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本文拟结合热力学基本原理,探究磷酸体系的解离平衡与水解平衡的关联.
25℃、100kPa下相关物质的ΔfGθm数据参见表1[1].
磷酸的解离平衡
磷酸解离反应如下式(1)、(2)及(3)所示.
H3PO4(aq)→H2PO4-(aq) + H+(aq) (1)Kθa1
H2PO4-(aq) → HPO42-(aq) + H+(aq) (2)Kθa2
HPO4-(aq) → PO43-(aq) + H+(aq) (3)Kθa3
1.1 磷酸的逐级解离常数(Kθa1、Kθa2及Kθa3)热力学计算
对于式(1),依热力学基本原理可得:
ΔrGθm,1=ΔfGθm(H2PO4-,aq)+ΔfGθm(H+,aq)-ΔfGθm(H3PO4,aq)
=-1130.39+0-(-1142.65)
=12.26(kJ·mol-1)
另对于式(1)有如下等式成立:ΔrGθm=-RT·lnKθa1
整理上式可得:lnKθa1=-ΔrGθm/(RT)
=-12.26×103/(8.314×298.15)
=-4.9459
解之得:Kθa1=7.11×10-3
同理对于式(2),依热力学基本原理可得:
ΔrGθm,2=ΔfGθm(HPO42-,aq)+ΔfGθm(H+,aq)-ΔfGθm(H2PO4-,aq)
=-1089.26-(-1130.39)
=41.13(kJ·mol-1)
lnKθa2=-ΔrGθm/(RT)
=-41.13×103/(8.314×298.15)
=-16.5926
解之得:Kθa2=6.22×10-8
对于式(3),依热力学基本原理可得:
ΔrGθm,3=ΔfGθm(PO43-,aq)+ΔfGθm(H+,aq)-ΔfGθm(HPO42-,aq)
=-1018.8-(-1089.26)
=70.46(kJ·mol-1)
lnKθa3=-ΔrGθm/(RT)
=-70.46×103/(8.314×298.15)
=-28.4248
解之得:Kθa3=4.52×10-13
2.磷酸系统的水解
水的解离反应参见式(4):
H2O(l)→H+(aq)+OH-(aq) (4)Kθw
对于式(4),依热力学基本原理可得:
ΔrGθm,4=ΔfGθm(OH-,aq)+ΔfGθm(H+,aq)-ΔfGθm(H2O,l)
=-157.24-(-237.13)
=79.89(kJ·mol-1)
lnKθw=-ΔrGθm/(RT)
=-79.89×103/(8.314×298.15)
=-32.2290
解之得:Kθw=1.01×10-14
磷酸系统水解反应如下式(5)、(6)及(7)所示:
PO43-(aq)+H2O(l)→HPO42-(aq)+OH-(aq) (5)Kθh1
HPO42-(aq)+H2O(l)→H2PO4-(aq)+OH-(aq) (6)Kθh2
H2PO42-(aq)+H2O(l)→H3PO4(aq)+OH-(aq) (7)Kθh3
观察式(5)、(6)及(7)可得:
(5)=(4)-(3)
(6)=(4)-(2)
(7)=(4)-(1)
根据多重平衡原则可得:
Kθh1=Kθw/ Kθa3 (8)
Kθh2=Kθw / Kθa2 (9)
Kθh3=Kθw /Kθa1 (10)
将Kθw、Kθa1、Kθa2及Kθa3数据依次代入式(8)、(9)及(10)可得:
Kθh1=1.01×10-14/(4.52×10-13)=2.23×10-2
Kθh2=1.01×10-14/(6.22×10-8)=1.62×10-7
Kθh3=1.01×10-14/(7.11×10-3)=1.42×10-12
3. 结论
对于H3PO4 溶液,Kθa1=7.11×10-3,H3PO4 不发生水解,溶液显酸性;
对于H2PO4-溶液,Kθa2>> Kθh3,H2PO4-主要发生解离,溶液显酸性;
对于HPO42-溶液,Kθa3<<Kθh2,HPO42-主要发生水解,产生OH-,溶液显碱性;
对于PO43-溶液,Kθh1=2.23×10-2,PO43-不发生解离,溶液显碱性.
参考文献
[1]Lide D R. CRC Handbook of Chemistry and Physics. 89th ed, Chemical Co, 2008,17:2688
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