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

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In summary: K_{sp} = [Pb^{2+}(aq)][I^{-}(aq)]^{2}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. 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.
  • #1
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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  • #2
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 [tex] PbI_{2}[/tex]:

[tex] PbI_{2} \longrightarrow Pb^{2+} + 2 I^{-}[/tex]

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

The same counts for micromols [tex]I^{-}[/tex]

The third one you need to use the law of Lambert-Beer: [tex]A =\epsilon\cdot c\cdot l[/tex]. From the given : Absorbance of a 1.000 mM Iodide soln = .775, you can calculate [tex]\epsilon[/tex] And then with this found [tex]\epsilon[/tex] you can calculate [tex] [I^{-}]_{eq}[/tex]. 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 [tex]K_{sp}[/tex]
 
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  • #3
sdekivit said:
The third one you need to use the law of Lambert-Beer: [tex]A =\epsilon\cdot c\cdot l[/tex]. From the given : Absorbance of a 1.000 mM Iodide soln = .775, you can calculate [tex]\epsilon[/tex] And then with this found [tex]\epsilon[/tex] you can calculate [tex] [I^{-}]_{eq}[/tex]. 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 [tex]K_{sp}[/tex]


Thank you sooo much! :rofl:

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!
 
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  • #4
rachelle said:
I should use Ksp = [Pb]^2 right?



yes, but watch your notations !

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

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.
 
  • #5
sdekivit said:
yes, but watch your notations !

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

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!
 
  • #6
sdekivit said:
first: your reaction equation is not completely correct. Dissolving [tex] PbI_{2}[/tex]:

[tex] PbI_{2} \longrightarrow Pb^{2+} + 2 I^{-}[/tex]

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

The same counts for micromols [tex]I^{-}[/tex]

The third one you need to use the law of Lambert-Beer: [tex]A =\epsilon\cdot c\cdot l[/tex]. From the given : Absorbance of a 1.000 mM Iodide soln = .775, you can calculate [tex]\epsilon[/tex] And then with this found [tex]\epsilon[/tex] you can calculate [tex] [I^{-}]_{eq}[/tex]. 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 [tex]K_{sp}[/tex]


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...
 

1. What is the purpose of calculating Pb(II) and I- at equilibrium in the Ksp pre-lab?

The purpose of calculating Pb(II) and I- at equilibrium in the Ksp pre-lab is to determine the solubility product constant (Ksp) of a given compound. This constant represents the equilibrium constant for the dissociation of a sparingly soluble salt into its ions in a solution. By calculating the concentrations of Pb(II) and I- at equilibrium, we can then use the Ksp equation to solve for the Ksp value.

2. What is the process for calculating Pb(II) and I- concentrations at equilibrium?

The process for calculating Pb(II) and I- concentrations at equilibrium involves first setting up an ICE (Initial, Change, Equilibrium) table to track the changes in concentration of the ions. The initial concentration of the compound is known, and the change in concentration is determined by the stoichiometry of the dissociation reaction. The equilibrium concentration can then be calculated using the known Ksp value and the concentrations of the ions at equilibrium.

3. How do temperature and pH affect the solubility product constant?

The solubility product constant (Ksp) is affected by both temperature and pH. Generally, an increase in temperature will lead to an increase in solubility and therefore a higher Ksp value. On the other hand, changes in pH can affect the formation of complex ions, which can decrease the solubility of a compound and therefore decrease the Ksp value.

4. Can the solubility product constant be used to predict the precipitation of a compound?

Yes, the solubility product constant (Ksp) can be used to predict the precipitation of a compound. If the ion product (Q) is greater than the Ksp value, the solution is supersaturated and precipitation will occur. If Q is less than the Ksp value, the solution is undersaturated and no precipitation will occur. If Q is equal to the Ksp value, the solution is at equilibrium and no precipitation will occur.

5. How is the Ksp value related to the molar solubility of a compound?

The Ksp value and the molar solubility of a compound are directly related. The molar solubility represents the maximum amount of a compound that can dissolve in a given solvent, and can be calculated using the Ksp value. The higher the Ksp value, the more soluble the compound is and the higher its molar solubility will be.

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