Question

Post-Lab Assignment: pH and Buffers 1. A buffer is prepared from a weak acid with a Ka - 7.1 x 104 and its conjugate base. a. What pH would provide maximum buffer capacity? b. What would be the buffer range for this acid? (Your answer should show the lowest and the highest pH that would provide a reasonably effective buffer.) 2. If you were provided with a 0.1 M solution of an unknown weak acid and a 0.1 M solution of its conjugate base, what procedure would you follow to determine the pka? 3. Nitrous acid, HNO2, has a Ka = 4.5 x 104. You are asked to prepare a buffer having a pH of 3.60 using the following solutions: 0.10 M HNO2 and 0.10 M NaNO2. How many milliliters of NaNO2 solution should you add to 20.0 mL of 0.10 M HNO2 to make the buffer?

Answer 1

1. Kete for better understanding het us take an example of Che WoH (weak acid) . . Cenjugate Base. ICCOUH) (Chor"; +HT , Kac 7:18104 u ķa = [CH, Cool] [HT] weak acid -- Сен, с огон) Аллергое Curstant of an acid Kax ([cy Coon). Cht) \tcң сw"] Taking log on both the sides we get, log (it) a looke + logo che con - long. In 23- loy ke - log e I concor pH e pka + log / Concoro ) ng TCH WHY < Pla Katleg [ conjugate Base] , (weak aced] / 12 - Henderson Hasselbatch I equation for acidic buffer Nowo, for maximum bafper capacity, [conjugate bree] - [ weak acild? MI DK a + log (1) = pka toa pka = -log (7.1 X 15") = 3.15)

to from the Henderson - Hasselbatch equation, pH = pka + loy_cycou] [cenjugate Base] ) "[ącout] - [weak Acid)) operating Buffer zone u Ic[Choo - The Coon) <10.} Range of Buffer to, Put [ch coo] cus.com to win pH e pka +676) pit pra - lub [ch cool] [CM Cour] We get pH = pkat leglo proti n pH pH range of buffer - pku-l 3:15-1 42-15) phati 3,15+1=14,15) "manimum bi . "maximum buffer capacity [cenjugate Base}=0.1m E 2 - [ weak acid] cool oscar . . . Ab t=0 CHCOOH S ORIM +H Ka seepoca U A cu 0-IM At t= teg ool-a a olta no [CH₂ cour][H] [chool]

Osaka [HT] - * = Kan -log [**]> - loqka | * * *) / М..., А., р found out. though it's been already disuessed that this is Pas Cendilicem of morenou na buffer capacity 3. het Volume (in me) of Nang solution be v. from Nano = H * No, ka = 4058104 ANG At=0 20x0.! 20 + V v Xol Tutv Also, given At ttes v 20X0. l 2o+V VXO - pH = 3.6 . yvܒܐ ☆ 2okovi 10-9, Floquet) - 3.6 [[ht] = 1072) 20+V DxV 20+V Kaz [NO] [HI] feixul x1036 = 4,5 xiuy [HNG] (20+ + 2oxor (20+) xl 4.5 xiy IN=35.83 mel do X