Ksp Pre-Lab Help: Calculate Pb(II) & I- at Equilibrium

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SUMMARY

This discussion focuses on calculating equilibrium concentrations and solubility product constant (Ksp) for the reaction of lead iodide (PbI2) dissociating into lead (Pb²⁺) and iodide ions (I⁻). Participants provided detailed steps for determining micromoles of Pb(II) and I⁻ in solution, utilizing the formula c = n/V for initial concentrations and the Lambert-Beer law (A = ε·c·l) for absorbance calculations. The correct notation for Ksp is emphasized as Ksp = [Pb²⁺][I⁻]², with a recommendation to apply the ICE (Initial, Change, Equilibrium) method for accurate equilibrium concentration determination.

PREREQUISITES
  • Understanding of chemical equilibrium concepts
  • Familiarity with the Lambert-Beer law for absorbance calculations
  • Knowledge of molarity and micromole calculations
  • Ability to apply ICE tables for equilibrium problems
NEXT STEPS
  • Learn how to calculate molar extinction coefficients using absorbance data
  • Study the application of ICE tables in chemical equilibrium
  • Explore the concept of solubility product constants (Ksp) in detail
  • Review the proper notation and units in chemical equations
USEFUL FOR

Chemistry students, laboratory technicians, and educators involved in analytical chemistry and equilibrium studies will benefit from this discussion.

rachelle
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Hi, I'm trying to do my prelab and I have no idea where to start... if someone can point me to the right direction, I will absolutely be grateful... :redface:

Given the following equilibrium:

PbI2 <----> Pb + 2I

You have a solution of the following:
4.77 ml of .00274M Lead Nitrate solution
4.35 ml of .00215M Iodide solution
5.00 ml of water

You measured the solution and get the following:
%T blank = 100.3 %
%T sample = 42.9 %
Absorbance of a 1.000 mM Iodide soln = .775

Calculate the following:
micromoles of Pb(II) originally put in solution
micromoles of I- originally put in solution
mM of I- at equilibrium
micromoles of I- in solution at equilibrium
micromoles of I- precipitated
micromoles of Pb(II) precipitated
micromoles of Pb(II) in solution at equilibrium
mM of Pb(II) in soluion at equilibrium
Ksp

The only ones I solved that were correct were the following:
total volume = 14.12 ml
absorbance of sample = .3688

Hope someone can help me... :cry:
 
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rachelle said:
Given the following equilibrium:

PbI2 <----> Pb + 2I

You have a solution of the following:
4.77 ml of .00274M Lead Nitrate solution
4.35 ml of .00215M Iodide solution
5.00 ml of water

You measured the solution and get the following:
%T blank = 100.3 %
%T sample = 42.9 %
Absorbance of a 1.000 mM Iodide soln = .775

The only ones I solved that were correct were the following:
total volume = 14.12 ml
absorbance of sample = .3688

first: your reaction equation is not completely correct. Dissolving PbI_{2}:

PbI_{2} \longrightarrow Pb^{2+} + 2 I^{-}

for the original micromols Pb^{2+} you have to use the formula c = \frac {n} {V} and you use the given: 4.77 ml of .00274M Lead Nitrate solution

The same counts for micromols I^{-}

The third one you need to use the law of Lambert-Beer: A =\epsilon\cdot c\cdot l. From the given : Absorbance of a 1.000 mM Iodide soln = .775, you can calculate \epsilon And then with this found \epsilon you can calculate [I^{-}]_{eq}. Only you need to know l: the length of the cuvet where the light goes through. Since it's a prelab, you'll probably have this parameter.

Then you have everything to go on calculating K_{sp}
 
Last edited:
sdekivit said:
The third one you need to use the law of Lambert-Beer: A =\epsilon\cdot c\cdot l. From the given : Absorbance of a 1.000 mM Iodide soln = .775, you can calculate \epsilon And then with this found \epsilon you can calculate [I^{-}]_{eq}. Only you need to know l: the length of the cuvet where the light goes through. Since it's a prelab, you'll probably have this parameter.

Then you have everything to go on calculating K_{sp}


Thank you sooo much! :smile:

Okay, I've solved everything correctly except for Ksp

I should use Ksp = [Pb]^2 right?

But which concentrations should I use? The ones at equilibrium? Do I need to do "ICE"? I'm not familiar with it, but I searched and they mentioned "ICE", not sure if it's applicable here though...

Thanks again!
 
Last edited:
rachelle said:
I should use Ksp = [Pb]^2 right?


yes, but watch your notations !

K_{sp} = [Pb^{2+}(aq)][I^{-}(aq)]^{2}

you already know the equilibrium concentrations and you need them here. As for your question about ICE: you already had to use it to get to the equilibriumconcentrations.
 
sdekivit said:
yes, but watch your notations !

K_{sp} = [Pb^{2+}(aq)][I^{-}(aq)]^{2}

you already know the equilibrium concentrations and you need them here. As for your question about ICE: you already had to use it to get to the equilibriumconcentrations.


Thank you, thank you, thank you~~sdekivit! I finally got it :rolleyes:

Also yes, I should learn how to use the notations correctly... I'll read up on how to use the notations here on the forum :smile:
Thank you again!
 
sdekivit said:
first: your reaction equation is not completely correct. Dissolving PbI_{2}:

PbI_{2} \longrightarrow Pb^{2+} + 2 I^{-}

for the original micromols Pb^{2+} you have to use the formula c = \frac {n} {V} and you use the given: 4.77 ml of .00274M Lead Nitrate solution

The same counts for micromols I^{-}

The third one you need to use the law of Lambert-Beer: A =\epsilon\cdot c\cdot l. From the given : Absorbance of a 1.000 mM Iodide soln = .775, you can calculate \epsilon And then with this found \epsilon you can calculate [I^{-}]_{eq}. Only you need to know l: the length of the cuvet where the light goes through. Since it's a prelab, you'll probably have this parameter.

Then you have everything to go on calculating K_{sp}


Hi, i couldn't help but wonder how you calculated the equilibrium concentration of I- from the molar extinction coefficient, E?? I have a very similar problem to Rachelle's post...