The Henderson–Hasselbalch equation shows the correlation of the pK_a and pH of the buffer solution.

The pK_a of a given acid, or the conjugate acid of the base, is constant and therefore, the only variable in the equation is the logarithm term which depends on the acid-base pair concentrations.

One application of this relationship is the possibility to predict whether the pH of the solution is going to be equal to or different than the pK_a of the acid. There are three scenarios here:

1) when [HA] = [A^–], 2) [HA] > [A^–], and 3) [HA] < [A^–].

 

1) When [HA] = [A^–], the logarithm becomes zero, and therefore, the pH = pK_a.

2) [HA] > [A^–]: If there is more acid than its conjugate base in the solution, then the pH will be smaller than the pKa (pH < pK_a) because the logarithm is a negative number. This also makes sense intuitively; more HA means more acid, and thus a lower pH.

3) [HA] < [A^–]: If the conjugate base is present in a larger quantity, the solution goes more basic because the logarithm now is greater than zero. So, in this case, pH > pK_a:

 

 

For example, calculate the pH of a solution consisting of 0.65 M NH₄Cl and 0.25 M NH₃. K_b (NH₃) = 1.8 x 10^-5.

So, right from the concentrations of the base (NH₃), and its conjugate acid, we can see that [HA] > [A^–], therefore the pH is going to be smaller than the pKa. Let’s confirm this by doing the calculations too.

If the pK_a of the acid is not given, you need to determine it from wither the K_a/K_b, or  pK_a/ pK_b relationship. Let’s use the pK_a/ pK_b relationship this time:

 

pK_b = -log K_b = -log 1.8 x 10^-5 = 4.75

pK_a + pK_b = 14

pK_a = 14 – pK_b = 14 – 4.75 = 9.25

 

And now that we know the pK_a and the concentrations of HA and A^–, we can calculate the pH using the Henderson–Hasselbalch equation:

 

\[{\rm{pH}}\; = \;{\rm{p}}{K_{\rm{a}}}\;{\rm{ + }}\;{\rm{log}}\frac{{{\rm{[N}}{{\rm{H}}_3}{\rm{]}}}}{{{\rm{[N}}{{\rm{H}}_4}{\rm{Cl]}}}}\;\]

 

\[{\rm{pH}}\; = \;9.25\; + \,{\rm{log}}\frac{{{\rm{0}}{\rm{.25}}\;M}}{{{\rm{0}}{\rm{.65}}\;{\rm{M}}}}\; = \;8.84\]

 

As expected, pH < pK_a (8.84 vs 9.25) because [HA] > [A^–].

 

An example when the concentration of the acid is smaller than the conjugate bases:

Calculate the pH of a solution containing 0.25 M HF and 0.55 M NaF. pK_a (HF) = 3.8.

This time we have the pK_a, so all we need to do is plug in the numbers:

 

\[{\rm{pH}}\; = \;{\rm{p}}{K_{\rm{a}}}\;{\rm{ + }}\;{\rm{log}}\frac{{{\rm{[}}{{\rm{F}}^{\rm{ – }}}{\rm{]}}}}{{{\rm{[HF]}}}}\]

 

\[{\rm{pH}}\; = \;{\rm{3}}{\rm{.8}}\;{\rm{ + }}\,{\rm{log}}\frac{{{\rm{0}}{\rm{.55}}\;{\rm{M}}}}{{{\rm{0}}{\rm{.25}}\;{\rm{M}}}}\;{\rm{ = }}\;{\rm{4}}{\rm{.14}}\]

 

And this confirms that if [HA] < [A^–] then pH > pK_a.

 

Check Also

• Buffer Solutions
• The Henderson–Hasselbalch Equation
• The pH of a Buffer Solution
• Preparing a Buffer with a Specific pH
• The Common Ion Effect
• Strong Acid–Strong Base Titrations
• Titration of a Weak Acid by a Strong Base
• Titration of a Weak Base by a Strong Acid
• Titration of Polyprotic Acids
• Buffer Solutions Practice Problems
• K_sp and Molar Solubility
• The Effect of a Common Ion on Solubility
• The Effect of pH on Solubility
• Will a Precipitate Form? K_sp and Q
• K_sp and Molar Solubility Practice Problems