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Chapter 15: Problem 14

When \(25.00 \mathrm{~mL}\) of unknown were passed through a Jones reductor, molybdate ion \(\left(\mathrm{MoO}_{4}^{2-}\right)\) was converted into \(\mathrm{Mo}^{3+}\). The filtrate required \(16.43 \mathrm{~mL}\) of \(0.01033 \mathrm{M} \mathrm{KMnO}_{4}\) to reach the purple end point. $$ \mathrm{MnO}_{4}^{-}+\mathrm{Mo}^{3+} \rightarrow \mathrm{Mn}^{2+}+\mathrm{MoO}_{2}^{2+} $$ A blank required \(0.04 \mathrm{~mL}\). Balance the reaction and find the molarity of molybdate in the unknown.

Short Answer

Expert verified
The molarity of molybdate ion in the unknown is 0.0339 M.

Step by step solution

01

Write and Balance the Chemical Equation

First, we need to balance the given chemical equation. The equation provided is:\[\mathrm{MnO}_{4}^{-} + \mathrm{Mo}^{3+} \rightarrow \mathrm{Mn}^{2+} + \mathrm{MoO}_{2}^{2+}\]Balancing the charges, we use stoichiometry to balance the reaction while considering both the change in oxidation states and charge. Initially, balance the oxygen and then balance the charges by adding appropriate numbers of electrons. This leads us to the balanced equation:\[2\mathrm{MnO}_{4}^{-} + 10\mathrm{Mo}^{3+} + 8H^+ \rightarrow 2\mathrm{Mn}^{2+} + 5\mathrm{MoO}_{2}^{2+} + 4H_2O\]
02

Calculate Moles of Permanganate Used

Calculate the moles of permanganate used, accounting for the blank:Given:- Volume of \(\mathrm{KMnO}_4\) used for the titration (corrected for the blank) is: \[16.43\,\mathrm{mL} - 0.04\,\mathrm{mL} = 16.39\,\mathrm{mL}\]- Molarity of \(\mathrm{KMnO}_4\) is \(0.01033\, M\).Convert volume to liters and calculate moles of \(\mathrm{KMnO}_4\) used:\[\text{Moles of } \mathrm{KMnO}_4 = 0.01033\, M \times \frac{16.39}{1000} = 1.694 \times 10^{-4}\, \text{moles}\]
03

Determine Moles of MoO4^{2-} in the Sample

Using the stoichiometry of the balanced equation, determine the moles of \(\mathrm{Mo}^{3+}\) that reacted with \(\mathrm{KMnO}_4\):From the balanced equation, 2 moles of \(\mathrm{MnO}_4^{-}\) react with 10 moles of \(\mathrm{Mo}^{3+}\), so the molar ratio is 1:5.Calculate the moles of \(\mathrm{Mo}^{3+}\) reacted:\[\text{Moles of } \mathrm{Mo}^{3+} = 1.694 \times 10^{-4}\, ext{moles of } \mathrm{KMnO}_4 \times 5 = 8.47 \times 10^{-4}\, ext{moles}\]
04

Calculate Molarity of Molybdate Ion in the Sample

The number of moles of \(\mathrm{Mo}^{3+}\) originally came from \(\mathrm{MoO}_{4}^{2-}\). Therefore, molarity of molybdate ion \(\mathrm{MoO}_{4}^{2-}\) in the unknown solution is calculated by dividing moles by the volume of solution in liters.Convert the sample volume from mL to L:\[25.00\, \mathrm{mL} = 0.02500\, \mathrm{L}\]Calculate molarity:\[\text{Molarity of } \mathrm{MoO}_{4}^{2-} = \frac{8.47 \times 10^{-4}\, \text{moles}}{0.02500\, \text{L}} = 0.0339\, M\]

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Key Concepts

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

Stoichiometry
Stoichiometry is the branch of chemistry that deals with the quantitative relationships between the substances involved in a chemical reaction. It involves using the coefficients of a balanced chemical equation to determine the proportions of reactants and products. In the context of titration analysis, stoichiometry helps us figure out how many moles of one substance react with a known quantity of another.
  • For the given exercise, a key stoichiometric calculation involved the conversion between moles of permanganate ions and molybdate ions.
  • The balanced chemical equation indicates a stoichiometric ratio between \(\mathrm{MnO}_{4}^{-}\) and \(\mathrm{Mo}^{3+}\), which is crucial for calculating moles accurately.
Understanding stoichiometry allows chemists to determine yields and the purity of substances produced in reactions.
Chemical Equation Balancing
Balancing a chemical equation is essential because it ensures that the law of conservation of mass and charge is upheld. Every chemical reaction must have the same number of each type of atom on both sides of the equation. For reactions, especially redox reactions in titration analysis, correct balancing also addresses changes in oxidation states and charges.In the provided exercise, the reaction between permanganate and molybdate ions needed proper balancing:
  • Ensure all atoms, including oxygen and hydrogen, are accounted for by adjusting coefficients appropriately.
  • Consider the charges to add the correct number of electrons transferred in the reaction to balance.
  • The original exercise provided a balanced equation that reflects both mass and charge balance: \[2\mathrm{MnO}_{4}^{-} + 10\mathrm{Mo}^{3+} + 8H^+ \rightarrow 2\mathrm{Mn}^{2+} + 5\mathrm{MoO}_{2}^{2+} + 4H_2O\]\
This balance ensures that the calculations for stoichiometry and other analyses, such as molarity, are accurate.
Molarity Calculation
Molarity is a measure of the concentration of a solute in a solution, expressed in moles per liter. Calculating molarity is vital in titration analysis because it allows us to determine how much of a substance is present in a given volume of solution.In our exercise, the molarity calculation follows using:
  • The volume of \(\mathrm{KMnO}_4\) solution used (adjusted for a blank) and its known concentration to find moles of \(\mathrm{KMnO}_4\).
  • The stoichiometric ratio from the balanced equation indicates how many moles of another reactant (\(\mathrm{Mo}^{3+}\)) have reacted.
  • This leads to the calculation of molarity of \(\mathrm{MoO}_{4}^{2-}\) by dividing the moles of \(\mathrm{Mo}^{3+}\) by the volume of the unknown solution: \[\frac{8.47 \times 10^{-4}\, \text{moles}}{0.02500\, \text{L}} = 0.0339\, M\]\leading to the concentration of the molybdate ion.
Redox Reaction
Redox reactions, or oxidation-reduction reactions, are essential in many chemical processes. They involve the transfer of electrons between substances, leading to changes in oxidation states. In chemistries like titration analysis using permanganate, understanding the redox nature of reactions is fundamental.For this exercise:
  • Permanganate ion \(\mathrm{MnO}_{4}^{-}\) acts as an oxidizing agent, while \(\mathrm{Mo}^{3+}\) is reduced during the reaction.
  • Electrons are transferred, where \(\mathrm{MnO}_{4}^{-}\) is reduced to \(\mathrm{Mn}^{2+}\), and \(\mathrm{Mo}^{3+}\) changes to \(\mathrm{MoO}_{2}^{2+}\).
  • Proper balancing of this redox reaction is necessary not just for mass and charge, but to correctly interpret the electron transfer occurring.
Mastering redox reactions and their balancing in titration helps chemists accurately quantify and characterize reactants in a mixture.

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Most popular questions from this chapter

Calcium fluorapatite \(\left(\mathrm{Ca}_{10}\left(\mathrm{PO}_{4}\right)_{6} \mathrm{~F}_{2}, \mathrm{FM} 1008.6\right)\) laser crystals were doped with chromium to improve their efficiency. It was suspected that the chromium could be in the \(+4\) oxidation state. 1\. To measure the total oxidizing power of chromium in the material, a crystal was dissolved in \(2.9 \mathrm{M} \mathrm{HClO}_{4}\) at \(100^{\circ} \mathrm{C}\), cooled to \(20^{\circ} \mathrm{C}\), and titrated with standard \(\mathrm{Fe}^{2+}\), using \(\mathrm{Pt}\) and \(\mathrm{Ag} \mid \mathrm{AgCl}\) electrodes to find the end point. Chromium oxidized above the \(+3\) state should oxidize an equivalent amount of \(\mathrm{Fe}^{2+}\) in this step. That is, \(\mathrm{Cr}^{4+}\) would consume one \(\mathrm{Fe}^{2+}\), and each atom of \(\mathrm{Cr}^{6+}\) in \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) would consume three \(\mathrm{Fe}^{2+}\) : $$ \begin{aligned} \mathrm{Cr}^{4+}+\mathrm{Fe}^{2+} & \rightarrow \mathrm{Cr}^{3+}+\mathrm{Fe}^{3+} \\ \frac{1}{2} \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+3 \mathrm{Fe}^{2+} & \rightarrow \mathrm{Cr}^{3+}+3 \mathrm{Fe}^{3+} \end{aligned} $$ 2\. In a second step, the total chromium content was measured by dissolving a crystal in \(2.9 \mathrm{M} \mathrm{HClO}_{4}\) at \(100^{\circ} \mathrm{C}\) and cooling to \(20^{\circ} \mathrm{C}\). Excess \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\) and \(\mathrm{Ag}^{+}\)were then added to oxidize all chromium to \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\). Unreacted \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\) was destroyed by boiling, and the remaining solution was titrated with standard \(\mathrm{Fe}^{2+}\). In this step, each \(\mathrm{Cr}\) in the original unknown reacts with three \(\mathrm{Fe}^{2+}\). $$ \begin{aligned} \mathrm{Cr}^{x+} & \stackrel{\mathrm{S}_{2} \mathrm{O}_{8}^{2-}}{\longrightarrow} \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \\ \frac{1}{2} \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+3 \mathrm{Fe}^{2+} & \rightarrow \mathrm{Cr}^{3+}+3 \mathrm{Fe}^{3+} \end{aligned} $$ In step \(1,0.4375 \mathrm{~g}\) of laser crystal required \(0.498 \mathrm{~mL}\) of \(2.786 \mathrm{mM}\) \(\mathrm{Fe}^{2+}\) (prepared by dissolving \(\mathrm{Fe}\left(\mathrm{NH}_{4}\right)_{2}\left(\mathrm{SO}_{4}\right)_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) in \(2 \mathrm{M} \mathrm{HClO}_{4}\) ). In step \(2,0.1566 \mathrm{~g}\) of crystal required \(0.703 \mathrm{~mL}\) of the same \(\mathrm{Fe}^{2+}\) solution. Find the average oxidation number of \(\mathrm{Cr}\) in the crystal and find the total micrograms of \(\mathrm{Cr}\) per gram of crystal.

A \(50.00-\mathrm{mL}\) sample containing La \(^{3+}\) was treated with sodium oxalate to precipitate \(\mathrm{La}_{2}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\), which was washed, dissolved in acid, and titrated with \(18.04 \mathrm{~mL}\) of \(0.006363 \mathrm{M} \mathrm{KMnO}_{4}\). Calculate the molarity of \(\mathrm{La}^{3+}\) in the unknown.

A \(3.026-\mathrm{g}\) portion of a copper(II) salt was dissolved in a \(250-\mathrm{mL}\) volumetric flask. A \(50.0-\mathrm{mL}\) aliquot was analyzed by adding \(1 \mathrm{~g}\) of KI and titrating the liberated iodine with \(23.33 \mathrm{~mL}\) of \(0.04668 \mathrm{M}\) \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\). Find the weight percent of \(\mathrm{Cu}\) in the salt. Should starch indicator be added to this titration at the beginning or just before the end point?

Write balanced half-reactions in which \(\mathrm{MnO}_{4}^{-}\)acts as an oxidant at (a) \(\mathrm{pH}=0\); (b) \(\mathrm{pH}=10\); (c) \(\mathrm{pH}=15\).

The Kjeldahl analysis in Section 10-8 is used to measure the nitrogen content of organic compounds, which are digested in boiling sulfuric acid to decompose to ammonia, which, in turn, is distilled into standard acid. The remaining acid is then back-titrated with base. Kjeldahl himself had difficulty in 1880 discerning by lamplight the methyl red indicator end point in the back titration. He could have refrained from working at night, but instead he chose to complete the analysis differently. After distilling the ammonia into standard sulfuric acid, he added a mixture of \(\mathrm{KIO}_{3}\) and KI to the acid. The liberated iodine was then titrated with thiosulfate, using starch for easy end-point detection-even by lamplight. \({ }^{31}\) Explain how the thiosulfate titration is related to the nitrogen content of the unknown. Derive a relationship between moles of \(\mathrm{NH}_{3}\) liberated in the digestion and moles of thiosulfate required for titration of iodine.
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