Volumetric Analysis

Chemical analysis plays a vital role in modern industrialized society. Quantitative analysis involves the estimation of an accurate amount of constituents of a substance. Qualitative analysis involves the detection and identification of constituents of a substance.

Volumetric analysis

The volumetric analysis deals with measurements of the volume of solutions involved in chemical reactions, leading to the determination of the amount of constituents in a given unknown solution.

The gravimetric analysis deals with the measurement of the volumes of solution and weight of precipitate obtained in chemical reactions which ultimately leads to the amount of constituents present in a given unknown solution.

Equivalent weight

Equivalent weight of an element is defined as the number of parts by weight of it which combines or displaces directly or indirectly 1.008 parts by weight of hydrogen, 8 parts by weight of oxygen, 35.5 parts by weight of chlorine or its equivalent.
Illustration:

1.\ Mg + H_{2}SO_{4} \rightarrow MgSO_{4} + H_{2}

2 parts by wt. of H₂ are displaced by 24 parts by wt. of Mg.
1 part by wt. of H₂ are displaced by 12 parts by wt. of Mg.
So, the equivalent weight of Mg = 12.

2.\ C + O_{2} \rightarrow CO_{2}

32 parts by wt. of O₂ combine with 12 parts by wt. of C
8 parts by wt. of O₂ combine with 3 parts by wt. of C
So, the equivalent weight of C = 3.

3.\ 2Al + 3Cl_{2} \rightarrow 2AlCl_{3}

213 parts by wt. of Cl₂ combine with 54 parts by wt. of Al
35.5 parts by wt. of Cl₂ combine with 9 parts by wt. of Al
So, the equivalent weight of Al = 9.

The equivalent weight of Mg is 12. It means that 12 parts by weight of Mg displace 1.008 parts by weight of hydrogen, 12 parts by weight of Mg combine with 8 parts by weight of oxygen and 12 parts by weight of Mg combine with 35.5 parts by weight of chlorine or one equivalent of any other element.

Gram equivalent weight

The equivalent weight expressed in gram is called gram equivalent weight.
Eg: one gram equivalent weight of oxygen is 8 gm.

No.\ of\ gram\ equivalent\ =\ \frac{Weight\ of\ substance}{Equivalent\ weight}

Q. How many gram equivalents are present in 480 gm of Calcium?
Solution:

\begin{align*} No.\ of\ gram\ equivalent\ &=\ \frac{Weight\ of\ substance}{Equivalent\ weight}\\ &= \frac{480}{20}\ = 24 \end{align*}

Variable equivalent weight

The elements having variable valency shows variable equivalent weight.
Examples:

i. In red oxide of copper (Cu₂O):
16 parts by wt. of O₂ combines with 127 parts by wt. of Cu
8 parts by wt. of O₂ combines with 63.5 parts by wt. of Cu
Eq. wt. of Cu in red oxide = 63.5

ii. In black oxide of copper (CuO)
16 parts by wt. of O₂ combines with 63.5 parts by wt. of Cu
8 parts by wt. of O₂ combines with 31.75 parts by wt. of Cu
Eq. wt. of Cu in black oxide = 31.75

Relation beween atomic weight, equivalent weight and valency

Let atomic weight, equivalent weight and valency of an element be A, E and V respectively. The valency of an element is the number of hydrogen atoms that combine with one atom of that element.
So,

V atoms of Hydrogen combine with one atom of an element
V x 1.008 parts by wt. of Hydrogen combine with A parts by wt. of element
1.008 parts by wt. of Hydrogen combine with A/V parts by wt. of the element.

Equivalent\ weight\ =\ \frac{Atomic\ weight}{Valency}

Eq. wt. of Fe in Fe⁺⁺ = 56/2 = 28
Eq. wt of Fe in Fe⁺⁺⁺ = 56/3 = 18.66

Equivalent weight of compounds

1. Equivalent weight of acid

Eq.\ wt.\ of\ acid\ =\ \frac{Molecular\ weight\ of\ acid}{Basicity}

Basicity is the number of replaceable hydrogen atoms present in a molecule of acid.
For HCl, eq. wt. = 36.5/1 = 36.5
For H₂SO₄, eq. wt. = 98/2 = 49
However eq. wt. of an acid depends upon the chemical reaction.

i.\ Na_{2}CO_{3} + 2H_{2}SO_{4}\ \rightarrow\ 2NaHSO_{4} + H_{2}O + CO_{2}\\ ii.\ Na_{2}CO_{3} + H_{2}SO_{4}\ \rightarrow\ Na_{2}SO_{4} + H_{2}O + CO_{2}

In reaction (i), the basicity of H2SO4 is 1. So eq. wt. is 98/1 = 98.
In reaction (ii), the basicity of H₂SO₄ is 2. So eq. wt. is 98/2 = 49.

2. Equivalent weight of base

Eq.\ wt.\ of\ base\ =\ \frac{Molecular\ weight\ of\ base}{Acidity}

Acidity is the number of replaceable hydroxyl groups present in a molecule of a base or twice the number of oxygen atoms per molecule of a base as an oxide.
For NaOH, eq. wt. = 40/1 = 40
For Ca(OH)₂, eq.wt. = 74/2 = 37
For FeO, eq. wt. = 71.84/2 = 35.92
For Fe₂O₃, eq. wt. = 159.7/6 = 26.61
However, equivalent weight of base depends upon the chemical reaction.

i.\ Ca(OH)_{2} + HCl \rightarrow Ca(OH)Cl + H_{2}O\\ ii.\ Ca(OH)_{2} + 2HCl \rightarrow CaCl_{2} + 2H_{2}O

In rxn. i, eq. wt. of Ca(OH)₂ = 74/1 = 74
In rxn. ii, eq. wt. of Ca(OH)₂ = 74/2 = 37

3. Equivalent weight of salt

Eq.\ wt.\ of\ salt\ =\ \frac{Molecular\ weight\ of\ salt}{\underset{+ve\ or\ -ve\ charges}{Total\ number\ of}}

For Na₂CO₃,
Na₂CO₃ → 2Na⁺⁺ + CO3^—
eq. wt. = 106/2 = 53

For Ca₃(PO₄)₂,
Ca₃(PO₄)₂ → 3Ca⁺⁺ + 2PO₄^—
eq. wt. = 310/6 = 51.67

4. Equivalent weight of oxidizing or reducing agent

i. Equivalent weight of HNO₃

2H \overset{+5}{N}O_{3}\ \rightarrow\ H_{2}O + 2 \overset{+4}{N}O_{2} + [O]\\ Eq.\ wt.\ of\ HNO3\ =\ 63/1\ =\ 63

ii. Equivalent weight of KMnO4
a. In acidic medium

\begin{align*} 2K \overset{+7}{Mn}O_{4}\ +\ 3H_{2}SO_{4} &\rightarrow\ K_{2}SO_{4} + 2 \overset{+2}{Mn}SO_{4} \\ &\ \ \ \ \ + 3H_{2}O + 5[O]\\ Eq.\ wt.\ of\ KMnO_{4} &= 158/5 = 31.6 \end{align*}

b. In neutral medium

\begin{align*} 2K \overset{+7}{Mn}O_{4}\ +\ 3H_{2}SO_{4} &\rightarrow\ 2 \overset{+4}{Mn}O_{2} + 2KOH+3[O]\\ Eq.\ wt.\ of\ KMnO_{4} &= 158/5 = 31.6 \end{align*}

c. In alkaline medium

\begin{align*} 2K \overset{+7}{Mn}O_{4} + KOH &\rightarrow\ 2K_{2} \overset{+6}{Mn}O_{4} + H_{2}O+[O]\\ Eq.\ wt.\ of\ KMnO_{4} &= 158/1 = 158 \end{align*}

Alternatively,

\begin{align*}i.\ HCl &\rightarrow H^{+} + Cl^{-}\\ Eq. \ wt.\ of\ HCl &= Eq.\ wt.\ of\ H^{+} + eq. wt. of Cl^{-}\\ &= 1/1 + 35.5/1\\ &= 36.5 \end{align*}

\begin{align*} ii.\ H_{2}SO_{4} &\rightarrow 2H^{+} + SO_{4}^{--}\\ Eq.\ wt.\ of\ H_{2}SO_{4} &= Eq.\ wt.\ of\ H^{+} + eq.\ wt.\ of\ SO_{4}^{--}\\ &= 1/1 + 96/2\\ &= 49 \end{align*}

\begin{align*} iii.\ NaOH &\rightarrow Na^{+} + OH^{-}\\ Eq.\ wt.\ of\ NaOH &= Eq.\ wt.\ of\ Na^{+} + eq.\ wt.\ of\ OH^{-}\\ &= 23/1 + 17/2\\ &= 40 \end{align*}

\begin{align*} iv.\ Ca(OH)_{2} &\rightarrow Ca^{++} + 2OH^{-}\\ Eq.\ wt.\ of\ CaOH_{2} &= Eq.\ wt.\ of\ Ca^{++} + eq.\ wt.\ of\ OH^{-}\\ &= 40/2 + 17/2\\ &= 37 \end{align*}

\begin{align*} iv.\ Na_{2}CO_{3} &\rightarrow 2Na^{+} + CO_{3}^{--}\\ Eq.\ wt.\ of\ Na_{2}CO_{3} &= Eq.\ wt.\ of\ 2Na^{+} + eq.\ wt.\ of\ CO_{3}^{--}\\ &= 23/1 + 60/2\\ &= 53 \end{align*}

Equivalent weight of some compounds