- Joined
- Jun 15, 2011
- Messages
- 91
- Reaction score
- 0
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.....
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.....