1
2 4/12 4/19 4/27 5/10 5/17 5/24 5/31 (20 ) (20 ) (10 ) (50 )
3 (mole fraction) X = (mol) (mol) i n 1, n 2,, n x N i X i = n i = n i n 1 + n 2 + + n x N
4 (molarity, M) 1 dm 3 ( L) (mol) (mol/l) = 1 L M x (g) y (ml) c = (x / M) y 1000
5 (molality, m) 1 kg (mol/kg) = (mol) 1 kg M x (g) y (g) m = (x / M) y 1000
6 ( wt%, w/w%) 100 g (%) = (g) X 100 (g)
7 ( V%, V/V%) 100 ml (ml) (%) = (ml) (ml) X 100
8 N 2 28.01 78.088 75.527 9x10 6 O 2 32.00 20.949 23.143 8x10 3 Ar 39.94 0.93 1.282 CO 2 44.01 0.03 0.0456 50-200 CO 28.01 1x10-5 1x10-5 0.1 Ne 20.18 1.8x10-3 1.25x10-3 He 4.00 5.24x10-4 7.24x10-5 3x10 7 CH 4 16.05 1.4x10-4 7.25x10-5 12 Kr 83.7 1.14x10-4 3.30x10-4 N 2O 44.02 5x10-5 7.6x10-5 120 H 2 2.02 5x10-5 3.48x10-6 6-8 O 3 48.0 2x10-6 3x10-6 0.1-0.3 H 2O 18.02 0.03
9 / ( wt/v%) 100 ml / (%) = (g) (ml) X 100
10 ppm() ppm = part per million = 10-6 1 mg 1 ppm = 1 L 1 g/ml ex. 1 L Cu 2+ 1 mg C Cu 2+ = 1 mg 1 L = 1 mg 1 L (1 kg/l) = 10 3 g 10 3 g =10 6 =1 ppm
11 ppb() ppm 1/1000 ppb = part per billion = 10-9 1 ppb = 1 µg 1 L 1 g/ml ex. 1 L Pb 2+ 1 µg C Pb 2+ = 1µg 1 L = 1µg 1 L (1 kg /L) = 10 6 g 10 3 g =10 9 =1 ppb
12 (normality, eq/l, N) 1 L.. (eq) (eq/l) = 1 L
13 1 (eq) = 1 mol H + 1 eq HCl = 1mol HCl 1 eq H 2 SO 4 = (1/2) mol H 2 SO 4 1 eq NaOH = 1 mol NaOH
14
15 Na + NaCl H 2 O Cl - Na + + SO 2-4 NaSO - 4
16 Na + K + Mg 2+ Ca 2+ 98 99 89 89 MSO 4 2 1 10 10 MHCO 3 1 1 Cl - SO 2-4 HCO - 3 CO 2-3 100 39 81 8 NaX 37 11 16 MgX 20 7 44 CaX 4 4 21 KX 1 Mg 2 CO 3 7 MgCaCO 3 4
17 (activity) a i = f i C i f i (activity coefficient) C i i (ionic strength) 1 2 µ = C i i C i Z i 2 Z i
18, f i, µ
19 Debye and Hückel (1923) log f i = AZ 2 i µ 1+ Bα µ (μ 0.2) A = 0.509 mol -1/2 dm 3/2 (25 ) B = 3.29 x 10 7 cm -1 mol -1/2 dm 3/2 (25 ) α (cm) 1 >> Bα µ log f i = 0.509 Z i 2 µ (μ 0.01) Davies (1938), Pitzer (1975)
20
21 f ± A m B n f Am B n f ± = = f A m f B n m +n ex. f NaCl = f Na + f Cl = f ± (m +n ) ( f A m f B n ) = f ± 2 f A f B A B f CaCl2 = f Ca 2+ f Cl 2 = f ± 3
22 [ ] 0.50 M Na 2 SO 4 [ ] Na 2 SO 4 2Na + + SO 4 2-0.50 M Na 2 SO 4 2 x 0.50 M Na + 0.50 M SO 4 2- Na + +1 SO 4 2- -2 µ = 1 ( 2 2 0.50 (+1)2 + 0.50 ( 2) 2 ) = 1 2 ( 1.0 + 2.0) =1.5
23 [ ] 0.0400 M HCl H + Cl - [ ] BαH + 3.0 Cl - 1.0. µ = 1 ( 2 0.0400 (+1)2 + 0.0400 ( 1) 2 ) = 1 2 0.512 0.0400 log f H + = 1+ 3.0 0.0400 = 0.0640 f H + = 0.863 0.512 0.0400 log f Cl = 1+1.0 0.0400 = 0.0853 f Cl = 0.822 ( 0.0400 + 0.0400) = 0.0400
24 ex. NaCl, NaOH, HCl NaCl Na + + Cl - ex. CH 3 COOH, H 2 S, NH 3 CH 3 COOH CH 3 COO - + H +
25 E = R I E (V) I (A) R (Ω) R = ρ S ρ (Ωm) S
26 (, ) S (Ω -1 ) K = 1 S ρ = κ κ () [S m -1 or S cm -1 ] (S/ ) () S
27 ( ) 1 ( ) Λ = κ C = 104 κ C [S m 2 mol 1 ] [S cm 2 mol 1 ]
28 Λ (S cm 2 mol -1 ) 400 300 200 100 Λ 0 Λ 0 Λ 0 NaOH AgNO 3 CH 3 COOH HCl 0 0.5 1.0 1.5 ( C ) 1/2 (mol L -1 ) 1/2 Λ 0 Λ = Λ 0 - k C 1/2
29 ( ) Λ Λ 0 - k C 1/2 Λ 0
30 Kohlraush Λ 0 Λ 0 = λ + + λ - ex. Λ 0 (NaCl) = λ Na+ + λ Cl- Λ 0 (HCl) = λ H+ + λ Cl- Λ 0 (CH 3 COONa) = λ Na+ + λ CH3COO-
31 [ ] [ ] Λ 0 (CH 3 COOH) = λ H+ º + λ CH3COO- º Λ 0 (CH 3 COOH) = λ H+ + λ Cl- + λ Na+ +λ CH3COO- - Λ 0 (NaCl) = Λ 0 (HCl) + Λ 0 (CH 3 COONa) - Λ 0 (NaCl) = 426.1 + 91.0-126.5 = 390.6 (S cm 2 mol -1 )
32 1 ( ) = ex. Ca 2+ = ( )/2
33
34 N 1252 m 0 1 2 km (700 m)
35 µ H + 2.14 x 10-2 Cl - 71.1 Na + 121 NO 3-70.5 + NH 4 40.3 2- SO 4 95.3 (µs cm -1 ) K + 4.04 - HCO 3 473 Mg 2+ 154 CO 3 2-2.46 Ca 2+ 389
36 (µs cm -1 ) = { λ H + [H + ] + λ NH + [NH + 4 4 ] + λ Na + [Na + ] + λ K + [K + ] + +λ Mg 2+ [Mg 2+ ] + λ Ca 2+ [Ca 2+ ] + λ Cl [Cl ] + λ NO3 [NO 3 ] +λ SO4 2 [SO 4 2 ] + λ HCO3 [HCO 3 ] + λ CO3 2 [CO 3 2 ]} 10 3 = 80.0 µs cm -1 λ [S cm 2 eq -1 ] [ ] [µeq L -1 ]