word格式
平衡浓度/ mol·dm-3 0.20 - x x 0.20 + x x = 6.4 ? 10-5 pH = pKa2 = 4.19
? (3)C2O24
43
(1) 氨水的[H+]
[OH?] =Kb?c?18.?10?5?010.= 1.3 ? 10-3 (mol·dm-3) [H+] =
1?10?14= 7.7 ? 10-12 (mol·dm-3 ) ?313.?10 (2) 加入 5.35 g NH4Cl后，溶液的[H+] [OH?] = Kb·
c碱c酸= 1.8 ? 10-5 ?
010.= 1.8 ? 10-5 (mol·dm-3) 010.10. [H+] =
?10?14= 5.6 ? 10-10 (mol·dm-3) ?518.?10 (3) 加入NH4Cl前后氨水的解离度 ① 加入NH4Cl前NH3·H2O的解离度
13.?10?13 ? = = 1.3 %
010. (或? =(18.?10?5)/010.= 1.3 % ② 加入NH4Cl后，NH3·H2O的解离度
18.?10?5 ? = = 0.018 %
010. 44
? HSO4H+ +SO4 K2 = 1.26 ? 10-2
2? 平衡浓度/ mol·dm-3 0.10 - x x x
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word格式
x2 = 1.26 ? 10-2
010.?x [H+] = x = 3.0 ? 10-2 mol·dm-3 pH = -lg 3.0 ? 10-2 = 1.52
Kw10.?10?14? Kh == 7.9 ? 10-13 故水解可忽略 ?2K2126.?10 45
c(Al3+) = 7.5 / (375 ? 0.20) = 0.10 (mol·dm-3 ) Al(NO3)3·9H2O溶于水发生水解 Al3+ + H2O
Al(OH)2+ + H+
平衡浓度/ mol·dm-3 0.10-x x x Kh由两个平衡求得： Al3+ + OH? +) H2O Al3+ + H2O
Al(OH)2+ K稳 = 1.9 ? 109 H+ + OH? Al(OH)2+ + H+
Kw = 10-14
Kh = K稳·Kw = 1.9 ? 109 ? 10-14 = 1.9 ? 10-5
x2 = 1.9 ? 10-5 x = 1.4 ? 10-3
010.?x c(Al(OH)2+) = 1.4 ? 10-3 mol·dm-3 c(H+) = 1.4 ? 10-3 mol·dm-3
c(OH?) = 1.0 ? 10-14 / (1.4 ? 10-3) = 7.1 ? 10-12 (mol·dm-3) c(Al3+) ? 0.10 mol·dm-3 46
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word格式
0.25?100= 0.17 (mol·dm-3)
1500.35?50 cs= [HPO42?] == 0.12 (mol·dm-3)
150? ca= [H2PO4] =
pH = pK2 - lgca017. = -lg 6.3 ? 10-8 - lg= 7.05 cs012. 加入NaOH后，c'a = (0.17 ? 150 - 0.10 ? 50) / 200 = 0.10 (mol·dm-3) c's = (0.12 ? 150 + 0.10 ? 50) / 200 = 0.12 (mol·dm-3) pH = pK2 - lg (c'a / c's) = 7.20 + 0.079 = 7.28 47
S2? + H2O
HS? + OH?
平衡浓度/ mol·dm-3 0.15 - x x x
Kx2 = Kb1=w= 1.0 解得 x = 0.13 mol·dm-3
K2015.?x HS? + H2O
H2S + OH?
平衡浓度/ mol·dm-3 0.13 - y y 0.13+y
Ky?(013.?y)= Kb2=w= 1.0 ? 10-7， 解得 y = 1.0 ? 10-7
015.?yK1 因为[Na+] = 0.15 ? 2 = 0.30 (mol·dm-3) [S2? ] = 0.15 - 0.13 = 0.02 (mol·dm-3) [HS? ] = [OH? ] = 0.13 (mol·dm-3) [H2S] = 1.0 ? 10-7 (mol·dm-3 ) [H+] =
Kw[OH]?= 7.7 ? 10-14 (mol·dm-3)
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2? CO3+ H2O
?HCO3+ OH?
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word格式
平衡浓度/ mol·dm-3 0.20 - x x x
Kx2 = Kb1 =w= 1.8 ? 10-4
K2020?x x =Kb1?c= 0.0060 mol·dm-3
? HCO3 + H2O
H2CO3 + OH?
Kb2 =
Kw= 2.3 ? 10-8 K1[OH?][H2CO3]?[HCO3] Kb2 =
? [H2CO3] (因为[OH?] = [HCO3])
? 所以[H2CO3] = Kb2 = 2.3 ? 10-8 mol·dm-3
? 所以[Na+] = 0.20 ? 2 = 0.40 (mol·dm-3) [HCO3] = [OH?] = 0.0060 mol·dm-3 ?-3-8-3 [CO23] = 0.19 mol·dm [H2CO3] = 2.3 ? 10 mol·dm
[H+] =
10.?10?14== 1.7 ? 10-12 (mol·dm-3) ?0.0060[OH]Kw 49
混合后：[OH?] = 0.20 ? 75/100 = 0.15 (mol·dm-3 ) [H2C2O4] = 0.40 ? 25/100 = 0.10 (mol·dm-3)
?-3
NaOH和H2C2O4反应后，[C2O24] = 0.050 mol·dm
-3 [HC2O?4] = 0.050 mol·dm
?[HC2O4]?[C2O24] (1) 混合溶液中[H+] = K (2) [Ca2+] =
2
= K2 = 6.4 ? 10-5mol·dm-3
2.0?10.= 2.0 ? 10-2 (mol·dm-3)
100?10.?-2-3
[Ca2+][C2O24] = 2.0 ? 10 ? 0.050 ? 1.0 ? 10 > Ksp
所以有CaC2O4? 生成
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