In volumetric (titrimetric) analysis, the accurate measurements of volumes of reacting acids and bases are taken out and employed to find out how much of the acid or base is present in the sample which is being analyzed. Acid-base (neutralization) reactions are employed in volumetric analysis as they fulfill the given conditions.
1) They are simple reactions which can be deduced via simple chemical equations.
2) Their reactions are complete and are in equal or stoichiometric proportions (that is, mole ratio).
3) Their reactions are immediate particularly whenever both are in solution.
4) There is a marked change in the pH of the solution at equivalence point.
5) Indicators are available to the signal that is the end-point of the reaction.
Types of Solutions:
Solutions employed in titrimetric analysis can be categorized into two kinds: the saturated and unsaturated solutions. The saturated solution is one which includes at a particular temperature, as much solute as it can hold in the presence of surplus solute. The saturated solution is in dynamic equilibrium having the surplus solid (Solute) example:
Ca(OH)2 (s) ↔ Ca2+ (aq) + 2 OH- (aq)
The unsaturated solution is one which includes less solute than it can hold at particular temperature. The concentration of a solution is deduced in mol dm3 that is the number of moles of solute in l dm3 of solution. The concentration of a saturated solution of a solute in a solvent provides the solubility of that solute in the solvent at particular temperature. Solutions employed in titrimetric analysis are generally unsaturated solutions.
For acids or bases which are very soluble in water, example: Ca(OH)2, titration of their saturated solutions having standard acid or alkali is frequently used to find out their solubility.
A standard solution is the solution of known concentration. It might be made by dissolving an identified weight of a pure sample of the substance in a known volume of solution. The solution so made is often termed to as the primary standard solution.
10.6g of pure anhydrous Na2CO3 was dissolved in 250 cm3 of solution. Compute its concentration in mol dm-3 (Na2CO3 = 106)
Mole of Na2CO3 = m/M = 10.6/106 = 0.1
0.1 mole of Na2CO3 dissolves in 250 cm3
0.1 x (1000/250) mole will dissolves in 1000 cm3
= 0.40 mol
Concentration = 0.40 mol dm-3
A few substances can't be getting in the pure form example: NaOH as it absorbs moisture (that is, hygroscopic) and CO2 from the atmosphere. A solution of NaOH prepared should be titrated having an acid of known concentration before its exact concentration is known. Such a solution whose concentration is acknowledged precisely after a titration with the other solution of known concentration is termed as a secondary standard solution.
Primary standard substances:
A primary standard solution is made by employing a pure sample of the substance. We will as well remember that sodium hydroxide can't be getting in very pure form. Substances which can be obtained in very pure form and are employed for making primary standard solutions are termed as primary standard substances.
Apart from being getting pure a primary standard substance should as well satisfy the given conditions.
1) Simple to get dry and purify if essential.
2) Be unaltered in air throughout weighing maintain its composition, not hygroscopic nor oxidized via air or influenced via CO2.
3) Its reaction by other reagents should be stoichiometric and instantaneous.
4) Be readily soluble.
Making a standard solution:
Step 1: Get the solute and solvent in extremely pure form. If there is requirement, purify and dry.
Step 2: Weigh out having a good chemical balance, the computed weight of the solute needed.
Step 3: Dissolve solute in small quantity of solvent in a beaker transfer to the volumetric flask.
Step 4: By a wash bottle, rinse the beaker numerous times by distilled water (that is, solvent) and add washings to the volumetric flask.
Step 5: Prepare the volumetric flask to the mark with solvent and shake well and then stopper.
The steps are symbolized in the description below:
Fig: Preparing a standard solution
The equivalence points and the end-point of a titration:
For a particular acid-base reaction there is for all time an equivalent amount of acid which will just neutralize an amount of base or vice-versa. The point throughout the titration whenever this equivalent amount bas been added is the equivalence point. For illustration: 25.0 cm3 of 0.10 mol dm-3 is titrated against 0.10 mol dm-3 HCl. The equivalence point is 25.0 cm3 as the acid and base are of equivalent concentration.
The main aim of adding indicator is to notice this equivalence point. The point at which the indicator changes color is the end point. The endpoint and the equivalence point will be similar or very close whenever the choice of indicator is correct. If not the indicator might give signal of end-point whenever the reaction is far away from-the equivalence point. This will lead to error in the determination of the acid or base concentration.
Acid-base indicators are organic compounds which behave as weak acid and bases. They encompass different colors in acid and alkaline solutions.
Table: Common acid-base indicators
Color change (Acid/Alkaline)
pH range of color range
5.0 - 8.0
3.1 - 4.4
8.3 - 10.0
4.2 - 6.3
As you can observe from the table, there is a range of pH over which the color range takes place for each and every indicator. A knowledge of this and the relative strengths of the acid and base titrated let for a good choice of indicator for the acid-base reaction.
Neutralization curves and choice of indicator:
It is a plot of pH against the volume of base throughout a titration.
Fig: Neutralization curves
The above is the Neutralization curve for a strong acid or strong base titration. The change in pH close to the equivalence point allows us to choose an indicator which will provide the least titration error. The curve exhibits that near the equivalence point the rate of change of pH is extremely fast. For the strong acid-strong base reaction any of the indicators in the table will give a sharp end point as only one drop or two of alkali causes the pH to move over the entire vertical part of the curve.
For the weak acid-strong base titration, (that is, ethanoic acid and sodium hydroxide) the hydrolysis of the salt forms the solution to be alkaline at equivalence point. Phenolphthalein is the choice of indicator as its color change is in the alkaline range. Whenever a strong acid is titrated against a weak base, at equivalence point the salt hydrolysis forms the solution acidic. Methyl orange or methyl red is the appropriate indicators for such reactions.
The utilization of sodium trioxocarbonate (IV) in acid-base titration:
Sodium trioxocarbonate is the salt of a weak acid and strong base. In aqueous solution, the salt is hydrolyzed.
Na2CO3 (aq) + H2O (l) ↔ 2 Na+ (aq) + 2OH- (aq) + H2CO3 (aq)
As OH- reacts to provide water, the equilibrium reaction moves in the forward direction till all the sodium carbonate has hydrolyzed. The reaction can go in portions.
Na2CO3 (aq) + HCl (aq) → NaCl (aq) + NaHCO3 (aq)
NaHCO3 (aq) + HCl (aq) → NaCl (aq) + H2CO3 (aq)
At equivalence point due to the weak acid H2CO3 the solution is weakly acidic.
H2CO3 ↔ 2H+ CO32-
Methyl orange is employed as indicator for the titration.
Computations in Volumetric Analysis:
Burette reading (final)
Burette reading (initial)
Volume of HCl used
The result from HCl/NaOH titration is represented in the table. The acid concentration is 0.10 mol dm3 and the sodium hydroxide solution consists of 1.10 g NaOH in 250 cm3 of solution. Employ the above titration outcome to compute the percentage purity of the NaOH sample employed. The pipette volume is 25.0 cm3
Equation of the reaction:
NaOH (aq) + HCl (aq) → NaCl (aq) + H2O (l)
Mole ratio NaOH: HCl = 1:1
Mole of HCl = (MHCl x VHCl)/1000
M: Concentration (mol dm3)
V: Average titre
V = (22.10 + 22.20 + 22.20)/3 = 22.17 cm3
Take an average of titre values that are in 0.2 cm3. These titres are known as concordant tins.
Mole of HCl = (0.10 x 22.17)/1000 = 2.217 x 10-3
Mole of NaOH = 2.217 x 10-3 as the mole ratio = 1:1
Mole NaOH = (MNaOH x VNaOH)/1000
MNaOH = (1000 x 2.217 x 10-3)/25 = 0.08867 = 0.0887 (3 sig. figs.)
Computation of concentration in g dm-3
Concentration (g dm-3) = Concentration mole dm-3 x molar mass
= .0887 x 40 (molar mass of NaOH)
= 3.548 g
= 3.55 g (3 significant figures) of pure NaOH.
The solution includes 1.10g/250cm3.
Concentration (g dm-3) of impure NaOH = (1.10 x 1000)/250 = 4.40 g
The percent purity = (mass of pure/mass of impure) x 100
= (3.55/4.40) x 100
= 80.7 %
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