Gen chem solubility and ksp- Destroyer specific!

[virtualmaster999]

Hi everyone!

Quick question about molar solubility and Ksp. So I thought that when you are solving for molar solubility/Ksp, you would incorporate the coefficient AND the exponent. For example...In Gen chem destroyer #126:

You have Fe (OH)2 so it would dissociate into Fe2+ and 2OH-, and OH- would be (2x)^2

Now for precipitation reaction, or whenever you are given a concentration of a common ion, you drop the coefficient.

When a question asks, "solubility of X in Y", does this always incorporate both the coefficient and the exponent? I'm trying to determine the fastest way to do these kinds of problems.

Anything helps. Thanks in advance!

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[cacajuate]

You only drop the molar solubilty (x) of the common ion but you still keep the exponent. If I'm understanding what you are saying

 

[virtualmaster999]

You only drop the molar solubilty (x) of the common ion but you still keep the exponent. If I'm understanding what you are saying

Ohhhh ok! So that applies for all situations where a common ion concentration is given: you drop the coefficient, but keep the exponent? Since the common ion was just "x", nothing changes. Is that right?

 

[cacajuate]

I'm not really too sure why you keep saying coefficient but you keep the concentration of the common ion and drop the X from the ice chart (which can have a coefficient.)


I'll let orgoman or feralis explain it much clearer.

Did you subscribe to chads videos? They are really worth the money and he explains these types of problems really well.

 

[virtualmaster999]

I'm not really too sure why you keep saying coefficient but you keep the concentration of the common ion and drop the X from the ice chart (which can have a coefficient.)


I'll let orgoman or feralis explain it much clearer.

Did you subscribe to chads videos? They are really worth the money and he explains these types of problems really well.

Yeah I used Chads. What I mean by coefficient is if you have say Fe(OH)2 going to Fe2+ and 2OH-. If no concentration is given for either, you have (x) (2x)^2, where the 2 is the coefficient referring to the ice chart (which is why I say coefficient to skip the process of doing the ice chart). So if, let's say, OH= 2*10^-2. you'd have (x)(2*10^-2)^2 and not (x)(2(2*10^-2))^2. Do you see what I mean?

 

[cacajuate]

Yeah I used Chads. What I mean by coefficient is if you have say Fe(OH)2 going to Fe2+ and 2OH-. If no concentration is given for either, you have (x) (2x)^2, where the 2 is the coefficient referring to the ice chart (which is why I say coefficient to skip the process of doing the ice chart). So if, let's say, OH= 2*10^-2. you'd have (x)(2*10^-2)^2 and not (x)(2(2*10^-2))^2. Do you see what I mean?

Typically they are going to give Fe2+ as the common ion, as in FeCl2. Which if [FE]=2*10^2 then the first equation you put would be correct. I would assume that if they did give a common ion with a 2x concentration, like Fe(OH)2 and the common ion was OH, then it would be like your second equation.

I searched through the whole destroyer and didn't find any that gave a common ion in a 2x concentration though.

 

[cdp1210]

Common ion problems are not the same as precipitation reactions. For common ion problems the ratio of products is important hence why you need to keep the coefficient when solving for the Ksp. For precipitation reactions, you are given a concentration of the two products so the ratio in which they are added doesn't matter, therefore you don't need a coefficient.

Just remember that you do not use the ICE chart for precipitation reactions because you are not dissolving the reactant solid (FeCl2 in your case) like you do when calculating the molar solubility in common ion problems. Instead you are adding the products and determining whether Q>Ksp or Q<Ksp.

 

[virtualmaster999]

Common ion problems are not the same as precipitation reactions. For common ion problems the ratio of products is important hence why you need to keep the coefficient when solving for the Ksp. For precipitation reactions, you are given a concentration of the two products so the ratio in which they are added doesn't matter, therefore you don't need a coefficient.

Just remember that you do not use the ICE chart for precipitation reactions because you are not dissolving the reactant solid (FeCl2 in your case) like you do when calculating the molar solubility in common ion problems. Instead you are adding the products and determining whether Q>Ksp or Q<Ksp.

Hmm Im a bit confused. Ok so lemme see if I got this:

1) For Ksp/Molar solubility: If you have something like MX3 dissociating-----> M + 3X, then you have (M)(3X)^3
-If you are given a concentration, such as [X]=5, you have (M)(5)^3

2) For Precipitation, you will be given both concentrations. Say x=5, m=5
You drop the coefficients, in this case, 1 and 3. So you have (5)(5)^3

Basically what I'm trying to get at is this: For molar solubility/Ksp problems, you may be given a common ion concentration. For THAT concentration, you do not include the coefficient, just the exponent. For precipitation problems, you will be given both concentrations, so you drop the coefficient in both. Is this right? I'm just trying to piece this all together and understand when coefficients are used and when they're not.

 

[cdp1210]

2) For Precipitation, you will be given both concentrations. Say x=5, m=5
You drop the coefficients, in this case, 1 and 3. So you have (5)(5)^3

Yup that looks like you understand that part for the precipitation reaction.

1) For Ksp/Molar solubility: If you have something like MX3 dissociating-----> M + 3X, then you have (M)(3X)^3
-If you are given a concentration, such as [X]=5, you have (M)(5)^3

Right, but make sure you understand it's technically like (M)(5+3X)^3, but you disregard the 3X because it is so small. By doing the ICE chart for this you would have the initial(I)=5 and the change (C)=3X so your equilibrium (E)=5+3X. Since X is going to be very small, adding it to 5 will be insignificant so you will ignore it and just be left with (M)(5)^3.

Here is one of the questions from chad's quizzes with a solution explanation to see if that helps clear anything up:
What is the molar solubility of BaF2 in 0.1M NaF (Ksp,BaF2 = 3.2x10-8)?
BaF2(s) → Ba2+(aq) + 2F-(aq)
Ksp = [Ba2+][F-]2 = (x)(0.1+2x)2
But x (the molar solubility) will be small and so 2x will be negligible when added to 0.1 and so the expression reduces to the following:
Ksp = (x)(0.1)2 = 3.2x10-8

 

[virtualmaster999]

Yup that looks like you understand that part for the precipitation reaction.


Right, but make sure you understand it's technically like (M)(5+3X)^3, but you disregard the 3X because it is so small. By doing the ICE chart for this you would have the initial(I)=5 and the change (C)=3X so your equilibrium (E)=5+3X. Since X is going to be very small, adding it to 5 will be insignificant so you will ignore it and just be left with (M)(5)^3.

Here is one of the questions from chad's quizzes with a solution explanation to see if that helps clear anything up:
What is the molar solubility of BaF2 in 0.1M NaF (Ksp,BaF2 = 3.2x10-8)?
BaF2(s) → Ba2+(aq) + 2F-(aq)
Ksp = [Ba2+][F-]2 = (x)(0.1+2x)2
But x (the molar solubility) will be small and so 2x will be negligible when added to 0.1 and so the expression reduces to the following:
Ksp = (x)(0.1)2 = 3.2x10-8

Awesome, thanks for the explanation!