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 _______________ µmol
b) µmol of I- originally put in solution _______________ µmol
c) Total volume of solution _______________ mL
d) Absorbance of sample _______________
e) Millimolarity of I- at equilibrium _______________ 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

given tha Volume of Lead Nitrate solution 1.88 mL
also Original concentration of Lead Nitrate solution 0.00465 M

a) Initial moles of lead nitrate solution = (1.88/1000)*0.00465= 0.000008742 mol =8.742 micro moles

b) Volume of Iodide ion solution 4.57 mL
Initial moles of iodide solution = 4.57*0.00308/1000 =1.47X10-5moles = 14.0756 micomoles

c) Total volume of solution = volume of lead nitrate solution + volume of iodide solution + volume of water = 1.88+ 4.57+ 4.5 ml = 10.95ml

d) absorbance of sample can be calculated from A= 2-log (%T) = 2- log(48.8) =0.3116

e) given that absorbance is 0.558 for one millimolar solution

Since absorbance is proportional to concentration

A = KC

Where K is constant

0.558 =K *1

K=0.558

The equation becomes A= 0.558C

given the absorbance of the sample= 0.3116

0.3116 =0.558C

c= 0.3116/0.558 =0.558423 Mill molar =5.58423X10-4 milli molar

Iodine in 10.5 ml of final solution = 5.58423X10-4 *10.5/1000 =5.86344X10-6 moles =5.86344 micro moles

molarity of iodine at equilibrium = 5.86344 10-6*10.5/1000 =6.1566X10-8 molar =6.1566 X10-5 mill molar

Iodine that got precipitated origianl moles- moles at equilibrium = (14.0756- 5.86344 =8.2412 micromoles

Lead preciptied as per stoichiometery ( PbI2----. Pb+2 + 2I-) =1/2 Iodime that got precipitated= 8.2412/2 micromoles =4.1206 micromoles

micromoles of lead initially present= 8.742 micromoles

lead precipiated= 4.1206 mircomoles

pb(II)present at equilibrium= initial lead- lead precipitated= 8.742-4.1206= 4.6214 micromoles

molarity of pb(II)= 4.6214*10-6 * 10.5/1000 =4.85X10-8 M= 4.85X10-8*103millMolar =4.85X10-5 milli molar

Ksp= [Pb] [I-]2 = 4.85X10-5 mill molar* (6.1566X10-5)2 milli molar 2=183.439X10-15 = 0.183X10-12

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.