# 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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