In: Chemistry
Data given -
PbI2(s) <===> Pb2+(aq) + 2I-
(aq) ( Ksp = [Pb2+][I-]2 )
5.0mL 0.0120 M Pb(NO3)2
2.0 mL 0.0300 M KI
3.0 mL 0.20 M KNO3
Total volume = 10 mL
Absorbance of solution = 0.280
[ I- ] in moles/liter at equilibrium = 4.2
CALCULATE -
a) The initial no. of moles Pb2+ Answer
b) The initial no. of moles I-
c) No. of moles I- at equilibrium
d) No. moles I- precipitated
e) No. moles Pb2+ precipitated
f) No. moles Pb2+ at equilibrium
g) [Pb2+] at equilibrium
h) Ksp PbI2
Answers of a to f are all raised to 10-5 (show calculations)
a) The initial no. of moles Pb2+
mol of Pb+2 = M1*V1 = 0.012 * 5*10^-3 = 0.00006 mol of Pb+2
b) The initial no. of moles I-
mol of I- = M2*V2= 0.03 *2*10^-3 = 0.00006 mol of I-
c) No. of moles I- at equilibrium
in equilibrium:
[ I- ] in moles/liter at equilibrium = 4.2
V =10 mL = 10*10^-3 = 0.01 L
mol = MV = 4.2*10^-5 *0.01 = 4.2*10^-7 mol of I-
d) No. moles I- precipitated
difference --> (0.00006 - 4.2*10^-7) =0.00005958 mol of I-
e)
ratio is 2 mol of I- = 1 mol of PB+2 --< 1/2*0.00005958 = 0.00002979 mol of Pb+2
f)
Pb+2 in eq =0.00006 - 0.00002979 = 0.00003021 mol of Pb+2
g)
[Pb+2] = mol/V = (0.00003021)/(10*10^-3) = 0.003021 M
h)
Ksp = [Pb+2][I-]^2 = (0.003021)(4.2*10^-5)^2
Ksp = 5.329*10^-12