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fill out the table. pKa=-log(Ka)
Acid pk, Base Ko K 1.4 x 102 1 HSO, N 1.0 x 10-12 H2SO3 HSO4 CH3C(O)COOH HF 3 2.8 x 10 4 3.17 1.5 x 10-11 F CH3COO 5 1.8 x 10
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Answer #1

- pka p2.4 10-19 10~9 1.4 ХО x (0-13 1. 25 ka 2.0.1 we know that ка. К – -19 and pkk - 14 о до -| со ка Now lets fill the gi5. Acid: CH₃COOH Base CH₃Coo 10-14 ka= 1.8x lo 05 Kb ka 1on14 plea= -log 1.8 X1055 -log 1:8 + 5 1.8810-S 0 -0.265 t - 15.558Base. HS? looly Ra= lonly kg = 1 x 107 1,1X10-7 (given) a g Llory gronxio 1.8810-5 8. Acid: HS ky 4.09xlo plea lorg a. alog 92 х 10-9 12. Neid e HiO₂ 3cs e“, (03 2 с looly lonry K, - Ка? Eь 2. Х lory ( 4 ivin) uПbx 10 РЕА. -| се ч. 16 x o-1 11 -| og

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