Transcript [MIn - ]/[HIn 2

考古題 99-5
Relative error : ( Ksp’ – Ksp ) / Ksp x 100%
Ksp = (γCd2+ x [Cd2+])2 x ( γFe(CN6)4- x [Fe(CN)64-])
考古題 98-6
Relative error : ( Ka’ – Ka ) / Ka x 100%
Ka‘ = (γH3O+[H3O+] • γNO2-[NO2-]) / (γHNO2[HNO2])
考古題 99-1
b) For the electrolyte AmBn: γ± = mean activity coefficient
= (γAm γBn )1/(m+n)
2
c)
- log γ X 
0 . 51 Z X
1  3 . 3 X


Z=3
γM : activity coefficient of the species M3+
ZM : charge on the species M3+
μ : ionic strength of the solution
αM : effective diameter of the hydrated ion M3+ in nanometers
考古題 98-4
2
a)
- log γ X 
0 . 51 Z X

1  3 . 3 X

b) Shielding effect , explain it
μ = ½ (0.015 x 32 + 0.045 x 12) = 0.09
ZA : # of charge  A+
ZB : # of charge  B+
αA  effective diameter of the hydrated ion A
αB  effective diameter of the hydrated ion B
Example 17-1
Calculate the molar Y4- concentration in a 0.0200M
EDTA solution buffered to a pH of 10.00
試求下列溶液的[Y4-]，0.0200 M EDTA 溶液，pH值為10.00
Example 17-2
Calculate the equilibrium concentration of Ni2+ in a
solution with an analytical NiY2- concentration of
0.0150M at pH (a) 3.0 and (b) 8.0
計算在不同pH值下，0.0150M的[NiY2-]溶液中有多少[Ni2+]
利用conditional formation constant 解題
Example 17-3
Calculate the concentration of Ni2+ in a solution that
was prepared by mixing 50.0mL of 0.0300M Ni2+ with
50.00mL of 0.05M EDTA. The mixture was buffered
to a pH of 3.0
將50.0mL，0.0300M Ni2+與50.00mL，0.05M EDTA溶液混
合，並將混合液的pH值調整至3.0。試計算Ni2+的濃度
Example 15-35
Construct a titration curve for 50.00 mL of 0.015
M Fe2+ with 0.0300M EDTA in a solution buffered
to pH 7.0. Calculate pFe values after the addition
of 0.00, 10.00, 24.00, 25.00, 26.00, and 30.00 mL
of titrant.
試求各滴定過程的[Fe2+] , 並繪製滴定曲線圖
1)當量點前 
直接計算[Fe2+]的剩餘量
2)當量點時
由Kf與[Fe2+]的關係式;其中CT=[Fe2+],解[Fe2+]
3)當量點後
由Kf與[Fe2+]的關係式;其中CT為EDTA的剩餘量,解[Fe2+]
考古題 98-7
考古題 99-7
考古題 99-3
[MIn-]/[HIn2-] ≥ 10
~red, before end point (pM small)
[MIn-]/[HIn2-] ≤ 1/10 ~blue, after end point (pM large)
試導出溶液顯色為紅色時,pPb與Ka1,Ka2,Kf之間的關係式
(當pH=9)
考古題 98-1
試導出溶液為藍色時,pPb與Ka1,Ka2,Kf之間的關係式(當
pH=9)
[MIn-]/[HIn2-] ≤ 1/10 ~blue, after end point (pM large)
加入過量EDTA,
再用Mg2+ or Zn2+ 滴定剩餘的EDTA;
溶液由BlueRed
Excess EDTA  blue
blue
pCa
red
Volume of EDTA (ml)
加入過量Mg-EDTA(MgY2-),
再加入EDTA滴定;
溶液由RedBlue
Kf : CaY2- > MgY2-
pCa
Volume of EDTA (ml)
考古題 99-8
A 0.3284-g sample of brass(containing lead, zinc, copper, and tin)
Pb2+ , Zn2+ , Cu2+ , Sn2+ (total : 0.3284g)
The sparingly soluble SnO2 ‧4H2O was removed by filtration, and the combined filtrate
and washings were then diluted to 500.0 mL.
Removed Sn2+
A 10.00-mL aliquot was then suitably buffered; titration of this aliquot required 37.56
mL of 0.002500 M EDTA.
Titrated  (Pb2+ , Zn2+ , Cu2+) 37.56mL x 0.0025M EDTA x (500/10) mL
The copper in a 25.00-mL aliquot was masked with thiosulfate; the solution was then
titrated with 27.67 mL of EDTA solution
Mask  Cu2+
Titrated  (Pb2+ , Zn2+) 27.67mL x 0.0025M EDTA x (500/25) mL
Cyanide ion was used to mask the copper and zinc in a 100-mL aliquot; 10.80 mL of the
EDTA solution were needed to titrate this aliquot
Mask  Zn2+ , Cu2+
Titrated  (Pb2+) 10.80mL x 0.0025M EDTA x (500/100) mL
Titrated  (Pb2+ , Zn2+ , Cu2+) 37.56mL x 0.0025M EDTA x (500/10) mL = 4.695 mmol
Titrated  (Pb2+ , Zn2+) 27.67mL x 0.0025M EDTA x (500/25) mL = 1.384 mmol
Titrated  (Pb2+) 10.80mL x 0.0025M EDTA x (500/100) mL = 0.135 mmol
(Zn2+) 1.384 – 0.135 = 1.2485 mmol
(Cu2+) 4.695 – 1.384 = 3.3115 mmol
(Cu2+)%  3.3115 x 0.06355 / 0.3284 g x 100% = 64.08%
(Zn2+)%  1.2485 x 0.06539 / 0.3284 g x 100% = 24.86%
(Pb2+)%  0.1350 x 0.2072 / 0.3284 g x 100% = 8.518%
(Sn2+)%  100% - (Cu2+)%- (Zn2+)% - (Pb2+)% = 2.54%
Exercise
Chromel is an alloy composed of nickel, iron, and chromium. A 0.6472-g sample was
dissolved and diluted to 250.0 mL. When a 50.00 mL aliquot of 0.05182 M EDTA was
mixed with an equal volume of the diluted sample, all three ions were chelated and a 5.11
mL back titration with 0.06241 M copper(II) was required. The chromium in a second 50.0
mL aliquot was masked through the addition of hexamethylene-tramine; titration of the
Fe and Ni required 36.28 mL of 0.05182 M EDTA. Iron and chromium were masked with
pyrophosphate in a third 50.0 mL aliquot, and the nickel was titrated with 25.91 mL of the
EDTA solution. Calculate the percentages of nickel, chromium, and iron in the alloy.
1. Chromel is an alloy composed of nickel, iron, and chromium
2. A 0.6472-g sample was dissolved and diluted to 250.0 mL
3. When a 50.00 mL aliquot of 0.05182 M EDTA was mixed with an equal volume of the diluted
sample, all three ions were chelated and a 5.11 mL back titration with 0.06241 M copper(II) was
required
4. The chromium in a second 50.0 mL aliquot was masked through the addition of
hexamethylenetramine; titration of the Fe and Ni required 36.28 mL of 0.05182 M EDTA
5. Iron and chromium were masked with pyrophosphate in a third 50.0 mL aliquot, and the
nickel was titrated with 25.91 mL of the EDTA solution.