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Review of Diprotic Acid Titration. Analysis of 0.181g Carbonate / Bicarbonate Sample. # mol HCO 3 - = (0.02630 - 0.00740 L) x (0.1000 M) x (1 mol HCO 3 - / 1 mol HCl) = 0.001890 mol HCO 3 - # mole CO 3 2- = (0.00740 - 0.00060 L) x (0.1000 M) x (1 mol CO 3 2- / 1 mol HCl) - PowerPoint PPT Presentation
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Review of Diprotic Acid Titration
Analysis of 0.181g Carbonate / Bicarbonate Sample # mol HCO3
- = (0.02630 - 0.00740 L) x (0.1000 M) x (1 mol HCO3- / 1 mol HCl)
= 0.001890 mol HCO3
-
# mole CO3
2- = (0.00740 - 0.00060 L) x (0.1000 M) x (1 mol CO32- / 1 mol HCl)
= 0.000680 mol CO3
2-
# mole HCO3
- in original mixture
= # mol HCO3- - # mol HCO3
- produced in titration to Phen. End Pt.
= # mol HCO3- - # mol CO3
2-
= 0.001890 mol - 0.000680 mol = 0.001210 mol HCO3
-
#g NaHCO3 = (0.001210 mol) x (84.007 g/mol) = 0.102 g #g Na2CO3 = (0.000680 mol) x (105.9885 g/mol) = 0.072 g wt% NaHCO3 = (0.102g / 0.181g) x 100 = 56.3 % wt % Na2CO3= (0.072g / 0.181g) x 100 = 39.8 %
Complexometric Titration
Typical Complexing Agents
EDTA
Ca-EDTA Titration
0.00 mL Titrant Added
Titration Reaction
EDTA Ca2+ CaEDTA2+
end = 0 M = 0.00500 M = 0 M
Equilibrium Reaction
EDTA Ca2+ CaEDTA2+
Initial 0 0.0500 0
[Ca2+] = 0.00500 M
Concentration
pCa = - log(0.00500) = 2.30
25.00 mL Titrant Added (Equivalence Point)
Titration Reaction
EDTA Ca2+ CaEDTA2+
start (0.0100M)(0.02500L) (0.0050M)(0.05000L)
= 0.000250 mol = 0.000250 mol = 0 mol
end 0 mol 0 mol 0.000250 mol
(0.000250mol)/(0.07500L)= 0 M = 0 M = 0.00333M
Equilibrium Reaction
EDTA Ca2+ CaEDAT2+
Initial 0 0 0.00333
Change + x + x - x
Equilibrium x x ~0.00333
1.75 x 1010 = (0.00333) / x2
x = [Ca+] = 4.36 x 10-7 MConcentration
pCa = - log(4.36 x 10-7) = 6.36
50.00 mL Titrant Added
Titration Reaction
EDTA Ca2+ CaEDTA2+
start (0.0100M)(0.05000L) (0.0050M)(0.05000L)
= 0.000500 mol = 0.000250 mol = 0 mol
end 0.000250 mol 0 mol 0.000250 mol
(0.000250mol)/(0.1000L) (0.000250mol)/(0.1000L)= 0.00250 M = 0 M = 0.00250M
Equilibrium Reaction
EDTA Ca2+ CaEDAT2+
Initial 0.00250 0 0.00250
Change + x + x - x
Equilibrium ~0.00250 x ~0.00250
1.75 x 1010 = (0.00250) / (0.00250) xx = [Ca+] = 5.71 x 10-11 M
Concentration
pCa = - log(5.71 x 10-11) = 10.24
Indicators