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Chemistry infolab reagents and resources
The preparation of titration standards and solutions

Acid-base, oxidation-reduction, iodine, silver, thiosulphate etc...

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Acid / Base Titrations

Ammonium sulphate
(NH4)2SO4. FW = 132.14, Eq. = 66g/l.
250ml 0.05M = 1.325g = 0.1N
Adipic acid
HO2C(CH2)4CO2H, FW = 146.4, Eq. = 73.03g/l
250ml 0.05M = 1.83g = 0.1N
Barium hydroxide
Ba(OH)2.8H2O, FW = 315.48, Eq. = 157.5g/l
250ml 0.05M = 3.94g = 0.1N
Benzoic acid
C6H5COOH, FW = 122.12, Eq. = 61g/l
250ml 0.05M = 1.52g = 0.1N
Calcium carbonate
CaCaCO3, FW = 100.00, Eq. = 50g/l
250ml 0.05M = 1.25g = 0.1N
Furroic acid
FW = 112.08, Eq. = 112g/l
250ml 0.1M = 2.8g = 0.1N
Hydrochloric acid
HCl, FW = 36.5, Density = 1.2
1M = 83mls = 1N (Use 86mls)
250ml 0.1M = 2mls = 0.1N
1 liter 0.1M soln = 8.6mls = 0.1N
Oxalic acid
H2C2O4.2H2O, FW = 126.07, Eq. = 63g/l
250ml 0.05M = 1.575g = 0.1N
Potassium hydrogen phthalate
KH(C8H4O4), FW = 204.23, Eq. =204g/l
250ml 0.1M = 5.105g = 0.1N
Potassiun hydrogen iodate
KH(IO3)2, FW = 389.92, Eq. 73.07g/l
250ml 0.1M = 9.75g = 0.1N
Sodium carbonate
Na2CO3, FW = 106, Eq. =53g/l
250ml 0.05M = 1.325g = 0.1N
Sodium hydroxide
NaOH, FW = 40, Eq. = 40g/l
250ml 0.1M = 1.0g = 0.1N
1 liter 0.1M soln = 4g = 0.1N
Sodium oxalate
Na2C2O4, FW = 134.00, Eq. =134g/l
250 0.1M = 3.35g = 0.1N
Sodium tetraborate
NaB4O7.10H2O, FW = 381.37, Eq. =190g/l
250ml 0.05M = 4.762g =0.1N
Succinic acid
HO2CCH2CH2CO2H, FW = 118.09, Eq. = 59.045g/l
250ml 0.05M = 1.475g = 0.1N
Sulphamic acid
H2NSO3H, FW = 97.09, Eq. = 99.09g/l
250ml 0.10M = 2.425 = 0.1N
Sulphuric acid
H2SO4, FW = 98.08, Density = 1.8
1M = 56mls = 2N
250ml 0.05M = 0.7mls = 0.1N
1 liter 0.05M = 2.8mls = 0.1N (use 3 mls)
Tris (hydroxymethyl)-aminomethane
H2N.C(CH2OH)3, FW = 121.14, Eq. = 121g/l
250ml 0.1M = 3.023g =0.1N

Titrations with Permanganate

Ammonium oxalate
(NH4)2C2O4, FW = 124, Eq. = 62g/l
250ml 0.05M = 1.55g = 0.1N
Ferrous ammonium sulphate
FeSO4(NH4)2SO4.6H2O, FW – 392, Eq. 392g/l
250ml 0,1M = 9.8g = 0.1N (add a few drops of conc. H2SO4 to clear)
Hhdrogen peroxide
H2O2, FW = 34. Eq. = 17g/l
10 Vol. = 3%
20 Vol. = 6% = 1.8M
100 Vol. = 30%
250ml 0.05M = 12.50ml of 20 Vol. = 0.1N
1 liter 0.05M = 50ml of 20 Vol. = 0.1N (add 2 mls conc. H2SO4 per liter soln)
1 liter 0.05M = 10mls of 100 Vol.
1 liter 0.05M =35mls of 30 Vol.
Volume strength = 11.2/34 x strongth of peroxide in g/l
Iron allum (ferric ammonium sulphate)
FeNH4(SO4)2.12H2O, FW = 482.19, Eq. =241g/l
250ml 0.05M = 6.02g = 0.1N

Oxalic acid (anhyd.)
H2C2O4, FW = 90, Eq. = 45g/l
250ml 0.05M =1.125g = 0.1N
Potassium permanganate
KMnO4, FW =158.04, Eq. ==31.6g/l
250ml 0.02M = 0.79g = 0.1N (dissolve in hot water)
1 liter 0.02M = 3.16g = 0.1N (boil to dissolve crystals, then dilute to 1liter)
Sodium nitrite
NaNO2, FW = 69.00, Eq. = 34.56/l
250ml 0.05M = 0.86g = 0.1N (use 1g)

Sodium oxalate
Na2C204, FW = 234, Eq. = 67g/l
250ml 0.05M = 1.675g = 0.1N

Titrations with Dichromate

Potassium dichromate
K2Cr2O7, FW = 294, Eq. = 49g/l
250ml 0.0167M = 1.22g = 0.1N
1 liter 0.0167M = 4.90g = 0.1N

Potassium iodate
KIO3, FW = 214, Eq. = 35.73g/l
250ml 0.0167M = 0.892g = 0.1N
Tin, Sn,
FW = 119.0, Eq. = 59.5g/l
Tin (11) chloride use metallic tin dissolved in conc. HCl
250ml 0.05M = 1.49g = 0.1N

Spathic iron ore
FeCO3, FW = 116, Eq. = 116g/l
250ml 0.1M = 2.9g = 0.1N

Titrations with iodine and Thiosulphate

Sodium thiosulphate
Na2S2O3.5H2O, FW = 248.18, Eq. = 248g/l
250ml 0.1M = 6.2g = 0.1N
1 liter 0.1M = 24.8g = 0.1N
Potassium iodide
KI, FW = 166, Eq. =166g/l
250ml 0.1M = 4.15g = 0.1N
1liter = 0.1M 16.6g = 0.1N
Potassium iodate
KIO3, FW = 214 Eq. = 35.73g/l
250ml 0.0167M = 0.892g = 0.1N
Iodine, I2
FW = 253.18, Eq. = 127g/l
1 liter 0.05M = 12.7g = 0.1N
(use 13g Iodine and add 25g KI
Copper sulphate
CuSO4.5H2O, FW = 249.68, Eq. = 249.68g/l
250ml 0.1M = 6.24g = 0.1N
Note: add anhy. NaCO3 untill a slight permanent bluish colour is formed. Add a little acetic acid to get a clear blue colour, then make up to the mark.
Copper sulphate, anhydrous
CuSO4, FW = 159.6, Eq. = 159.6g/l
250ml 0.1M = 4g =0.1N
Bleaching powder
CaOCL2, FW = 145, Eq. = 72,5g/l
250ml 0.05M = 1.81g = 0,1N
(use 2.5g)
Sodium sulphite
Na2SO3.5H2O, FW = 126.04, Eq. = 63g/l
250ml 0.05M = 1.57g = 0.1N
Signature: Dhanlal De Lloyd, Chem. Dept, The University of The West Indies, St. Augustine campus
The Republic of Trinidad and Tobago.
Copyright: delloyd2000© All rights reserved.