Analytical chemistry Trinidad & 

Tobago Lab resources
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250ml Volumetric analysis standards and solutions

Preparation of 0.1M and 0.1N Volumetric analysis titration Standards


This page gives info on the weight of salt required to make up 250 ml volumetric analysis standards for titration against acid base, permanganate, dichromate, iodine, sodium thiosulphate and silver nitrate. The weight was calculated according to the reaction process of the particular determination. Concentrations are expressed in molarity, normality and equivalents.


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On this page:
Acid / base titrations
Oxidation / Reduction
permanganate and dichromate
Iodometry
iodine and thiosulphate
Precipitation titrations
silver nitrate

Acid / Base Titrations

Example for the info on this page:

Sodium carbonate
Na2CO3, FW = 106, Eq. =53g/l
250ml 0.05M = 1.325g = 0.1N

In this example, the FW or Formula weight (Molecular weight) of the salt is 106. Eq. is the equivalent weight which is 53 grams per liter. And 1.325 grams dissolved up to mark in 250 ml volumetric flask will give a 0.05 M standard solution or a 0.1N standard solution.

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
CaCO3, 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


5Na2C2O4 + 2KMnO4 + 8H2SO4 = 10CO2 + 2MnSO4 + 5Na2SO4 + 8H2O
Reaction of Potassium permanganate and Sodium oxalate
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


K2Cr2O7 + 6Fe(NH4)2(SO4)2 + 7H2O4 = 3Fe2(SO4)3 + Cr2(SO4)3 + K2SO4 + 6(NH4)2SO4 + 7H2O
Reaction of Potassium dichromate and Ammonium ferrous sulphate
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


I2 + 2Na2S2O3 = Na2S4O6 + 2NaI
Reaction of Iodine and Sodium 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

Titrations with Silver Nitrate


1. AgNO3 + NaCl = AgCl = NaNO3 and K2CrO4 + 2AgNO3 = Ag2CrO4 + 2KNO3
2. AgNO3 + KSCN = AgSCN + KNO3 and AgSCN + Fe3+ indicator = [FeSCN]+2
Silver nitrate
AgNO3, FW = 170, Eq. =170g/l
250ml 0.1M = 4.25g = 0.1N

Sodium chloride
NaCl, FW = 58.5, Eq. = 58.5g/l
250ml 0.1M = 1.4625g = 0.1N
Potassium chromate
K2CrO4, FW = 194.20, Eq. =32.4g/l
250ml 0.0167M = 0.81g = 0.1N
1 liter 0.0167M = 3.24g = 0.1N
(add a few mls dil. H2SO4 to clear)
Ammonium thiocyanate
NH4CNS, FW = 76.12, Eq. = 76.12g/l
250ml 0.1M = 1.9g = 0.1N
1 liter 0.1M = 7.6g = 0.1N
Potassium thiocyanate
KCNS, FW = 97.18, Eq. = 97g/l
250ml 0.1M = 2.4g = 0.1N (use 3g)
1 liter 0.1M = 9.74g = 0.1N (use 12g)


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.