What is the role of sodium and chloride ions in this experiment 9:51File DetailsCHEM 121L - ALL SPRING 20...OBJECTIVES:To explore the stoichiometry of a chemical equationTo perform a gravimetric analysisTo run a double replacement reactionTo work with solution molaritiesTo 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 chlorideYou will run this reaction in the lab and recover and weigh the white calcium phosphate that isformed. In general, this approach to studying a reaction that forms a solid is called gravimetricanalysis.Each of the two reactants is an ionic compound in aqueous solution. This means that when youfirst draw the samples of sodium phosphate and calcium chloride from the reagent bench for oneof 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 ReagentExcess ReagentWhite SolidIons in solutionFor the reaction you will mix the two solutions, as you mix these solutions you intermingle theions present and the anions and cations' have the opportunity to switch partners. This is theessence of a double replacement reaction. If one of the new pairings of ions is able to combineto form a solid or a gas or something that will show a net change, then a reaction will occur. Inthis 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 ofsolid NazPO4 in a 100.0 mL volumetric flask to make the 0.100 M Na3PO4 solution. You candetermine the amount of solid you need by multiplying the volume in liters times the molarity toAnions are negative ions. Cations are positive ionsA Precipitate is simply a solid product that forms in a reaction. Any evidence of a solid forming, such ascloudiness, is evidence of precipitation. The solid doesn't have to fall to the bottom of the container to be aprecipitate.Page 2 of 8Experiment 6get the amount of moles then use the molar mass to determine the amount of solid you shouldweigh out.Moles = V(L) X MolarityandMass to weigh out = Moles x Molar MassNext you will mix 30.0 mL of 0.100 M Na3PO4 with 20.0 mL of a CaCl2 solution of unknownmolarity. The molarity of your calcium chloride solution will be low enough that the calciumchloride will function as the limiting reactant in this reaction and the sodium phosphate willremain in excess when the reaction is complete. The amount of CaCl2 present in your 20.0 mLof solution will control the amount of calcium phosphate precipitate formed. By determining themass and then the moles of precipitate present, it will be possible to calculate the number ofmoles CaCl2 present in the original limiting reactant solution of unknown concentration by usingthe stoichiometry of the chemical equation. Then using the definition of molarity, you cancalculate the molarity of your unknown CaCl2 solution.Molarity of calcium chloride solution = Moles of calcium chloride in your sampleLiters of solution in your sample( PreviousNext 15DashboardCalendarТo DoNotificationsInbox

Question

What is the role of sodium and chloride ions in this experiment 9:51File DetailsCHEM 121L - ALL SPRING 20...OBJECTIVES:To explore the stoichiometry of a chemical equationTo perform a gravimetric analysisTo run a double replacement reactionTo work with solution molaritiesTo 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 chlorideYou will run this reaction in the lab and recover and weigh the white calcium phosphate that isformed. In general, this approach to studying a reaction that forms a solid is called gravimetricanalysis.Each of the two reactants is an ionic compound in aqueous solution. This means that when youfirst draw the samples of sodium phosphate and calcium chloride from the reagent bench for oneof 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 ReagentExcess ReagentWhite SolidIons in solutionFor the reaction you will mix the two solutions, as you mix these solutions you intermingle theions present and the anions and cations' have the opportunity to switch partners. This is theessence of a double replacement reaction. If one of the new pairings of ions is able to combineto form a solid or a gas or something that will show a net change, then a reaction will occur. Inthis 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 ofsolid NazPO4 in a 100.0 mL volumetric flask to make the 0.100 M Na3PO4 solution. You candetermine the amount of solid you need by multiplying the volume in liters times the molarity toAnions are negative ions. Cations are positive ionsA Precipitate is simply a solid product that forms in a reaction. Any evidence of a solid forming, such ascloudiness, is evidence of precipitation. The solid doesn't have to fall to the bottom of the container to be aprecipitate.Page 2 of 8Experiment 6get the amount of moles then use the molar mass to determine the amount of solid you shouldweigh out.Moles = V(L) X MolarityandMass to weigh out = Moles x Molar MassNext you will mix 30.0 mL of 0.100 M Na3PO4 with 20.0 mL of a CaCl2 solution of unknownmolarity. The molarity of your calcium chloride solution will be low enough that the calciumchloride will function as the limiting reactant in this reaction and the sodium phosphate willremain in excess when the reaction is complete. The amount of CaCl2 present in your 20.0 mLof solution will control the amount of calcium phosphate precipitate formed. By determining themass and then the moles of precipitate present, it will be possible to calculate the number ofmoles CaCl2 present in the original limiting reactant solution of unknown concentration by usingthe stoichiometry of the chemical equation. Then using the definition of molarity, you cancalculate the molarity of your unknown CaCl2 solution.Molarity of calcium chloride solution = Moles of calcium chloride in your sampleLiters of solution in your sample( PreviousNext >15DashboardCalendarТo DoNotificationsInbox

Accepted Answer

A spectator ion is an ion that exists as a reactant and a product in a chemical equation. and it…