In this lesson, you will learn about isoelectric points. We will review pH and pKa, explore the definition of isoelectric points with use of examples, and then describe how to calculate the value knowing pKa values.
Introduction to pH and pKa
Every solution has a pH value, which simply describes how many hydrogen ions are in the solution. This can be measured with the use of a pH meter. This value also tells us if the solution is acidic or basic. Values less than 7 are acidic, values at 7 neutral, and values greater than 7 are basic. The pH of a solution also informs us about what will occur to compounds when they are put into the solution. Some compounds have functional groups that can give up a proton or can accept a proton.
Their ability depends on the pH of the solution and the acid disassociation constant, or pKa as we’ll be calling it from now on, of the group.For now, and as an example, let’s consider this figure on screen where there are two compounds represented: compound A, phenol, and compound B, 4-chlorophenol.
At low pH values both functional groups, the ammonia, or NH3+, and carboxylic acid, or COOH group, are both protonated. As the pH increases the COOH group becomes deprotonated first, as its pKa value is low. Then after the pH value is increased to a basic pH the NH3+ group is deprotonated.Notice that the charge of the overall glycine compound changes in different pH values.
At low pH, or in acidic solution, the overall compound has a positive charge, due to the NH3+ group. At moderate pH, or close to neutral, the overall compound is neutral, due to the loss of proton and subsequent minus charge on the COO– group and positive charge on the NH3+ group. At high pH, or in basic solution, the overall compound is negative, as both groups have lost their protons, leaving a neutral charge on the NH2 group and negative charge on the COO– group.
The isoelectric point is the pH at which the molecule contains no net electric charge. This is represented visually by the middle image of gylcine in this figure. The compound in this form is known as a zwitterion, or the form of the molecule that is neutral. In our example of glycine, this occurs at a pH of 5.97, when the COO– group has a negative charge and the NH3+ group has a positive charge. The isoelectric point is often shortened to pH(I), or pI.
Calculating the Isoelectric Point
Calculating pI is relatively simple, especially when there are only two groups with pKa values. The equation is simply this:
Let’s apply this equation to glycine. The pKa value of the NH3+ group was 9.60, and the pKa value of the COOH group was 2.34. Therefore: pIgylcine = (9.
60 + 2.34) / 2 = 5.97. This means that at a pH of 5.
97, we would expect the zwitterion to be present. In other words, the glycine would be neutral at this pH.Sometimes a compound has more than two groups with pKa values.
When this occurs, it is appropriate to use the two values of the pK’s that are closest together. This represents the two functional groups that are the most similar to each other.This figure depicts an amino acid, arginine, that has three different functional groups with pKa values.
The two groups that are the most similar are the two NH groups. It should also be noted that these pKa values are also the most similar to each other. Therefore the pI for arginine could be calculated as follows: pIarginine = (9 + 12.5) / 2 = 10.
Let’s review what we’ve learned. The pKa of a functional group in a molecule describes the pH where 50% of the functional group will be deprotonated. Some molecules have multiple functional groups that are acidic or basic and, therefore, have multiple pKa values. When this occurs, the isoelectric point (pI), which is the pH at which the molecule contains no net electric charge, can be calculated.
This is the pH of the solution that is required to have the molecule be neutral, completely in the zwitterion form. The equation for the isoelectric point is this:
Also remember that when a compound has more than two pKa values, you should choose the two values that are closest in value and use those in the formula. A good example of this would be the arginine example we solved earlier, in which there were three values.