Please help!!! When 35.0ml. of 0.05M Pb(N0_3)_2 are mixed with 15.0mL of 0.02M K
ID: 957558 • Letter: P
Question
Please help!!! When 35.0ml. of 0.05M Pb(N0_3)_2 are mixed with 15.0mL of 0.02M KI, a yellow precipitate of PbI_2(s) forms. How many moles of Pb^2+ are initially present? How many moles of I^- are initially present? The concentration of I^- is found by analysis to be 1.75 times 10^-3M at equilibrium. How many moles of I^- arc present in the solution (which has a total volume of 60mL)? How many moles of I^- are in the precipitate? How many moles of Pb^2+ arc in the precipitate? How many moles of Pb^2+ are left in solution? What is the concentration of Pb^2+ left in the solution at equilibrium? Calculate K_sp of PbI_2 from parts (c) and (g).Explanation / Answer
a) Number of mols of Pb2+ = (0.05 mol/L)(0.035 L) = 0.00175 mol
b) Number of mols of I- = (0.015 L)(0.02 mol/L) = 0.0003 mol
c) Number of mols of I- = (1.75*10-3 mol/L)(0.060 L) = 1.05*10-4 mol
d) Number of mols of I- in precipitate = 0.000195 mol
e) Number of mols of Pb2+ = 0.000195 mol * (1/2) = 0.0000975 mol
f) Number of mols of Pb2+ left = 0.00175 mol - 0.0000975 mol = 0.0016525 mol
g) [Pb2+] = 0.0016525 mol / 0.060 L = 0.0275 mol/L
h) Ksp = [Pb2+][I-]2 = ( 0.000195 mol/0.060)[ 0.0000975 mol/0.060]2 = 8.58*10-9