# 0.4 Gram equivalent concept  (Page 3/4)

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## Normality(n)

Normality is a measure of concentration of solution. It compares solute in terms of gram equivalents to the volume of solution in litres.

$\text{Normality(N)}=\frac{\text{geq}}{{V}_{L}}$

Its unit is geq/ litres. In case we consider volume in ml i.e. cc, then the expression of normality is given as :

$⇒\text{Normality(N)}=\frac{\text{geq}}{{V}_{CC}}X1000=\frac{\text{milli-gram equivalent}}{{V}_{CC}}=\frac{\text{meq}}{{V}_{CC}}$

If volume of a solution of known normality is known, then number of gram equivalents or milli-gram equivalents of solute are obtained as :

$⇒\text{geq}=N{V}_{L}$

$⇒\text{meq}=N{V}_{CC}$

It is evident from the definition of normality that normality of a given bulk solution and that of a sample taken from it are same. For example, if we take 10 cc of sulphuric acid from 1 litre of 0.2N sulphuric acid solution, then normality of sample taken is also 0.2N. However, geq or meq will be different as the amount of solutes are different.

$\text{meq in bulk solution}=0.2X1000=200$

$\text{meq in sample}=0.2X10=2$

Problem : Find volume of 0.2 N solution containing 2.5 meq of solute.

Solution : The gram equivalent is given by :

$\text{meq}=N{V}_{CC}$

${V}_{CC}=\frac{\text{meq}}{N}=\frac{2.5}{0.2}=12.5\phantom{\rule{1em}{0ex}}cc$

Problem : How many grams of wet NaOH containing 10% water are required to prepare 1 litre of 0.1 N solution.

Solution : We can determine amount of NaOH for preparation of 1 litre of 0.1 N solution by using normality relation :

$geq=NV=0.1X1=0.1$

$⇒geq=\frac{g}{E}=\frac{xg}{{M}_{O}}$

The molecular weight of NaOH is 23+16+1= 40. Its valence factor is 1. Hence,

$⇒g=\frac{\text{geq}X{M}_{O}}{x}=\frac{0.1X40}{1}=4\phantom{\rule{1em}{0ex}}gm$

But 100 gms of wet NaOH contain 88 gm NaOH. Applying unitary method, 4 gm of NaOH is present in wet NaOH given by :

$⇒\text{Mass of wet NaOH}=\frac{4X100}{88}=4.55\phantom{\rule{1em}{0ex}}gm$

Problem : Borax ( $N{a}_{2}{B}_{4}{O}_{7},10{H}_{2}O$ ) is a strong base in aqueous solution. Hydroxyl ions are produced by ionic dissociation as :

$2N{a}^{+}+{B}_{4}{O}_{7}{}^{2-}+7{H}_{2}O\to 4{H}_{3}B{O}_{3}+2N{a}^{+}+2O{H}^{-}$

A solution of borax is prepared in water. The solution is completely neutralized with 50 ml of 0.1N HCl. Find the mass of borax in the solution. Consider atomic weight of boron 24 amu.

Solution : Applying gram equivalent concept,

$⇒\text{meq of borax}=\text{meq of 50 ml 0.1N HCl}$

$⇒\text{meq of borax}=NV=0.1X50=5$

Now, gram equivalent is connected to mass as :

$⇒meq=\frac{1000g}{E}=\frac{1000xg}{{M}_{O}}$

$⇒g=\frac{meqXE}{1000x}=\frac{1000xg}{{M}_{O}}$

Molecular weight of borax is 2X23 + 4X24 + 7X16 + 7X18 = 380. The valence factor of borax is 2 as one molecule of borax gives 2 hydroxyl ions. Putting values, we have,

$⇒g=\frac{5X380}{1000X2}=0.95\phantom{\rule{1em}{0ex}}gm$

## Relation between normality and molarity

Normality is defined as :

$\text{Normality(N)}=\frac{\text{geq}}{{V}_{L}}$

Substituting for gram equivalent,

$⇒\text{Normality(N)}=\frac{\text{geq}}{{V}_{L}}=\frac{x{M}_{O}}{{V}_{L}}$

where “x” is valance factor. Now, molarity is defined as :

$M=\frac{{M}_{O}}{{V}_{L}}$

Note that we have used the convention that “M” represents molarity and “ ${M}_{O}$ ” or subscripted symbol " ${M}_{N{H}_{3}}$ " represents molecular weight. Now, combining two equations, we have :

$⇒N=xM$

$\text{Normality}=\text{valance factor}X\text{Molarity}$

## Combining solutions of different normality

Let us consider two solutions : 100 ml of 1 N ${H}_{2}S{O}_{4}$ and 50 ml of 0.5 N ${H}_{2}S{O}_{4}$ . What would be the normality if these two volumes of different normality are combined?

In order to determine normality of the final solution, we need to find total milli-gram equivalents and total volume. Then, we can determine normality of the final solution by dividing total gram equivalents by total volume. Here,

${\text{meq}}_{1}={N}_{1}{V}_{1}=1X100=100$

${\text{meq}}_{2}={N}_{2}{V}_{2}=0.5X50=25$

Total meq is :

$⇒\text{meq}={\text{meq}}_{1}+{\text{meq}}_{2}=100+25=125$

Now, important question is whether we can combine volumes as we have combined milli-gram equivalents. Milli equivalents can be combined as it measures quantity i.e. mass of solutes, which is conserved. Can we add volumes like mass? Density of 1N solution has to be different than that of 0.5N solution. We need to account for the density of the individual solution. It means that given volumes can not be added. However, concentrations of these solutions are low and so the difference in densities of solutions. As a simplifying measure for calculation, we can neglect the minor change in volume of the resulting solution. With this assumption,

$⇒\text{Total volume}={V}_{1}+{V}_{2}=100+50=150\phantom{\rule{1em}{0ex}}ml$

The normality of the resulting solution is :

$⇒N=\frac{125}{150}=\frac{5}{6}=0.83$

## Normality of combination of different acids

Normality is based on the concept of gram-equivalent, which, in turn, depends on equivalent weights. The equivalent weight, on the other hand, is obtained by dividing molecular weight by basicity. The basicity, in turn, is equal to numbers of furnishable hydrogen ions. We may conclude that gram equivalent is amount of acid per furnishable hydrogen ion. Thus, we can see combination of acids (like sulphuric and hydrochloric acids) in terms of their ability of furnishing hydrogen ions. What it means that we can add gram equivalents of two acids to find total gram equivalents and find normality of resulting solution.

We can extend the concept to basic solution as well.

Problem : 500 ml of 0.5 N HCl is mixed with 20 ml of ${H}_{2}S{O}_{4}$ having density of 1.18 gm/cc and mass percentage of 10. Find the normality of resulting solution assuming volumes are additative.

Solution : Here, we first need to find normality of ${H}_{2}S{O}_{4}$ solution, whose density and mass percentage are given. Now, molarity of the solution is given as :

$⇒M=\frac{10yd}{{M}_{O}}=10X10X\frac{1.18}{98}=118/98=1.2M$

Normality of sulphuric acid solution is :

$⇒N=xM=2X1.2=2.4N$

We can now determine total gram equivalents and total volumes to determine the normality of resulting solution.

$⇒{\text{meq}}_{1}={N}_{1}{V}_{1}=0.5X500=250$

$⇒{\text{meq}}_{2}={N}_{2}{V}_{2}=2.4X20=48$

Total meq is :

$⇒\text{meq}={\text{meq}}_{1}+{\text{meq}}_{2}=250+48=298$

$⇒\text{Total volume}={V}_{1}+{V}_{2}=500+20=520\phantom{\rule{1em}{0ex}}ml$

The normality of the resulting solution is :

$⇒N=\frac{298}{520}=0.57$

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