The correctness of the assumption in the title can be intuitively acknowledged by considering the precise interpretation of the correlation between charges and ionisation states as follows. According to the HendersonHasselbalch equation (Equation 3.14), the amount of charge on a group is proportional to the distance of the actual pH from the pK_{a} of the group. Thus, the net charge will be zero (i.e. the degree of deprotonation of the carboxylic acid will equal that of the protonation of the amino group) if the pH is at an equal distance from the pK_{a} values of these two groups.
We can prove this assumption by calculating the pH at which the amount of charges carried by the carboxylic acid and amino groups is identical, i.e. at which the degree of ionisation of these groups is the same. Using Equation 3.14, we can calculate the ratio of the concentrations of the ionised and nonionised (i.e. charged and uncharged) forms of an ionisable group ([COO^{}]/[COOH] or [NH_{2}]/[NH_{3}^{+}]). According to the mole concept, this ratio is identical to the ratio of the number of groups in the ionised and nonionised forms. This, in turn, is identical to the fraction of the form of a group adopting the charged state, while this fraction is identical to the number of ionised groups relative to the nonionised ones. From this, the degree of ionisation of a group (i.e. the amount of charges it possesses) can be calculated. (See the procedure of such a calculation below in Demo calculation 3.4.2.) This tight correspondence between the charge on an ionisable group and the ratio of concentrations of the two forms of such a group simplifies the subsequent charge calculations because we can use the ratio of concentrations of the two forms of a group instead of its charge. According to this, we are looking for the pH at which

(3.16) 
The concentration ratios on the two sides of the equation can be obtained from the HendersonHasselbalch equation (Equation 3.14) applied to the carboxylic acid and amino groups. As the desired pH will be the pI, we can write the following equations:

(3.17) 
and

(3.18) 
From these:

(3.19) 
and

(3.20) 
and

(3.21) 
Substituting into Equation 3.16:

(3.22) 
that is:

(3.23) 
from which:

(3.24) 
Thus, indeed, the pI is the average of the two pK_{a} values.
In the case of glycine it is:

(3.25) 
Let us check by calculation whether the charge on the two ionisable groups is identical. Now Equation 3.16 will look as follows:

(3.26) 
and

(3.27) 
From these:

(3.28) 
and

(3.29) 
That is,

(3.30) 
and

(3.31) 
In the case of the carboxylic acid group, this result means that the fraction of the groups in the charged state is 4.27×10^{3}fold relative to those in the uncharged state. In the case of the amino group, this fraction is 1/2.34×10^{4 }= 4.27×10^{3}, i. e. the same as above. Thus, indeed, the net charge of glycine will be zero in a pH 5.97 solution. This pH is its isoelectric point. Notably, fraction values above a thousand mean that the charge of either ionisable group differs only negligibly from one, the charge at complete ionisation. This is natural since the pI differs more than three pH units from both pK_{a} values, which means that its is sufficiently basic for the complete ionisation of the carboxylic acid group and, at the same time, too acidic for the deprotonation of the amino group. Thus, the zero net charge is achieved by the molecule having a full negative and a full positive charge. In the next example, in which the pI will be much closer to the pK_{a} values, we will see how a zero net charge is achieved at fractional charges (partial ionisation).
Before we set out to prove the above statement, it is worth considering what result can be expected. It is easily acceptable that the following conditions must hold at the pI value: (i) It is not possible for all three ionisable groups (the two carboxylic acids and the amino group) to be completely ionised because this would give a net charge of 1 to aspartic acid. (ii) The negative charges on the two carboxylic acid groups (that on the α–carbon atom, COOH_{(α)}, and that in the side chain, –COOH_{(R)}) must together neutralise the positive charge on the amino group. (iii) For this reason, the combined charge of the two carboxylic acids cannot be larger than one, which implies a pI value at which the charge of both carboxylic acid groups is only fractional. (iv) Considering the data in Figure 3.2 (pK_{a(α) }= 1.88 and pK_{a(R) }= 3.65) as well as the meaning of pK_{a} (above), this is such a low pH where the amino group is completely protonated and thus it carries a full positive charge (see the interpretation of the previous pI calculation). Thus, the combined charge of the two carboxylic acid groups must be one at the pH of pI. We can formulate this condition as follows:

(3.32) 
(In writing Equation 3.32 we again exploited the reduction allowed by the unambiguous correspondence between the charge on an ionisable group and the ratio of concentrations of that group’s two forms.)
Since in this case both groups are of the same type (ionise by forming an anion), their different pK_{a} values do not allow a pH at which they carry the same amount of charge so that their combined charge is one. (This is contrary to the case of two groups that ionise by forming different (positive and negative) ions, in which case the same amounts of opposite charges neutralise each other—see the previous example.) At pH > 3.65 (cf. Equation 3.12), the degree of ionisation of both carboxylic acid groups would be larger than 50 %, which would result in a combined charge larger than one. For the same reason, at pH < 1.88 the degree of ionisation of neither carboxylic acid group would reach 50 %, and thus the combined charge would be less than one. Thus, the pI must be between the two pK_{a} values. In addition, at this pH it must hold that as much the charge of one group is short of 0.5, the charge of the other must exceed 0.5 to the same extent. (In other words, the absolute values of the charges must fall in equal “distance” from 0.5 on the two “sides” of it.) This can happen only if the pK_{a }values are in equal distance from the pI. Therefore, the pI will be the average of the two pK_{a }values. Now we will prove this.
As in Equation 3.17, we can write for both carboxylic acid groups that

(3.33) 
From this, the ratio of concentrations of the ionised and nonionised forms, which is proportional to the amount of charges, will be

(3.34) 
Therefore, we can transform Equation 3.32 as follows:

(3.35) 
From which:

(3.36) 
Thus indeed:

(3.37) 
Substituting the pK_{a} values we get the pI value of aspartic acid:

(3.38) 
As a check, we can calculate the charges on the two carboxylic acid groups. After substituting the appropriate values into Equation 3.34 we will get:

(3.39) 
and

(3.40) 
which are

(3.41) 
and

(3.42) 
Due to the identity of the ratio of the concentrations and the ratio of charged and uncharged groups, we can state that the fraction of charged α–carboxylic acid groups is 7.67fold relative to the uncharged ones, while this fraction is only 0.13fold in the case of the carboxylic acid groups in the side chain. As these fractions reflect the number of charged groups relative to the uncharged ones (see above), we can determine the degree of ionisation—which corresponds to the charge possessed by an ionisable group—by dividing the number of charged groups by the total number of groups. In the case of the α–carboxylic acid, for each nonionised group there are 7.67 ionised ones. Therefore:

(3.43) 
Thus the α–carboxylic acid group has 0.885 charges. The case is just the opposite for the side chain carboxylic acid, in which there are 1/0.13 = 7.67 nonionised groups for each ionised one. Therefore:

(3.44) 
Thus the side chain carboxylic acid group has 0.115 charges. Indeed, the two carboxylic acid groups contribute unequally to the total negative charge of one and, as much the share of the α–carboxylic acid is in excess of 0.5 (0.8850.5=0.385), this will be the same as the extent to which the share of the side chain carboxylic acid (0.50.115=0.385) is short of 0.5.
Let us calculate the pI of a protein that consists of 152 amino acids and has the following amino acid composition:
H_{2}NGly(150 amino acids)GlyCOOH, containing the following amino acids with ionisable residues: 2Arg, 3Lys, 2Tyr, 1Cys, 5His, 3Asp.
As there are many ionisable groups in such a large molecule, it is obvious that the calculation of pI will be more a complex procedure. Therefore the procedures employed above are suitable only for an approximate estimation of pI. In some very rare cases, however, the calculation can follow that of the pI of aspartic acid (Demo calculation 3.4.2), and will not even be more difficult than that. In these cases the expected pI must be close to a pK_{a} so that, at the same time, it is at least two pH units away from all the other pK_{a }values. In order to determine whether this criterion is met, an estimation must be performed concerning the expected value of the pI. To this end, we must calculate the net charge of the protein at several different pH values. In order to facilitate these calculations, we can construct a charge calculation table listing the amino acids possessing ionisable residues in the order of their increasing pK_{a} values. In the case of our protein, this table will look as follows (Figure 3.5):
In order to reach our goal more rapidly, prior to the calculations it worth considering the pH range into which the pI is expected to fall. As a conclusion from the reasoning employed at the pI calculation of aspartic acid (Demo calculation 3.4.2), the pI will fall in the acidic or basic range if the number of negatively or positively ionising groups exceeds that of the other, respectively. The more the ratio is shifted towards one type of group, the more the pI will be shifted towards the pH range of the pK_{a} of these groups. In our protein, the number of groups of the two types is the same. Therefore, it is appropriate to start the calculation of charge at pH 7.0.
Calculation of the charge at pH 7.0
The charge of the α–amino group:
According to Equation 3.18:

(3.45) 
After substitution:

(3.46) 
From this, following the earlier calculations:

(3.47) 
According to the above considerations regarding the connection between the ratio of concentrations and the fraction of charges, for every charged group there are 0.0025 uncharged ones. Based on this and according to Equation 3.43, the degree of ionisation, i.e. the amount of charge will be:

(3.48) 
This value differs negligibly from one. Therefore, we can take it as a full charge on the α–amino group. During approximate calculations we can always apply such rounding whenever the difference between the pH and the pK_{a} is more than two pH units. This is allowed because in these cases the equilibrium of ionisation is shifted so much towards one of the forms of the group that the difference of the charge from one or zero will only be a few thousandths. Applying this simplification, without detailed calculations we can assign a full charge to the residues of aspartic acids, lysines and arginines and to the α–carboxylic acid, as well as zero charge to tyrosine residues.
Following the calculations above, the ratio of forms of the cysteine residue will be

(3.49) 
From this the charge of cysteine:

(3.50) 
While the above ratio regarding histidine residues will be

(3.51) 
and thus their charge:

(3.52) 
By using the values of charges calculated above—and also taking into account the number of groups—we can fill the pH = 7.0 column of Figure 3.5. Summing up the charges, we can see that the charge of our protein at this pH will be 1.593. Thus, as it follows from the relationship between the net charge and the pH (see above), the pI must be below 7.0.
Continuing the estimation of the expected value of pI, we can now calculate the net charge of the protein at pH 6.0. As the minimally expected twounit difference between the pH and the pK_{a} values again holds for Asp, Lys, Arg and Tyr residues as well as for the α–carboxylic acid and α–amino groups, their charge can be taken the same as at pH 7.0. In contrast, the charges of cysteine and histidine residues will change. The former will decrease essentially to zero (because now the distance of pH from the pK_{a} will be more than two units), while the latter will increase to +0.5 because now pH = pK_{a} and, by the definition of pK_{a}, here the degree of ionisation will be 50 %. Filling the pH = 6.0 column of Figure 3.5 and summing up the charges, we get that the net charge of the protein is +0.5. Thus, this pH is below the pI. Therefore, the pI must be between 6.0 and 7.0.
If we now assume (with some reason) that the pI will be close to 6, we can expect that even this value will differ at least two pH units also from the pK_{a} of cysteine. Thus the criterion set above seems to be met, and its significance can now be understood. At several tenths of pH units above 6, the charge of every group—including cysteine—will only negligibly differ form that at pH 6.0. Therefore we can try to calculate the exact pI by employing the same logic that was used during the pI calculation of aspartic acid. Thus, at pH = pI, the net charge of the protein will be 2 without the charge of histidines. Therefore, the reasoning applied in writing Equation 3.32 will in this case appear as the following: we will look for the pH at which the combined charges of histidines can neutralise the 2 charge resulting from the other ionisable groups. That is,

(3.53) 
thus

(3.54) 
Writing the expression of pI as in Equation 3.18 and performing the appropriate substitutions we get:

(3.55) 
Thus the isoelectric point we look for is 6.4. After the calculation and summation of the charge on every ionisable group, we must get zero. Performing this control calculation (see the pH = 6.4 column in Figure 3.5) indeed yields zero net charge.