to convert \(\mathrm{C}_{19} \mathrm{H}_{16} \mathrm{O}_{4}\) to \(\mathrm{CHI}_{3}\). This treatment gives one mole of \(\mathrm{CHI}_{3}\) for every mole of \(\mathrm{C}_{19} \mathrm{H}_{16} \mathrm{O}_{4}\) that was initially present in the sample. The iodine in \(\mathrm{CHI}_{3}\) is then precipitated as \(\mathrm{AgI}(\mathrm{s})\) by treatment with excess \(\mathrm{AgNO}_{3}(\mathrm{aq}):\) $$\begin{aligned} \mathrm{CHI}_{3}(\mathrm{aq})+3 \mathrm{AgNO}_{3}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow & \longrightarrow 3 \mathrm{AgI}(\mathrm{s})+3 \mathrm{HNO}_{3}(\mathrm{aq}) &+\mathrm{CO}(\mathrm{g}) \end{aligned}$$ If \(0.1386 \mathrm{g}\) solid \(\mathrm{AgI}\) were obtained, then what is the percentage by mass of warfarin in the sample analyzed?

A \(110.520 \mathrm{g}\) sample of mineral water is analyzed for its magnesium content. The \(\mathrm{Mg}^{2+}\) in the sample is first precipitated as \(\mathrm{MgNH}_{4} \mathrm{PO}_{4},\) and this precipitate is then converted to \(\mathrm{Mg}_{2} \mathrm{P}_{2} \mathrm{O}_{7},\) which is found to weigh 0.0549 g. Express the quantity of magnesium in the sample in parts per million (that is, in grams of \(\mathrm{Mg}\) per million grams of \(\mathrm{H}_{2} \mathrm{O}\) ).

Chile saltpeter is a natural source of \(\mathrm{NaNO}_{3}\); it also contains \(\mathrm{NaIO}_{3} .\) The \(\mathrm{NaIO}_{3}\) can be used as a source of iodine. Iodine is produced from sodium iodate in a two-step process occurring under acidic conditions: \(\begin{aligned} \mathrm{IO}_{3}^{-}(\mathrm{aq})+\mathrm{HSO}_{3}^{-}(\mathrm{aq}) & \longrightarrow \mathrm{I}^{-}(\mathrm{aq}) +\mathrm{SO}_{4}^{2-}(\mathrm{aq}) \end{aligned} \quad\) ( not balanced) \(\mathrm{I}^{-}(\mathrm{aq})+\mathrm{IO}_{3}^{-}(\mathrm{aq}) \longrightarrow\) \(\mathrm{I}_{2}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \quad(\text { not balanced })\) In the illustration, a 5.00 L sample of a \(\mathrm{NaIO}_{3}(\mathrm{aq})\) solution containing \(5.80 \mathrm{g} \mathrm{NaIO}_{3} / \mathrm{L}\) is treated with the stoichiometric quantity of \(\mathrm{NaHSO}_{3}\) (no excess of either reactant). Then, a further quantity of the initial \(\mathrm{NaIO}_{3}(\mathrm{aq})\) is added to the reaction mixture to bring about the second reaction. (a) How many grams of NaHSO \(_{3}\) are required in the first step? (b) What additional volume of the starting solution must be added in the second step?

Complete and balance these half-equations. (a) \(\mathrm{SO}_{3}^{2-} \longrightarrow \mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) (acidic solution) (b) \(\mathrm{HNO}_{3} \longrightarrow \mathrm{N}_{2} \mathrm{O}(\mathrm{g})\) (acidic solution) (c) \(\mathrm{Al}(\mathrm{s}) \longrightarrow \mathrm{Al}(\mathrm{OH})_{4}^{-}\) (basic solution) Indicate whether oxidation or reduction is involved.

In your own words, define or explain the terms or symbols \((\mathrm{a}) \rightleftharpoons(\mathrm{b})[] ;(\mathrm{c})\) spectator ion; (d) weak acid.
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