The Henderson–Hasselbalch equation shows the correlation of the p*K*_{a} and pH of the buffer solution.

The p*K*_{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 p*K*_{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 = p**.

*K*_{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 < p**) because the logarithm is a negative number. This also makes sense intuitively; more HA means more acid, and thus a lower pH.

*K*_{a}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 > p**

*K*_{a}_{:}

**For example, **calculate the pH of a solution consisting of 0.65 *M *NH_{4}Cl and 0.25 *M *NH_{3}. *K*_{b} (NH_{3}) = 1.8 x 10^{-5}.

So, right from the concentrations of the base (NH_{3}), 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 p*K*_{a} of the acid is not given, you need to determine it from wither the *K*_{a}/*K*_{b, }or p*K*_{a}/ p*K*_{b} relationship. Let’s use the p*K*_{a}/ p*K*_{b} relationship this time:

p*K*_{b }= -log *K*_{b} = -log 1.8 x 10^{-5} = 4.75

p*K*_{a} + p*K*_{b} = 14

**p K_{a}** = 14 – p

*K*

_{b }= 14 – 4.75 =

**9.25**

And now that we know the p*K*_{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 < p K_{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. p*K*_{a} (HF) = 3.8.

This time we have the p*K*_{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 > p**

*K*_{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