Question

A buffer solution contains 0.202 M C₆H₅NH₃Br and 0.445 M C₆H₅NH₂ (aniline). Determine the pH change when 0.114 mol HClO₄ is added to 1.00 L of the buffer.

pH after addition − pH before addition = pH change =

Answer 1

The oth Buffer solution of a buffer soluti tion can be calulated by ang Henderson- Hasselbalch equation, - pH = pkatlog EBase] A taidy pha & clog ka Concentration of base (Base] [Acidy ka a concentration of Acid dissociation constant. - acid - 6 Aniline (CGHE NH₂) is a weak base. With kb value - 4.3x10-10 . (6 HG WH, +4T - CHO NH3 KaxKb = kW acid kaa 16 14 4-381010 - betone addition of Alloy = 2.325 x 105 According to Henderson-Hosselbalch equation, gi = pua loa flog ? [ Co HG NH 2 a thog IG HJÁng] SEGHE NH 2] 2 0.445m - pH = -log (2.325 x165)+log 10.4 a) pH 2 4.6335+ 0.343 pH conjugate 3251 445 ([Cat5 Min3] 20-2022 - pH a 4.9765 pH before addition = (4.9765 0202 Spolarity = Volume of moles solution in ď pH-after addition of steloy Moles of GHE NH2 in 11002 = 0445mole XIL? solution = 0.445 mole moles of C6H5NH3. in l.roL 0.202mole, au Slution = 0202 mole moles of teloy added ally mole The neutralization reaction combe expressed as 1C6 fable) Cats NH₂ can tat can > CGHTHz can Initially mole 0.445 billy 01202 change – 0.114 -ally tolly Equilibrium 0.831 o 1L x 12 Molar concentrations, 0.316 0.331 mole I Co H5NH₂] = L«Ну, ми ) . 20.332m 0316 mole 14 = 0.316m

1. According to Henderson - Hasselbalch equation, рч = p<a +1° С (авья», - 3 pH = -log (2.3254105) +log / 0.331 306A133 + 109 (0:3316) pH = 4.6335 0.316 +0:02014" pH = 4.6536 pH after addition - Tu pH after addition - pH after addition - & pH change = pH her pt before addition - 4.6536 - 4.9765 - - 0.3228 Ove sign means pH decreases. П