Question

note: the beginning of this video is the same as the video from the previous assignment, but the end is different.
at the half - equivalence point of a weak acid - strong base titration, ph =
pkapkbpohka
question 2
for a weak acid - strong base titration, the ph at the equivalence point will be greater than 7 because
the acid has been converted to its conjugate basethere is more base than acidthe temperature has changed

Explanation:

• Question 1:
• In a weak acid - strong base titration, at the half - equivalence point, the Henderson - Hasselbalch equation \(pH = pK_a+\log\frac{[A^-]}{[HA]}\) is used. At the half - equivalence point, \([A^-]=[HA]\), so \(\log\frac{[A^-]}{[HA]} = 0\), and \(pH=pK_a\).
• Question 2:
• In a weak acid - strong base titration (\(HA+OH^-\to A^- + H_2O\)), at the equivalence point, the acid \(HA\) is completely neutralized and converted to its conjugate base \(A^-\). The conjugate base \(A^-\) undergoes hydrolysis (\(A^-+H_2O

ightleftharpoons HA + OH^-\)), producing \(OH^-\) ions. So the solution is basic (\(pH>7\)). At the equivalence point, the amount of base added is stoichiometrically equivalent to the amount of acid initially present (not more base than acid), and temperature change is not the reason for \(pH>7\) in this context.

Answer:

• Question 1: \(pKa\)
• Question 2: the acid has been converted to its conjugate base