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Question: Nitric acid (70%) has a specific gravity of 1.42g/ml. Find the normality and molarity of the acid....

Nitric acid (70%) has a specific gravity of 1.42g/ml. Find the normality and molarity of the acid.

Explanation

Solution

Normality is defined as the gram equivalent weight of the solute present in one liter of solution. Molarity is defined as the number of moles of solute particles present in one liter of solution. The relationship between molarity and normality is expressed by the following equation
Normality ×\times equivalent mass = molarity ×\times molar mass

Complete step by step solution:
Here in this question the density of nitric acid solution is given that is specific density = 1.42g/ml
The formula for normality is =number  of  equivalencevolume  in  liter\dfrac{number\;of\;equivalence}{volume\;in\;liter}
Equivalent weight can be calculated by the formula given below.
So the formula is equivalent weight = molecular  massvalency  factor\dfrac{molecular\;mass}{valency\;factor}
Here in the above formula valency factor is equal to the number of replaceable hydrogen ions present. So here in nitric acid only one hydrogen ion is there. This can be found by dissociating the nitric acid. So the equation for dissociation is given below
HNO3H++NO3HN{{O}_{3}}{{H}^{+}}+N{{O}_{3}}^{-}. This shows that there is only one replaceable hydrogen ion.
So now we will calculate the molecular mass,
The molecular mass for nitric acid is = 1×14×16(3)=63g/mol1\times 14\times 16(3)=63g/mol
Now the equivalent weight = 631=63g/mol\dfrac{63}{1}=63g/mol
Now the density of nitric acid = 1.42g/mL (given)
The density of nitric acid in liter will be= 1.42×1000=1420g/L1.42\times 1000=1420g/L
Now we will calculate the mass percentage of nitric acid in liter =1420×70100=1420\times \dfrac{70}{100}= 994g
Now we will calculate the number of equivalence of nitric acid = weight  of  compoundmolecular  mass=99463=15.777\dfrac{weight\;of\;compound}{molecular\;mass}=\dfrac{994}{63}=15.777
The formula for normality is =number  of  equivalencevolume  in  liter\dfrac{number\;of\;equivalence}{volume\;in\;liter}
Now substituting the values in above formula we get= 15.7771\dfrac{15.777}{1}= 15.777N
So the normality of nitric acid is 15.777N.
Now we will calculate the molarity by the following formula
Normality ×\times equivalent mass = molarity ×\times molar mass
Molarity=Normality×equivalent  massmolar  mass\dfrac{Normality\times equivalent\;mass}{molar\;mass}
Normality= 15.777N
Equivalent mass= 63g/mol
Molar mass= 63g/mol
Substituting the values in the above relation we get
15.777×6363=15.777M\dfrac{15.777\times 63}{63}=15.777M

Therefore, the molarity of nitric acid is 15.777M.

Note: We can calculate the normality for acidic solution by following formula
Normality= molarity ×\times basicity15.777×6363=15.777M\dfrac{15.777\times 63}{63}=15.777M
Basicity is the number of hydrogen ions that can be given by the acid molecule.
We can calculate normality for basic solution by following formula
Normality= molarity ×\times acidity
Acidity is defined as the number of hydroxyl ions that can be given by base molecule.