# When to square variable or not in Ksp, Keq, Kp

Discussion in 'DAT Discussions' started by pandalove89, Aug 7, 2011.

1. ### pandalove89 7+ Year Member

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Okay so this is one of the examples where I don't understand when to square the variable or not with Ksp, Keq, or Kp problems.

This is in the 2011 DAT Destroyer General Chemistry Portion:

For question number 86 it asks:
The Ksp of Fe(OH)2 at 25 degrees celsius is 1.6*10^-14 . What is the solubility of Fe(OH)2 in .025 M FeCl2. (an obvious common ion problem)

So you would set up the equation like:

Ksp = [Fe^2+][OH-]^2
1.6 *10^-14 = [.025][2x]^2
Then 1.6*10^-14 = [.025][4x^2]

For question number 101 it asks:

Consider the following: 2A (g) + B (g) <-- --> 2C (g)
.50 atm of A and .20 atm of B are placed in a flask at 300 K. At equilibrium, the total pressure is found to be .60 atm. Calculate Kp for the reaction.

So you set up an ice chart and get:

.50 -2x + .20 - x + 2x = .60

so x = .10 atm

however, in the equation it is K = [C]^2 / [A]^2 = [.2]^2 / [.3]^2[.1]

my question is how come in the [C] area it wouldn't be set up like the question in 86. So instead of [C]^2 it wouldn't be 2[C]^2 or [4C^2]

I always get caught up with this.....

2. ### JustwantDDS DrAMG 7+ Year Member

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hmm i cant answer your question- but for some reason im blanking as to why its -2x and -x in the problem..........in the ice table
i totally dont remember

3. ### JustwantDDS DrAMG 7+ Year Member

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i think its because for 86 the ksp equation will be ksp= [Fe2+][OH-]2 but since we dont have a value for OH- we put 2x since 2OH- are produced for every 1 Fe2+

but for the next problem - the K is the same set up K= [c]^2/ [A]^2 but we use to ICE to set it up............ ? uh

4. OP

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bump.

5. ### RalphMouth 7+ Year Member

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We do set it up the same in problem one as in problem 2, the Ksp is Ksp = [Fe^2+][OH-]^2.

In the second example you're calculating the K value given the concentrations (or rather the concentrations you solved for), which equals K = [C]^2 / [A]^2. These setups are similar.

The difference is in problem 1, we are trying to solve for the unknown concentrations, therefore must assign a variable in order to get us the answer. In that example, it is solubility of Fe (OH)2 assigned as x. If 1 mole of Fe OH 2 is soluble, 2 moles of OH 2 are soluble, therefore OH concentration = 2x. Therefore [OH-]^2 = (2x)^2

In the second problem we don't use variables at all in the Kp equation because we have the actual concentrations already from the first part of the problem involving the ICE chart. Just plug it into the equation

For further clarification in the first example, we could represent x to be the OH- concentration. In that case, the solubility would actually be 1/2 x. Its honestly totally arbitrary when you solve for the variable. But most people set it up the way you describe.

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Last edited: Aug 7, 2011
6. ### JustwantDDS DrAMG 7+ Year Member

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How come we dont solve for C like this = (2* 0.2)^2 its just 0.2^2 but when we solve for molar solubility of lets say FeOH2 for ex, its like 2x^2 which becomes 4x^2 and we set x= ksp and solve

7. ### RalphMouth 7+ Year Member

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Because we're not solving for C, we're plugging a value for C into the equation.

By definition Kp= [C]^2/ ([A]^2 ), you figured out the concentration already, so the number is the concentration, not 2*number, therefore it is not (2[C])^2, it's simply [C]^2.

A simple example:

If I told you at equilibrium [C] is 2 M, [A] is 1 M, is 1 M. You would for the K calculation simply take 2^2 for the numerator, not (2*2)^2. It's not different for the original problem. Just because you figured out the concentration of C using a variable, doesn't mean you use 2x...they're 2 different calculations.

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Last edited: Aug 8, 2011
8. ### csulapredental

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same q

same question! why is it -2x and -x???? why negative i don't get it

9. ### csulapredental

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why is it a negative on ice chart

10. ### DMikes 2+ Year Member

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This is from 2011.

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