THE HENDERSON–HASSELBALCH EQUATION
We have learned that the pH of a solution depends on the concentration of
H^{+} (more correctly called H_{3}O^{+}) ions present
in the solution. We have also learned that weak acids don’t fully dissociate
into ions when placed in water. How much they ionize, or how “acid”
they can make the water in which they dissolve, is different for each weak
acid. The pK_{a} is a measure of the weak acid’s ability to ionize.
Thus, pH and pK_{a} seem like very similar terms, and it is important
not to get them confused. Remember, pH is a direct measure of the H^{+}
concentration, while pK_{a} represents a particular acid’s ability
to ionize (dissociate into ions, including H^{+}).
The Henderson–Hasselbalch equation
However, since the extent of ionization of a weak acid (the pK_{a})
influences the final concentration of H^{+} ions (the pH) of the solution,
there must be a relationship between pH and the pK_{a} of a weak acid.
This relationship is given by the Henderson–Hasselbalch
equation:
pH 
= 
pK_{a} 
+ 
log 
[base] 

[acid] 

Where “base” represents the conjugate base, and “acid”
the conjugate acid, of a conjugate acid–base pair (for a review of conjugate
acid–base pairs see section 5). Note that there is a way for the pH and
the pK_{a} of a solution to be the same: the term “log [base]/[acid]”
must be equal to zero! When does that happen? If we solve the equation
log [base]/[acid] = 0
[base]/[acid] = 1
[base] = [acid]


Thus, only in the rare circumstance when the molar concentrations of the
conjugate acid and conjugate base in a solution are equal, the pH = pK_{a}.
Below is a graph of a weak acid titration, which is also a graph of the HendersonHasselbalch
equation, showing how pH changes as acid or base is added. Note that when
the pH = pK_{a}, there are equivalent amounts of the conjugate acid
and base. It is also important to note that just because the amounts of
conjugate acid and base are the same at the pK_{a}, the pH is not
7 (neutral) at this point. Remember, pH is a function of the the concentration
of H_{3}O^{+}, not the conjugate acid, and because weak acids
do not ionize completely, these concentrations will not be the same.

Buffer titration/Henderson–Hasselbalch graph: how the
pH of a buffered solution changes as acid or base is added. 
It is important to remember that the pH of a weak acid solution can be calculated,
using the K_{a} value of the weak acid.
Example 4
What is the pH when 10 mL of 1 M acetic acid is added to pure water so that the final volume is 100 mL? The K_{a} for acetic acid is 1.74 x 10^{–5}.
The reaction that occurs when acetic acid dissociates in water is
CH_{3}COOH + H_{2}O CH_{3}COO^{–} + H_{3}O^{+}
It is given that 10 mL of 1 M acetic acid is diluted to 100 mL in water. Thus,
the initial acetic acid concentration is
(0.01 liter)(1 mol/liter) / (0.100 liter total volume)
= 0.1 M acetic acid.
Once the acetic acid is diluted in water, the dissociation reaction begins
and continues until it reaches equilibrium. We can use a table to organize
what happens to all the species involved in the reaction:
ACETIC
ACID DISSOCIATION TABLE 

CH_{3}COOH 
CH_{3}COO^{–} 
H_{3}O^{+} 
Initial concentration 
0.1 M 
0 
0 
Equilibrium concentration 
0.1 M – [CH_{3}COOH] 
[CH_{3}COO^{–}] 
[H_{3}O^{+}] 

We also know that the K_{a} of acetic acid is 1.74 x 10^{–5} . Using the K_{a} expression for acetic acid:
K_{a}

= 
[CH_{3}COO^{–}][ H_{3}O^{+}] 

[CH_{3}COOH]

Plugging in the known K_{a} value, as well as the
equilibrium concentrations of each species:
1.74 x 10^{–5} 
= 
[CH_{3}COO^{–}][
H_{3}O^{+}] 

0.1 – [CH_{3}COO^{–}] 
But since one H_{3}O^{+} is produced for every CH_{3}COO^{–}
ion,
[CH_{3}COO^{–}] =
[ H_{3}O^{+}]
and the expression becomes
1.74 x 10^{–5} 
= 
[ H_{3}O^{+}][
H_{3}O^{+}] 

0.1 –
[ H_{3}O^{+}] 
This is of course now just an algebra problem, where we have an quadratic equation
in the form of
1.74 x 10^{–5} 
= 
x^{2} 

0.1 – x^{2}

But don’t drag out that quadratic formula just yet! Thankfully there is
a shortcut that we can use. Remember that weak acids dissociate very poorly in
water, which means that the final CH_{3}COO^{–} concentration
will be very small compared to the 0.1 M acetic acid that we started with. This
means that the equilibrium concentration of CH_{3}COOH will be very close
to 0.1 M. This changes the K_{a} expression to
1.74 x 10^{–5}

= 
[CH_{3}COO^{–}][H_{3}O^{+}] 

0.1

or
1.74 x 10^{–5}

= 
[H_{3}O^{+}][H_{3}O^{+}] 

0.1

1.74 x 10^{–6} = [H_{3}O^{+}]^{2}
[H_{3}O^{+}] = 1.32 x 10^{–3}
Using the pH equation:
pH = – log [H_{3}O^{+}]
pH = – log (1.32 x 10^{–3})
pH = 2.88

