pK (pK_w, pK_a, pK_b) 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:

  K_w = 10⁻¹⁴ = [H⁺_(aq)][OH^-]

  pK_w = 14.00 = pH + pOH

• For acids:

  ⚛ the stronger the acid the larger its K_a but the smaller its pK_a

  ⚛ the weaker the acid the smaller its K_a but the larger its pK_a

• For bases:

  ⚛ the stronger the base the larger its K_b but the smaller its pK_b

  ⚛ the weaker the base the smaller its K_b but the larger its pK_b

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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:

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!)

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:

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 10^{log value}):

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 (10^{log value}) and hence the solution to your original mathematical equation:

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:

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 = 10^−pK

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

• pK_w for the self-dissociation of water :

  pK_w = −log₁₀K_w

  K_w = 10^-pK_w

• pK_a for the dissociation of a weak acid:

  pK_a = −log₁₀K_a

  K_a = 10^-pK_a

• pK_b for the dissociation of a weak base:

  pK_b = −log₁₀K_b

  K_b = 10^-pK_b

pK_w

You've probably been using pK_w 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, K_w, also known as the ion-product for water, is:

K_w = [H⁺_(aq)][OH^-_(aq)]

or for the equivalent proton-transfer reaction,

K_w = [H₃O⁺_(aq)][OH^-_(aq)]

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

K_w = [H⁺_(aq)][OH^-_(aq)] = 10^-14

or for the equivalent proton-transfer reaction,

K_w = [H₃O⁺_(aq)][OH^-_(aq)] = 10^-14

We then define the value of pK_w as the negative logarithm of K_w:

pK_w = −log₁₀K_w

which at 25°C is:

pK_w = −log₁₀K_w = −log₁₀(10^-14) = 14

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

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

pK_a

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 (K_a) as given below:

or, for the equivalent proton-transfer reaction:

and pK_a is defined as −log₁₀K_a:

pK_a = −log₁₀K_a

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

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

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

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

If you need to convert a pK_a value into a K_a value, then:

K_a = 10^−pK_a

pK_b

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 (K_b) as shown below:

or, for the equivalent proton-transfer reaction:

And pK_b is defined as −log₁₀K_b

pK_b = −log₁₀K_b

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

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

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

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

If you need to convert a pK_b value into a K_b value, then:

K_b = 10^−pK_b

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 pK_a 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:

   Flask	Acid	concentration of 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 pK_a value than a stronger acid.

4. Determine which acid has the largest pK_a:

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

   6.93 (HB_(aq)) > 5.76 (HC_(aq)) > 4.02 (HA_(aq))

   HB_(aq) is the weakest acid.

5. Is your answer plausible?

   Have you answered the question that was asked?
   Yes, we have identified the weakest acid.

   Is your answer reasonable?
   Try another approach to determine which is the weakest acid.
   For example, convert each pK_a 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 K_a.
   The acid with the smallest K_a 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 pK_a 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 pK_a = 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?

   HC₄H₇O_2(aq) is a monoprotic acid.

   [HC₄H₇O_2(aq)] = 0.100 mol L^-1

   pK_a = 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:

   HC₄H₇O_2(aq) ⇋ H⁺_(aq) + C₄H₇O₂^-_(aq)

   Write the equation to calculate the value of the acid dissociation constant, K_a using pK_a:

   K_a = 10^-pK_a

   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, K_a using pK_a:

   K_a = 10^-pK_a = 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

5. Is your answer plausible?

   Have you answered the quuestion that was asked?
   Yes, we have calculated the pH of the solution.

   Is your answer reasonable?
   Work backwards, use this value for pH and the concentration of butanoic acid given to calculate K_a and hence pK_a:
   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

   
   pK_a = −log₁₀K_a = −log₁₀(1.51 × 10^-5) = 4.82
   Since this pK_a 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