1 (172)

Prévia do material em texto

Principles of Instrumental Analysis, 6th ed. Chapter 24
 
 6
24-9. 1 mol As2O3 = 2 mol HAsO3
2- = 2 mol I2 = 4 mol e–
 Q = 0.1276 C/s × (11 min × 60 s/min + 54 s) = 91.106 C 
 Mass As2O3 = 2 3 2 31 mol As O 197.8 g As O91.106 C 
96485 C/mol e 4 mol e mol− −× × = 0.04669 
 %As2O3 = 0.04669 g 100
6.39
× = 0.731 
24-10. The equivalent mass of an acid is that mass that contains one mole of titratable H+. 
 1 F 1 equiv HA0.0441 C/s 266 s 
96485 C F
× × × = 1.216 × 10–4 equiv HA 
 equiv mass = 4
0.0809 g HA = 665 g/equiv
1.216 10 equiv HA−×
 
24-11. 1 mol C6H5NH2 = 3 mol Br2 = 6 mol e–
 Q = (3.76 – 0.27) min × 60 s/min × 1.00 × 10–3 C/s = 0.2094 C 
 mol C6H5NH2 = 6 5 21 mol C H NH1 mol e0.2094 C 
96485 C 6 mol e
−
−× × = 3.617 × 10–7 
 mass C6H5NH2 = 7 6
6 5 2
6 5 2
93.128 g 3.617 10 mol C H NH 10 μg/g
mol C H NH
−× × × 
 = 33.7 μg 
24-12. The spreadsheet is attached.