In the previous post, we talked about the pH of buffer solutions and the two approaches to calculating it.

Let’s now discuss how to prepare a buffer with a given pH. So, we need to choose a proper acid/base pair with certain concentrations. When choosing the acid, remember that the acid should have a pK_a as close to the desired pH value as possible.

This can be seen from the Henderson–Hasselbalch equation.

 

 

When [A^–] = [HA], pH = pK_a since the log term is equal to zero.

For example, if we need a buffer with pH = 4.75, an acid with a pK_a close to 4.75 would be a great candidate.

The pK_a of acetic acid is 4.75, and if a buffer solution contains, for example , 0.2 M CH₃CO₂^– and 0.2 M CH₃CO₂H, the pH will be equal to the pKa:

 

\[{\rm{pH}}\; = \;{\rm{p}}{K_{\rm{a}}}\;{\rm{ + }}\;{\rm{log}}\frac{{{\rm{[C}}{{\rm{H}}_{\rm{3}}}{\rm{C}}{{\rm{O}}_{\rm{2}}}^{\rm{ – }}{\rm{]}}}}{{{\rm{[C}}{{\rm{H}}_{\rm{3}}}{\rm{C}}{{\rm{O}}_{\rm{2}}}{\rm{H]}}}}\]

 

\[{\rm{pH}}\; = \;4.75\; + \,\cancel{{{\rm{log}}\frac{{{\rm{0}}{\rm{.2}}\;{\rm{M}}}}{{{\rm{0}}{\rm{.2}}\;{\rm{M}}}}}}\; = \,4.75\]

 

So, how do we determine the ratio of A^–/HA if the pH is different than the pK_a of the acid with the closest value?

For example, let’s say we need a buffer solution with pH = 4.0.

The closest match is going to be the acetate buffer and to determine the [CH₃CO₂^–]/[CH₃CO₂H] ratio required to prepare it, we need to rearrange the Henderson-Hasselbalch equation to obtain an expression for [CH₃CO₂^–] and [CH₃CO₂H]:

 

\[{\rm{pH}}\; = \;{\rm{p}}{K_{\rm{a}}}\;{\rm{ + }}\;{\rm{log}}\frac{{{\rm{[C}}{{\rm{H}}_{\rm{3}}}{\rm{C}}{{\rm{O}}_{\rm{2}}}^{\rm{ – }}{\rm{]}}}}{{{\rm{[C}}{{\rm{H}}_{\rm{3}}}{\rm{C}}{{\rm{O}}_{\rm{2}}}{\rm{H]}}}}\]

 

Therefore,

 

\[{\rm{log}}\frac{{{\rm{[C}}{{\rm{H}}_{\rm{3}}}{\rm{C}}{{\rm{O}}_{\rm{2}}}{\rm{Na]}}}}{{{\rm{[C}}{{\rm{H}}_{\rm{3}}}{\rm{C}}{{\rm{O}}_{\rm{2}}}{\rm{H]}}}}\;\; = \;{\rm{pH}}\; – \;{\rm{p}}{K_{\rm{a}}}\; = \;4.0\; – \;4.75\; = \; – 0.75\]

 

The antilog of this expression is:

 

antilog (-0.75) = 10^-0.75 = 0.18

 

\[{10^{\left( {{\rm{log}}\frac{{[{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{C}}{{\rm{O}}_{\rm{2}}}{\rm{Na}}]}}{{[{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{C}}{{\rm{O}}_{\rm{2}}}{\rm{H}}]}}} \right)}}\;\; = \;{10^{\left( { – 0.75} \right)}}\]

 

\[\frac{{{\rm{[C}}{{\rm{H}}_{\rm{3}}}{\rm{C}}{{\rm{O}}_{\rm{2}}}{\rm{Na]}}}}{{{\rm{[C}}{{\rm{H}}_{\rm{3}}}{\rm{C}}{{\rm{O}}_{\rm{2}}}{\rm{H]}}}}\; = \;0.18\]

 

This means that, for example, if the solution contains 1.0 M CH₃CO₂H, the concentration of CH₃CO₂^– would have to be 0.18 M.

 

Check Also

• Buffer Solutions
• The Henderson–Hasselbalch Equation
• The pH of a Buffer Solution
• The Common Ion Effect
• The pH and pK_a Relationship
• 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