Need help Experiment 26 Data and Calculations The Solubility

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Experiment 26 Data and Calculations: The Solubility Product of Ba(lOsh The calculations here are similar to those in Experiment 23, and are outlined in the Experimental Procedure section. To use a spreadsheet, set up a table like the one below. Test Tube Number mL 0.0350 M KIO, 23S mL 0.0200 M Ba(NO) Total volume in mL Absorbance of soln. Lar 110,-I in soln. Processing the Data Initial moles Ba Initial moles 10, Moles 10, in equil. soln. (12 mL) Moles IO precipitated (continued on following page)

Solution

calculation with experiment #1 )

moles IO3- in equilibrium state=mol of [IO3-] in solution [Ba(IO3)2(s) <---->Ba2+ +2IO3-]

      =(1.2*10^-3 mol/L *total volume of solution)

=(1.2*10^-3 mol/L * 0.012L)=1.44*10^-5 mol

mol of IO3- precipitatd=initial mol-moles IO3- in equilibrium solution=(3.5*10^-5 mol) - (1.44*10^-5 mol)=2.06*10^-5 mol

mol Ba2+ precipitated=1/2*(2.06*10^-5)=1.03*10^-5mol

mol Ba2+ in equilibrium=initial mol of Ba2+ - mol Ba2+ precipitated=1.0*10^-4 mol-1.03*10^-5mol=0.897*10^-4 mol

[ Ba2+] in equilibrium=0.897*10^-4 mol/0.012L=0.00747mol/L

ksp (Ba(IO3)2)=[Ba2+]eq[IO3-]eq^2=(0.00747mol/L)(1.2*10^-3 mol/L)^2=1.076*10^-8 M^2