𝐸𝑖𝑛𝑑 = 𝐸𝐴 = 𝐸𝐴𝑜 −
𝐸 = 𝐸𝐵 = 𝐸𝐵𝑜 −
[𝐴𝑟𝑒𝑑 ]
0.05 916
log
[𝐴𝑜𝑥 ]
𝑛𝐴
[𝐵𝑟𝑒𝑑 ]
0.05 916
log
[𝐵𝑜𝑥 ]
𝑛𝐵
𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑢𝑛𝑟𝑒𝑎𝑐𝑡𝑒𝑑 𝐹𝑒 3+
[𝐹𝑒
=
𝑡𝑜𝑡𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒
[𝐹𝑒 3+ ]𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 ∙ 𝑉𝐹𝑒 3+ − 2 ∙ [𝐴𝐴] ∙ 𝑉𝐴𝐴
=
𝑉𝐹𝑒 3+ + 𝑉𝐴𝐴
3+ ]
3+
𝑚𝑜𝑙𝑒𝑠
𝑜𝑓
𝑟𝑒𝑎𝑐𝑡𝑒𝑑
𝐹𝑒
[𝐹𝑒 2+ ] =
𝑡𝑜𝑡𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒
2 ∙ [𝐴𝐴] ∙ 𝑉𝐴𝐴
=
𝑉𝐹𝑒 3+ + 𝑉𝐴𝐴
𝐸 = {𝐸𝐴𝑜 −
[𝐴𝑟𝑒𝑑 ]
0.05 916
log
} − 0.197
[
]
𝑛𝐴
𝐴𝑜𝑥
[𝐹𝑒 2+ ]
0.05 916
𝐸 = {0.767 −
log
} − 0.197
[𝐹𝑒 3+ ]
𝑛𝐴
[𝐹𝑒
3+ ]
(0.020 𝑀)(10 𝑚𝐿) − 2(0.0100 𝑀)(2.0 𝑚𝐿)
=
10 𝑚𝐿 + 2.0 𝑚𝐿
0.20 𝑀 ∙ 𝑚𝐿 − 0.040 𝑀 ∙ 𝑚𝐿
=
12.0 𝑚𝐿
0.16 𝑀 ∙ 𝑚𝐿
=
12.0 𝑚𝐿
1
=
≈ 0.0133
75
[𝐹𝑒
2+ ]
2(0.0100 𝑀)(2.0 𝑚𝐿)
=
10 𝑚𝐿 + 2.0 𝑚𝐿
0.040 𝑀 ∙ 𝑚𝐿
=
12.0 𝑚𝐿
1
=
≈ 0.0033
300
1
]
0.05 916 𝑉
300
𝐸 = {0.767 𝑉 −
log
} − 0.197 𝑉
1
1
[ ]
75
1
𝐸 = {0.767 𝑉 − (0.05 916 𝑉) log } − 0.197
4
𝐸 = {0.767 𝑉 − (0.05 916)(−0.60206) 𝑉} − 0.197 𝑉
[
𝐸 = {0.767 + 0.036}𝑉 − 0.197 𝑉
𝐸 = 0.803 𝑉 − 0.197 𝑉
𝐸 = 0.606 𝑉
[𝐹𝑒
3+ ]
(0.020 𝑀)(10 𝑚𝐿) − 2(0.0100 𝑀)(5.0 𝑚𝐿)
=
10 𝑚𝐿 + 5.0 𝑚𝐿
0.20 𝑀 ∙ 𝑚𝐿 − 0.10 𝑀 ∙ 𝑚𝐿
=
15.0 𝑚𝐿
0.10 𝑀 ∙ 𝑚𝐿
=
15.0 𝑚𝐿
=
1
≈ 0.0066 𝑀
150
[𝐹𝑒
2+ ]
2(0.0100 𝑀)(5.0 𝑚𝐿)
=
10 𝑚𝐿 + 5.0 𝑚𝐿
0.10 𝑀 ∙ 𝑚𝐿
=
15.0 𝑚𝐿
1
=
≈ 0.0066
150
1
]
0.05 916 𝑉
150
𝐸 = {0.767 𝑉 −
log
} − 0.197 𝑉
1
1
[
]
150
𝐸 = {0.767 𝑉 − (0.05 916 𝑉) log 1} − 0.197 𝑉
[
𝐸 = {0.767 𝑉 − (0.05 916)(0) 𝑉} − 0.197 𝑉
𝐸 = {0.767 + 0}𝑉 − 0.197 𝑉
𝐸 = 0.767 𝑉 − 0.197 𝑉
𝐸 = 0.570 𝑉
𝐸 = 𝐸𝐴𝑜 − 𝐸𝑟𝑒𝑓
𝐸𝑖𝑛𝑑 = 𝐸𝑡𝑖𝑡𝑟𝑎𝑛𝑡 + 𝐸𝑡𝑖𝑡𝑟𝑎𝑛𝑑 = 𝑛𝐴 𝐸𝐴 + 𝑛𝐵 𝐸𝐵
𝐸𝑖𝑛𝑑 = 𝑛𝐴 𝐸𝑒𝑞 + 𝑛𝐵 𝐸𝑒𝑞
𝐸𝑖𝑛𝑑 = (𝑛𝐴 + 𝑛𝐵 )𝐸𝑒𝑞
𝐸𝑖𝑛𝑑 = 𝐸𝑒𝑞
(𝑛𝐴 + 𝑛𝐵 )𝐸𝑒𝑞 = 𝐸𝐴 + 𝐸𝐵
(𝑛𝐴 + 𝑛𝐵 )𝐸𝑒𝑞 = 𝐸𝐴 + 𝐸𝐵
𝑛𝐴 𝐸𝐴 = 𝑛𝐴 𝐸𝐴𝑜 − 0.05 916 log
[𝐴𝑟𝑒𝑑 ]
[𝐴𝑜𝑥 ]
𝑛𝐵 𝐸𝐵 = 𝑛𝐵 𝐸𝐵𝑜 − 0.05 916 log
𝑛𝐴 𝐸𝑒𝑞 + 𝑛𝐵 𝐸𝑒𝑞 = 𝑛𝐴 𝐸𝐴𝑜 − 0.05 916 log
[𝐵𝑟𝑒𝑑 ]
[𝐵𝑜𝑥 ]
[𝐴𝑟𝑒𝑑 ]
[𝐵𝑟𝑒𝑑 ]
+ 𝑛𝐵 𝐸𝐵𝑜 − 0.05 916 log
[𝐴𝑜𝑥 ]
[𝐵𝑜𝑥 ]
𝑛𝐴 𝐸𝑒𝑞 + 𝑛𝐵 𝐸𝑒𝑞 = 𝑛𝐴 𝐸𝐴𝑜 + 𝑛𝐵 𝐸𝐵𝑜 − 0.05 916 log
[𝐴𝑟𝑒𝑑 ]
[𝐵𝑟𝑒𝑑 ]
− 0.05 916 log
[𝐴𝑜𝑥 ]
[𝐵𝑜𝑥 ]
𝑛𝐴 𝐸𝑒𝑞 + 𝑛𝐵 𝐸𝑒𝑞 = 𝑛𝐴 𝐸𝐴𝑜 + 𝑛𝐵 𝐸𝐵𝑜 − 0.05 916 {log
[𝐴𝑟𝑒𝑑 ]
[𝐵𝑟𝑒𝑑 ]
+ log
}
[𝐴𝑜𝑥 ]
[𝐵𝑜𝑥 ]
𝑛𝐴 𝐸𝑒𝑞 + 𝑛𝐵 𝐸𝑒𝑞 = 𝑛𝐴 𝐸𝐴𝑜 + 𝑛𝐵 𝐸𝐵𝑜 − 0.05 916 {log
[𝐴𝑟𝑒𝑑 ][𝐵𝑟𝑒𝑑 ]
}
[𝐴𝑜𝑥 ][𝐵𝑜𝑥 ]
𝑛𝐴 𝐸𝑒𝑞 + 𝑛𝐵 𝐸𝑒𝑞 = 𝑛𝐴 𝐸𝐴𝑜 + 𝑛𝐵 𝐸𝐵𝑜 − 0.05 916(log 1)
𝑛𝐴 𝐸𝑒𝑞 + 𝑛𝐵 𝐸𝑒𝑞 = 𝑛𝐴 𝐸𝐴𝑜 + 𝑛𝐵 𝐸𝐵𝑜 − 0.05 916(0)
𝑛𝐴 𝐸𝑒𝑞 + 𝑛𝐵 𝐸𝑒𝑞 = 𝑛𝐴 𝐸𝐴𝑜 + 𝑛𝐵 𝐸𝐵𝑜 − 0
𝑛𝐴 𝐸𝑒𝑞 + 𝑛𝐵 𝐸𝑒𝑞 = 𝑛𝐴 𝐸𝐴𝑜 + 𝑛𝐵 𝐸𝐵𝑜
(𝑛𝐴 + 𝑛𝐵 )𝐸𝑒𝑞 = 𝑛𝐴 𝐸𝐴𝑜 + 𝑛𝐵 𝐸𝐵𝑜
(𝑛𝐴 + 𝑛𝐵 )𝐸𝑒𝑞 = 𝑛𝐴 𝐸𝐴𝑜 + 𝑛𝐵 𝐸𝐵𝑜
𝑛𝐴 𝐸𝐴𝑜 + 𝑛𝐵 𝐸𝐵𝑜
𝐸𝑒𝑞 =
(𝑛𝐴 + 𝑛𝐵 )
𝐸𝑒𝑞 = 𝐸𝐴 = 𝐸𝐴𝑜 −
[𝐴𝑟𝑒𝑑 ]
0.05 916
log
[𝐴𝑜𝑥 ]
𝑛𝐴
𝐸𝑒𝑞 = 𝐸𝐵 = 𝐸𝐵𝑜 −
[𝐵𝑟𝑒𝑑 ]
0.05 916
log
[𝐵𝑜𝑥 ]
𝑛𝐵
𝑛𝐴 𝐸𝑒𝑞 = 𝑛𝐴 𝐸𝐴𝑜 − 0.05 916 log
[𝐴𝑟𝑒𝑑 ]
[𝐴𝑜𝑥 ]
𝑛𝐵 𝐸𝑒𝑞 = 𝑛𝐵 𝐸𝐵𝑜 − 0.05 916 log
[𝐵𝑟𝑒𝑑 ]
[𝐵𝑜𝑥 ]
𝐸 = 𝐸𝑖𝑛𝑑 − 𝐸𝑟𝑒𝑓
𝐸 = 𝐸𝑒𝑞 − 𝐸𝑟𝑒𝑓
𝐸=
𝑜
𝐴
𝑜
𝐵
𝑛𝐴 𝐸 + 𝑛𝐵 𝐸
(𝑛𝐴 + 𝑛𝐵 )
− 𝐸𝑟𝑒𝑓
[𝐹𝑒 2+ ]
𝐸𝑒𝑞 = 0.767 𝑉 − 0.05 916 log
[𝐹𝑒 3+ ]
[𝐴𝐴]
0.05 916
2𝐸𝑒𝑞 = 2 (0.390 𝑉 −
log
)
[𝑑𝑒ℎ𝑦𝑑𝑟𝑜][𝐻 + ]2
2
[𝐴𝐴]
2𝐸𝑒𝑞 = 0.780 𝑉 − 0.05 916 𝑉 log
[𝑑𝑒ℎ𝑦𝑑𝑟𝑜][𝐻 + ]2
[𝐹𝑒 2+ ]
[𝐴𝐴]
𝐸𝑒𝑞 + 2𝐸𝑒𝑞 = (0.767 𝑉 + 0.780 𝑉) − 0.05 916 log
−
0.05
916
log
[𝐹𝑒 3+ ]
[𝑑𝑒ℎ𝑦𝑑𝑟𝑜][𝐻 + ]2
3𝐸𝑒𝑞 = 1.547 𝑉 − 0.05 916 {log
[𝐹𝑒 2+ ]
[𝐴𝐴]
+ log
}
3+
[𝐹𝑒 ]
[𝑑𝑒ℎ𝑦𝑑𝑟𝑜][𝐻 + ]2
3𝐸𝑒𝑞 = 1.547 𝑉 − 0.05 916 {log
[𝐹𝑒 2+ ][𝐴𝐴]
}
[𝐹𝑒 3+ ][𝑑𝑒ℎ𝑦𝑑𝑟𝑜][𝐻 + ]2
3𝐸𝑒𝑞 = 1.547 𝑉 − 0.05 916 {log
1
}
[𝐻 + ]2