The correct answer is option 'A', i.e., the pOH of the solution will decrease in Kolbe's electrolytic synthesis. Let's understand why this is the case:

Kolbe's electrolytic synthesis is a process used to prepare alkanes from carboxylic acids by electrolysis. It involves the electrolysis of a concentrated solution of a carboxylic acid using an inert electrode, such as platinum.

During the electrolysis process, the carboxylic acid molecules are oxidized at the anode, while hydrogen gas is evolved at the cathode. The overall reaction can be represented as follows:

2R-COOH → R-R + 2CO2 + 2H2O

Here, R represents the alkyl group.

Now, let's analyze the changes in the solution during Kolbe's electrolytic synthesis and how they affect the pOH of the solution:

1. Increase in pH:
During the electrolysis, carboxylic acids are oxidized at the anode to form carbon dioxide gas and water. As a result, the concentration of carboxylic acid decreases in the solution, leading to a decrease in the concentration of hydronium ions (H3O+). This decrease in hydronium ion concentration causes an increase in the pH of the solution.

2. Decrease in pOH:
pOH is the negative logarithm of the hydroxide ion concentration (OH-) in a solution. Since the concentration of hydronium ions decreases during Kolbe's electrolytic synthesis, the concentration of hydroxide ions increases. As a result, the pOH of the solution decreases.

3. Formation of alkanes:
The carboxylic acids are reduced at the cathode to form alkanes. This reduction process consumes hydronium ions from the solution, further reducing their concentration. Therefore, the pH of the solution increases, leading to a decrease in pOH.

In summary, during Kolbe's electrolytic synthesis, the concentration of hydronium ions decreases, leading to an increase in pH and a decrease in pOH. Therefore, the correct answer is option 'A', i.e., the pOH of the solution will decrease.