reagent boules for varying pro- In this experiment out thesc solutions different
ID: 509902 • Letter: R
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reagent boules for varying pro- In this experiment out thesc solutions different conditions, with their iodate portions of the we will precipitate solid Baao., under several for we can reagents used. We the andr the prepared, equilibrium, and from those values and the way the mixtures were the concentrations of the barium and iodate ions, and the value in your test Label five medium (18 x 150 with labels or by noting their location mm) dry test tubes, either To Test Tube 1. carefully 3, 4. and 5 mL reagent casiest Tubes 2, 3, pipet 1.00 mL so M Kuo, 2, for these additions. to 1.00. In way to do and 5, respectively. graduated pipet level fall and Tube 2, this is to to the o 00 your the pipet in Test Tube 1, let the fall to 600, the level You let level. With let in Tube 4 the level fall to 3.00, thereby adding 2 the Tube. In Tube 3. looo) pipet let it go to 10.00. Refill ml. to Tube 5 let it fall to pipet Finally, tube 6, Then, to each of the test the pipet to the 5.00 level and in 5-mL each 5, tubes add 500 mL of M Bacwo,,, using your final volume in and 2 mL 0.0200 The should be 12.00 distilled water into TestTubes 1 through 5, respectively. 26.1. of mL The compositions of the summarized in Table mixtures are Table 20.1 volumes of Reagents used in Precipitating Baool Test Tube 0.0350 M KIO 0.0200 M Ba(NO, with a stirring rod stireach selution for,o seconds up and downaswellas swirling the todo paper towel before proceeding to the next solution Touch the wall of the tube occasionally as you settles Look at each tube. In most of them you should see a white precipitate, which slowly crystalline out. It will be most noticeable in the higher number tubes. should be present as a light cloudiness in Test Tubes 1 and 2. There will only be auny amount of precipitate, only a few mg, so the amount is not impressive. It is most easily seen if you look through the tube toward the light, where you may observe tiny particles, very slowly settling out. The occurs slowly, so give time. slowly rotate the tubes so that Put a cork in each of the tubes in which is a and and help bring the system to the liquid goes from end to end twenty times or so to ensure thorough mixing (For any tubes in which you observe only a clear liquid, and no solid, repeat the stinring operation. scratching the walls of the tube as you stir, which may get things staned. If that still doesn't do the job, touch your stirring rod to the sample of pure solid barium iodate that we have in the lab and put the rod in the reluc- tant liquid. Insert a cork stopper in the tube and shake with vigor. Sooner or later you will get the cloudiness Repeat the rotation step, tuming each tube end to end to do what you can to make the liquid homoge- neous and bring it to equilibrium with the precipi You can't oversti, so be Give minutes with this. It will be time well spent. precipitate particles settle out completely, Finally, let the test tubes stand for 15 minutes to let the solid leaving a clear solution above the solids.Explanation / Answer
The initial concentrations of IO3- is given by the initial volume and concentration of KIO3 added to each test tube. The final concentration (equilibrium concentration) of KIO3 in each test tube is obtained from the absorbance data.
The final volume of the equilibrium solutions = 12 mL.
Remember that 1 mL = 0.001 L.
Construct the following table.
Test tube
Volume of IO3- added (mL)
Moles of IO3- added initially
Use the relation
Number of moles = (concentration in mol/L)*(volume added in L)
Moles of IO3- present at equilibrium = (concentration of IO3- at equilibrium)*(volume of solution in L)
Change in number of moles of IO3-, = (moles of IO3- initially added) – (moles of IO3- present at equilibrium)
1
1
3.5*10-5
1.296*10-5
2.204*10-5
2
2
7.0*10-5
1.524*10-5
5.476*10-5
3
3
1.05*10-4
1.272*10-5
9.228*10-5
4
4
1.40*10-4
1.344*10-5
1.2656*10-4
5
5
1.75*10-4
1.488*10-5
1.6012*10-4
Write down the dissociation of Ba(IO3)2 as below:
Ba(IO3)2 (s) <=====> Ba2+ (aq) + 2 IO3- (aq)
Look at the 1:2 nature of dissociation of the compound; the moles of Ba2+ will be equal to half the moles of IO3- at equilibrium.
Construct the next table.
Test tube
Volume of Ba2+ added (mL)
Moles of Ba2+ added = (volume of Ba2+ in L)*(concentration of Ba2+ in mol/L)
Moles of Ba2+ present at equilibrium = ½*(moles of IO3- present at equilibrium)
Change in number of moles of Ba2+, = (moles of Ba2+ initially added) – (moles of Ba2+ present at equilibrium)
1
5
1.75*10-4
6.48*10-6
1.6852*10-4
2
5
1.75*10-4
7.62*10-6
1.6738*10-4
3
5
1.75*10-4
6.36*10-6
1.6864*10-4
4
5
1.75*10-4
6.72*10-6
1.6828*10-4
5
5
1.75*10-4
7.44*10-6
1.6756*10-4
Next fill out the table for mixture 1.
[Ba2+]
[IO3-]
Initial = (moles added)/(total volume of solution in L)
[total volume of solution = 12 mL = 0.012 L]
(1.75*10-4 mole)/(0.012 L) = 0.01458 M
(3.5*10-5 mole)/(0.012 L) = 2.9167*10-3 M
= (moles reacted)/(total volume of solution in L)
(1.6852*10-4 mole)/(0.012 L) = 0.01404 M
(2.204*10-5 mole)/(0.012 L) = 1.8367*10-3 M
Equilibrium = (moles at equilibrium)/(total volume of solution)
(6.48*10-6 mole)/(0.012 L) = 5.4*10-4 M
(1.296*10-5 mole)/(0.012 L) = 1.08*10-3 M
Ksp
The expression for Ksp is given as
Ksp = [Ba2+][IO3-]2 = (5.4*10-4)*(1.08*10-3)2 = 6.29856*10-10 6.3*10-10
Test tube
Volume of IO3- added (mL)
Moles of IO3- added initially
Use the relation
Number of moles = (concentration in mol/L)*(volume added in L)
Moles of IO3- present at equilibrium = (concentration of IO3- at equilibrium)*(volume of solution in L)
Change in number of moles of IO3-, = (moles of IO3- initially added) – (moles of IO3- present at equilibrium)
1
1
3.5*10-5
1.296*10-5
2.204*10-5
2
2
7.0*10-5
1.524*10-5
5.476*10-5
3
3
1.05*10-4
1.272*10-5
9.228*10-5
4
4
1.40*10-4
1.344*10-5
1.2656*10-4
5
5
1.75*10-4
1.488*10-5
1.6012*10-4
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