Question

Salt Hydrolysis: Hydrolysis at the Endpoint of a Titration

Consider the following equation:
NH₃ (aq) + HBr (aq) → NH₄Br (aq) + H₂O (l)

Here the neutralization reaction is creating ammonium bromide.

If you add enough strong acid to react with all of the weak base, all that will remain will be ammonium bromide and water. The conjugate acid from this salt will hydrolyze in solution to determine the pH at this point.

Weak Acid	Ka	Weak Base	Kb
CH3COOH (Acetic Acid)	1.8 X 10-5	NH3 (Ammonia)	1.76 X 10-5
C6H5COOH (Benzoic Acid)	6.5 X 10-5	C6H5NH2 (Aniline)	3.9 X 10-10
CH3CH2CH2COOH (Butanoic Acid)	1.5 X 10-5	(CH3CH2)2NH (Diethyl amine)	6.9 X 10-4
HCOOH (Formic Acid)	1.8 X 10-4	C5H5N (Pyridine)	1.7 X 10-9
HBrO (Hypobromous Acid)	2.8 X 10-9	CH3CH2NH2 (Ethyl amine)	5.6 X 10-4
HNO2 (Nitrous Acid)	4.6 X 10-4	(CH3)3N (Trimethyl amine)	6.4 X 10-5
HClO (Hypochlorous Acid)	2.9 X 10-8	(CH3)2NH (Dimethyl amine)	5.4 X 10-4
CH3CH2COOH (Propanoic Acid)	1.3 X 10-5
HCN (Hydrocyanic Acid)	4.9 X 10-10

If 23.00 mL of a 0.74 M solution of NH₃ required 8.80 mL of the strong acid to completely neutralize the solution, what was the final pH at the endpoint?

Answer 1