In the Fischer esterification reaction, a carboxylic acid reacts with an excess of alcohol in acidic conditions to form an ester
1 answer:
Answer:
See explanation and image attached
Explanation:
Esterification is a reaction of an alkanol and an alkanoic acid to yield an ester and water as products. It is analogous to inorganic neutralization reaction.
The process may be acid catalyzed. As mentioned in the question, the process of ester formation between pentanoic acid and n‑propanol first involves the carbonyl in going through an sp3 hybridized intermediate before returning to sp2 hybridized state in the product.
The image of the neutral form of this intermediate as well as the final structure of the ester are both shown in the image attached to this answer.
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Answer:
7.3
Explanation:
By Henderson Hasselbalch equation we can calculate the pH or the pOH of a solution by its pKa. Remember that , and pKa = -logKa. Ka is the equilibrium constant of the acid.
Henderson Hasselbalch equation :
Where [HA] is the concentration of the acid, and is the concentration of the anion which forms the acid.
So, acid X, has two ionic forms, the carboxyl group and the other one. First, we have 0.1 mol/L of the acid, in 100 mL, so the number of moles of X
n1 = (0.1 mol/L)x(0.1 L) = 0.01 mol
When it dissociates, it forms 0.005 mol of the carboxyl group and 0.005 mol of the other group. Assuming same stoichiometry.
Adding NaOH, with 0.1 mol/L and 75 mL, the number of moles of will be
n2 = (0.1 mol/L)x(0.075 L) = 0.0075 mol
So, the 0.0075 mol of reacts with 0.005 mol of carboxyl, remaining 0.0025 mol of
, which will react with the 0.005 mol of the other group. So, it will remain 0.0025 mol of the other group.
The final volume of the solution will be 175 mL, but both concentrations (the acid form and ionic form) have the same volume, so we can use the number of mol in the equation.
Note that, the number of moles of the acid form is still 0.01 mol because it doesn't react!
So,
6.72 = pKa - log 4
pKa - log4 = 6.72
pKa = 6.72 + log4
pKa = 6.72 + 0.6
pKa = 7.3