18 0 10 2 0 ca 2 y cay cca eq 001 ca

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Unformatted text preview: 1 + 0.1 − K sp 3.2 × 10 10 + = = 3.9 × −9 [ Ag ] = 10 − [Cl ] 0.082 pAg = 8.41 θ = 0.5 pAg = 8.01 θ =1 + − [ Ag ] = [Cl ] = K sp 1 1 = × .49 = 4.74 pK sp 9 2 2 c 0 + (θ − 1) 0.1×1.5 − 1) ( θ = 1.5 [ Ag+ ] = Ag = = 0.02 1+θ 1 + 1.5 pAg = 1.70 pAg = θ pAg θ pAg 0.050 0.100 0.200 0.400 0.500 0.600 0.800 0.900 0.990 8.45 8.41 8.31 8.12 8.01 7.89 7.54 7.21 6.19 0.999 1.000 1.001 1.010 1.100 1.20 1.30 1.50 2.00 5.24 4.74 4.25 3.30 2.32 2.04 1.88 1.70 1.48 pAg ~ θ 作图 注:θ = 0.999 9 8 7 0 cCl − (1 − 0.999) − = + [ Ag+ ] [Cl ] 1 + 0.999 θ = 1.001 6 pAg 5 4 3 2 1 0.0 0.5 1.0 θ 1.5 2.0 + [ Ag ] = c 0 + Ag (1.001 − 1) 1 + 1.001 + [Cl − ] 〈例〉 pH = 12 , c Ca = cY = 0.02 mol/L 时 Y 滴定 Ca , 计算θ = 0.5, 1.0, 1.5 时的 pCa。 β = 10.7 CaY lg 解:无副反应, Ca + Y 0 cCa (1 − θ ) 0.02 ×1 − 0.5) ( 2+ θ = 0.5 = = = 6.67 × −3 [Ca ] 10 1+θ 1 + 0.5 pCa = 2.18 0 θ = 1.0 2+ 0 + [Ca 2 ] = [Y], [CaY ] = cCa .eq = 0.01 + [Ca 2 ] = [CaY ] β CaY = 0.01 = 4.47 × −7 10 10.7 10 pCa = 6.35 或 pCa eq = 1 1 β + pC ) = (10.7 + 2) = 6.35 (lg 2 2 θ = 1.5 0 cCa 0.02 ×1.5 − 1) ( [Y] = , [CaY] = cCa = 1 + 1.5 1+θ 0 cCa [CaY] = 1 + θ = 1 = 1 [Ca ]= = 10−10.4 0 β[ ] cY (θ − 1) β (θ − 1) 1010.7 (1.5 − 1) Y β 1+θ pCa = 10.40 12 10 pCa ~ θ 作图 8 pCa 6 4 2 0 0.0 0.5 1.0 θ 1.5 2.0 二.滴定误差与滴定突跃 1.滴定误差:终点与化...
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This note was uploaded on 08/18/2012 for the course CHE 322 taught by Professor Liuyu during the Spring '12 term at Nanjing University.

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