Question
How do we know when x is negligible when calculating equilibrium concentrations?
Answers
Jess
Do you divide the concentration given in the question by the Kb or Ka value or is that incorrect?
DrBob222
Actually you don't REALLY know if x is small enough to neglect. My advice has always been to work the problem neglecting x. When you get the answer you decide if x can be neglected; i.e., you see if you should have done it the long way. Here is an example:
Calculate the (H^+) and (Ac^-) in a 0.1 M solution of acetic acid(HAc)
......................HAc --> H^+ + Ac^-
I.......................0.1.........0..........0
C.......................-x............x..........x
E...................0.1-x...........x..........x
Ka = 1.8E-5 = (x)(x)/1.8E-5
x^2 = 1.8E-6. See I've neglected the x and done it the easy way.
x = 1.3E-3 or 0.0013
Now you decide if this is OK or if you should have done it the long way.
There are two ideas out there. One is the 5% rule; the other is the 10% rule. Take your pick. I prefer the 10% rule but that's what I used when I was a student which was a LONG TIME AGO when Ka values were not much better than 10%. Now days Ka values are better than that and 5% may be justified. Anyway, using the 10% rule you do this.
The answer is 0.0013 M for (H^+)
So 10% of the 0.1 is 0.1 x 0.1 = 0.01. The answer when neglecting x is 0.0013 which is less than the 10% so the shortcut is OK and no need to redo it the long way with the quadratic. Hope this is what you needed.
Calculate the (H^+) and (Ac^-) in a 0.1 M solution of acetic acid(HAc)
......................HAc --> H^+ + Ac^-
I.......................0.1.........0..........0
C.......................-x............x..........x
E...................0.1-x...........x..........x
Ka = 1.8E-5 = (x)(x)/1.8E-5
x^2 = 1.8E-6. See I've neglected the x and done it the easy way.
x = 1.3E-3 or 0.0013
Now you decide if this is OK or if you should have done it the long way.
There are two ideas out there. One is the 5% rule; the other is the 10% rule. Take your pick. I prefer the 10% rule but that's what I used when I was a student which was a LONG TIME AGO when Ka values were not much better than 10%. Now days Ka values are better than that and 5% may be justified. Anyway, using the 10% rule you do this.
The answer is 0.0013 M for (H^+)
So 10% of the 0.1 is 0.1 x 0.1 = 0.01. The answer when neglecting x is 0.0013 which is less than the 10% so the shortcut is OK and no need to redo it the long way with the quadratic. Hope this is what you needed.
DrBob222
oops. BIG typo.
Here is what I typed.
Ka = 1.8E-5 = (x)(x)/1.8E-5
It should have been this.
Ka = 1.8E-5 = (x)(x)/(0.1-x)
Then neglecting x we get x^2 = 1.8E-6 etc.
Here is what I typed.
Ka = 1.8E-5 = (x)(x)/1.8E-5
It should have been this.
Ka = 1.8E-5 = (x)(x)/(0.1-x)
Then neglecting x we get x^2 = 1.8E-6 etc.