Ka : Solving Weak Acid Problems: Part One
http://dbhs.wvusd.k12.ca.us/webdocs/AcidBase/Ka-Solving1.html
Copyright © 1998 by John L. Park
Important note: all constants refered to: Kc, Kw, Ka, and Kb are temperature-dependant. All discussions are
assumed to be at 25 °C, i.e. standard temperature.
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 for two reasons:
a) you memorized a short list of strong acids. (You did, didn't you?) Everything else is weak.
b) The Ka value of 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-on 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 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.
Ka : Solving Weak Acid Problems: Part Two
Important note: all constants refered to: Kc, Kw, Ka, and Kb are temperature-dependant. All discussions are
assumed to be at 25 °C, i.e. standard temperature.
Here's a second typical weak acid problem:
What is the pH of a 0.300 M solution of benzoic acid? Ka = 6.46 x 10¯5
These are the important equations:
HBz + H2O <==> H3O+ + Bz¯
Ka = ( [H3O+] [Bz¯] ) / [HBz] = 6.46 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 oneto-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:
6.46 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 = 1.938 x 10¯5
Taking the square root (of both sides!!), we get:
x = 4.40 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 2.356.
Checking for 5% we find we had 1.47%, which is acceptable.
Ka : Solving Weak Acid Problems: Part Three
Important note: all constants refered to: Kc, Kw, Ka, and Kb are temperature-dependant. All discussions are
assumed to be at 25 °C, i.e. standard temperature.
Here's a third typical weak acid problem:
What is the pH of a 0.250 M solution of cacodylic acid? Ka = 6.4 x 10¯7
You may have noticed that the solutions in parts one and two were exactly the same. Both ended up with this:
x = square root of (Ka times starting acid concentration)
When you're doing the 'drop subtract x' thing, this above equation always works for weak acids.
I better add a cautionary note here: the equation works for weak MONOPROTIC acids. However, the study of
diprotic acids is not touched on in high school chemistry nor really in Advanced Placement, so we're safe for
the time being.
So the solution to the above problem is:
x = square root of (6.4 x 10¯7 times 0.250) = 4.0 x 10¯4
From this, the pH = 3.40
Checking the 5% rule, we get 0.16% error.
Here's a fourth example. The 5% rule fails.
What is the pH of a 0.150 M solution of nitrous acid, HNO2? Ka = 4.6 x 10¯4
x = square root of (4.6 x 10¯4 times 0.150) = 8.31 x 10¯3
Checking the 5% rule:
(8.31 x 10¯3 / 0.15) x 100 we get 5.54% error.
In order to get a correct answer, we must turn to the quadratic method. In other words, we cannot ignore the
'subtract x' portion in the denominator.
The equation to use is as follows (I left off the sub a on the K):
x = [-K + sq. root (K2 + 4KC)] / 2
The C stands for the starting concentration of the acid.
The solution is left to the reader.
Period ______________
Name ________________________
Ka Problems, from Merrill Chemistry Workbook
SHOW YOUR WORK
1. Determine the Ka for 0.00100M acetic acid (HCH3COO) with [H3O+1] = 1.27 x 10-4
HCH3COO(aq) + H2O(l) 
 H3O+1(aq) + CH3COO-1(aq)
2. Determine the Kb for 0.0070M aqueous ammonia (NH3) with [OH-1] = 3.46 x 10-4
NH3(g) + H2O(l)  NH4+1(aq) + OH-1(aq)
3. Determine the Ka for 0.100M hydrogen cyanide (HCN) solution with [H3O+] = 7.85 x 10-6
HCN(aq) + H2O(l) 
 H3O+1(aq) + CN-1(aq)
4. Determine the [H3O+] concentration in a 0.045M HIO solution. Ka = 2.29 x 10-11. What is the pH? Also
show you’ve checked the 5% rule.
HIO(aq) + H2O(l) 
 H3O+1(aq) + IO-1(aq)
5. Determine the molar concentration of all substances in a 1.000L volumetric flask filled with a water solution
containing 1.000 moles of acetic acid, HCH3COO. The Ka = 1.75 x 10-5. See #1 for the chemical equation.
Also show you’ve checked the 5% rule.