### Ka : The acid ionization constantFifteen Examples

Some discussion.

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.

Basic Information

(1) Weak acids are less than 100% ionized in solution.
(2) Acetic acid (formula = HC2H3O2) is the most common weak acid example used by instructors.
(3) Another way to write acetic acid's formula is CH3COOH.
(4) A common abbreviation for acetic acid is HAc, where Ac¯ refers to the acetate polyatomic ion.

The following equation describes the reaction between acetic acid and water:

HAc + H2O ⇌ H3O+ + Ac¯

Note that it is a reaction that comes to a state of equilibrium.

The equilibrium constant for this reaction is written as follows:

 [H3O+] [Ac¯] Kc = ––––––––––– [HAc] [H2O]

However, in pure liquid water, [H2O] is a constant value. To demonstrate this, consider 1000 mL of water with a density of 1.00 g/mL. This 1.00 liter (1000 mL) would weigh 1000 grams. This mass divided by the molecular weight of water (18.0152 g/mol) gives 55.5 moles. The "molarity" of this water would then be 55.5 mol / 1.00 liter or 55.5 M.

The solutions studied in introductory chemistry are so dilute that the "concentration" of water is unaffected. So 55.5 molar can be considered to be a constant if the solution is dilute enough.

Moving [H2O] to the other side gives:

 [H3O+] [Ac¯] Kc [H2O] = ––––––––––– [HAc]

Since the term Kc [H2O] is a constant, let it be symbolized by Ka, giving:

 [H3O+] [Ac¯] Ka = ––––––––––– [HAc]

This constant, Ka, is called the acid ionization constant. It can be determined by experiment and each acid has its own unique value. For example, acetic acid's value is 1.77 x 10¯5.

From the chemical equation above, it can be seen that H3O+ and Ac¯ concentrations are in the molar ratio of one-to-one. This will have an important consequence as we move into solving weak acid poblems.

By the way, you stand a relatively decent chance of seeing this:

 [H+] [Ac¯] Ka = ––––––––––– [HAc]

It has the exact same meaning as the one that uses [H3O+]. Exact.

Comment: the type of problem discussed in this tutorial covers only monoprotic acids. The techniques for di- and triprotic are more complex and will not be covered.

More comment: if you don't know the formula of the monoprotic weak acid, that's OK. Simply use HA as the formula. It does not matter what the anion portion is, it only matters that the acid is weak and monoprotic.

Last comment: the first five examples calculate the pH when given the Ka. Starting at #6, I have examples that give the pH (and other data) and ask you to calculate the Ka. Asking you to calculate the Ka on the test while having only taught in class how to calculate the pH is a very popular tactic.

Example #1: What is the pH of a 0.100 M solution of acetic acid? Ka = 1.77 x 10¯5

Some facts of importance before the solution:

(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 a weak acid is small (10¯5 is a typical value). Strong acids have very large Ka values (105 or 107 being typical values).
(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.

Comment on 1b: This means Ka values of strong acids are billions and trillions of times larger than Ka values of weak acids. This means the reactions for strong acids essentially go completely to the right. In other words, strong acids (and bases) are 100% ionized in solution.

Solution:

Warning: long solution, some detailed explanation

1) From above, here is the Ka expression for acetic acid:

 [H3O+] [Ac¯] Ka = ––––––––––– [HAc]

2) 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

3) 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¯, both being created as one HAc molecule dissociates (see coefficients of the balanced chemical equation above).

The end result of this is that [H3O+] will equal [Ac¯] in the solution.

4) So now, we have all but one value in our equation:

 (x) (x) 1.77 x 10¯5 = –––––– [HAc]

All we have to do is figure out [HAc] and we can calculate an answer to 'x.'

5) 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. That mean this is the final set-up:

 (x) (x) 1.77 x 10¯5 = –––––––– 0.100 − x

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.

6) It turns out that 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% of the answer using the quadratic, we can use approximate techniques to get an answer. The 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 ≈ 0.100

7) We now write a new equation:

 (x) (x) 1.77 x 10¯5 = –––––––– 0.100

8) We move the 0.100 to the other side to get:

x2 = 1.77 x 10¯6

9) Taking the square root (of both sides!!), we get:

x = 1.33 x 10¯3 M

10) Take note of two things:

(a) The Ka value is unitless, but x is a molarity. Some instructors insist on units for the Ka.
(b) Square root both sides. I have had students square root the x2, but not the other side. Weird, but true.

11) We finish by determining the pH:

pH = −log [H3O+]

pH = −log 1.33 x 10¯3

pH = 2.876

12) The final step has to do with checking for 5%. The formula is:

([H3O+] / [HAc]o) (100) < 5%

Where [HAc]o is the starting concentration of the acid.

In our case, we had 1.33%, which is acceptable.

There is a brief discussion of the 5% Rule here.

Example #2: What is the pH of a 0.300 M solution of benzoic acid? Ka = 6.46 x 10¯5

Solution:

1) Write the chemical reaction and the Ka expression:

HBz + H2O ⇌ H3O+ + Bz¯

 [H3O+] [Bz¯] Ka = ––––––––––– [HBz]

Bz¯ refers to the benzoate ion. It is completely unimportant to the pH calculation what its formula is.

2) We want to determine the [H3O+]. So we have:

[H3O+] = x

as well as

[Bz¯] = x

Remember, there is a one-to-one molar ratio between H3O+ and Bz¯, created as each HBz molecule dissociates.

This leads to the fact that [H3O+] equals[Bz¯]

3) 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. Therefore, we write this:

[HBz] = 0.300 − x.

4) Next is our 'dropping the subtract x' trick:

0.300 − x ≈ 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.

5) We now have our fully filled-in equation:

 (x) (x) 6.46 x 10¯5 = –––––– 0.300

6) And we proceed with the solution:

x2 = 1.938 x 10¯5

x = 4.40 x 10¯3 M

7) Determine the pH:

pH = −log 4.40 x 10¯3 = 2.356

8) Apply the 5% rule:

(x / [HA]o) (100)

(4.40 x 10¯3 / 0.300) (100) = 1.47%

Using the approximation is valid.

Example #3: What is the pH of a 0.250 M solution of cacodylic acid? Ka = 6.4 x 10¯7

Solution:

1) You may have noticed that the solutions in Example #1 and #2 were exactly the same. Both ended up with this:

x = $\sqrt{\mathrm{\left(Ka\right) \left(\left[HA\right]o\right)}}$

where [HA]o is the starting concentration of the acid.

When you're doing the 'drop subtract x' approximation, this equation just above always works for MONOPROTIC weak acids. Diprotic and triprotic acids have different solving techniques. Those techniques will not be covered.

2) Solve the example:

x = $\sqrt{\mathrm{\left(6.4 x 10¯7\right) \left(0.250\right)}}$

x = 4.0 x 10¯4 M

pH = 3.40

Checking the 5% rule, we get 0.16%. Using the approximate technique is valid.

Example #4: What is the pH of a 0.150 M solution of nitrous acid, HNO2? Ka = 4.6 x 10¯4

Solution:

1) Solve the example:

x = $\sqrt{\mathrm{\left(4.6 x 10¯7\right) \left(0.150\right)}}$

x = 8.3 x 10¯3 M

This leads to a pH of 2.08

2) Check to see if the approximation is valid:

(8.31 x 10¯3 / 0.150) (100) = 5.54%

The approximate technique fails. We must use the quadratic equation.

3) There is a specialized form of the quadratic that can be used at this point. It's only for these type of weak acid problems. Here it is:

 −K + $\sqrt{\mathrm{K2+ 4KC}}$ x = –––––––––––––– 2

x = 0.016 M

pH = 1.80

Example #5: Calculate the pH of a 0.036 M solution of HF. The Ka is 4.5 x 10¯4

Solution:

In this example, the 5% rule fails. The solution is given in this video.

The presentation is weird, but the chemistry is 100% correct. He uses the general formula for the quadratic, which is a bit different than the specialized one I used just above.

You might want to try the solution on your own before looking at the video. The 5% rule gives 11% and the pH with the quadratic is 2.42.

Example #6: A 0.120 M solution of a generic weak acid (HA) has a pH of 3.26. Determine the Ka.

Notice that a generic weak acid is used, symbolized by the formula HA. No one cares what the specific acid is because the technique to be explained works for all weak acids. What happens is that some teachers will use the name of a specific weak acid while others go the generic route.

Solution

1) Write the dissociation equation for the acid:

HA + H2O ⇌ H3O+ + A¯

2) Write the Ka expression:

 [H3O+] [A¯] Ka = ––––––––––– [HA]

3) Our task now is to determine the three concentrations on the right-hand side of the Ka expression since the Ka is our unknown.

(a) We will use the pH to calculate the [H3O+]. We know pH = −log [H3O+], therefore [H3O+] = 10¯pH
[H3O+] = 10¯3.26 = 5.4954 x 10¯4 M

I've kept a couple guard digits; I'll round off the final answer to the proper number of significant figures.

(b) From the dissociation equation, we know there is a 1:1 molar ratio between [H3O+] and [A¯]. Therefore:

[A¯] = 5.4954 x 10¯4 M

(c) The final value, [HA] is given in the problem. In the example being discussed, 0.120 M is the value we want. Some teachers will use 0.120, while others would say to subtract the 5.4954 x 10¯4 value from 0.120 first. Let's do both.

(i) Without subtraction
 (5.4954 x 10¯4) (5.4954 x 10¯4) Ka = –––––––––––––––––––––––––– 0.120

Ka = 2.52 x 10¯6

(ii) With subtraction

 (5.4954 x 10¯4) (5.4954 x 10¯4) Ka = –––––––––––––––––––––––––– (0.120 − 5.4954 x 10¯4)

Ka = 2.53 x 10¯6

This all, of course, harkens back to the 5% rule. In the end, you do what your teacher recommends. So, ask your teacher if you're not sure.

Example #7: A 0.128 M solution of uric acid (HC5H3N4O3) has a pH of 2.39. Calculate the Ka of uric acid.

Discussion:

In this example, I'll use a real acid and numbers that lead to the actual Ka for the acid.

Before that, a comment: one reason teachers might tend to avoid real substances in this type of question is that you can just look up the answers on the Internet. For example, I used this page to get the Ka for the following problem.

Another reason for using the generic acid formula of HA is that this avoids the need to constantly write a somewhat complex formula for the anion portion of the weak acid. In fact, as I do this problem, I will write Ur¯ for the anion portion of uric acid (the C5H3N4O3¯).

Solution:

1) Determine [H3O+]:

10¯pH = 10¯2.39 = 4.0738 x 10¯3 M

2) Determine [Ur¯]:

[Ur¯] = [H3O+] = 4.0738 x 10¯3 M

3) I will use 0.128 M for [HUr]

4) Determine the Ka:

 (4.0738 x 10¯3) (4.0738 x 10¯3) Ka = ––––––––––––––––––––––––– 0.128

Ka = 1.30 x 10¯4

Example #8: HC9H7O4 (MW = 180. g/mol) is prepared by dissolving 3.60 g into a 1.00 L solution. The pH of this solution was determined to be 2.60. What is the Ka?

Solution

1) The only thing different from #6 and #7 is that you must calculate the molarity:

3.60 g / 180. g/mol = 0.0200 mol

0.0200 mol / 1.00 L = 0.0200 mol/L

2) Then apply the usual technique:

 (2.5119 x 10¯3) (2.5119 x 10¯3) Ka = ––––––––––––––––––––––––– 0.0200

Ka = 3.15 x 10¯4

Example #9: A student prepared a solution of salicylic acid (a monoprotic weak acid, MW = 138.123 g mol¯1) and measured the pH of the solution to be 2.430. Then, she evaporated 100.0 mL of the solution and collected 0.220 g of dry salicylic acid. What is the Ka of salicylic acid?

Solution:

1) Write the dissociation equation for salicylic acid, HSal:

HSal + H2O ⇌ H3O+ + Sal¯

2) Write the Ka expression for HSal:

 [H3O+] [Sal¯] Ka = ––––––––––– [HSal]

3) Use the pH to determine [H3O+] and [Sal¯]:

[H3O+] = 10¯pH = 10¯2.20 = 3.71535 x 10¯3 M (some guard digits)

We know from the stoichiometry of the chemical equation that [Sal¯] = [H3O+]

4) Determine molarity of HSal:

0.220 g / 138.123 g mol¯1 = 1.59278 x 10¯3 mol

1.59278 x 10¯3 mol / 0.100 L = 1.59278 x 10¯2 M

5) Calculate Ka:
 (3.71535 x 10¯3) (3.71535 x 10¯3) Ka = –––––––––––––––––––––––––– 1.59278 x 10¯2

Ka = 8.67 x 10¯4

Comment: I did some searching the the Ka of salicylic acid and found some striking differences in reported values. For example, I found "Quantitative Chemical Analysis" (by Daniel C. Harris) on Google Books. On page 183, he reports the pKa of salicylic acid to be 2.97; from that we see the Ka to be 1.07 x 10¯3, almost a 20% difference.

This is not to say any one number is wrong, just that various sources give different values for the Ka of salicylic acid.

Example #10: An aqueous solution containing 8.40 g/L of Cyanoacetic acid (abbreviated HCya, formula is CH2CNCOOH) has a pH of 1.77. What is the value of the Ka?

Solution:

1) Write the dissociation equation for HCya:

HCya + H2O ⇌ H3O+ + Cya¯

2) Write the Ka expression for HCya:

 [H3O+] [Cya¯] Ka = ––––––––––– [HCya]

3) Use the pH to determine [H+] and [Cya¯]:

[H3O+] = 10¯pH = 10¯1.77 = 1.69824 x 10¯2 M (I kept some guard digits)

We know from the stoichiometry of the chemical equation that [Cya¯] = [H3O+]

4) Determine molarity of HCya:

8.40 g / 85.0618 g mol¯1 = 9.875 x 10¯2 mol

1.59278 x 10¯3 mol / 1.00 L = 9.875 x 10¯2 M

5) Calculate Ka:

 (1.69824 x 10¯2) (1.69824 x 10¯2) Ka = –––––––––––––––––––––––––– 9.875 x 10¯2

Ka = 2.92 x 10¯3

Example #11: What is the Ka of a monoprotic acid in which an 0.0100 M solution exerts an osmotic pressure of 200.0 torr at 25.0 °C

Solution:

1) Determine the van 't Hoff factor:

(200.0 / 760.0) = i (0.0100) (0.08206) (298.0)

i = 1.07614 (I'll carry some guard digits.)

2) Determine the concentrations of all three species in solution:

a) We can assign concentrations using one unknown:
[H+] = [A¯] = x
[HA] = 0.0100 − x
b) The sum of all three concentrations is 0.0107614 M (from 0.0100 M times the van 't Hoff factor). Therefore:
x + x + (0.0100 − x) = 0.0107614 M
x = 0.0007614 M
3) Calculate the Ka:
Ka = ([H+] [A¯]) / [HA]

Ka = [(0.0007614) (0.0007614)] / (0.01 − 0.0007614)

Ka = 6.27 x 10¯5

A personal note about this problem: I presented the solution to a problem very much like this to a study group of about 25 high school chemistry teachers. (There were about 50 in the month-long workshop, so two study groups.) One teacher complained about my solution, pointing to a "problem" in the van't Hoff equation. He pointed out that the liters in the molarity comes from a solution-based theory and that the liters in the gas constant came from a gas-based theory. His claim was that those two units could not cancel.

Quite frankly, I didn't know what to do. This teacher did not know that one of van 't Hoff's contributions was to apply the ideal gas equation to solutions, showing that the same principles govern both gases and solutions. My personal struggle was to not hold him up to ridicule in front of his peers. I finally said we could discuss this later and proceeded with the solution.

Here's the funny part: the problem that I was solving was from Steven Zumdahl's general chemistry textbook (the second edition, it was about 1992, if I remember correctly) and Steve was leading the study group. When I came to the board, he moved to the back of the room, and wound up laughing quietly at my discomfort. He later came over to me and "apologized" for laughing at me, but I had not taken offense.

That other teacher and I talked after the workshop, but I was unable to convince him of his error. I later saw him talking to Steve, but I don't remember if it was about my presentation.

Example #12: A 0.150 M solution of acetic acid (shorthand formula = HAc) is found to be 1.086% dissociated. What is the Ka?

Solution:

The equation for the dissociation of HAc is:

HAc + H2O ⇌ H3O+ + Ac¯

The Ka expression is:

Ka = ([H3O+] [Ac¯]) / [HAc]

We need to know the three values on the right.

1) We find [H3O+] using the concentration and the percent dissociation:

(0.01086) (0.15) = 1.629 x 10¯3 M

2) For the [Ac¯], we use the 1:1 stoichiometry from the above equation to find:

[Ac¯] = [H+] = 1.629 x 10¯3 M

3) For the [HAc], we use the concentration given. We could subtract 1.629 x 10¯3 from it, but this is usually not done. Warning: you might have a teacher who does ask you to do it. Check to make sure.

4) Now, we insert into the Ka expression and solve:

x = [(1.629 x 10¯3) (1.629 x 10¯3)] / 0.15

x = 1.77 x 10¯5

Example #13: A generic weak acid (use HA for the formula) has a concentration of 0.200 M and is 1.235% dissociated. Determine the Ka.

Solution:

Please note the use of a generic weak acid in the question. The solution technique for this type of problem works for almost all weak acids. In an introductiory course, the technique works for all acids, since the more unusual cases are reserved for a more advanced course. Hence, the use of a generic acid.

1) [H3O+]:

(0.200 M) (0.01235) = 2.47 x 10¯3 M

2) [A¯]:

Equals [H3O+] of 2.47 x 10¯3 M by 1:1 stoichiometry

3) [HA]:

0.200 M

4) Insert into the Ka expression:

x = [(2.47 x 10¯3) (2.47 x 10¯3)] / 0.200

x = 3.05 x 10¯5

Example #14: 2.55 g of a generic weak acid (HA) with a molar mass of 85.0 g/mol is dissolved in 250.0 g of water. The freezing point of the resulting solution is −0.227 °C. Calculate the Ka for this acid.

Solution:

1) Calculate the molality (Reminder: molality is mol solute/kg solvent):

(2.55 g/85.0 g/mol) / 0.250 kg = 0.120 m

2) Determine the van 't Hoff factor:

Δt = i Kf m

0.227 °C = (i) (1.86 °C/m) (0.12 m)

i = 1.017025

3) Make a key assumption: the density of the solution is 1.00 g/mL. Given the dilute nature of the solution, this is a fairly safe assumption. If more exactness is required, then the density of the solution would have to be measured. The assumed density allows us to use 0.120 M for the concentration of the solution.

4) Determine the concentration of [H3O+], [A¯], and [HA]:

[H3O+] = x
[A¯] = x
[HA] = 0.120 − x

[H3O+] + [A¯] + [HA] = (0.120 M) (1.017025) = 0.122043 M

x + x + (0.120 − x) = 0.122043

x = 0.002043 M

5) By the way, the percent dissociation can also be determined from the van 't Hoff factor:

i = 1 + % dissociation. (see here)

% dissociation = 0.017025

% dissociation = [H3O+]/[ HA]o

0.017025 = x / 0.120

x = 0.002043 M

6) Determine the Ka:

x = $\sqrt{\mathrm{\left(Ka\right) \left(\left[HA\right]o\right)}}$

0.002043 = $\sqrt{\mathrm{\left(Ka\right) \left(0.120\right)}}$

0.000004173849 = (Ka) (0.120)

Ka = 3.48 x 10¯5

7) If you use 0.120 − x in the Ka calculation, you obtain 3.54 x 10¯5 for the answer. The two answers differ by 1.7%, falling well within the 5% rule for determining if the approximate answer is valid.

Bonus Example #1: How many liters of a 0.54 M monoprotic weak acid (call it HA) should be added to 685,000 L of water (pH 7.00) to reach the pH to 5.20? Ka = 7.1 x 10¯4

Solution:

1) Use the Ka expression to determine the concentration of HA needed to obtain a pH of 5.20.

 [0.0000063096] [0.0000063096] 7.1 x 10¯4 = –––––––––––––––––––––––––– [HA]

[HA] = 5.6072 x 10¯8 M

Note: 0.0000063096 M came from 10¯5.20

2) Use M1V1 = M2V2 to determine how much of the 0.54 M solution to be added.
(0.54 ) (x) = (5.6072 x 10¯8) (685000 + x)

0.54x = 0.03840795 + (5.6072 x 10¯8)x

0.539999943928x = 0.03840932

x = 0.071 L (to two sig figs, based on the pH of 5.20)

Note that the pH of the water does not play a role in the calculation.

Bonus Example #2: At 25 °C the enthalpy change, ΔH°, for the ionization of trichloroacetic acid is +6.3 kJ/mol and the entropy change, ΔS°, is +0.0084 kJ/mol-K. What is the pKa of trichloroacetic acid?

Solution:

1) The equations to use are these:

ΔG° = ΔH° − TΔS°

ΔG° = −RT ln K

2) Therefore:

ΔH° − TΔS° = −RT ln K

3) Remember this:

R = 8.31447 has the units J/mol-K, not kJ/mol-K

4) Substituting:

6300 − (298) (8.4) = − (8.31447) (298) ln K

3796.8 / −(8.31447) (298) = ln K

ln K = −1.532381450328817

K = 0.216

pKa = 0.664

5) This page gives a Ka value of 0.23 rather than the 0.216 calculated in this problem. To make it closer, we could round off the 0.216 to 0.22, but I'll just leave it the way it is.