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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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Redox Reaction
In a redox reaction, electrons are transferred between two substances. The substance that donates electrons is oxidized, while the one that accepts electrons is reduced. In the titration provided, the oxalate ion (\(\mathrm{C_{2}O_{4}^{2-}}\)) donates electrons and gets oxidized, whereas the permanganate ion (\(\mathrm{MnO_{4}^{-}}\)) accepts electrons and is reduced. This transfer of electrons results in the conversion of \(\mathrm{C_{2}O_{4}^{2-}}\) to \(\mathrm{CO_{2}}\) gas, and \(\mathrm{MnO_{4}^{-}}\) to \(\mathrm{Mn^{2+}}\) ions. Redox reactions are vital in understanding numerous chemical processes such as energy production. These reactions are characterized by changes in oxidation states, highlighting the importance of keeping track of electron flow.
Stoichiometry
Stoichiometry involves calculating the relative quantities of reactants and products involved in a chemical reaction. It is based on the law of conservation of mass, where the amount of each element in the reactants equals that in the products. In this exercise, stoichiometry is used in the balanced chemical equation \(5 \mathrm{C}_{2} \mathrm{O}_{4}^{2-} + 2 \mathrm{MnO}_{4}^{-}\), which indicates that 5 moles of oxalate react with 2 moles of permanganate. From stoichiometry, after finding moles of \(\mathrm{KMnO}_{4}\), it relates directly to determine moles of \(\mathrm{Na}_{2}\mathrm{C}_{2} \mathrm{O}_{4}\) through a ratio of \(\frac{5}{2}\). The ability to convert between moles using molar ratios is essential to determine the amounts required for reactions.
Molarity Calculation
Molarity is a measure of the concentration of a solute in a solution. It is defined as the number of moles of solute per liter of solution (\(\mathrm{M} = \text{moles/L}\)). Calculating molarity gives an idea of how much solute is present in the solution. In this exercise, we calculate the molarity of sodium oxalate in the initial 5.00 mL sample, determined as \(0.27621\, \mathrm{M}\).To find it, use the formula \(M = \frac{n}{V}\), where \(n\) is moles of solute and \(V\) is volume in liters. With this data, you can deduce further calculations or conversions needed to solve related problems.
Chemical Equations Balancing
Balancing chemical equations is crucial for accurately representing chemical reactions. It involves making sure the number of atoms of each element is the same on both sides of the equation, which preserves mass adherent to the law of conservation of mass. For the initial unbalanced equation, \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-} + \mathrm{MnO}_{4}^{-} \rightarrow \mathrm{Mn}^{2+} + \mathrm{CO}_{2}+\ldots\), we end up with the balanced equation given in Step 1:\[5 \mathrm{C}_{2} \mathrm{O}_{4}^{2-} + 2 \mathrm{MnO}_{4}^{-} + 16 \mathrm{H}^{+} \rightarrow 10 \mathrm{CO}_{2} + 2 \mathrm{Mn}^{2+} + 8 \mathrm{H}_{2} \mathrm{O}\]Prediction and understanding of this balancing provide accuracy not only in academic scenarios but also in lab settings where exact ratios are required for experiments.

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Most popular questions from this chapter

A compound contains only Fe and O. A \(0.2729 \mathrm{g}\) sample of the compound was dissolved in \(50 \mathrm{mL}\) of concentrated acid solution, reducing all the iron to \(\mathrm{Fe}^{2+}\) ions. The resulting solution was diluted to \(100 \mathrm{mL}\) and then titrated with a \(0.01621 \mathrm{M} \mathrm{KMnO}_{4}\) solution. The unbalanced chemical equation for reaction between \(\mathrm{Fe}^{2+}\) and \(\mathrm{MnO}_{4}^{-}\) is given below. \(\begin{aligned} \mathrm{MnO}_{4}^{-}(\mathrm{aq})+& \mathrm{Fe}^{2+}(\mathrm{aq}) \longrightarrow \mathrm{Mn}^{2+}(\mathrm{aq})+\mathrm{Fe}^{3+}(\mathrm{aq}) \quad(\text { not balanced }) \end{aligned}\) The titration required \(42.17 \mathrm{mL}\) of the \(\mathrm{KMnO}_{4}\) solution to reach the pink endpoint. What is the empirical formula of the compound?

Explain why these reactions cannot occur as written. (a) \(\mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{MnO}_{4}^{-}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq}) \longrightarrow\) \(\mathrm{Mn}^{2+}(\mathrm{aq})+\mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(1)\) (b) \(\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq})+\mathrm{Cl}_{2}(\mathrm{aq}) \longrightarrow\) \(\mathrm{ClO}^{-}(\mathrm{aq})+\mathrm{O}_{2}(\mathrm{g})+\mathrm{H}^{+}(\mathrm{aq})\)

Balance these equations for disproportionation reactions. (a) \(\mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow \mathrm{Cl}^{-}+\mathrm{ClO}_{3}^{-}\) (basic solution) (b) \(\mathrm{S}_{2} \mathrm{O}_{4}^{2-} \longrightarrow \mathrm{S}_{2} \mathrm{O}_{3}^{2-}+\mathrm{HSO}_{3}^{-}\) (acidic solution)

Phosphorus is essential for plant growth, but an excess of phosphorus can be catastrophic in aqueous ecosystems. Too much phosphorus can cause algae to grow at an explosive rate and this robs the rest of the ecosystem of oxygen. Effluent from sewage treatment plants must be treated before it can be released into lakes or streams because the effluent contains significant amounts of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\mathrm{HPO}_{4}^{2-}\). (Detergents are a major contributor to phosphorus levels in domestic sewage because many detergents contain \(\mathrm{Na}_{2} \mathrm{HPO}_{4}\) ) A simple way to remove \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\mathrm{HPO}_{4}^{2-}\) from the effluent is to treat it with lime, \(\mathrm{CaO}\) which produces \(\mathrm{Ca}^{2+}\) and \(\mathrm{OH}^{-}\) ions in water. The \(\mathrm{OH}^{-}\) ions convert \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\mathrm{HPO}_{4}^{2-}\) ions into \(\mathrm{PO}_{4}^{3-}\) ions and, finally, \(\mathrm{Ca}^{2+}, \mathrm{OH}^{-}\), and \(\mathrm{PO}_{4}^{3-}\) ions combine to form a precipitate of \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH}(\mathrm{s})\) (a) Write balanced chemical equations for the four reactions described above. [Hint: The reactants are \(\mathrm{CaO}\) and \(\mathrm{H}_{2} \mathrm{O} ; \mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\left.\mathrm{OH}^{-} ; \mathrm{HPO}_{4}^{2-} \text { and } \mathrm{OH}^{-} ; \mathrm{Ca}^{2+}, \mathrm{PO}_{4}^{3-}, \text { and } \mathrm{OH}^{-} .\right]\) (b) How many kilograms of lime are required to remove the phosphorus from a \(1.00 \times 10^{4}\) L holding tank filled with contaminated water, if the water contains \(10.0 \mathrm{mg}\) of phosphorus per liter?

Balance these equations for redox reactions in basic solution. (a) \(\mathrm{MnO}_{2}(\mathrm{s})+\mathrm{ClO}_{3}^{-} \longrightarrow \mathrm{MnO}_{4}^{-}+\mathrm{Cl}^{-}\) (b) \(\mathrm{Fe}(\mathrm{OH})_{3}(\mathrm{s})+\mathrm{OCl}^{-} \longrightarrow \mathrm{FeO}_{4}^{2-}+\mathrm{Cl}^{-}\) (c) \(\mathrm{ClO}_{2} \longrightarrow \mathrm{ClO}_{3}^{-}+\mathrm{Cl}\) (d) \(\mathrm{Ag}(\mathrm{s})+\mathrm{CrO}_{4}^{2-} \rightarrow \mathrm{Ag}^{+}+\mathrm{Cr}(\mathrm{OH})_{3}(\mathrm{s})\)

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