发布时间 : 星期一 文章物化作业课后答案更新完毕开始阅读812e9830ee06eff9aef807f5
?m? ( Cl-)= ?m? (KCl) ??m? (K+ ) = 76.28×10-4
S · m2· mol-1
(2) ?m? (Na+ )= ?m? (NaCl)×t (Na+ ) =50.07×10-4
S · m2/mol
?m?(Cl-) = ?m?(NaCl) ? ?m?(Na+) = 76.38×
10-4S · m2/mol
习题5.5
解:(1)首先从已知条件计算出电导池常数 Kcell = κ ·R = 0.1412S · m-1×484.0Ω = 68.34 m-1 由Kcell及NaCl水溶液的电阻值计算不同浓度时的电导率数值,再由电导率和浓度计算相应的摩尔电导率,计算公式和所得结果如下:
κ = Kcell / R ; ?m = κ /c
c/ mol · dm-3 0.0005 0.0010 0.0020
0.0050
κ / S · m-1 0.006264 0.01244
0.02465 0.06054
?m / S · m2· mol-1 0.01253 0.01244 0.01233 0.01211
以 ?m对 c1/2作图得一直线,把直线外推到c → 0时,得截距0.01270 S · m2/ mol。根据公式5-9可知所得截距就是 ?m?的值。
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习题5.6
解: ?m= κ /c= Kcell/(R·C) = 36.7/(2220×0.01×103 ) = 1.65×10 -3 S · m 2· mol-1
?m? (HAc) = ?m? (H+) +?m? (Ac-)= 390.7×10
-4S
· m2 · mol-1
α = ?m/?m? = 0.0422
习题5.7
解: ?m? (H2O) = ?m? (HAc)+ ?m? (NaOH)??m? (NaAc)= 5.478×10-2 S · m2 / mol
?m(H2O) = κ /c ,
???m?????C??mm
c (H+) = c (OH-) = α · c (H2O) =
5.80?10?6??1.059×10-7mol ??m0.05478?· dm-3
K w = c (H+)/c?· c (OH-) /c? =1.12×10-14
习题5.8
解 :I?1?miZi2?1(0.01?0.001?22?0.012)= 0.013 mol · kg-1
22在水溶液中298K时A =0.509,所以
lg?? = ? A|z+z-|(I/m?)1/2=?0.509×2×1 ×
0.0131/2= ? 0.116
?? = 0.765
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习题5.9 画出用BaCl2滴定Li2SO4时电导变化的示意图。
112?1BaCl2?Li??SO4?BaSO4(s)?Li??Cl? 22212?SO4?Cl?,因此滴定终点前体系的电2 由于导电能力
导下降,在滴定终点体系的电导达到最小,终点后导电离子增加,电导又开始增大。 ?
V
习题5.10
解:I = 1(2×0.0810×0.01)=8.1×10-4 mol · kg-1
2lg?? = ? A|z+z-|(I/m?)1/2=?0.509×1×1×(8.1×
10-4)1/2= ? 0.01449
?? =0.967
a ( H+) = a (A-) = ??×0.0810×0.01 = ??× 8.1×10-4
mol/kg
a (HA) = ??(0.01-0.00081) = ??×0.00919 mol/kg
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K??aH?aA?aHA???,H?mH?m????,A?mHAm?mA???,HA?42m??0.967?(8.1?10)?6.9?10?5
0.00919习题5.11
解:由5-13式得 ?(盐) = 0.01482 – 0.00015 = 0.01467 S/m
?m? (盐) =( 2×59.46 +2×80.0) ×10-4 =
0.02789 Sm2/mol
C = ?(盐)/?m? (盐) = 0.01467/0.02789 =
0.5260 mol/m3 = 5.26×10-4 mol/L
M = 5.26×10-4×183.7 = 0.0966 g/L
习题5.12
解:由5-21式得 I =0.5×(0.3×12+0.1×22+0.1×12) =0.4 mol · kg-1
第六章 可逆电池电动势
习题6.1 (1)解:(-) Cu (s) → Cu2+(a Cu2+) + 2e
(+) 2Ag+ (a Ag+) + 2e→ 2Ag (s)
电池反应: Cu (s) + 2Ag+ (a Ag+) → Cu2+ ( a Cu2+) +
2Ag(s)
(2)解: (-) H2 ( pH2) → 2H+ (a H+) +2 e (+) 2Ag (a Ag+) + 2e → 2Ag (s)
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