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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 10^{2} 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 log_{10} 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 10^{log value}):
Mathematical operation 

using logarithms (base 10) 

10^{log} 
(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 × 10^{8} 
Similarly, if you wanted to square a number you multiply its log_{10} by 2, but if you want to find the square root you divide its log_{10} by 2, then you reverse the process to find the antilog (10^{log value}) and hence the solution to your original mathematical equation:
Mathematical operation 

using logarithms (base 10) 

10^{log} 
(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 log_{10} 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 "log_{10} 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 log_{10}s when we want to multiply numbers, subtract log_{10}s when we want to divide one number by another, multiply log_{10}s by 2 to square a number, and, divide a log_{10} 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 log_{10} value into a base 10 number because first we must remember to multiply our log_{10} value by −1:
pK = −1 × log_{10}K
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 = −log_{10}K
 and
K = 10^{−pK}
The equilibrium constants you will often found tabulated as pK values are:
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 = −log_{10}[H^{+}_{(aq)}] (or alternatively, pH = −log_{10}[H_{3}O^{+}_{(aq)}])
pOH = −log_{10}[OH^{}_{(aq)}]
So where does the 14.00 come from?
It comes from the value for the equilibrium constant when water selfdissociates (selfionises or autodissociates):
H_{2}O_{(l)} ⇋ H^{+}_{(aq)} + OH^{}_{(aq)}
or, the equivalent protontransfer reaction:
2H_{2}O_{(l)} ⇋ H_{3}O^{+}_{(aq)} + OH^{}_{(aq)}
For which the expression for the equilibrium constant, K_{w}, also known as the ionproduct for water, is:
K_{w} = [H^{+}_{(aq)}][OH^{}_{(aq)}]
or for the equivalent protontransfer reaction,
K_{w} = [H_{3}O^{+}_{(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 protontransfer reaction,
K_{w} = [H_{3}O^{+}_{(aq)}][OH^{}_{(aq)}] = 10^{14}
We then define the value of pK_{w} as the negative logarithm of K_{w}:
pK_{w} = −log_{10}K_{w}
which at 25°C is:
pK_{w} = −log_{10}K_{w} = −log_{10}(10^{14}) = 14
Remember the magic of using logarithms we discussed above?
If you want to multiply two numbers you add their logarithms ....
K_{w} = 
[H^{+}_{(aq)}] 
× 
[OH^{}_{(aq)}] 
−log_{10}K_{w} = 
−log_{10}[H^{+}_{(aq)}] 
+ 
−log_{10}[OH^{}_{(aq)}] 
pK_{w} = 
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}
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 protontransfer reaction in water:
HA_{(aq)} + H_{2}O_{(l)} ⇋ H_{3}O^{+}_{(aq)} + A^{}_{(aq)}
We can write an expression for the equilibrium constant for this acid dissociation (K_{a}) as given below:
K_{a} = 
[H^{+}_{(aq)}][A^{}_{(aq)}] [HA_{(aq)}] 
or, for the equivalent protontransfer reaction:
K_{a} = 
[H_{3}O^{+}_{(aq)}][A^{}_{(aq)}] [HA_{(aq)}] 
and pK_{a} is defined as −log_{10}K_{a}:
pK_{a} = −log_{10}K_{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}):
acid strength 
acid 
formula 
K_{a} 
Trend in K_{a} 
pK_{a} =log_{10}K_{a} 
Trend in pK_{a} 
weakest 
ammonium ion

NH_{4}^{+} 
5.6 × 10^{10} 
smaller K_{a} 
9.25 
largest pK_{a} 
↓ 
boric acid 
H_{3}BO_{3} 
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

C_{2}H_{5}COOH 
1.3 × 10^{5} 
↓ 
4.89 
↑ 
↓ 
acetic acid (ethanoic acid) 
CH_{3}COOH 
1.8 × 10^{5} 
↓ 
4.74 
↑ 
↓ 
benzoic acid 
C_{6}H_{5}COOH 
6.4 × 10^{5} 
↓ 
4.19 
↑ 
↓ 
lactic acid

HC_{3}H_{5}O_{3} 
1.4 × 10^{4} 
↓ 
3.85 
↑ 
↓ 
formic acid (methanoic acid) 
HCOOH 
1.8 × 10^{4} 
↓ 
3.74 
↑ 
↓ 
nitrous acid 
HNO_{2} 
7.2 × 10^{4} 
↓ 
3.14 
↑ 
↓ 
hydrofluoric acid 
HF 
7.6 × 10^{4} 
↓ 
3.12 
↑ 
strongest (of these weak acids) 
chlorous acid 
HClO_{2} 
1.1 × 10^{2} 
larger K_{a} 
1.96 
smallest pK_{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^{−pKa}
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 protontransfer reaction:
B_{(aq)} + H_{2}O_{(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:
K_{b} = 
[B^{+}_{(aq)}][OH^{}_{(aq)}] [BOH_{(aq)}] 
or, for the equivalent protontransfer reaction:
K_{b} = 
[BH^{+}_{(aq)}][OH^{}_{(aq)}] [B_{(aq)}] 
And pK_{b} is defined as −log_{10}K_{b}
pK_{b} = −log_{10}K_{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}):
base strength 
base 
formula 
K_{b} 
Trend in K_{b} 
pK_{b} =log_{10}K_{b} 
Trend in pK_{b} 
weakest 
phosphine 
PH_{3} 
1 × 10^{14} 
smallest K_{b} 
14.00 
largest pK_{b} 
↓ 
hydroxylamine 
NH_{2}OH 
9.1 × 10^{9} 
↓ 
8.04 
↑ 
↓ 
ammonia 
NH_{3} 
1.8 × 10^{5} 
↓ 
4.74 
↑ 
strongest (of these weak bases) 
methanamine (methylamine) 
CH_{3}NH_{2} 
4.4 × 10^{4} 
largest K_{b} 
3.36 
smallest pK_{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^{−pKb }