4.5 Equivalent Weight

  Stoichiometry

Number of parts by mass of an element which reacts or displaces from a compound $1.008$ parts by mass of hydrogen, $8$ parts by mass of oxygen and $35.5$ parts by mass of chlorine, is known as the equivalent weight of that element e.g.

\[2Mg + {O_2}\xrightarrow{{}}2MgO\]

$48g$ $32g$

$12g$ $8g$

$\therefore $ $32\ g$ of ${O_2}$ reacts with $48\ g$ of $Mg$

$\therefore $ $8\ g$ of ${O_2}$ = $\frac{{48 \times 8}}{{32}}$ $=$ $12\ g$

$\therefore $ Equivalent weight of $Mg$ $=$ $12$

Similarly,

\[Zn + {H_2}S{O_4}\xrightarrow{{}}ZnS{O_4} + {H_2}\]

$65.5g$ $2 \times 1.008$

$\because $ $2 \times 1.008\ g$ of hydrogen is displaced by $65.5\ g$

$\therefore $ $1.008\ g$ of $H$ = $\frac{{65.5}}{{2 \times 1.008}}$ $ = 32.75g$

$\therefore $ Equivalent weight of $Zn$ = $\frac{{65.5}}{2}$ $ = 32.75$

\[Al + \frac{3}{2}C{l_2}\xrightarrow{{}}AlC{l_3}\]

$32g$ $\frac{3}{2} \times 71g$ $\because $ $111.5\ g$ chlorine reacts with $27\ g$ of $Al$

$\therefore $ $35.5\ g$ chlorine reacts with $\frac{{27 \times 35.5}}{{111.5}} = 9.0\;g$ of $Al$

$\therefore $ Equivalent weight of aluminium $ = \frac{{27}}{3} = 9.0$

As we can see from the above examples that equivalent weight is the ratio of atomic weight and a factor (say $n-$factor) which is in above three cases is their respective valencies. So, equivalent

$$Weight =\frac{{atomic{\text{ weight}}}}{{n - factor}}$$

In a similar way, the equivalent weight of acid/base is the ratio of molecular weight and the basicity/acidity and for oxidizing agents and reducing agent it is the ratio of molecular weight and the number of moles of electrons gained or lost.

So in case of acid/base the $n-$factor is basicity/acidity (i.e. number of dissociable $H^+$ ions/number of dissociable $OH^–$ ion and in case of oxidizing agent/reducing agent, $n-$factor is number of moles of electrons gained/lost per mole of oxidizing agent/reducing agent. Therefore, in general, we can write.

$$Equivalent\ weight (E)=\frac{{Atomic{\text{ or molecualr weight}}}}{{n - factor}}$$

$$No.\ of\ equivalents\ of\ solute =\frac{{Wt}}{{Eq.{\text{ wt}}{\text{.}}}} =\frac{W}{E} =\frac{W}{{M/n}}$$

$$No.\ of\ equivalents\ of\ solute\ = No.\ of\ moles\ of\ solute \times n-factor$$

And also $$Normality = n-factor \times molarity\ of\ solution$$