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# 001 3 1 y 0001 c m eq 1 1 0001 10 001

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Unformatted text preview: p 沉淀滴定 TE = − 络合滴定 TE = sin 键 2 K sp c D ,eq sinh( 2.303∆pT ) 2 (c D,eq β ) 1 sinh( 2.303∆pD ) 2 〈例〉0.1 mol/L AgNO3 滴定 0.1 mol/L NaCl,当 pAgep = 6.00 时， 求滴定误差。 − − [ Ag]eq = K sp = 3.2 × 10 = 1.79 × 5 10 10 解： pAgeq = 4.75 ∆pAg = 6.00 − 4.75 = 1.25 0.1 = 0.05 cCl − ,eq 2 − − 1.79 × 5 (10 1.25 − 101.25 ) 10 = = −6.3 × 10 −3 = −0.63% TE 0.05 − 2 × .79 × 5 1 10 =− × .25) = −6.3 × −3 sinh( 2.303 1 10 或 TE 0.05 = 0.02 mol/L EDTA 滴定同浓度 Ca2+，当 pCa,ep = 8.85 〈例〉 时，计算滴定误差。 1 1 解：pCa eq = 2 (lg β + pCCaeq ) = 2 (10.7 + 2) = 6.35 ∆pCa = 8.85 − 6.35 = 2.50 TE = − 10 2.50 − 10 2.50 (10 10.7 或 TE = −2 × 10 ) 1 2 10 2.5 = 4.35 = 10 −1.85 = 0.014 = 1.4% 10 2 × 2.5) = 10 −1.85 = 1.4% sinh(2.303 4.35 10 2. 滴定突跃 定义 θ在 0.999 ~ 1.001 之间时的 Δ pI 值来衡量滴定突跃 1:1 型反应滴定突跃范围 求 ∆pI 0.001 滴定突跃大小 ∆ pI eq ± pI 0.001 2∆pI 0.001 2[ D]eq TE = ± sinh( 2.303∆pI ) 出发 我们从 cD,eq sinh( 2.303∆pI 0.001 ) = ± 令 y= 0.001× cD,eq 2[D]eq 0.001× cD,eq 2[D]eq 则 sinh −1 ( y ) = 2.303 ∆pI 0.001 ⇒ ∆pI 0.001 = 用计算器按 亦可用下式计算 y Inv hyp sin sinh −1 ( y ) 2.303 键 sinh −1 ( y ) = ln( y + y 2 + 1) ∆pI 0.001 = lg( y + y...
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