i need help for lead molarities. J and K PbI2 = Pb+2 + 2 I- You mix the followin

Question

i need help for lead molarities. J and K

PbI2 = Pb+2 + 2 I-
You mix the following:
Volume of Lead Nitrate solution 1.88 mL
Original concentration of Lead Nitrate solution 0.00465 M
Volume of Iodide ion solution 4.57 mL
Original concentration of Iodide ion solution 0.00308 M
Volume of distilled water 4.50 mL
You measure the %T of the blank and sample at 525 nm. Assume Beer's Law, absorbance is proportional to concentration.
%T of blank 99.5 %
%T of sample 48.8 %
Absorbance of a 1.000 MilliMolar Iodide solution 0.558
Note: This constant is only for this pre-lab! In the experiment you will use a "standard plot!"
Calculate [to 3 decimal places, except c)]:
a) µmol (micromoles) of Pb(II) originally put in solution ___8.742____________ µmol
b) µmol of I- originally put in solution ________14.0756_______ µmol
c) Total volume of solution ______10.95_________ mL
d) Absorbance of sample ______.309_________
e) Millimolarity of I- at equilibrium _________.554______ mM
f) µmol of I- in solution at equilibrium _________6.07______ µmol
g) µmol of I- precipitated ________8.0056_______ µmol
h) µmol of Pb(II) precipitated __________4.0028_____ µmol
i) µmol of Pb(II) in solution at equilibrium ___________4.74____ µmol
j) Millimolarity of Pb(II) in solution at equilibrium _______________ mM
k) Calculated Ksp _______________ x 10-12

Explanation / Answer

j) millimolarity of Pb(II) = micromoles /total volume of solution in mL

Millimolarity = 4.74 / 10.95 = 0.432 millimolar

k) The Ksp = [Pb+2][I-]^2

so concentration of Pb+2 = 0.432 millimolar = 0.432 X 10^-3

concentration of I- = 0.554 millimolar = 0.554 X 10^-3

So Ksp = 0.432 X 10^-3 X (0.554 X 10^-3) ^2 = 0.13258 X 10^-9 = 132.58 X 10^-12

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