How do we know when x is negligible when calculating equilibrium concentrations?
3 answers
Do you divide the concentration given in the question by the Kb or Ka value or is that incorrect?
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.
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.
2 answers