Question

Part E: pH pH pH pH Average Volume Trial 1 Trial 2 Trial 3 5.0 0 5 5.0। 5.1 .0 । 5.1 5.1 5.2 5.3 [5.0। 5.1 5.1 5.1 5.2 5.3 [5.2 5.3 5.3 5.4 5.5 16. 16.5 17.0 | 17.5। 18.0 18.5 19.0 19.5 | 20.0 20.5 21.0 | 21.5 | 22.0 22.5 23.0 23.5 24.0 24.5 | 5.6 5.7 [5.3 [5.4 | 5.5 | 5.6 | 5.7 5.4 5.5 5.6 5.7 5.9 5.1 5.1 5.2 5.2 5.3 5.4 5.5 5.6 5.7 | 6.0 6.3 7.2 10.9 | 10.9 10.9 | 10.9 10.9 10.9 10.9 10.9 6.0 6.2 | 6.8 | 10.6 10.9 10.9 । 10.9 । | 109, 10.9 10.9 10.9 10.9 | 6.0 । 6.5 10.0 10.9 10.9 । 10.9 10.9 10.9 । 10.9 10.9 । 10.9 6.5 9.2 10.9 10.9 10.9 10.9 10.9 10.9 10.9 10.9 25.0 25.5 | 26.0

calculations for:

concentration of the unknown acid and values for pKa and Ka.

Answer 1

2 DATE: PAGEN 20. toooooooooo 6 17 18 19 20 21 22 23 24 2+ Volunu ol NaOH 19 find conc. of unknown acid concentration of NaOH should be mention in question. Since you have not given the complete Question, I am considering it to be 0.1 M NÃO H] at equivalence pt moles of base added = moles of acid we started with Volume of base reqwed to reach equivalence point - 21.5 mL M= n n = M XV = 0.1X21.5 = .00215 1000 moles = 0.00215 x1000 = 0.0869 . M= n 25

PAGE NG b) Apka & Kad It hay equivalence point prapka Volume = 21.5 = 10.75 pH valu corresponding to this volume is not given You should have calculated pH around 10.15 mL volume also