Chemistry: Buffers

Buffers

Your blood is a very slightly basic solution. Let's consider what happens when you drink a big bottle of soda (pH ~ 3):

• The acid in the soda neutralizes the small quantity of base in your blood.
• The remaining acid in the soda causes the pH of your blood to decrease rapidly to a pH of about 5.
• Because the enzymes in your body don't work well at a pH of 5, you die a horrible death.

Okay, maybe that's not what happens when you drink a soda. However, if there wasn't something in your blood that stabilizes its pH, it's exactly what would happen.

As it turns out, your blood, like many solutions, is a buffer. Buffers are solutions consisting of a weak acid and its conjugate base—these solutions resist changes in pH when either acid or base is added to it.

Let's say that we have a buffered solution that contains acetic acid as its weak acid and sodium acetate as its conjugate base. If we were to add some hydrochloric acid to this solution, the sodium acetate would react with it by the following process:

• HCl + NaC₂H₃O₂ ⇔ C₂H₃O₂H + NaCl

As you can see, the very strong HCl that's been added to the solution has been converted to acetic acid, which is a weak acid. Because weak acids cause a much smaller disruption in pH than strong acids, the pH of the solution will decrease much less than if it contained no sodium acetate.

Likewise, if we were to add sodium hydroxide to this solution, the acetic acid would react to it by the following process:

• NaOH + C₂H₃O₂H ⇔ NaC₂H₃O₂ + H₂O

Because the strong base NaOH has been converted to the weak base sodium acetate, the pH of the solution won't rise nearly as much as if the acetic acid weren't present in the first place.

We can determine the pH of a buffered solution by using the Henderson-Hasselbalch equation:

• pH = - log K_a + log ^[base]⁄_[acid]

where K_a is the acid dissociation constant of the weak acid in the buffered solution.

Example: What's the pH of a solution that contains 0.100 M acetic acid and 0.200 M sodium acetate? K_a (C₂H₃O₂H) = 1.75 × 10^-5.

Solution: Using the Henderson-Hasselbalch equation, we find that:

• pH = - log K_a + log ^[base]⁄_[acid]
• pH = - log(1.75×10^-5) + log ^[0.200]⁄_[0.100]
• pH = 4.76 + 0.301
• pH = 5.06