# One mole of ${{P}_{2}}{{O}_{5}}$ undergoes hydrolysis as

${{P}_{2}}{{O}_{5}}+{{H}_{2}}O\to {{H}_{3}}P{{O}_{4}}$

The normality of the phosphoric acid formed is (The volume of solution is 1L)

(a) 2

(b) 12

(c) 24

(d) 4

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**Hint:**First balance the equation. With balancing the number of moles can be calculated. Molarity can be calculated by dividing the number of moles with the volume of the solution. Normality is equal to molarity multiplied to the basicity of the compound.

**Complete answer:**

Molarity: The molarity of the solution is defined as the number of moles of the solute present per liter. It is represented by the symbol, M.

\[Molarity=\dfrac{\text{Moles of the solute}}{\text{Volume of the solution}}\]

Moles of the solute can be calculated by:

\[Moles=\dfrac{\text{mass of the solute}}{\text{molar mass of the solute}}\]

The normality of the solution is defined as the number of grams equivalent of the solute present in liters. It is represented by the symbol, N.

\[Normality=\dfrac{\text{Gram equivalent of the solute}}{\text{Volume of the solution}}\]

First, let us balance the given equation:

${{P}_{2}}{{O}_{5}}+{{H}_{2}}O\to {{H}_{3}}P{{O}_{4}}$

After balancing the equation will be:

$2{{P}_{2}}{{O}_{5}}+6{{H}_{2}}O\to 4{{H}_{3}}P{{O}_{4}}$

And the volume of the solution is 1 L. (Given)

So, from the balanced equation we can that there are 4 moles of phosphoric acid in 1 L of solution.

From this we can calculate the molarity of the solution:

\[Molarity=\dfrac{\text{Moles of the solute}}{\text{Volume of the solution}}\]

So, by putting the values in the above equation:

\[Molarity=\dfrac{\text{Moles of the solute}}{\text{Volume of the solution}}=\dfrac{4}{1}=4\]

The molarity is 4.

With molarity, we can find the normality of the solution.

Normality of the acid = molarity x basicity.

Phosphoric acid has basicity 3 because of the presence of hydrogen ions.

So, normality of the phosphoric acid will be:

$normality=\text{ 4 x 3 = 12 N}$

Hence, the normality is 12 N.

**So, the correct answer is “Option B”.**

**Note:**The balancing of the equation must be right. The molarity of the acid can be converted into normality by multiplying it with basicity. The molarity of the base can be converted into normality by multiplying it with acidity.