Problem 125 - A Complete Resource Book in Chemistry for JEE Main

Question

The equivalent mass of phosphoric acid \(\left(\mathrm{H}_{3} \mathrm{PO}_{4}\right)\) in the reaction, \(\mathrm{NaOH}+\mathrm{H}_{3} \mathrm{PO}_{4} \longrightarrow \mathrm{NaH}_{2} \mathrm{PO}_{4}+\mathrm{H}_{2} \mathrm{O}\) is (a) 98 (b) 89 (c) 49 (d) 58

Step-by-Step Solution

Answer

The equivalent mass of phosphoric acid in the reaction is 98 g/equiv.

Step 1: Identify Molar Mass of Phosphoric Acid

First, determine the molar mass of phosphoric acid, \( \mathrm{H}_3\mathrm{PO}_4 \). It comprises three hydrogen atoms, one phosphorus atom, and four oxygen atoms. The atomic masses are approximately: H = 1, P = 31, O = 16.

Step 2: Calculate the Molar Mass

Using the atomic masses, calculate the molar mass of \( \mathrm{H}_3\mathrm{PO}_4 \): \[m = 3(1) + 1(31) + 4(16) = 3 + 31 + 64 = 98 ext{ g/mol}.\]

Step 3: Determine the Basicity of Phosphoric Acid

Identify the number of moles of hydrogen ions \((\mathrm{H}^+)\) phosphoric acid can donate in this reaction. Since only one \( \mathrm{NaOH} \) is neutralized by \( \mathrm{H}_3\mathrm{PO}_4 \), the reaction involves the loss of just one \( \mathrm{H}^+ \) ion from \( \mathrm{H}_3\mathrm{PO}_4 \).

Step 4: Calculate the Equivalent Mass

The equivalent mass of an acid is calculated by dividing its molar mass by its basicity. For \( \mathrm{H}_3\mathrm{PO}_4 \) in this reaction, its basicity is 1, therefore its equivalent mass is \[\frac{98}{1} = 98\text{ g/equiv}.\]

Step 5: Select the Correct Answer

Based on the calculated equivalent mass, the answer is: (a) 98 g/equiv.

Key Concepts

Phosphoric Acid

Phosphoric acid, with a chemical formula of \( \mathrm{H}_3\mathrm{PO}_4 \), is a slightly viscous, colorless liquid. It's commonly used in fertilizers and food flavoring.

• Phosphoric acid consists of three hydrogen atoms, one phosphorus atom, and four oxygen atoms.
• The chemical structure allows it to behave as a triprotic acid, meaning it can donate three hydrogen ions \( (\mathrm{H}^+) \).

This gives it the flexibility to participate in various chemical reactions, primarily with bases, making it crucial in understanding acid-base reactions.

Acid-base Reaction

An acid-base reaction refers to a process where an acid and a base neutralize each other to form water and a salt. In the context of our chemical reaction: \[ \mathrm{NaOH} + \mathrm{H}_3\mathrm{PO}_4 \longrightarrow \mathrm{NaH}_2\mathrm{PO}_4 + \mathrm{H}_2 \mathrm{O} \]

• Here, sodium hydroxide \((\mathrm{NaOH})\), a strong base, reacts with phosphoric acid, neutralizing one of its hydrogen ions, resulting in the formation of sodium dihydrogen phosphate \((\mathrm{NaH}_2\mathrm{PO}_4)\) and water.
• This reaction is a typical example of a monoprotic reaction because only one hydrogen ion is donated by the acid, despite it having the capacity to donate more.

Basicity

Basicity in the context of acids refers to the number of hydrogen ions \((\mathrm{H}^+)\) an acid can release when dissolved in water. Phosphoric acid is unique due to its triprotic nature:

• It can donate up to three hydrogen ions \((\mathrm{H}^+)\), which makes its potential basicity three, under complete neutralization scenarios.
• However, in the specified reaction, only one \((\mathrm{H}^+)\) ion is lost, defining its basicity as one in this particular circumstance.

This understanding is pivotal when determining the acid's equivalent mass, as the calculation involves dividing the molar mass by its basicity.

Molar Mass

The molar mass of a compound is the mass of one mole of its molecules and is usually expressed in grams per mole \((\text{g/mol})\). To find this for phosphoric acid \((\mathrm{H}_3\mathrm{PO}_4)\), one must add the atomic masses of all constituent atoms:

• Each hydrogen \((\mathrm{H})\) atom has an atomic mass of about 1, so for three hydrogens: \(3 \times 1 = 3 \).
• The phosphorus \((\mathrm{P})\) atom contributes 31.
• Each oxygen \((\mathrm{O})\) atom contributes 16, and with four oxygens: \(4 \times 16 = 64 \).

Summing them gives a total molar mass of 98 g/mol for \(\mathrm{H}_3\mathrm{PO}_4\). This is crucial for calculating its equivalent mass in reactions.

Chemical Equation Analysis

Analyzing chemical equations involves understanding how substances interact during a reaction. The given reaction, \( \mathrm{NaOH} + \mathrm{H}_3\mathrm{PO}_4 \longrightarrow \mathrm{NaH}_2\mathrm{PO}_4 + \mathrm{H}_2 \mathrm{O} \), represents a single hydrogen ion transfer:

• The reaction identifies the conversion of one reactant equivalent, \(\mathrm{H}_3\mathrm{PO}_4\), through the donation of a hydrogen ion, into specific products.
• This highlights the concept of equivalent mass, which is the molar mass divided by the number of hydrogen ions exchanged, important for determining the stoichiometry and predicting reaction outcomes.

A thorough understanding of such interactions allows chemists to calculate and predict the amounts of substances consumed and produced.