Question

Supplemental Questions - Titration Lab 4. Explain why the initial pH is not the same for all the acids while it is nearly the same when the titration has been carried a few ml. past the final equivalence point. Your answer should include appropriate proton transfer NIE(s). 5. The buffer range for a given weak acid/conjugate base pair is pk, 1 Explain why 6. A weak triprotic acid, H.Z has the following K, values: K -1.4 x 104 K.9.7 x 10.2 K, -5.3 x 10 a. Write the NIE associated with each K, value. b. Of the four weak acid species: H,Z, H,Z, HZ.Z'; which two could be used to prepare a buffer of pH = 3.55? Explain. c. A buffer is made by pouring 250.0 mL of 1.688 M Na, HZ into 350.0 mL of 1.455 M Z Cit. Calculate the pH of this buffer solution. d. Calculate the pH if the original buffer solution is diluted to 1200 mLs.

Answer 1

bestial, pH of different aside is not same due to differen concentration o f obtained from dissociation of rud. eg. In hie, it is completely dissociated and in Me lors, une of tht is less than the are of the city as it is partially conised. In case of tixrason with a strong base on from the base reacts with the it from acid and form water. the on fusson strong base pushes the dissocia ogn of weak acid in eroard. direction and produses voore and more tt. This pH near equivalence point is same for all acids.

Buffer nge can be evaluated wing the following ын рид + оо. | Толх] Jacid Since is [salt] s 10 10 and -1 cost log Esalt] s acid] l. pka- 1 <pka t ka t log Isalt Tracid] log Isalt) s pkat pka-1 s pH s pkärl e This makes the buffer range (pka-l , pratl] for H₂ 2 ka, = 1.4810-4 3 pkai = 3.85 ka = 9.78 lot 2 pk az 26.01. kl2 = 5.3 x 10 9 3 pka, 8.27. Ž H + H₂2' ka, Ž H+ + H22- ~ Ht + 23- a kaz H₂ Z t₂ H22 1 I lacid To make a buffer of pH - 3.55 , 32 and H2 2 ů used as we know for buffer H & pray + log Isole pkan + log [salt] = 3.85 + log [bal] [acid] d log [H22 -3.85 +3.55 I [H₂2] [122] = 0.5 [H34 L

250 me 1.688 M Nalle 350 ml 1.485 M 422H + 23 pka = 8.27 - pka + log [23] ( TH2?] ( 350 X1.955 8.27 t tog 250 x 1.68% TpH = 8.35 8.55 of the buffer is diluted to 1200 me, then pH of the buffer remains same il. 8.35 as on dilution only the volume and conch of 23 and 422 changes but the moles of them remains same