内容发布更新时间 : 2025/5/6 13:05:49星期一 下面是文章的全部内容请认真阅读。
0.059lg
CFe3?CFe2? = E电池 – ?0Fe3+/Fe2+ + 0.24 = 0.44 – 0.68 + 0.24 = 0
CFe3+/CFe2+ = 1.0
?sp= (1.44 + 0.68)/2 = 1.06 V E电池 = 0.76 = ?Fe3+/Fe2+ = –0.24 ?Fe3+/Fe2+ = 0.76 + 0.24 = 1.00 V 与化学计量点相差0.06V
17.国产231型玻璃电极是适于测量高pH值的锂玻璃电极。若KPotH+,Na=1×10–15, 这意味着提供相同电位时,溶液中允许的Na+浓度是H+浓度的多少倍?若Na+浓度为1.0mol/l时,测量pH=13的溶液,所引起的相对误差是多少? 解:Kpoti,j?i[H?]???10?15,即Na+浓度为H+浓度的1015倍时,产生相同的电位响应,故: ??j[Na]?i1.0×100% = 10–15 ×?13 = 1.0% ?j10相对误差 = Kpoti,j×18.用液膜电极测定Ca2+, ?Ca2+ = 0.001mol/l,如果溶液中存在Mg2+,且?Mg2+ = 0.01mol/l,由此所产生的相对误差是
多少?电池的电动势将如何变化(Kpot Ca2+,Mg2+ = 0.01)? 解:相对误差 = Kpoti,j?j?ini/nj(0.012/2)?100%?0.01??10%
0.001如有Mg2+存在:??Ca2+ = ?Ca2+ + KpotCa2+,Mg2+ = 0.001 + 0.01 × 0.01 = 0.0011mol/l Mg2+干扰所产生的电池电动势变化为: ?? = ?? – ? =
0.059?'Ca2?0.0590.0011lg?lg0.001 = 1.22 × 10–3 V 2?Ca2?219 当一个电池用0.010mol/l的氟化物溶液矫正氟离子选择电极时,所得读数为0.101V;用3.2×10–4mol/l氟溶
液校正所得度数为0.194V。如果未知浓度的氟溶液校正所得的读数为0.152V,计算未知溶液的氟离子浓度? 解:0.104 = K? – Slg(0.010) = K? + 2S 0.194 = K? – Slg(3.2×10–4) = K? + 3.49S 解得: S = 0.060 K? = –0.016V 0.152 = –0.016 – 0.060lg[F–] lg[F–] =
?(0.152?0.016) = –2.8 [F–] = 1.6×10–3mol/l
0.06020.用下列电池桉直接电位法测定C2O42–
(–)Ag?AgCl(S), KCl(饱和)║C2O42–(x/mol/l), Ag2C2O4(S)?Ag(+) 在250C时测得电池电动势为0.402V。计算未知溶液的C2O42–浓度。(已知?Ag/AgCl = 0.1988V, Ksp,Ag2C2O4 = 3.5×10–11, ?0Ag+/Ag = 0.7995V)
解:(1) E = ?Ag2C2O4/Ag – ?AgCl ?Ag2C2O4/Ag = ?0Ag2C2O4/Ag –
0.059lg[C2O42–] 20.059lg[C2O42–] 2因为?Ag+/Ag = ?0Ag+/Ag + 0.059lg[Ag+] 且 Ksp,Ag2C2O4 = [Ag+]2 [C2O42–] 所以?Ag+/Ag = 0.7995 + 0.059lg
Ksp[C2O4]2? = 0.7995 + 0.059lg
Ksp=
当[ C2O42–] = 1mol/l时,该电位即?0Ag2C2O4/Ag,所以?0Ag2C2O4/Ag = 0.7995 + 0.059lgKsp = 0.4910V (2)当KAg2C2O4/Ag = 0.4910 – </