Quantitative determination of components in solution containing
Sodium Hydroxide, Sodium Carbonate, and Sodium Hydrogen Carbonate
Two of these three components can exist in an appreciable amount in any solution because reaction
with acids (neutralisation titration) will eliminate the third. The analysis of such mixtures can be
realised at any technique:
1) direct titration (Warder’s method) or
2) back-titration (Winkler’s method).
The analysis of such mixtures requires two titrations: one with an alkaline-range indicator, such as
phenolphthalein, and the other with an acid-range indicator, such as methylorange.
1. Direct titration the mixture of Sodium Hydroxide and Sodium Carbonate
The composition of the solution can be deduced from the relative volumes of acid needed to titrate
equal volumes of the sample.
Titration with hydrochloric acid to a phenolphthalein end point (Vphth):
NaOH + HCl = NaCl + H2O
Na2CO3 + HCl = NaHCO3 + NaCl
Titration with hydrochloric acid to a methyl orange end point (Vmo):
NaHCO3 + HCl = NaCl + H2O + CO2
Components in sample
Relationship between Vphth and Vmo in the
titration of an equal volume of sample
Vphth = Vmo
Vphth = 1/2Vmo
Vphth > 1/2Vmo
NaOH
Na2CO3
NaOH + Na2CO3
1. Load a burette with 0,1 N hydrochloric acid solution.
2. To conical flask pour in 10 ml aliquot of sample (VS).
3. Introduce 2-3 drops of phenolphthalein solution, and titrate with 0.1 N hydrochloric acid solution
until the solution lost the red colour. Read the burette mark (Vphth).
4. To conical flask pour in 10 ml aliquot of sample (VS).
5. Introduce 2-3 drops of methyl orange solution, and titrate with 0,1 N hydrochloric acid solution
until the solution just begin to change colour from yellow to red. Read the burette mark (Vmo).
6. Repeat titration also two times. Calculate the median volumes of used hydrochloric acid solution.
7. Calculate sodium hydroxide and sodium carbonate percentage in sample:
xNa2CO3 =
xNaOH =
2(Vmo − Vphth ) ⋅ N HCl ⋅ meq Na2CO3
VS
⋅100%
[Vphth - (Vmo − Vphth )] ⋅ N HCl ⋅ meq NaOH
VS
⋅100%
2
2. Back-titration the mixture of Sodium Carbonate and Sodium Hydrogen Carbonate
The technique takes advantage of the limited solubility of barium carbonate in neutral and basic
solutions. The presences of solid barium carbonate not hamper end-point detection in this method.
Carbonate and hydrogen carbonate ions can be accurately determined in mixtures by first
titration both ions with standard acid to an end point with an acid-range indicator as methyl orange.
The hydrogen carbonate in a second aliquot is converted to carbonate by the addition of a known
excess of standard base. An unmeasured excess of neutral barium chloride solution to precipitate
barium carbonate, following which the hydroxide ion residue is titrated to a phenolphthalein end
point.
Titration with hydrochloric acid to a methyl orange end point (Vmo):
NaHCO3 + HCl = NaCl + H2O + CO2
Na2CO3 + 2HCl = 2NaCl + H2O + CO2
Conversion the hydrogen carbonate ion to carbonate ion in a second aliquot:
NaHCO3 + NaOH = Na2CO3 + H2O
excess
Precipitation of carbonate ion with barium chloride:
Na2CO3 + BaCl2 = BaCO3↓ + 2NaCl
Titration with hydrochloric acid to a phenolphthalein end point (Vphth):
NaOH + HCl = NaCl + H2O
residue
1. Load a burette with 0.1 N hydrochloric acid solution.
2. To conical flask pour in 10 ml aliquot of sample (VS).
3. Introduce 2-3 drops of methyl orange solution, and titrate with 0.1 N hydrochloric acid solution
until the solution just begins to change colour from yellow to red. Read the burette mark (Vmo).
4. To conical flask pour in 10 ml aliquot of sample (VS).
5. Add 10 ml of 0.1 N sodium hydroxide solution (VNaOH).
6. Add 10 ml of 10 % barium chloride solution.
7. Introduce 2-3 drops of phenolphthalein solution, and titrate with 0.1 N hydrochloric acid solution
until the solution lost the red colour. Read the burette mark (Vphth).
8. Repeat titration also two times. Calculate the median volumes of used hydrochloric acid solution.
9. Calculate sodium hydrogen carbonate and sodium carbonate percentage in sample:
xNaHCO3 =
xNa2CO3 =
M.m. Na2CO3 = 105.989;
(VNaOH ⋅ N NaOH − Vphth ⋅ N HCl ) ⋅ meq NaHCO3
VS
⋅100%
[Vmo ⋅ N HCl - (VNaOH ⋅ N NaOH − Vphth ⋅ N HCl )] ⋅ meq Na2CO3
⋅ 100%
VS
M.m. NaHCO3 = 84.007