Need any help Experiment 26 Data and Calculations The Solubi
Need any help!
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
