Question

The normal pH range for blood plasma is 7.35-7.45. Under these conditions, would you expect the carboxyl group of lactic acid \(\left(\mathrm{p} K_{\mathrm{a}} 3.08\right)\) to exist primarily as a carboxyl group or as a carboxylic anion? Explain.

Short answer

Answer: In the normal pH range for blood plasma, the carboxyl group of lactic acid primarily exists as a carboxylic anion (COO-).

Step-by-step solution

Understand the Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is used to describe the relationship between the pH, pKa, and the concentration of acidic and basic forms of a molecule. It is given by: pH = pKa + log10([A-]/[HA]) Where pH is the pH of the solution, pKa is the acid dissociation constant, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the acid. In our case, HA represents the carboxyl group (COOH) and A- represents the carboxylic anion (COO-).

Calculate the pH - pKa value

We need to determine the difference between the pH value and pKa value to understand the dominating form of the carboxyl group. To simplify the calculation, we will consider the average pH value of the blood plasma, which is (7.35+7.45)/2 = 7.4. The difference is: Difference = pH - pKa = 7.4 - 3.08 = 4.32

Evaluate the pH - pKa difference

The pH - pKa difference is 4.32, which is greater than 0. According to the Henderson-Hasselbalch equation, if pH > pKa, then log10([A-]/[HA]) > 0, which means the concentration of the carboxylic anion ([A-], or COO-) is higher than the concentration of the carboxyl group ([HA], or COOH). When blood plasma has a pH of 7.4, the carboxyl group of lactic acid will primarily exist as the carboxylic anion (COO-) rather than the undissociated carboxyl group (COOH).