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Chapter 15: Problem 17

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.

Short Answer

Expert verified
The molarity of \(\text{La}^{3+}\) in the solution is \(0.003830 \text{ M}\).

Step by step solution

01

Write the Balanced Chemical Reaction

First, write out the balanced chemical reaction for the reaction between the oxalate ion \( \text{C}_2\text{O}_4^{2-} \) and the permanganate ion \( \text{MnO}_4^{-} \):\[2\text{MnO}_4^{-} + 5\text{C}_2\text{O}_4^{2-} + 16\text{H}^{+} \rightarrow 2\text{Mn}^{2+} + 10\text{CO}_2 + 8\text{H}_2\text{O}\]
02

Calculate Moles of KMnO4 Used

Use the volume and molarity of the KMnO4 solution to find the moles of KMnO4 used in the titration:\[\text{moles KMnO}_4 = 0.01804 \text{ L} \times 0.006363 \text{ M} = 1.1489512 \times 10^{-4} \text{ moles}\]
03

Relate Moles of KMnO4 to Moles of C2O42-

From the balanced equation, 2 moles of \( \text{MnO}_4^{-} \) react with 5 moles of \( \text{C}_2\text{O}_4^{2-} \). Therefore, we can calculate the moles of \( \text{C}_2\text{O}_4^{2-} \):\[\text{moles C}_2\text{O}_4^{2-} = 1.1489512 \times 10^{-4} \times \frac{5}{2} = 2.872378 \times 10^{-4} \text{ moles}\]
04

Relate Moles of C2O42- to Moles of La3+

La^3+ is precipitated as \(\text{La}_2(\text{C}_2\text{O}_4)_3\), so use the stoichiometry from the formula: 3 moles of \(\text{C}_2\text{O}_4^{2-}\) react with 2 moles of \(\text{La}^{3+}\):\[\text{moles La}^{3+} = 2.872378 \times 10^{-4} \times \frac{2}{3} = 1.914919 \times 10^{-4} \text{ moles}\]
05

Calculate Molarity of La3+ in the Solution

Now, find the molarity of \(\text{La}^{3+}\) in the original 50.00 mL solution using the moles of \(\text{La}^{3+}\) calculated:\[\text{Molarity of La}^{3+} = \frac{1.914919 \times 10^{-4} \text{ moles}}{0.05000 \text{ L}} = 0.003830 \text{ M}\]

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Key Concepts

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

Stoichiometry
Stoichiometry is the cornerstone of calculating and understanding chemical reactions. It involves using balanced chemical equations to determine the relationships between reactants and products in a chemical reaction. In our exercise, stoichiometry helps us understand how many moles of one substance are required to react with a given number of moles of another substance.
For instance, we look at the balanced reaction between permanganate ions (\( \text{MnO}_4^- \)) and oxalate ions (\( \text{C}_2\text{O}_4^{2-} \)). The reaction is represented as follows:
  • 2 moles of permanganate ions react with 5 moles of oxalate ions.
  • These reactants then produce carbon dioxide (\( \text{CO}_2 \)), manganese ions (\( \text{Mn}^{2+} \)), and water (\( \text{H}_2\text{O} \)).
By using stoichiometry, we can relate the amount of one reactant or product to another, which is crucial for determining the amount of \( \text{La}^{3+} \) present in the sample. Stoichiometry is vital because it allows precise quantitative predictions in chemical reactions.
Titration
Titration is an analytical technique used to determine the concentration of an unknown solution by reacting it with a solution of known concentration. In this problem, we are titrating a dissolved precipitate that contains \( \text{La}^{3+} \) ions with a \( \text{KMnO}_4 \) solution.
The steps of titration include:
  • Adding the titrant (here, \( \text{KMnO}_4 \)) until the reaction is complete. The point at which this occurs is called the equivalence point.
  • Using the volume and molarity of the titrant, we can calculate the number of moles of the titrant used.
In our exercise, 18.04 mL of 0.006363 M \( \text{KMnO}_4 \) was required to reach the equivalence point with the oxalate present, which was initially precipitated from \( \text{La}^{3+} \) by forming \( \text{La}_2(\text{C}_2\text{O}_4)_3 \). This provides us important data to calculate the number of \( \text{C}_2\text{O}_4^{2-} \) ions present, which in turn helps determine the concentration of \( \text{La}^{3+} \) in the sample.
Chemical Reactions
Chemical reactions involve the transformation of substances through the breaking and forming of chemical bonds. In the context of quantitative analysis, these reactions are used to determine the composition and concentration of components in a mixture.
In the provided exercise, the chemical reactions play a pivotal role in finding the molarity of \( \text{La}^{3+} \) ions. This analysis involves:
  • Initially, \( \text{La}^{3+} \) ions react with oxalate ions to form a precipitate \( \text{La}_2(\text{C}_2\text{O}_4)_3 \).
  • Once precipitated, it is redissolved in acid and reacts again during titration with \( \text{KMnO}_4 \), involving a redox reaction where \( \text{MnO}_4^- \) oxidizes the \( \text{C}_2\text{O}_4^{2-} \) ions.
The balanced chemical equation provided is fundamental because it ensures that the stoichiometric calculations are based on the correct proportion of reactants. Understanding these reactions is key to accurately determining the concentration of \( \text{La}^{3+} \), highlighting the interrelationship between reactions in practical scenarios.

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Most popular questions from this chapter

(a) Potassium iodate solution was prepared by dissolving \(1.022 \mathrm{~g}\) of \(\mathrm{KIO}_{3}\) (FM 214.00) in a 500-mL volumetric flask. Then \(50.00 \mathrm{~mL}\) of the solution were pipetted into a flask and treated with excess \(\mathrm{KI}(2 \mathrm{~g})\) and acid \(\left(10 \mathrm{~mL}\right.\) of \(\left.0.5 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\right)\). How many moles of \(\mathrm{I}_{3}^{-}\)are created by the reaction? (b) The triiodide from part (a) reacted with \(37.66 \mathrm{~mL}\) of \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) solution. What is the concentration of the \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) solution? (c) A 1.223-g sample of solid containing ascorbic acid and inert ingredients was dissolved in dilute \(\mathrm{H}_{2} \mathrm{SO}_{4}\) and treated with \(2 \mathrm{~g}\) of \(\mathrm{KI}\) and \(50.00 \mathrm{~mL}\) of \(\mathrm{KIO}_{3}\) solution from part (a). Excess triiodide required \(14.22 \mathrm{~mL}\) of \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) solution from part (b). Find the weight percent of ascorbic acid (FM 176.13) in the unknown. (d) Does it matter whether starch indicator is added at the beginning or near the end point in the titration in part (c)?

Would tris \(\left(2,2^{\prime}\right.\)-bipyridine \()\) iron be a useful indicator for the titration of \(\mathrm{Sn}^{2+}\) with Mn(EDTA) \(^{-}\)? (Hint: The potential at the equivalence point must be between the potentials for each redox couple.)

Two possible reactions of \(\mathrm{MnO}_{4}^{-}\)with \(\mathrm{H}_{2} \mathrm{O}_{2}\) to produce \(\mathrm{O}_{2}\) and \(\mathrm{Mn}^{2+}\) are Scheme 1: \(\mathrm{MnO}_{4}^{-} \rightarrow \mathrm{Mn}^{2+}\) \(\mathrm{H}_{2} \mathrm{O}_{2} \rightarrow \mathrm{O}_{2}\) Scheme 2: \(\mathrm{MnO}_{4}^{-} \rightarrow \mathrm{O}_{2}+\mathrm{Mn}^{2+}\) \(\mathrm{H}_{2} \mathrm{O}_{2} \rightarrow \mathrm{H}_{2} \mathrm{O}\) (a) Complete the half-reactions for both schemes by adding \(\mathrm{e}^{-}\), \(\mathrm{H}_{2} \mathrm{O}\), and \(\mathrm{H}^{+}\)and write a balanced net equation for each scheme. (b) Sodium peroxyborate tetrahydrate, \(\mathrm{NaBO}_{3} \cdot 4 \mathrm{H}_{2} \mathrm{O}\) (FM 153.86), produces \(\mathrm{H}_{2} \mathrm{O}_{2}\) when dissolved in acid: \(\mathrm{BO}_{3}^{-}+2 \mathrm{H}_{2} \mathrm{O} \rightarrow\) \(\mathrm{H}_{2} \mathrm{O}_{2}+\mathrm{H}_{2} \mathrm{BO}_{3}^{-}\). To decide whether Scheme 1 or Scheme 2 occurs, students at the U.S. Naval Academy \({ }^{30}\) weighed \(1.023 \mathrm{~g} \mathrm{NaBO}_{3}\). \(4 \mathrm{H}_{2} \mathrm{O}\) into a \(100-\mathrm{mL}\) volumetric flask, added \(20 \mathrm{~mL}\) of \(1 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\), and diluted to the mark with \(\mathrm{H}_{2} \mathrm{O}\). Then they titrated \(10.00 \mathrm{~mL}\) of this solution with \(0.01046 \mathrm{M} \mathrm{KMnO}_{4}\) until the first pale pink color persisted. How many \(\mathrm{mL}\) of \(\mathrm{KMnO}_{4}\) are required in Scheme 1 and in Scheme \(2 ?\) (The Scheme 1 stoichiometry was observed.)

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.

Nitrite \(\left(\mathrm{NO}_{2}^{-}\right)\)can be determined by oxidation with excess \(\mathrm{Ce}^{4+}\), followed by back titration of the unreacted \(\mathrm{Ce}^{4+}\). A \(4.030-\mathrm{g}\) sample of solid containing only \(\mathrm{NaNO}_{2}\) (FM 68.995) and \(\mathrm{NaNO}_{3}\) was dissolved in \(500.0 \mathrm{~mL}\). A \(25.00-\mathrm{mL}\) sample of this solution was treated with \(50.00 \mathrm{~mL}\) of \(0.1186 \mathrm{M} \mathrm{Ce}^{4+}\) in strong acid for \(5 \mathrm{~min}\), and the excess \(\mathrm{Ce}^{4+}\) was back- titrated with \(31.13 \mathrm{~mL}\) of \(0.04289 \mathrm{M}\) ferrous ammonium sulfate. $$ \begin{aligned} 2 \mathrm{Ce}^{4+}+\mathrm{NO}_{2}^{-}+\mathrm{H}_{2} \mathrm{O} & \rightarrow 2 \mathrm{Ce}^{3+}+\mathrm{NO}_{3}^{-}+2 \mathrm{H}^{+} \\ \mathrm{Ce}^{4+}+\mathrm{Fe}^{2+} & \rightarrow \mathrm{Ce}^{3+}+\mathrm{Fe}^{3+} \end{aligned} $$ Calculate the weight percent of \(\mathrm{NaNO}_{2}\) in the solid.
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