Titration

A process of determining the concentration of an unknown solution by finding out the exact volume of it
required to react completely a known volume of standard solution is called titration.

Acidimetry

It is the process of determining the concentration of an acidic solution with the help of a standard alkali solution by using a suitable indicator.

Alkalimetry

It is the process of determining the concentration of alkali solution with the help of a standard acid solution by using a suitable indicator.

Indicator

It is a chemical substance with indicates the completion of a reaction by changing its color. eg. phenolphthalein, methyl orange, etc.

End point

The completion of a reaction is indicated by an indicator is called endpoint.

Equivalence point

A point at which the equivalent quantity of titrant is added to titrand is called equivalent point. it is a theoretical endpoint. The difference between endpoint and equivalence point is called titration error.

Neutral point

A point at which the pH of the resulting solution is 7 is called the neutral point.

Titrant

It is the solution that is taken in a burette.

Titrand

It is the solution that is taken in a titration flask.

Acid-base titration

The titration between acid and base is called an acid-base titration. In this titration, a neutralization reaction takes place.

HCl + NaOH \rightarrow NaCl + H_{2}O

Redox titration

The titration between the oxidizing agent and reducing agent is known as redox titration. In this titration, a redox reaction takes place.

Iodimetric and Iodometric titration

Iodine is a mild oxidizing agent and its solution undergoes titration in two ways:
a. Iodimetric titration(Iodimetry):
The direct titration of standard iodine solution with a reducing agent is called Iodimetry. eg. titration of iodine with sodium thiosulphate (hypo).

\underset{Hypo}{2Na_{2}S_{2}O_{3}} + I_{2}\ \xrightarrow{titration} \underset{\underset{tetrathionate}{Sodium}}{Na_{2}S_{4}O_{6}} + 2NaI

b. Iodometric titration (Iodometry):
The titration in which the prepared iodine from KI by means of some chemical reaction is titrated against hypo is called iodometry. It is used to determine the concentration of the oxidizing agents. eg. titration of CuSO₄ and hypo.

2CuSO_{4} + KI\ \rightarrow\ 2K_{2}SO_{4} + Cu_{2}I_{2} + I_{2}\\ 2Na_{2}S_{2}O_{3} + I_{2}\ \xrightarrow{titration} Na_{2}S_{4}O_{6} + 2NaI

Precipitation or argentometric titration

The titration in which a precipitate is formed as a result of the reaction between the reacting solutions is called precipitation titration. eg. titration of silver nitrate and sodium chloride. It is called argentometric titration because silver nitrate solution is used for precipitation.

\underset{(aq.)}{AgNO_{3}} + \underset{(aq.)}{NaCl}\xrightarrow{titration} AgCl\downarrow + NaNO_{3}

Complexometric titration

The titration between the metal ion and complexing agent in which the formation of a soluble complex takes place is called complexometric titration. The metal ion is called the central atom and the complexing agent is called a ligand. eg. titration between calcium or magnesium ions with EDTA (ethylene diamine tetra acetic acid).

Ca^{++}+EDTA\rightarrow EDTA\ complex

Types of indicators

On the basis of use in volumetric analysis, indicators are classified into two groups:

• Internal indicator
• External indicator

i. Internal indicator

The indicators which are added to one of the two solutions during titration to find the endpoint are called an internal indicators. Based upon the nature of the reaction, they are divided into three classes:

a. Acid-base indicator: These are used in acid-base titrations. They change their color at definite pH. eg. methyl orange, phenolphthalein, methyl red, litmus paper, etc.

b. Adsorption indicator: They can change the color of the precipitate by getting adsorbed in it. In iodometry and iodimetry, 1% freshly prepared starch solution is used as an indicator. eg. organic dyes like fluorescin and eosin.

c. Self indicator: They themselves act as an indicator during titration. In the titration of oxalic acid with KMnO₄, the end point is marked by the appearance of pink color. So, KMnO₄ acts as self indicator.

ii. External indicator

The indicator which is not added to solutions but used externally is called an external indicator. It is placed outside on a reference porcelain basin. eg. in the titration of K₂Cr₂O₇ and ferrous ammonium sulphate solution, K₃[Fe(CN)₆] is used as an external indicator.

Normality equation

According to the law of equivalence, substances react in their equivalent proportion. This means that a certain number of gram equivalent of acid neutralizes the same number of gram equivalent of alkali.
Number of gram equivalent of acid = number of gram equivalent of alkali
We know that,

Normality= \frac{No.\ of\ gram\ eq.}{Vol\ of\ sol^{n}\ in\ litre}

So, Number of Gram equivalent = volume of solution x normality

At the equivalence point,
Number of gram equivalent of acid = Number of gram equivalent of alkali
Normality of acid volume of acid = Normality of alkali volume of alkali
S_AV_A = S_BV_B or,
N₁V₁ = N₂V₂

This equation can be applied in all reactions between any two solutions and also for dilutions of the solution.

Principles of volumetric analysis

The principle of volumetric analysis is based on the law of chemical equivalence which can be stated as,” substances always react to their equivalent proportion”. Followings are the law of volumetric analysis:

1. Acid and alkali solution of same normality always neutralize in equal volume.
   10 ml of N/10 acid = 10 ml of N/10 alkali

2. Equivalent volume is inversely proportional to its normality
   100 ml of N/10 acid = 100 N/10 ml of 1N acid
   = 10 ml of 1N acid

3. When an acid is completely neutralised by a base, the normality equation holds good.
   V_AS_A = V_BS_B
   
   V_A = Volume of acid S_A = Normality of acid
   V_B = Volume of base S_B = Normality of base

• If two different acid solutions are mixed, the resulting solution will be: V_MS_M = V₁S₁ + V₂S₂

• If acid and alkali are mixed, the resulting solution will be: V_MS_M = V₁S₁ – V₂S₂

Titration curve and selection of indicator

The pH of a solution depends upon the [H⁺] concentration. Greater the [H⁺] ion, the lower is its pH value and vice versa. When a certain volume of acid is titrated by a base, the [H⁺] decreases for each turn addition of a base, corresponding pH of a solution increases. If a graph is plotted between pH and volume of base added, the steep rise in pH occurs at the endpoint. The curve thus obtained is called the titration curve. An indicator whose pH range falls within this interval of pH shows a sharp colour change and it can be used to determine the endpoint. Different types of acid-base titration are given below:

i. Titration between a strong acid and strong base:
In this case, the shape of the curve is ABCD. The steep rise occurs between pH 3-11. The pH range of both methyl orange and phenolphthalein falls within this interval of pH. So, both indicators can be separately used to determine the endpoint for this titration.

ii. Titration between a strong acid and weak base:
In this case, the shape of the curve is EFCD. The steep rise occurs between pH 3-8. The pH range of methyl orange falls within this interval of pH. So, this indicator is used to determine the endpoint for this titration.

iii. Titration between weak acid and strong base:
In this case, the shape of the curve is ABGH. The steep rise occurs between pH 6-11. The pH range of phenolphthalein falls within this interval of pH. So, this indicator is used to determine the endpoint for this titration.

iv. Titration between weak acid and weak base:
In this case, the shape of the curve is EFGH. The steep rise occurs between pH 6-8. The pH range of no indicator falls within this interval of pH. So, titration cannot be carried out.

Indicator and their colour change and pH range

Solved Examples

1. 2 gm of CaCO3 neutralizes 50 cc of HCl solution. What is the normality of HCl?
Solution:
2 gm of CaCO₃ = 50 cc of HCl
So,

\begin{align*} V_{B}S_{B} &= V_{A}S_{A}\\ V_{B} \times \frac{W}{V_{B}} \times \frac{1000}{50}&=50 \times S_{A}\\ \frac{2 \times 1000}{50} &= 50 \times S_{A}\\ \therefore S_{A}&= 0.8N \end{align*}

2. To a 150 cc solution of NaOH, 10 cc of N/2 HCl was added and for complete neutralization, it requires 60 cc of N/12 HCl. Calculate the normality of NaOH?
solution:
150 cc of NaOH = 10 cc of N/2 HCl + 60 cc of N/12 HCl
150 cc of NaOH = 5 cc of 1N HCl + 5 cc of 1N HCl
150 cc of NaOH = 10 cc of 1N acid
150 x S_B = 10 x 1N
S_B = N/15

3. What volume of NaOH containing 96 gm/lt is required to neutralize completely 1 lt of N/10 H₂SO₄?
solution:
x volume of NaOH containing 96 gm/lt = 1000 cc of N/10 H₂SO₄

\begin{align*} V_{B}S_{B} &= V_{A}S_{A}\\ x \times \frac{gm/lit}{eq.\ wt.}&= 1000 \times N/10\\ x \times \frac{96}{40}&= 100\\ \therefore x &= 41.667\ cc. \end{align*}

4. 10 cc of N/10 HCl is neutralized by 50 cc of NaOH. What volume of this alkali is required to prepare 1000 cc of N/100 solution?
solution:

\begin{align*} 10\ cc\ of\ N/10\ HCl &= 50cc\ of\ NaOH\\ 10 \times N/10 &= 50 \times S_{B}\\ or,\ S_{B}&=N/50\\ Again,\\ V_{conc} = ?,\ \ S_{conc} = N/50, & \ V_{dil} = 1000\ cc, \ S_{dil} = N/100\\ V_{conc} \times N/50 &= 1000 \times N/100\\ \therefore V_{conc}&=500\ cc. \end{align*}

5. 5.3 gm of impure Na₂CO₃ was added in 100 cc of 1N HCl . The residual acid required 10 cc of N/2 NaOH. Calculate the % of Na₂CO₃ in the sample.
solution:
Let w gm be the pure wt. of Na₂CO₃
w gm of Na₂CO₃ + 10 cc of N/2 NaOH = 100 cc of 1N HCl
w gm of Na₂CO₃ = 100 cc of 1N HCl – 5 cc of 1N NaOH
w gm of Na₂CO₃ = 95 cc of 1N HCl

\begin{align*} V_{B}S_{B} &= V_{A}S_{A}\\ V_{B} \times \frac{W}{V_{B}} \times\frac{1000}{53}&=95 \times 1N\\ w &= 5.035 gm\\ \%\ of\ Na_{2}CO_{3} &=\frac{5.035}{5.3} \times 100 = 95\% \end{align*}

6. A sample of sulphuric acid having specific gravity 1.51 contains 60.65% of pure H₂SO₄ by weight. What volume of this acid would be required to furnish 1 lt. of 1N H₂SO₄?
solution:

\begin{align*}Normality &= \frac{\% \times sp.\ gr. \times 10}{eq.\ wt.}\\ &=\frac{60.65 \times 1.51 \times 10}{49}=18.69N\\ again,\\ x\ ml\ of\ 18.69N &= 1000\ ml\ of\ 1N\ H_{2}SO_{4}\\ x \times 18.69 &= 1000 \times 1\\ x &= 53.5\ ml \end{align*}

7. 1.84 gm of a mixture of CaCO₃ and MgCO₃ were heated with 50 ml of 0.97N HCl solution. The excess of acid required 17 ml of N/2 NaOH for neutralization. Calculate the percentage composition of the mixture.
solution:
Amount of acid used by CaCO₃ and MgCO₃ = (50 x 0.97) – (17 x 1/2) = 40 ml
CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O
MgCO₃ + 2HCl → MgCl₂ + CO₂ + H₂O

Let amount of CaCO₃ be x and MgCO₃ be 1.84-x
100 gm of CaCO₃ = 73 gm of HCl
x gm of CaCO₃ = (73/100) x gm of HCl
84 gm of MgCO₃ = 73 gm of HCl
1.84-x gm of MgCO₃ = (73/84) (1.84-x) gm of HCl
1000 ml of 1N HCl = 36.5 gm
40 ml of 1N HCl = 1.46 gm
According to question,

\begin{align*} \frac{73x}{100}+\frac{73}{84}(1.84-x) &=1.46\\ or,\ x &=1\\ \%\ of\ CaCO_{3}&=\frac{1}{1.84}\times 100=54.34\%\\ \%\ of\ MgCO_{3}&=100-54.34\% = 45.65\% \end{align*}