Preparing a Buffer — Overview & Methods

Buffer solutions are extremely useful in both biology and chemistry, so it is very important to know how to prepare one!

A buffer is designed to resist changes in the pH value of a solution when small amounts of a different acid or base are introduced to the solution. Maintaining a certain pH range is important, because if the pH is too high or too low, certain reactions will either not take place at all or do so very slowly. In fact, preparing a buffer solution is often one of the first tasks required of a new lab tech. While considered tedious or sometimes difficult, there are a few different methods that can be used.

Recall that buffer solutions are able to resist changes in pH because of the acid-base conjugates they contain that neutralize $H^{+}$ and $OH^{-}$ ions. These solutions consist of weak acids or bases along with the salt of that acid or base.

The first step in preparing a buffer solution is to select an appropriate buffer system for your desired pH range:

$$\begin{array}{r|l} Buffer System & pH Range \\ \hline Carboxylic Acid & 3-6\\ Phosphate & 6.5-7.5 \\ Borate & 8.5-10\\ Amino Acid & Varies \end{array}$$

Once you have selected a buffer system, it is time to glove up and put your safety goggles on! We will discuss four different methods that you could follow to prepare your buffer:

1. Start with the weak acid and its conjugate base by dissolving the acid form of the buffer in about three-fifths of the volume of water required to obtain the final solution volume, and mix the two solutions until your desired pH is achieved, using a pH probe to monitor. When the pH is correct, dilute the solution to the final desired volume.

1. Start with the weak acid and add strong base until the required pH is achieved, diluting to the final desired volume at the end.

1. Start with the weak base and add strong acid until the required pH is achieved, diluting to the final desired volume at the end.

Note: Dilution will not have any significant effect on pH

1. Determine the exact amount of acid and conjugate base needed to make a buffer of a certain pH by using the Henderson-Hasselbalch equation:

$pH = pK_{a} + log(\frac{|A^{-}|}{|HA|})$

where,

-pH is the concentration of [$H^{+}$]

-$pK_{a}$ is the acid dissociation constant,

• and [$A^{-}$] and [HA] are concentrations of the conjugate base and starting acid

In some situations, the Henderson-Hasselbalch approximation can be used instead, making the math a bit simpler:

$pH = pK_{a} + log[A^{-}]$

This approximation is only valid IF:

1. The conjugate acid/base concentration falls between 0.1 and 10

1. The buffer's molarity exceeds $K_{a}$ by a factor of 100+

The Henderson-Hasselbalch equation is used to mathematically relate a measurable pH of a solution to the $pK_{a}$ of an acid, recall that $pK_{a}$ is equal to $-logK_{a}$. With a provided pH and known $pK_{a}$, the equation outputs a logarithm of the ratio of acid and base needed to achieve a certain pH.

Let's work through an example to hopefully make sense of this!

Example:

Say you have 0.250 L of 0.300 M $HNO_{2}$. What volume of 6.00 M NaOH must you add to prepare a buffer of pH=4.00?

$HNO_{2} + OH^{-} \longrightarrow H_{2}O + NO^{-}_{2}$

Solution:

First, look up the $K_{a}$ value for nitrous acid and convert it into its $pK_{a}$ value

$K_{a}= 4.0 x 10^{-4}$, so

$pK_{a}= -log(4.0 x 10^{-4})= 3.40$

Now that you have a pH and a $pK_{a}$ value, fill in the Henderson-Hasselbalch equation

$pH = pK_{a} + log [base/acid]$

4.00 = 3.40 +log [base/acid]

From here, determine the moles of nitrous acid found in the buffer fill in the rest of the equation

(0.250 L $HNO_{2}$)(0.300 $\frac{mol}{L} HNO_{2})$= 0.0750 mol $HNO_{2}$

$4.00 = 3.40 + log (\frac{x}{0.0750-x})$

The 'x' is the amount of $NO^{-}_{2}$ that is produced when $HNO_{2}$ reacts with the $OH^{-}$ and also the amount of $HNO_{2}$ that reacted, in terms of moles. Since the reaction occurs with a 1:1 molar ratio, 'x' is the amount of NaOH that needs to be added to achieve the desired pH.

Next, solve using algebra!

$4.00 = 3.40 + log[\frac{x}{(0.0750 - x)}]$

$log [\frac{x}{(0.0750-x)}] = 0.6$

$\frac{x}{(0.0750 - x)}= 3.98$

$0.2985 - 3.98 x = x$

$0.2985 = 4.98 x$

$x = 0.05994$ moles of NaOH to be added

And last, convert to volume using stoichiometry!

$0.05994 mol NaOH \times \frac{1 L}{6.00 mol}= 0.00999 L$

Use two significant figures, based on the $K_{a}$ value

$0.00999 L \times \frac{1000 mL}{1 L} = 10 mL$

We have determined that 10 mL of 6.00 M NaOH must be added to 0.250 L of 0.300 M $HNO_{2}$ to prepare a buffer at pH=4.00!

(Video) 17.1 Buffers | General Chemistry

by Chad's Prep

In this section of the video, Chad explains buffers. He starts by defining what a buffer is. Then, he shows how to approximate a buffer's pH from its pKa values. Finally, he illustrates how to make a buffer. He also shares some common ways your teacher might trick you. This is a great video for understanding how to make buffers!