The reaction of MnO 4 – with oxalic acid (H 2 C 2 O 4 ) in acidic solution, yielding Mn 2+ and CO 2 gas, is widely used to determine the concentration of permanganate solutions. (d) A 1.200 g sample of sodium oxalate (Na 2 C 2 O 4 ) is dissolved in dilute H 2 SO 4 and then titrated with a KMnO 4 solution. If 32.50 mL of the KMnO 4 solution is required to reach the equivalence point, what is the molarity of the KMnO 4 solution?

Hello everyone today. We are being asked the following question. The concentration of permanganate in an acidic solution can be determined by titrate in with arsenic tie tri oxide A point 1762g of arsenic trioxide was dissolved and dilute acidic solution. Then the resulting solution was tight treated with the potassium Permanganate solution. A total of 28.5 ml of the potassium permanganate solution was used to reach the end point, calculate the concentration of this potassium permanganate solution. So the first thing we need to do is we need to balance these half reactions. So first, the first step in doing that is we need to balance all of the non hydrogen and non oxygen's. So for our first we'll label this one, we label this as our second one. We don't need to do much with our second one so we can move on to our our first one so we can move on to our second one. We see here that we have our arsenic try outside. We have to arsenic on the left, but we only have one on the right. So what we must do is we must add to as a coefficient to the right hand side. Now we can move on to step two and step two is to balance the oxygen's by adding water. So we're gonna do that for both stages here for our permanganate. We have four oxygen's on the left, meaning on the right, we must add four waters for arsenic trioxide truck sign, We have three oxygen's on the left and four on the right. We have zero hydrogen on the left and two on the right. So what we can do is we could add five waters on the left to give us an equal amount of oxygen's Now. We move onto the 3rd step. The third step is to after balancing the oxygen's, we must then balance our hydrogen by adding protons for our permanganate half reaction. We must add a total of eight protons because there are eight on the right and so that's going to balance that out for us. And for arsenic trioxide, we are essentially Going to have to add four to the right. And so why is that? Well, we have 10 hydrogen on the left, we only have six on the right. So what we need to do to get from 6 to 10 is to add four. And lastly we just have to balance the charges. And to balance charges, we simply add electrons. So on the left hand side for our permanganate we have a positive eight charge overall. And on the right we have a positive to charge. So to get from a positive 8 to positive two plus this negative charge here. So essentially have a positive seven charge on the left, we have to add five electrons to the left hand side to balance that out for our arsenic trioxide, we see that we have a charge of zero on the left hand side. So what we need to do on the right hand side, which has a charge of positive four is we need to add four electrons. So we're going to add four electrons to the right hand side. Now we have to balance these half reactions. We actually have to balance our electrons, So balance charges of the electrons and balance the electrons themselves. So we have five electrons for our permanganate and we have four electrons for arsenic, a common multiplier factor of four and five is 20. So what we need to do is we need to multiply our entire half reaction for the permanganate by four and our entire half reaction of the arsenic trioxide by five. And this is going to give us the following data from the following reaction. And so now we have to simplify these into one general equation. So we see here that we have 20 electrons on the left here and 20 on the right hand on the other equation. So we can cross out our electrons. What else can we cross out? Well, We see that we have 16 Water for our manganese on the right and we have 25 on the left for our arsenic. So what we can do as we can say that if we cross that out, we'll have nine remaining waters for our arsenic trioxide equation side. We can also do the same for our hydrogen. So we have 32 on the left here and 20 on the right for our other equation. What we can do is we can subtract 20 from both. And we'll be left with 12 hydrogen or 12 protons on the left. And so when we simplify these two Equations into one, we get four of our permanganate Plus five of our arsenic try oxen Plus our 12 remaining protons Plus our nine water. And that's going to give us our for Manganese two plus an hour, 10 arsenic acid here. Now, the next thing we must do as you must find the molar mass of our arsenic trioxide. So how are we going to do that? Well, based on the periodic table, we can look at the mother mass for one arsenic ion and we have two of them. So we're gonna take two multiplied by that mass, which is 74.92 g per mole. We're gonna multiply that by oxygen's molar mass and we have three of them. So that's three times 16 g per mole. We're gonna end up with 197. g per mole. Next we need to find out how many moles of permanganate we have. So we take our mass that we were given for our arsenic trioxide, which is 0.1762 g over arsenic trioxide. And we multiply by its molar mass, which says that we have one mole Over arsenic trioxide is equal to 197.84 32 g and then we multiply by our multiple ratio. Which according to our chemical reaction that we solved for here, we can say that for every five moles of our arsenic trioxide, we have four moles of per Megane. When our units cancel that, we're left with the 7.125 times 10 to the negative 4th malls of permanganate. We can then use this to solve for our polarity of permanganate and minorities and units of moles per liter. So we're gonna use the value that we just sold for 7.125 times 10 to negative fourth bowls of permanganate Over our volume which was 28.5 middle leaders. We must convert that to leaders. So we used the conversion factor that one. Middle leaders equal to 10 to the negative third leaders. And once our units cancel out, We're left with a value of 0. polarity. And so since the polarity of our permanganate is going to be equal to that of our potassium permanganate, we can say that the concentration of our potassium permanganate is going to be the same value .250 molar. And with that we have answered the question overall, I hope that this helped. And until next time

Ch.19 - Electrochemistry

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McMurry 8th Edition

Ch.19 - Electrochemistry

Problem 155d

Chapter 19, Problem 155d

The reaction of MnO₄^– with oxalic acid (H₂C₂O₄) in acidic solution, yielding Mn²⁺ and CO₂ gas, is widely used to determine the concentration of permanganate solutions. (d) A 1.200 g sample of sodium oxalate (Na₂C₂O₄) is dissolved in dilute H₂SO₄ and then titrated with a KMnO₄ solution. If 32.50 mL of the KMnO₄ solution is required to reach the equivalence point, what is the molarity of the KMnO₄ solution?

Verified step by step guidance

Write the balanced chemical equation for the reaction between MnO4- and oxalic acid (H2C2O4) in acidic solution. This will help in determining the stoichiometry of the reaction.

Convert the mass of sodium oxalate (Na2C2O4) to moles using its molar mass. This step is crucial as it provides the amount of substance which reacts with the permanganate solution.

Using the stoichiometry from the balanced equation, calculate the moles of KMnO4 needed to react completely with the moles of sodium oxalate.

Convert the volume of KMnO4 solution used in the titration from mL to liters to facilitate the calculation of molarity.

Calculate the molarity of the KMnO4 solution using the formula: Molarity = Moles of solute / Volume of solution in liters.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions based on the balanced chemical equation. It allows us to determine the relationships between the amounts of substances consumed and produced. In this case, understanding the stoichiometric ratios between KMnO4 and oxalic acid is essential for calculating the molarity of the KMnO4 solution.

Titration

Titration is a laboratory technique used to determine the concentration of a solute in a solution by reacting it with a solution of known concentration. In this scenario, the KMnO4 solution is titrated against sodium oxalate, and the volume of KMnO4 used at the equivalence point is crucial for calculating its molarity. The equivalence point indicates that the amount of titrant added is stoichiometrically equivalent to the amount of substance being analyzed.

Molarity

Molarity is a measure of concentration defined as the number of moles of solute per liter of solution. It is expressed in moles per liter (mol/L). To find the molarity of the KMnO4 solution, one must calculate the number of moles of sodium oxalate used and relate it to the volume of KMnO4 solution at the equivalence point, applying the stoichiometric relationships from the balanced equation.

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Given the following standard reduction potentials at 25 °C, (a) balance the equation for the reaction of H2MoO4 with elemental arsenic in acidic solution to give Mo3+ and H3AsO4 and (b) calculate E° for this reaction.

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Textbook Question

The reaction of MnO₄^– with oxalic acid (H₂C₂O₄) in acidic solution, yielding Mn²⁺ and CO₂ gas, is widely used to determine the concentration of permanganate solutions. (a) Write a balanced net ionic equation for the reaction.

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Textbook Question

The reaction of MnO₄^– with oxalic acid (H₂C₂O₄) in acidic solution, yielding Mn²⁺ and CO₂ gas, is widely used to determine the concentration of permanganate solutions. (b) Use the data in Appendix D to calculate E° for the reaction. (c) Show that the reaction goes to completion by calculating the values of ∆G° and K at 25 °C. (H₂C₂O₄) in acidic solution, yielding Mn²⁺ and CO₂ gas, is widely used to determine the concentration of permanganate solutions.

Open Question

Calculate the standard reduction potential for Ba2+ (aq) + 2 e- -> Ba(s) given that ∆G° = 16.7 kJ for the reaction Ba2+(aq) + 2Cl-(aq) -> BaCl2(s). Use any necessary data from Appendices B and D.

Textbook Question

A concentration cell has the same half-reactions at the anode and cathode, but a voltage results from different concentrations in the two electrode compartments. (b) A similar cell has 0.10 M Cu2+ in both compartments. When a stoichiometric amount of ethylenediamine (NH2CH2CH2NH2) is added to one compartment, the measured cell potential is 0.179 V. Calculate the formation constant Kf for the complex ion Cu(NH2CH2CH2CH2)22+. Assume there is no volume change.

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Textbook Question

Consider the redox titration (Section 4.13) of 120.0 mL of 0.100 M FeSO4 with 0.120 M K2Cr2O7 at 25 °C, assuming that the pH of the solution is maintained at 2.00 with a suitable buffer. The solution is in contact with a platinum electrode and constitutes one half-cell of an electrochemical cell. The other half-cell is a standard hydrogen electrode. The two half-cells are connected with a wire and a salt bridge, and the progress of the titration is monitored by measuring the cell potential with a voltmeter. (a) Write a balanced net ionic equation for the titration reaction, assuming that the products are Fe3+ and Cr3+.

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