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What is the role of sodium and chloride ions in this experiment

What is the role of sodium and chloride ions in this experiment

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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What is the role of sodium and chloride ions in this experiment 

9:51 File Details CHEM 121L - ALL SPRING 20... OBJECTIVES: To explore the stoichiometry of a chemical equation To perform a gravimetric analysis To run a double replacement reaction To work with solution molarities To determine the molarity of an unknown solution. INTRODUCTION: The reaction that is being explored in this lab is the following double replacement. 3 СаCl(aq) + 2 NasPO4(aq) > Саз(РО4)2(8) + 6 NaC{aq) calcium chloride + sodium phosphate → calcium phosphate + sodium chloride You will run this reaction in the lab and recover and weigh the white calcium phosphate that is formed. In general, this approach to studying a reaction that forms a solid is called gravimetric analysis. Each of the two reactants is an ionic compound in aqueous solution. This means that when you first draw the samples of sodium phosphate and calcium chloride from the reagent bench for one of your trials, each will contain an anion and a cation dissolved in water. 3 CaCl2(aq) + 2 Na3PO4(aq) → Ca3(PO4)2(s) + 6 NaCI(aq) Limiting Reagent Excess Reagent White Solid Ions in solution For the reaction you will mix the two solutions, as you mix these solutions you intermingle the ions present and the anions and cations' have the opportunity to switch partners. This is the essence of a double replacement reaction. If one of the new pairings of ions is able to combine to form a solid or a gas or something that will show a net change, then a reaction will occur. In this particular reaction, the pairing of Nat and Cl gives not net change, but the pairing of Ca2+ and PO4 forms an insoluble solid and a precipitate? Ca3(PO4)2 (s) forms. In this experiment, you will prepare a 0.100 M Na3PO4, by dissolving the appropriate amount of solid NazPO4 in a 100.0 mL volumetric flask to make the 0.100 M Na3PO4 solution. You can determine the amount of solid you need by multiplying the volume in liters times the molarity to Anions are negative ions. Cations are positive ions A Precipitate is simply a solid product that forms in a reaction. Any evidence of a solid forming, such as cloudiness, is evidence of precipitation. The solid doesn't have to fall to the bottom of the container to be a precipitate. Page 2 of 8 Experiment 6 get the amount of moles then use the molar mass to determine the amount of solid you should weigh out. Moles = V(L) X Molarity and Mass to weigh out = Moles x Molar Mass Next you will mix 30.0 mL of 0.100 M Na3PO4 with 20.0 mL of a CaCl2 solution of unknown molarity. The molarity of your calcium chloride solution will be low enough that the calcium chloride will function as the limiting reactant in this reaction and the sodium phosphate will remain in excess when the reaction is complete. The amount of CaCl2 present in your 20.0 mL of solution will control the amount of calcium phosphate precipitate formed. By determining the mass and then the moles of precipitate present, it will be possible to calculate the number of moles CaCl2 present in the original limiting reactant solution of unknown concentration by using the stoichiometry of the chemical equation. Then using the definition of molarity, you can calculate the molarity of your unknown CaCl2 solution. Molarity of calcium chloride solution = Moles of calcium chloride in your sample Liters of solution in your sample ( Previous Next > 15 Dashboard Calendar Тo Do Notifications Inbox
Transcribed Image Text:9:51 File Details CHEM 121L - ALL SPRING 20... OBJECTIVES: To explore the stoichiometry of a chemical equation To perform a gravimetric analysis To run a double replacement reaction To work with solution molarities To determine the molarity of an unknown solution. INTRODUCTION: The reaction that is being explored in this lab is the following double replacement. 3 СаCl(aq) + 2 NasPO4(aq) > Саз(РО4)2(8) + 6 NaC{aq) calcium chloride + sodium phosphate → calcium phosphate + sodium chloride You will run this reaction in the lab and recover and weigh the white calcium phosphate that is formed. In general, this approach to studying a reaction that forms a solid is called gravimetric analysis. Each of the two reactants is an ionic compound in aqueous solution. This means that when you first draw the samples of sodium phosphate and calcium chloride from the reagent bench for one of your trials, each will contain an anion and a cation dissolved in water. 3 CaCl2(aq) + 2 Na3PO4(aq) → Ca3(PO4)2(s) + 6 NaCI(aq) Limiting Reagent Excess Reagent White Solid Ions in solution For the reaction you will mix the two solutions, as you mix these solutions you intermingle the ions present and the anions and cations' have the opportunity to switch partners. This is the essence of a double replacement reaction. If one of the new pairings of ions is able to combine to form a solid or a gas or something that will show a net change, then a reaction will occur. In this particular reaction, the pairing of Nat and Cl gives not net change, but the pairing of Ca2+ and PO4 forms an insoluble solid and a precipitate? Ca3(PO4)2 (s) forms. In this experiment, you will prepare a 0.100 M Na3PO4, by dissolving the appropriate amount of solid NazPO4 in a 100.0 mL volumetric flask to make the 0.100 M Na3PO4 solution. You can determine the amount of solid you need by multiplying the volume in liters times the molarity to Anions are negative ions. Cations are positive ions A Precipitate is simply a solid product that forms in a reaction. Any evidence of a solid forming, such as cloudiness, is evidence of precipitation. The solid doesn't have to fall to the bottom of the container to be a precipitate. Page 2 of 8 Experiment 6 get the amount of moles then use the molar mass to determine the amount of solid you should weigh out. Moles = V(L) X Molarity and Mass to weigh out = Moles x Molar Mass Next you will mix 30.0 mL of 0.100 M Na3PO4 with 20.0 mL of a CaCl2 solution of unknown molarity. The molarity of your calcium chloride solution will be low enough that the calcium chloride will function as the limiting reactant in this reaction and the sodium phosphate will remain in excess when the reaction is complete. The amount of CaCl2 present in your 20.0 mL of solution will control the amount of calcium phosphate precipitate formed. By determining the mass and then the moles of precipitate present, it will be possible to calculate the number of moles CaCl2 present in the original limiting reactant solution of unknown concentration by using the stoichiometry of the chemical equation. Then using the definition of molarity, you can calculate the molarity of your unknown CaCl2 solution. Molarity of calcium chloride solution = Moles of calcium chloride in your sample Liters of solution in your sample ( Previous Next > 15 Dashboard Calendar Тo Do Notifications Inbox
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