ABSE_Buffers Lesson

Buffers

In some situations, a change in pH can cause extreme reactions to occur. For example, the pH of blood should be from 7.38-7.42. If it changes to 7 or 8, we would die. We must therefore have a mechanism to protect against even a slight change in pH. The chemical that does this is called a buffer. Buffers maintain pH, they do not make solutions neutral.

Buffer Components:

Buffers are made of two solutes, a conjugate pair containing a weak acid and a weak base.  

Here are some examples:

HC2H3O2 and NaC2H3O2

H2CO3 and HCO3 -

NH4 + and NH3 

pH of a Suffered Solution

The pH of a buffer depends on

  1. ka (or kb for a base)
  2. the ratio of [A-]/[HA]

Diluting a buffer does not change its pH because it does not change the ratio or the ka. 

Follow along with this video to learn how to calculate the pH of a buffer.

Note that you have a choice of what equation to use to begin a problem like th one above. You can either write the ka equation using the acid or the kb equation using the base.

There is a shortcut to calculate the pH of a buffer solution.   The basic ka expression can be rearranged to another very useful equation known as Henderson-Hasselbalch equation.

Henderson- Hasselbalch pH = pk +log [base]A/[acid] H

This equation can be used for buffers with the initial concentrations of the buffer components. This equation can be written for pOH as well, but it is not used as commonly.

Let's do the same problem done in the video above. But this time we will use the Henderson-Hasselbalch equation.  

What is the pH of a buffer solution containing 0.90 M HC2H3O2 and 1.10 M NaC2H3O2? pH = pk +log [A⁻]/[HA] pH = -log(1.8 x 10⁻⁵) + log (1.10/0.90) pH 4.74 + 0.08715 pH = 4.83

How Buffers Work

Now that you know how to calculate the pH of a prepared buffer, let's talk about how buffers actually work and how to prepare them. A buffer works by neutralizing any strong acid or base added. This allows the original pH to remain constant or to change only by a negligible amount. A buffer solution contains a large concentration of both members in a conjugate acid-base pair. The conjugate acid in the buffer will react to neutralize added base and the conjugate base in the buffer will react to neutralize added acid.

Consider a buffer of HA and A, where HA represents the weak acid component and A represents the weak base component of the buffer. What happens if a strong acid is added to this solution? The strong acid (shown below as H⁺) will be neutralized by one of the buffer components. It will react with the base component of the buffer (A⁻). H⁺+ A⁻→ HA Notice that this reaction is NOT an equilibrium reaction. The point is to remove the strong acid or base by reacting it completely.

Preparing a Buffer of a Given pH

Buffers are usually made so that the ratio of [A-]/[HA] is near 1. Look at the Henderson-Hasselbalch equation below. What do you notice when the ratio of [A-]/[HA] is exactly 1?

LaTeX: pH=pk_a+log \frac{[A^-]}{[HA]}pH=pka+log[A−][HA]

When this ratio is exactly 1, pH = pka. This is true any time [acid] = [base].

Buffers are also made to work around a certain pH. Here is how a buffer of a specified pH is prepared:

  1. Choose a weak acid with a pka near the desired pH. The general rule is that the desired pH = pka ± 1.
  2. Adjust the ratio of weak base to weak acid ([A-]/[HA]) to get the exact desired pH.

Example: Describe how to prepare a buffer solution of pH = 5. First we need to select a weak acid whose pk.=5±1. Look on the list of ka values in the back of your book or find one on the web. Acetic acid, HC₂H₃O₂, has a k₂ of 1.8 x 10-5. pk.log(1.8 x 10-5) pk₂ = 4.7 This is within the range of pH = 5 ± 1. Now, we need to adjust the ratio of [A]/[HA], which is [C₂H₂O₂-]/[HC2H3O2] in this example. (The specific compound paired with the acetic acid could be NaC2H3O2 or any other salt with a soluble cation and acetate as the anion.) To adjust the ratio, we start with the ka expression. ka=[H⁺][A⁺]/[HA] (Note that if you don't remember this expression, just write the balanced equation and use it to write the mass action expression.) Then, rearrange to solve for [A]/[HA]. ka/[H⁺]=[A⁺]/[HA] Now we can plug in the k.. We can use the desired pH to determine [H+]. pH = 5 = -log[H⁺] [H⁺'] = 10⁻⁵ 1.8 x 10⁻⁵/1 x 10⁻⁵ [A]/[HA] 1.8= [A⁺]/[HA] As long as the ratio between these concentrations is 1.8, their specific values could be any molarities. For example, you could use [NaC₂H₂O₂] = 1.8 M and [HC₂H₃O₂] = 1.0 M or [NaC₂H₂O₂] = 0.18 M and [HC₂H₃O₂] = 0.10 M or [NaC₂H₂O₂] = 0.432M and [HC₂H₃O₂] = 0.240 M These are just examples of solutions that have the ratio of 1.8:1.

By comparing the pH of a solution to the pKa of the acid in the solution, the concentration ratio between the acid and base forms of that acid (sometimes called the protonation state) can be determined. For example, if pH < pKa, the acid form has a higher concentration than the base form. If pH > pKa, the base form has a higher concentration than the acid form. Applications of this relationship include the use of acid-base indicators, the protonation state of protein side chains (including acids or proteins with multiple labile protons), and the pH required for acid-catalyzed reactions in organic chemistry. 

Buffer Capacity

There is a limit to the amount of strong base or strong acid that a buffer can neutralize. This limit is known as buffer capacity. Buffer capacity is the number of moles of base (or acid) that, when added to 1 liter of buffer solution, changes its pH by 1 unit. Buffer capacity is determined by the actual concentration (or number of moles) of each component present NOT by their ratio.

The more A-  and HA molecules available, the less of an effect the addition of a strong acid or strong base will have on the pH of a system. Consider the addition of a strong acid such as HCl. Initially, the HCl donates its proton (H+) the weak base (A-) through the reaction

A- + HCl → HA + Cl-

This changes the pH by lowering the ratio of [A-]/[HA], but as long as there is still a lot of A- present, the change in pH will be small. But if we keep adding HCl, the weak base A- will eventually run out. Once the A- is gone, any additional HCl will donate its proton to water (HCl + H2O ⇌  H3O+ + Cl-). This will dramatically increase the [H+] and so the pH drops.

Remember to work on the module practice problems as you complete each section of content.  

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