What will be the value for volume of potassium hydroxide in the titration of potassium hydroxide against oxalic acid?
Answers
The word titration comes from the Latin word "titulus", which means inscription or title. The French word title means rank. Therefore, Titration means the determination of concentration or rank of a solution with respect to water with a pH of 7.
The standard solution is usually added from a graduated vessel called a burette. The process of adding standard solution until the reaction is just complete is termed as titration and the substance to be determined is said to be titrated.
The reaction must be a fast one.It must proceed stoichiometrically.The change in free energy (ΔG) during the reaction must be sufficiently large for spontaneity of the reaction.There should be a way to detect the completion of the reaction.
It is a chemical reagent used to recognize the attainment of end point in a titration. After the reaction between the substance and the standard solution is complete, the indicator should give a clear colour change.
When a titration is carried out, the free energy change for the reaction is always negative.
That is, during the initial stages of the reaction between A & B, when the titrant A is added to B the following reaction takes place.
Equilibrium constant,
a = activity co-efficient.
Large values of the equilibrium constant K implies that the equilibrium concentration of A & B are very small at the equivalence point. It also indicates that the reverse reaction is negligible and the product C & D are very much more stable than the reactants A & B. Greater the value of K the larger the magnitude of the negative free energy change for the reaction between A & B. Since,
Where,
R = Universal gas Constant = 8.314 JK-1mol-1,
T = Absolute Temperature.
The reaction of the concentration of A & B leads to the reduction of the total free energy change. If the concentrations of A & B are too low the magnitude of the total free energy change becomes so small and the use of the reaction for titration will not be feasible.
Expressions of Concentration of Solutions:
The concentration or strength of solution means the amount of solute present in a given amount of the solution. The concentration may be expressed in physical or chemical units.
Normality (N): It is defined as number of gram equivalents of the solute present in 1 litre (1000mL.) of the solution. If W g of solute of equivalent weight E is present in V mL of the solution, the normality of the solution is given by: Molarity (M): It is defined as the number of moles of the solute present in 1 litre (or 1000 mL) of the solution. A one molar solution contains 1 mole of the solute dissolved in 1 litre of the solution. Molality (m): It is defined as the number of moles of solute dissolved in 1000 g of the solvent. One molal solution contains one mole of the solute dissolved in 1000 g of the solvent. Normal solution:
A solution containing one gram equivalent weight of the solute dissolved per litre is called a normal solution; e.g. when 40 g of NaOH are present in one litre of NaOH solution, the solution is known as normal (N) solution of NaOH. Similarly, a solution containing a fraction of gram equivalent weight of the solute dissolved per litre is known as subnormal solution. For example, a solution of NaOH containing 20 g (1/2 of g eq. wt.) of NaOH dissolved per litre is a sub-normal solution. It is written as N/2 or 0.5 N solution.
Formulae used in solving numerical problems on volumetric analysis;
Strength of solution = Amount of substance in g litre-1. Strength of solution = Amount of substance in g moles litre-1. Strength of solution = Normality × Eq. wt. of the solute = molarity × Mol. wt. of solute. Molarity = Moles of solute/Volume in litre. Number of moles = Wt.in g/Mol. wt = M × V (initial) = Volume in litres/22.4 at NTP (only for gases). Number of milli moles = Wt. in g × 1000/mol. wt. = Molarity × Volume in mL. Number of equivalents= Wt. in g/Eq. wt = x × No. of moles × Normality × Volume in litre (Where x = Mol. wt/Eq. wt). Number of mill equivalents (meq.) = Wt. in g × 1000 / Eq. wt = normality × volume in mL. Normality = x × No. of mill moles (Where x = valency or change in oxi. number). Normality formula, N1V1 = N2V2, (Where N1, N2 → Normality of titrant and titrate respectively, V1, V2 → Volume of titrant and titrate respectively). % by weight = Wt. of solvent/Wt. of solution × 100 .
A solution is a homogeneous mixture of two or more components, the composition of which may be changed. The substance which is present in smaller proportion is called the solute, while the substance present in large proportion is called the solvent.
i) The reaction between the titrant and titrate must be expressed.
ii) The reaction should be practically instantaneous.
iii) There must be a marked change in some physical or chemical property of the solution at the end point.
iv) An indicator should be available which should sharply define the end point.