Buffers (or buffer systems) are combinations of weak bases and weak acids, which when in the right proportions relative to each other, help maintain a tighter pH range even when exposed to additional amounts of acids or bases. They do this by effectively serving to neutralize excess acid or base that is introduced into the solution. Without a buffer system in the solution, small fluctuations in acid or base concentrations introduced into the solution could drastically change the pH.
There are natural buffer systems in the body such as the carbonic acid/bicarbonate buffer system which is one of the main systems that maintain blood pH. There are also buffer systems that are made in the clinical laboratory to help maintain specific pH ranges either to stabilize reagents or samples or to optimize chemical reactions used to quantify analytes. A common buffer system that can help maintain a broad range of clinically relevant pH setpoints is the phosphate buffer system. In this system, a weak acid in the form of potassium dihydrogen phosphate (KH2PO4) and a weak base in the form of dipotassium hydrogen phosphate (K2HPO4) are mixed in differing proportions to one another which helps set the buffering pH at which the overall solution will be maintained.
To determine the pH of these differing proportions of KH2PO4 and K2HPO4 in solution, the Henderson-Hasselbalch equation can be used.
The equation is as follows:
- pH = pKa + log([A-] / [HA])
- pKa is a unique dissociation constant for the buffer system. Each buffer has its own pKa.
- [A-] is the concentration of the conjugate base (or weak base) in moles.
- [HA] is the concentration of the conjugate acid (or weak acid) in moles.
If 1 liter of phosphate buffer was made in water containing 6.31 x 10-2 moles of K2HPO4 and 0.1 moles of KH2PO4, calculate the pH of the buffer (follow the steps below with the image).
- Step 1 requires filling in the calculation to solve for pH.
- The pKa for phosphate buffer is 7.2.
- The conjugate base concentration is 6.31 x 10-2 moles and the conjugate acid concentration is 0.1 moles.
- Step 2 requires completion of the division calculation within the parentheses.
- Step 3 requires taking the log of the number within the parentheses.
- Step 4 requires completion of the addition (or subtraction) of the equation to calculate the pH.
Another example would be if 1 liter of buffer was made in water containing a higher concentration of conjugate base such as 0.63 moles of K2HPO4 and maintaining the sample concentration of KH2PO4 at 0.1 moles, the calculated pH would come out to 8.
- pH = 7.2 + log(0.63/0.1) = 7.2 + 0.8 = 8
