Question

Part A You need to produce a buffer solution that has a pH of 5.14. You already have a solution that contains 10. mmol (millimoles) of acetic acid. How many millimoles of acetate (the conjugate base of acetic acid) will you need to add to this solution? The pk of acetic acid is 4.74 Express your answer numerically in millimoles. View Available Hint(s) V AED ROO? mmol acetate

Answer 1

@ pH = pka +log L COOH ] GS op 5.14 = 4.74 + log I CH3COO oo [CH3coo 10 ] . or log [CH3COo] =(5-14-4.74) = 0.40 or [CH3coo 100.40 10 = 2.51 or [CH3C00] = (0x2151) = 25.4 Ans => 25.1 mmol

2 ® 1.90x10 mL of 0.100 M = (1.90x1080.100) mmol = 19 mmol Now, ciz COOH + CH3COo = 19. CH3COOH = 19.- CH3200 The pH of the buffer pH = pka +log [CH3200 ] or 5 = 4.74 +10g CH3C002 ) 11 19-CH3Coo) orlog CH3200 0 [CH3COOH] 19- CH3 coo = 0.26 or : CH3cool 0.26 19.- Chs coo = 100126=1182 2.82 or CH3 Coo. = 1:82: (19- CH3600 ) or 2.82* CH3 COO = 1.82419 or CH3 Coo = (4.82x19) = 12:26 LA CH3COO = 12:26 mmol & CH3COOH =(19-12:26). = 6.74 minor Now, gml of 0.380M HU =(8x 0.380) mmol. Hy = 342 99Hal . CH3COO +Hul CH3COOH tau 3.42 mmol. He will react with 3.42 mmoli Cits cooto . produce 3.42 mmol CH3COOH. . Amount of CH3COOH (3.42 + 6.74) mmol. = 10.16 mmol. 19: 19 Hts.coo =(12:26 - 3:42) 99 = 8.84.99 :pH = pka tlog: [CH3200-] *P = pramog T CH3COOH 7. = 4.74. Hog (8.84) = 4068 -:45H = (4.68-5:00) = -0:32 Cans) 10.16