Weak Acid problems
The typical weak acid problem to solve in high school classes looks like this:
What is the pH of a 0.100 M solution of acetic acid? Ka = 1.77 x 10¯5
Some facts of importance:
1) you know this is a weak acid because the Ka value is small. Strong acids have very large Ka values, as in 105 or 107.
2) The solution technique explained below applies to almost all weak acids. The only things to change are the concentration and the Ka, if doing another acid.
You've seen these equations:
HAc + H2O <==> H3O+ + Ac¯
Ka = ( [H3O+] [Ac¯] ) / [HAc] = 1.77 x 10¯5
The key quantity we want is the [H3O+]. Once we have that, then the pH is easy to calculate. Since we do not know the value, let's do this:
[H3O+] = x
I hope that, right away, you can see this:
[Ac¯] = x
This is because of the one-to-one molar ratio between [H3O+] and [Ac¯] that is created as HAc molecules dissociate.
So now, we have all but one value in our equation:
1.77 x 10¯5 = {(x) (x)} / [HAc]
All we have to do is figure out [HAc] and we can calculate an answer to 'x.'
In this problem, the [HAc] started at 0.100 M and went down as HAc molecules dissociated. In fact, due to the one-to-one ratio, it went down by 'x' amount and wound up at an ending value of 0.100 - x.
So, here's the final set-up:
1.77 x 10¯5 = {(x) (x)} / (0.100-x)
Now, that is a quadratic equation and can easily be solved with the quadratic formula. However, there is a trick we can use to make our calculation easier.
You don't know it, but Ka values are very difficult to figure out. There's a whole bunch of variables that are difficult to control. The end result is that Ka are approximate and most are in error about ± 5%.
So that means, if we stay within 5%, we can use approximate techniques to get an answer. The major approximation occurs with '0.100 - x.' Since x is rather small, it will not change the value of 0.100 by much, so we can say:
0.100 - x approximately equals 0.100
We now write a new equation:
1.77 x 10¯5 = {(x) (x)} / 0.100
which is very easy to solve. We move the 0.100 to the other side to get:
x2 = 1.77 x 10¯6
Taking the square root (of both sides!!), we get:
x = 1.33 x 10¯3 M
Take note of two things:
1) The Ka value is unitless, but x is a molarity.
2) Square root both sides. I have had students square root the x2, but not the other side. Weird, but true.
We finish by taking the pH (the antilog) to get a final answer of 2.876.
The final comment has to do with checking for 5%. The formula is:
( [H3O+] / [HAc] ) x 100 < 5%
In our case, we had 1.33%, which is acceptable.
Example 2:
What is the pH of a 0.300 M solution of benzoic acid? Ka = 6.3 x 10¯5
These are the important equations:
HBz + H2O <==> H3O+ + Bz¯
Ka = ( [H3O+] [Bz¯] ) / [HBz] = 6.3 x 10¯5
Bz¯ refers to the benzoate ion. It is completely unimportant what its formula is.
As before, we want the [H3O+]. So we have:
[H3O+] = x
and
[Bz¯] = x
This is because of the one-to-one molar ratio between [H3O+] and [Bz¯] that is created as HBz molecules dissociate.
Remember, the [HBz] started at 0.300 M and went down as HBz molecules dissociated. In fact, due to the one-to-one ratio, it went down by 'x' amount and wound up at an ending value of 0.300 - x.
Next is our 'dropping the subtract x' trick:
0.300 - x approximately equals 0.300
We will check the validity of the trick after completing the calculation. If the approximation exceeds 5%, then we have to use the quadratic.
We now have our equation:
6.3 x 10¯5 = {(x) (x)} / 0.300
which is very easy to solve. We move the 0.300 to the other side to get:
x2 = 1.89 x 10¯5
Taking the square root (of both sides!!), we get:
x = 4.35 x 10¯3 M
Remember:
1) The Ka value is unitless, but x is a molarity.
2) Square root both sides. I have had students square root the x2, but not the other side. Weird, but true.
We finish by taking the pH to get a final answer of 2.36.
Checking for 5% we find we had 1.45%, which is acceptable.