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X 20- What is the dominant form of aspartic acid (a triprotic acid (H3A+)) at a pH of 2.50 (pka1 = 1.990, pka2 = 3.900,

Posted: Sun Apr 10, 2022 6:26 pm
by answerhappygod
X 20 What Is The Dominant Form Of Aspartic Acid A Triprotic Acid H3a At A Ph Of 2 50 Pka1 1 990 Pka2 3 900 1
X 20 What Is The Dominant Form Of Aspartic Acid A Triprotic Acid H3a At A Ph Of 2 50 Pka1 1 990 Pka2 3 900 1 (27.4 KiB) Viewed 46 times
X 20 What Is The Dominant Form Of Aspartic Acid A Triprotic Acid H3a At A Ph Of 2 50 Pka1 1 990 Pka2 3 900 2
X 20 What Is The Dominant Form Of Aspartic Acid A Triprotic Acid H3a At A Ph Of 2 50 Pka1 1 990 Pka2 3 900 2 (32.88 KiB) Viewed 46 times
X 20 What Is The Dominant Form Of Aspartic Acid A Triprotic Acid H3a At A Ph Of 2 50 Pka1 1 990 Pka2 3 900 3
X 20 What Is The Dominant Form Of Aspartic Acid A Triprotic Acid H3a At A Ph Of 2 50 Pka1 1 990 Pka2 3 900 3 (38.7 KiB) Viewed 46 times
X 20 What Is The Dominant Form Of Aspartic Acid A Triprotic Acid H3a At A Ph Of 2 50 Pka1 1 990 Pka2 3 900 4
X 20 What Is The Dominant Form Of Aspartic Acid A Triprotic Acid H3a At A Ph Of 2 50 Pka1 1 990 Pka2 3 900 4 (33.79 KiB) Viewed 46 times
X 20- What is the dominant form of aspartic acid (a triprotic acid (H3A+)) at a pH of 2.50 (pka1 = 1.990, pka2 = 3.900, pka3 = 10.002)? * D. H2A and HA- are equal O C.H2A O A. H3A+ and H2A are equal B. H3A+ Х Correct answer C. H2A
✓ 30. A 0.4376 g aspirin tablet was heated gently with 50.00 mL of 0.196 M sodium hydroxide solution. The aspirin reacts according to the equation: C6H4(OCOCH3)COOH(aq) + 2NaOH(aq) --C6H4OH)COONa(aq) + CH3COONa(aq) + H2O0) After cooling, the resulting solution was titrated against 0.298 M hydrochloric acid to determine the amount of unreacted sodium hydroxide. The initial burette reading was 1.00 ml while the end burette reading volume of 19.64 ml at the end point. Calculate: the percentage by mass %w/w. of aspirin in the tablet (Molecular weight of aspirin is 180g/mole) A. 87.31 B. 43,66 O C. 21.83 D. 10.91
✓ 31- An unknown amount of acid can often be determined by adding an excess of base and then "back-titrating the excess. A 0.3471g sample of a mixture containing oxalic acid in addition to an impurity is treated with 100.0 mL of 0.1000M NaOH. The excess NaOH is titrated with 20.00 mL of 0.200 M HC1. Find the mass % of oxalic acid. Mw of oxalic acid = 90.03 g/mol A. 77.81 B. 55.626 O c.22.19 D. 38.905
✓ 32- What volume of 2.811 M oxalic acid solution is needed to react to the equivalence point with a 5.090 g sample of material that is 92.10% NaOH? Oxalic acid is a diprotic acid. Mw NaOH 39.997 g/mol A. 20.85 ml T B. 41.70 ml O C. 83.41 ml O D. 13.33 ml