• D L 2 3 4 5. 6 7 8 1 TO m 2 2) 24 25 NaOH added (ml) 0 1 2 3 4 5 6 7 8 9 10 11 12 HER98528 14 15 16 19 20 21 22 23 PH

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D L 2 3 4 5 6 7 8 1 To M 2 2 24 25 Naoh Added Ml 0 1 2 3 4 5 6 7 8 9 10 11 12 Her98528 14 15 16 19 20 21 22 23 Ph 1 (23.53 KiB) Viewed 11 times

D L 2 3 4 5 6 7 8 1 To M 2 2 24 25 Naoh Added Ml 0 1 2 3 4 5 6 7 8 9 10 11 12 Her98528 14 15 16 19 20 21 22 23 Ph 2 (23.53 KiB) Viewed 11 times

D L 2 3 4 5 6 7 8 1 To M 2 2 24 25 Naoh Added Ml 0 1 2 3 4 5 6 7 8 9 10 11 12 Her98528 14 15 16 19 20 21 22 23 Ph 3 (36 KiB) Viewed 11 times

D L 2 3 4 5 6 7 8 1 To M 2 2 24 25 Naoh Added Ml 0 1 2 3 4 5 6 7 8 9 10 11 12 Her98528 14 15 16 19 20 21 22 23 Ph 4 (37.63 KiB) Viewed 11 times

• D L 2 3 4 5. 6 7 8 1 TO m 2 2) 24 25 NaOH added (ml) 0 1 2 3 4 5 6 7 8 9 10 11 12 HER98528 14 15 16 19 20 21 22 23 PH 2.99 3.65 3.75 4.06 4.08 4.19 4.27 4.38 4.48 4.57 4.67 4.77 4.89 5.04 5.18 5.34 5.54 6.03 10.97 11.64 11.81 11.97 12.06 12.09 pH vs. NaOH added (mL) 15 € D 2.99 365 375 406-408-430 427 438 448 457 467 477 499 504 510 534 5.54 605 10 M NaOH added (m) 10.97 11.64 1181 11 97 12.06 12.09
1. Initial concentration of unknown acid: Volume of NaOH at equivalence point: 18mL Concentration of NaOH : 3.00M Unknown acid: 20mL (according to lab manual, step # Base: n=Mx V n=3.00 mol/L 0.018L = 0.054mol Acid: M-0.054mol / 0.0200L = 2.7M 2. The pKa from the initial pH Initial pH= 2.99 pH= -log[H₂O*] →2.99 → [H3O+] = 10-2.9⁹ M M=0.00102M
Acid: M-0.054mol / 0.0200L = 2.7M 2. The pKa from the initial pH Initial pH= 2.99 pH= -log[H,0¹] →→2.99 → [H₂O¹] = 10299 M M=0.00102M I C E [HA] 2.7 -0.00102 2.69898 Ka-[H,O'][A-]/[HA] 0.00102-0.00102/2.69898 3.88 107 [H,0¹] 0 0.00102 0.00102 263 pKa= -log(Ka) → pKa= -log(3.88 10) = 6.41 3. The pka of weak acid from the two points in your buffer region 4. The % range of three pKa values | [A-] 0 0.00102 0.00102