 # pK (pKw, pKa, pKb) Chemistry Tutorial

## Key Concepts

• The equilibrium constant for a given reaction is given the symbol K
• pK is defined as:

pK = -log10K

• So we can convert pK values into K values using the relationship:

K = 10-pK

• Some special relationships between K and pK are given below:

Type of reaction Symbol for
Equilibrium Constant
Definition of pK Finding K using pK
Self-dissociation of water Kw pKw = −log10Kw Kw = 10−pKw
Dissociation of acids Ka pKa = −log10Ka Ka = 10−pKa
Dissociation of bases Kb pKb = −log10Kb Kb = 10−pKb

• For water at 25°C and atmospheric pressure:

Kw = 10−14 = [H+(aq)][OH-]

pKw = 14.00 = pH + pOH

• For acids:

⚛ the stronger the acid the larger its Ka but the smaller its pKa

⚛ the weaker the acid the smaller its Ka but the larger its pKa

• For bases:

⚛ the stronger the base the larger its Kb but the smaller its pKb

⚛ the weaker the base the smaller its Kb but the larger its pKb

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## Defining pK : and a quick review of logarithms (base 10)

For the general reaction:

aA + bB ⇋ cC + dD

The mathematical expression to calculate the equilibrium constant (K) for this reaction is:

 K = [C]c[D]d [A]a[B]b

For different reactions, even under the same conditions, the values of K can be enormously different, of the order of 102 down to 10-20

Now, in the bad old days, before the invention of digital devices, even before electronic calculators, calculations involving these numbers was tedious and prone to errors for those who weren't paying enough attention to what they were doing.

Just try dividing 1.64 × 10-3 by 7.39 × 10-12 without the aid of an electronic device!
And for an encore, try calculating the square root of 7.39 × 10-12 in your head !

One piece of equipment every student in these "dark ages" had access to was a book containing "log tables".
Students in these evil times became adept at using these "log tables" to find the logarithm of numbers like 1.64 × 10-3.
And this was useful because it simplified calculations so much that you could do them in your head!

If you want to multiply 1.64 × 10-3 by 7.39 × 10-12 you use your "log tables" to find the log10 of each number (you can use your electronic device instead if you don't have access to log tables, or if, like me, you've effectively forgotten how to use them!)

Number: logarithm (base 10)
1.64 × 10-3 -2.785
7.39 × 10-12 -11.131

If you want to multiply these two numbers we just add their logarithms, if we want to divide these two numbers we just subtract one logarithm from the other:

Mathematical operation   using logarithms (base 10)
(1.64 × 10-3) × (7.39 × 10-12)   (-2.785) + (-11.131) = -13.916
(1.64 × 10-3) ÷ (7.39 × 10-12)   (-2.785) - (-11.131) = 8.346

Then you could reverse the process to find the number corresponding to your logarithm using your "log book", that is, we find the antilog (you can use your electronic device since the reverse process, the antilog, is 10log value):

Mathematical operation   using logarithms (base 10)   10log
(1.64 × 10-3) × (7.39 × 10-12)   (-2.785) + (-11.131) = -13.916   1.213 × 10-14
(1.64 × 10-3) ÷ (7.39 × 10-12)   (-2.785) - (-11.131) = 8.346   2.218 × 108

Similarly, if you wanted to square a number you multiply its log10 by 2, but if you want to find the square root you divide its log10 by 2, then you reverse the process to find the antilog (10log value) and hence the solution to your original mathematical equation:

Mathematical operation   using logarithms (base 10)   10log
(1.64 × 10-3)2   (-2.785) × 2 = -5.57   2.692 × 10-6
√(1.64 × 10-3)   (-2.785) ÷ 2 = -1.3193   4.050 × 10-2

So, in the "old days" there was value in converting really big or really small numbers to log10 numbers because it simplified the maths (after you managed to find the numbers in the "log tables" ofcourse!).

Now you may have noticed that the range of K values we talked about above seems to fall mostly in the range below 0 rather than above 0.
So chemists defined their "p" values as -1 × the "log10 value" which results in a positive value most of the time for numbers chemists are interested in:

Number   logarithm (base 10)   -1 × logarithm (base 10)
1.64 × 10-3   -2.785   2.785
7.39 × 10-12   -11.131   11.131

This doesn't change the operations: we still add log10s when we want to multiply numbers, subtract log10s when we want to divide one number by another, multiply log10s by 2 to square a number, and, divide a log10 by 2 to find the square root of a number, however, it does introduce a small change into the reverse process when we want to turn our log10 value into a base 10 number because first we must remember to multiply our log10 value by −1:

pK = −1 × log10K

K = 10(−1 × pK)

And that's it really. The only reason we have pK values is that it makes the numbers nicer to play with!

Just remember:

• K is the value of the equilibrium constant for a given reaction at a specified temperature and pressure
• For the same reaction under the same conditions:

pK = −log10K

• and

K = 10pK

The equilibrium constants you will often found tabulated as pK values are:

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## pKw

You've probably been using pKw values for a very long time, but did you realise you were doing it?
Have you ever performed a calculation using the equation below?

pH + pOH = 14.00 (for aqueous solutions at 25°C)

You may recall that we define pH and pOH as follows:

pH = −log10[H+(aq)] (or alternatively, pH = −log10[H3O+(aq)])

pOH = −log10[OH-(aq)]

So where does the 14.00 come from?

It comes from the value for the equilibrium constant when water self-dissociates (self-ionises or auto-dissociates):

H2O(l) ⇋ H+(aq) + OH-(aq)

or, the equivalent proton-transfer reaction:

2H2O(l) ⇋ H3O+(aq) + OH-(aq)

For which the expression for the equilibrium constant, Kw, also known as the ion-product for water, is:

Kw = [H+(aq)][OH-(aq)]

or for the equivalent proton-transfer reaction,

Kw = [H3O+(aq)][OH-(aq)]

and, at 25°C, the value of this equilibrium constant is 10-14

Kw = [H+(aq)][OH-(aq)] = 10-14

or for the equivalent proton-transfer reaction,

Kw = [H3O+(aq)][OH-(aq)] = 10-14

We then define the value of pKw as the negative logarithm of Kw:

pKw = −log10Kw

which at 25°C is:

pKw = −log10Kw = −log10(10-14) = 14

Remember the magic of using logarithms we discussed above?
If you want to multiply two numbers you add their logarithms ....

 Kw = [H+(aq)] × [OH-(aq)] −log10Kw = −log10[H+(aq)] + −log10[OH-(aq)] pKw = pH + pOH 14 = pH + pOH

This is important.

If we know the pH of an aqueous solution at 25°C we can calculate its pOH:

pOH = 14.00 - pH

or, if we know its pOH we can calculate its pH:

pH = 14.00 - pOH

Similarly, if we know the solution's pH, we can calculate its hydrogen ion concentration in mol L-1:

[H+(aq)] = 10-pH

or, if we know the pOH we can calculate the concentration of hydroxide ions in mol L-1

[OH-(aq)] = 10-pOH

## pKa

A weak acid only partially dissociates (ionises) in aqueous solution, so undissociated acid molecules (HA(aq)) are in equilibrium with a certain concentration of hydrogen ions (H+(aq)) and anions (A-(aq)) as given by the general chemical equation below:

HA(aq) ⇋ H+(aq) + A-(aq)

or for the equivalent proton-transfer reaction in water:

HA(aq) + H2O(l) ⇋ H3O+(aq) + A-(aq)

We can write an expression for the equilibrium constant for this acid dissociation (Ka) as given below:

 Ka = [H+(aq)][A-(aq)] [HA(aq)]

or, for the equivalent proton-transfer reaction:

 Ka = [H3O+(aq)][A-(aq)] [HA(aq)]

and pKa is defined as −log10Ka:

pKa = −log10Ka

We can therefore calculate the value of pKa for any weak acid if we know the value of its acid dissociation constant (Ka):

acid strength acid formula Ka Trend in Ka pKa
=log10Ka
Trend in pKa
weakest ammonium ion
NH4+ 5.6 × 10-10 smaller Ka 9.25 largest pKa
boric acid H3BO3 5.8 × 10-10 9.24
hydrocyanic acid
(hydrogen cyanide)
HCN 6.3 × 10-10 9.20
hydrobromous acid
HOBr 2.4 × 10-9 8.62
hypochlorous acid HOCl 2.9 × 10-8 7.54
propanoic acid
C2H5COOH 1.3 × 10-5 4.89
acetic acid
(ethanoic acid)
CH3COOH 1.8 × 10-5 4.74
benzoic acid C6H5COOH 6.4 × 10-5 4.19
lactic acid
HC3H5O3 1.4 × 10-4 3.85
formic acid
(methanoic acid)
HCOOH 1.8 × 10-4 3.74
nitrous acid HNO2 7.2 × 10-4 3.14
hydrofluoric acid HF 7.6 × 10-4 3.12
strongest
(of these weak acids)
chlorous acid HClO2 1.1 × 10-2 larger Ka 1.96 smallest pKa

Can you see any patterns (or trends) in the data in the table?

• for weak acids, Ka << 1
• for weak acids, pKa > 1
• the weaker the acid, the smaller the value of Ka BUT the larger the value of pKa
• the stronger the acid, the larger the value of Ka BUT the smaller the value of pKa

So, if we are given two or more weak acids to compare, the weakest acid will have the smallest Ka but the largest pKa.
And the strongest weak acid will have the largest Ka but the smallest pKa

If you need to convert a pKa value into a Ka value, then:

Ka = 10−pKa

## pKb

A weak base only partially dissociates (ionises) in aqueous solution, so undissociated base molecules (BOH(aq)) are in equilibrium with a certain concentration of hydroxide ions (OH-(aq)) and cations (B+(aq)) as given by the general chemical equation below:

BOH(aq) ⇋ B+(aq) + OH-(aq)

or, for the equivalent proton-transfer reaction:

B(aq) + H2O(l) ⇋ BH+(aq) + OH-(aq)

We can write an expression for the equilibrium constant for the dissociation of the base (Kb) as shown below:

 Kb = [B+(aq)][OH-(aq)] [BOH(aq)]

or, for the equivalent proton-transfer reaction:

 Kb = [BH+(aq)][OH-(aq)] [B(aq)]

And pKb is defined as −log10Kb

pKb = −log10Kb

Therefore we can calculate the value of pKb for any weak base if we know the value of its base dissociation constant (Kb):

base strength base formula Kb Trend in Kb pKb
=log10Kb
Trend in pKb
weakest phosphine PH3 1 × 10-14 smallest Kb 14.00 largest pKb
hydroxylamine NH2OH 9.1 × 10-9 8.04
ammonia NH3 1.8 × 10-5 4.74
strongest
(of these weak bases)
methanamine
(methylamine)
CH3NH2 4.4 × 10-4 largest Kb 3.36 smallest pKb

Can you see patterns (or trends) in the data in the table?

• For weak bases, Kb << 1
• For weak bases, pKb > 1
• The weaker the base the smaller the value of Kb BUT the larger the value of pKb
• The stronger the base the larger the value of Kb BUT the smaller the value of pKb

If we are asked to compare two or more bases, the weakest base will have the smallest Kb but the largest pKb.
And the strongest base will have the largest Kb BUT the smallest pKb.

If you need to convert a pKb value into a Kb value, then:

Kb = 10−pKb

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## Worked Examples of Using pK Values

Question 1: Chris the Chemist has been given three stoppered flasks labelled A, B, and C.
Each flask contains 0.100 mol L-1 of the weak acids, HA(aq), HB(aq) and HC(aq), respectively.
The respective pKa values for the contents of each flask are as follows: 4.02, 6.93, 5.76
Which acid, HA(aq), HB(aq) or HC(aq), contains the weakest acid?

Solution:

(Based on the StoPGoPS approach to problem solving)

1. What have you been asked to do?

Identify the weakest acid.

2. What data (information) have you been given in the question?

Organise the data into a table for easy reference:

aqueous solution (mol L-1)
pKa
A HA(aq) 0.100 4.02
B HB(aq) 0.100 6.93
C HC(aq) 0.100 5.76

3. What is the relationship between what you have been given and what you need to find?

A weaker acid has a larger pKa value than a stronger acid.

4. Determine which acid has the largest pKa:

Arrange the acids in order from the largest to smallest values of pKa:

6.93 (HB(aq)) > 5.76 (HC(aq)) > 4.02 (HA(aq))

HB(aq) is the weakest acid.

Yes, we have identified the weakest acid.

Try another approach to determine which is the weakest acid.
For example, convert each pKa to a K(a)

Acid pKa Ka = 10-pKa
HA(aq) 4.02 9.55 × 10-5
HB(aq) 6.93 1.17 × 10-7
HC(aq) 5.76 1.74 × 10-6

The weakest acid is the acid that dissociates least, that is, the concentration of ions in solution will be least, therefore [H+][anion] will be smallest, so given all the solutions are the same concentration of monoprotic acid, the acid with the lowest concentration of ions will have the lowest value of Ka.
The acid with the smallest Ka value is HB(aq) (1.17 × 10-7 is smaller than either 1.74 × 10-6 or 9.55 × 10-5).
HB(aq) is the weakest acid.

Since this answer agrees with the answer we got by examining pKa values we are reasonably confident that our answer is plausible.

6. State the solution to the problem (identify the weakest acid):

Weakest acid is HB(aq)

Question 2: Butanoic acid, HC4H7O2, is a monoprotic organic acid.
For an aqueous solution of butanoic acid pKa = 4.82 (at 25°C).
Determine the pH of a 0.100 mol L-1 aqueous solution of butanoic acid.

Solution:

(Based on the StoPGoPS approach to problem solving)

1. What have you been asked to do?

Calculate pH

pH = ?

2. What data (information) have you been given in the question?

HC4H7O2(aq) is a monoprotic acid.

[HC4H7O2(aq)] = 0.100 mol L-1

pKa = 4.82 (at 25°C)

3. What is the relationship between what you have been given and what you need to find?

Write the equation for the dissociation of a monoprotic acid:

HC4H7O2(aq) ⇋ H+(aq) + C4H7O2-(aq)

Write the equation to calculate the value of the acid dissociation constant, Ka using pKa:

Ka = 10-pKa

Write an expression for the acid dissociation constant:

 Ka = [H+(aq)][C4H7O2-(aq)] [HC4H7O2(aq)]

And rearrange this expression to find [H+(aq)]

 Multiply both sides of the equation by [HC4H7O2(aq)] Ka[HC4H7O2(aq)] = [H+(aq)][C4H7O2-(aq)] Recognise that [H+(aq)] = [C4H7O2-(aq)], so Ka[HC4H7O2(aq)] = [H+(aq)]2 Assume that the acid dissociates only a tiny amount so that its concentration after dissociation does not change significantly, so √(Ka[HC4H7O2(aq)]) = [H+(aq)]

Calculate the pH of the solution:

pH = −log10[H+(aq)]

4. Substitute in the values and solve to find pH:

Write the equation to calculate the value of the acid dissociation constant, Ka using pKa:

Ka = 10-pKa = 10−4.82

Find [H+(aq)]

 [H+(aq)] = √(Ka[HC4H7O2(aq)]) [H+(aq)] = √(10−4.82 × 0.100) [H+(aq)] = √(1.51 × 10-6) [H+(aq)] = 1.23 × 10-3 mol L-1

Calculate the pH of the solution:

pH = −log10[H+(aq)] = −log10[1.23 × 10-3] = 2.91

Yes, we have calculated the pH of the solution.

Work backwards, use this value for pH and the concentration of butanoic acid given to calculate Ka and hence pKa:
pH = 2.91
[H+] = 10−pH = 10−2.91 = 1.23 × 10-3 mol L-1
HC4H7O2 ⇋ H+ + C4H7O2-

 Ka = [H+][C4H7O2-][HC4H7O2]

[H+] at equilibrium = [C4H7O2-]
Equilibrium concentration of HC4H7O2 ≈ 0.100 mol L-1 (assuming it dIssociates only a tiny amount)
 Ka = [H+]2[HC4H7O2] Ka = [1.23 × 10-3]2[0.100] Ka = 1.51 × 10-6[0.100] Ka = 1.51 × 10-5

pKa = −log10Ka = −log10(1.51 × 10-5) = 4.82
Since this pKa value agrees with that given in the question we are reasonably confident that the pH of the solution is 2.91

6. State the solution to the problem (pH of solution):

pH = 2.91

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