The fundamental basis of all titrations is the **law of equivalence.**

According to which at the end point of titration, **the volume of the two titrants reacted have same number of equivalents or milli equivalents.**

Equivalent of solute = Normality × Volume in litre

⇒ Meq. = Normaltiy × Volume in mL

Also Meq./1000 = Eq

⇒ Also moles of solute = Molarity × Volume in litre

⇒ Milli moles of solute = Molarity × Volume in ml

⇒ Milli moles /1000 = Moles

⇒ $\large M_{eq.} = \frac{w}{Equivalent \; wt.} \times 1000 $

$\large M_{eq.} = \frac{W}{M} \times n \times 1000 $

⇒ **For a general reaction**

aA + bB → mC + nD

Eq. of A = Eq. of B = Eq. of C = Eq. of D

Or , Meq. of A = Meq. of B = Meq. of C = Meq. of D

⇒ In a compound AxBy,

Eq. of AxBy = Eq. of A = Eq. of B

**For the chemical change**

$\large a A + b B \rightarrow m C \rightarrow^{d D} n E \rightarrow X $

Meq. of A used = Meq. of B used = Meq. of C formed = Meq. of D used = Meq. of E formed = Meq. of X formed.

⇒ That is equivalent or meq. of reactants react in equal number to give the same number of equivalent or meq. of the products separately.

Mole and millimole react according to the equation. It is, therefore, advisable to solve numerical problems by equivalent or meq. rather than using mole or milli mole.

For this purpose change molarity into normality.

While diluting a solution from one concentration to another

(i) M_{1}V_{1} = M_{2}V_{2}

(conc.) (dilute)

(ii) N_{1}V_{1} = N_{2}V_{2}

(conc.) (dilute)

For acid-base (neutralization reaction) or redox reaction

N_{1}V_{1} = N_{2}V_{2} is always true

But M_{1}V_{1} = M_{2}V_{2} (may or may not be true)

But M_{1} × n_{1} × V_{1} = M_{2} × n_{2} × V_{2}

(always true where n −terms represent n-factor).