PbI2 = Pb+2 + 2 I- You mix the following: Volume of Lead Nitrate solution 1.88 m

Question

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 _______________ µmol
g) µmol of I- precipitated _______________ µmol
h) µmol of Pb(II) precipitated _______________ µmol
i) µmol of Pb(II) in solution at equilibrium _______________ µmol
j) Millimolarity of Pb(II) in solution at equilibrium _______________ mM
k) Calculated Ksp _______________ x 10-12

Explanation / Answer

The solution is as follows,

a) µmol (micromoles) of Pb(II) originally put in solution 8.742 µmol
b) µmol of I- originally put in solution 14.076 µmol
c) Total volume of solution 10.95 mL
d) Absorbance of sample 0.312
e) Millimolarity of I- at equilibrium 0.001 mM
f) µmol of I- in solution at equilibrium 8.742 µmol
g) µmol of I- precipitated 8.742 µmol
h) µmol of Pb(II) precipitated 4.371 µmol
i) µmol of Pb(II) in solution at equilibrium 0.7984 µmol
j) Millimolarity of Pb(II) in solution at equilibrium 7.984 x 10^-4 mM
k) Calculated Ksp 2.05 x 10-12

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