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第八章 电解质溶液 答案
一、选择题 题号 答案
二、判断题 题号 1 答案 ×
三、计算题 1．[答]
2.21S?m-1?42-1?221?10 S?m?mol⑴．1．?m(NaOH)?? 3-3c0.1?10 mol?m1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 A C A B C B C C B A B B A D A A C B C A 2 × 3 √ 4 √ 5 √ 6 × 7 √ 8 × 9 √ 10 × ?0.56S?m-1?112?10?4 S?m2?mol-1 ⑵．?m(NaCl)?3-30.05?10mol?m-1(1.70?2?42-13?0.56)S?m⑶．?m(HCl)? ?399?10 S?m?mol3-3(0.1)?10mol?m32．[答] ?(AgCl)??(溶液)??(水)?1.82?10?4S?m-1
???-22-1??m(AgCl)??m(AgNO3)??m(HCl)-?m(HNO3)?1.38?10S?m?mol
c??(AgCl)?(AgCl)???1.32?10-2mol?m-3?1.32?10-5mol?dm-3?m(AgCl)?m(AgCl)
?1.89?10-3g?dm-33．[答] 计算公式： lg?? = (-A│z+z-│I1/2 ) I = 1/2?mBzB2
B 结果： AB m = 0.0001 ?? = 0.9883 m = 0.0005 ?? = 0.9741 AB2 m = 0.0001 ?? = 0.9601 m = 0.0005 ?? = 0.9132
AB3 m = 0.0001 ?? = 0.9175 m = 0.0005 ?? = 0.8248 4．[答] (1) ?m(KCl) = ?m(KNO3) + ?m(NaCl) - ?m(NaNO3) = 1.5×10-2 S·m2·mol-1 t+ = ?m,+/?m ?m(K+) = t+?m(KCl) = 7.5×10-3 S·m2·mol-1 ?m(Cl-) = t-?m(KCl) = 7.5×10-3 S·m2·mol-1 ?m(Na+) = ?m(NaCl) - ?m(Cl-) = 5.1×10-3 S·m2·mol-1
?m(NO?3) = ?m(KNO3) - ?m(K+) = 7.0×10-3 S·m2·mol-1
(2) R = ?×(l/A) = Kcell/k ?m = k/c ?m(HCl)/?m(NaCl) = [(k/c(HCl)]/[(k/c(NaCl)] = Rc(NaCl)/[Rc(HCl)] = 3.5 ?m(HCl) = 3.5?m(NaCl) = 4.41×10-2 S·m2·mol-1
5．[答] t(Ag+)?Q(Ag+)Q(总)?1??m(Ag+)m(Ag+)?0.4739 t(NO3-) = 1 - t(Ag+) = 0.5261 6．[答] ⑴??mm??c，????????0.09025 mc?m⑵CNH3OHC?2H52H5NH3?OH-
c(1-)
K??2c??(c$)?5.596?10?41
c(OH -) =(0.09025/16) mol·dm-3=5.641×10-3 mol·dm-3 7．[答]
(溶液)=(KCl)+
(ACl)=c(KCl)
m(KCl)+c(ACl)m(ACl)
≈
(KCl)[
m(K+) + m
(Cl -)]+
(ACl)[
m
(A+) + m
代入具体数据得：
m
(A+)=40×10-4 S·m2·mol-1
(Cl -)]
8．[答] Λm=κ/c=13.32/0.405×103 S·m2·mol-1 =13.32/(0.405×103)=328.9×10-4 S·m2·mol-1
K=CΛm2/[Λm∞(Λm∞-Λm)]=2.18×10-4
9．[答] κx=(l/A)(1/R)2.767×10-1=(l/A)(1/82.4);
∴电导池常数 (l/A)=22.8m -1 κ（K2SO4）=22.8×1/163=0.140S·m-1 Λm(K2SO4)=κ/c=0.028S·m2·mol-1 10．[答] ??Kcell13.7?1?1?m???2.092?10?2S?m?1 R6552.092?10?2?m??S?m2?mol?1?1.32?10?3S?m2?mol?1
c15.81???m(HAc)???m(H?)???m(Ac?)?(349.82?40.9)?10?4S?m2?mol?1 ?3.91?10?2S?m2?mol?1
?m1.32?10?3?????3.38?10?2 ?2?m3.91?10c215.81?2?2??(3.38?10)??c1 KC??1??1?3.38?10?2?1.87?10?5