You are at isoelectric point for a solution of Leucine. If the number of negatively charged carboxyl ate groups, -COO-, is 1x10^19, how many positively charged amino groups, NH3+, would there be?
Wouldn't that be 1 x 10^19 + charges so that the net charge would be zero?
To determine the number of positively charged amino groups, NH3+, at the isoelectric point of Leucine, we need to know the net charge of the Leucine molecule.
At the isoelectric point, the net charge of the molecule is zero. Leucine contains one negatively charged carboxylate group, -COO-. The carboxylate group, -COO-, carries a charge of -1.
Therefore, to achieve a net charge of zero, there must be one positively charged group present on Leucine. This group is the amino group, NH3+.
Thus, there would be 1 positively charged amino group, NH3+, in the Leucine molecule at its isoelectric point.
To determine the number of positively charged amino groups, NH3+, in the solution of Leucine at its isoelectric point, we need to consider the overall charge balance.
At the isoelectric point, the net charge on the molecule is zero. Since Leucine is an amino acid, it has both negatively charged carboxylate groups, -COO-, and positively charged amino groups, NH3+.
The formula of Leucine is:
NH2-CH(CH2)3-CH(COOH)-COOH
From the given information, we know that the number of negatively charged carboxylate groups, -COO-, is 1x10^19 (The carboxylate group carries -1 charge on it).
Now, let's consider the overall charge balance at the isoelectric point:
1. The amino group, NH2, has a +1 charge.
2. There are two carboxylate groups, -COO-, each carrying a -1 charge, resulting in a total charge of (-1) x 2 = -2.
To neutralize the negative charge, we need two positively charged amino groups, NH3+.
So, the number of positively charged amino groups, NH3+, in the solution would be 2.