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• Hey Leland, I´m not sure whats going on in AP chem right now. Can you help me with ion molarity?﻿
• I´d love to help. What´s the problem?﻿
• 250.ml of 5.53M ferrous nitrate is mixed with 250.ml of ﻿4.75M potassium phosphate. Calculate ion molarity.
• Alright so first you´re going to have to figure out your balanced equation.
• In this case its : 3Fe(NO3)2 (aq) + 2KPO4 (aq) →Fe3(PO4)2(s) + 6KNO3 (aq)﻿
• Then you want to determine which reactant is limiting. To do this you need to stoich it. Fe(NO3)2: 250.ml/1000ml x 1L/1L x 5.53mol / 3 mol x 1 mole Fe3(PO4)2 = .460 mol Fe3(PO4)2 made LR K3PO4: 250.ml/1000ml x 1L/1L x 4.75 mol / 2 mol x 1 mol Fe3(PO4)2 = .593 mol Fe3(PO4)2 made ER
• Now we know which ion is 0M. You can figure this out by looking at the limiting reactant and which ever ion is in both the LR and solid is 0M because, it all precipitated.﻿
• K+ and NO3- are the spectator ions 1.38 mol Fe(NO3)2/1mol x 2 mol NO3- = 2.76 mol/.5L = 5.52M NO3- 1.18 mol K3PO4/1mol x 3 mol K+ = 3.54 mol/.5L = 7.08M K=
• Last step is the easiest. First find how much precipitate was never made from our previous info. .593 mol - .460 mol = .113 mol Fe3(PO4)2 not made. .113 mol/1 mol x 2 mol = .266 mol PO43-/.5L = .532M PO43-﻿
• Now lets solve for our spectators. To do so you must combine volumes and find moles of both and stoich to moles of the sole ion.
• OH. It all makes sense now. Thanks for the help Leland you´re a hero.
• Anytime!
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