Question

A student must make a buffer solution with a pH of 1.50. Determine which weak acid is the best option to make a buffer at the specified pH. O propionic acid, Ka = 1.34 x 10-5, 3.00 M formic acid, Ka = 1.77 x 10-4, 2.00 M O acetic acid, Ka = 1.75 x 10-5,5.00 M O sodium bisulfate monohydrate, Ka = 1.20 x 10-2, 3.00 M Determine which conjugate base is the best option to make a buffer at the specified pH. O sodium sulfate decahydrate, Na, S04 · 10H20 sodium formate, HCOONa O sodium propionate, CH, CH, COONa Osodium acetate trihydrate, CH, COONa · 3H20

Answer 1

Salt acid 1.50 AV 1.92 Buffer solution with a pH of 1.50 pH = pkatlog salt) pka= -log(ka) =-109 (r.20x162) 1.505 1.92 + log ( salt) 58.92 moles of acid = MxV [salt] = 0.0038 moles Fo.lxoil = 0.01 mdes mans of solid conjugate base = Molesx molarman = 0.0038x322 =102236 511229 Acid = sodiumbisulfate monohydrate (NaHSOM HO ) Conjugale bave = sodium sulfate de cahydrale (ną som-toto )