The Fischer esterification mechanism is examined in the following two questions in the assignment. Part 1 involves MeOH addition to form the key tetrahedral intermediate. Part 2 will involve loss of H2O to form the ester. Part 1 of 2:

Fisher esterification is one of the ways of obtaining an ester, dissolving a carboxylic acid in boiling alcohol with a mineral acid catalyst. The mechanism is addition-elimination by activating the initial protonation of group C=O.

Explanation:

The mechanism begins with the protonation of the carbonyl group of the carboxylic acid, increasing the electrophilia of that group. Then methanol attacks the carbonyl group for the formation of a tetrahedral intermediate, which by proton exchange produces a tetrahedral intermediate that has a leaving group, water. The regeneration of the carbonyl group leads to the elimination of water and the formation of the ester. Finally, the proton exchange with the water molecule regenerates the acid catalyst.

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.