Problem 140 - A Complete Resource Book in Chemistry for JEE Main

Question

If 1 mole of \(\mathrm{H}_{3} \mathrm{PO}_{4}\) is reacted with 1 mole of \(\mathrm{X}(\mathrm{OH})_{2}\) as: \(\mathrm{H}_{3} \mathrm{PO}_{4}+\mathrm{X}(\mathrm{OH})_{2} \longrightarrow \mathrm{XHPO}_{4}+2 \mathrm{H}_{2} \mathrm{O}\) then (a) The equivalent weight of \(\mathrm{H}_{3} \mathrm{PO}_{4}\) is \(\frac{98}{3}\). (b) The equivalent weight of base is \(\frac{\text { Molecular mass }}{2}\) (c) 1 mole of \(\mathrm{X}(\mathrm{OH})_{2}\) more is required for complete neutralization of \(\mathrm{XHPO}_{4}\).

Step-by-Step Solution

Answer

Parts (b) and (c) are true; (a) is incorrect based on typical reaction donation.

Step 1: Understanding the Reaction

The reaction is given as: \(\mathrm{H}_3\mathrm{PO}_4+\mathrm{X}(\mathrm{OH})_2 \longrightarrow \mathrm{XHPO}_4+2 \mathrm{H}_2 \mathrm{O}\). In this reaction, \(\mathrm{H}_3\mathrm{PO}_4\) donates two protons (\(\mathrm{H}^+\)) to \(\mathrm{X}(\mathrm{OH})_2\), forming \(\mathrm{XHPO}_4\). Each \(\mathrm{X}(\mathrm{OH})_2\) neutralizes two protons.

Step 2: Calculating Equivalent Weight of \(\mathrm{H}_3\mathrm{PO}_4\)

The equivalent weight of an acid is its molar mass divided by the number of protons it donates. \(\mathrm{H}_3\mathrm{PO}_4\) has a molar mass of 98 g/mol and donates 2 protons in this reaction. Thus, its equivalent weight is \(\frac{98}{2} = 49\, \text{g/equiv}\). However, the problem states \(\frac{98}{3}\) which should correspond to a reaction donating 3 protons.

Step 3: Calculating Equivalent Weight of Base \(\mathrm{X}(\mathrm{OH})_2\)

For a base, the equivalent weight is the molar mass divided by the number of \(\mathrm{OH}^-\) ions it provides. Since \(\mathrm{X}(\mathrm{OH})_2\) provides 2 \(\mathrm{OH}^-\) ions, the equivalent weight of the base is \(\frac{\text{Molecular mass of } \mathrm{X}(\mathrm{OH})_2}{2}\). This matches the statement given in part (b).

Step 4: Analyzing the Need for More \(\mathrm{X}(\mathrm{OH})_2\)

Upon forming \(\mathrm{XHPO}_4\), one more \(\mathrm{X}(\mathrm{OH})_2\) molecule can react with \(\mathrm{XHPO}_4\), which has the potential to accept another \(\mathrm{OH}^-\) group, leading to the formation of \(\mathrm{X}_2\mathrm{HPO}_4\). Hence, 1 more mole of \(\mathrm{X}(\mathrm{OH})_2\) is needed for complete neutralization, supporting statement (c).

Key Concepts

Acid-Base Reactions

When it comes to chemistry, an important concept to grasp is acid-base reactions. These types of reactions occur when an acid and a base interact to produce water and a salt. In general, these reactions can be represented as:

• Acid + Base → Salt + Water

In the case of the provided exercise, phosphoric acid (\(\mathrm{H}_3\mathrm{PO}_4\)) is the acid that reacts with a hydroxide base, represented by \(\mathrm{X}(\mathrm{OH})_2\), to form a salt \(\mathrm{XHPO}_4\) and water \(2 \mathrm{H}_2 \mathrm{O}\).
Acid-base reactions are essential in many real-world applications, including the human body and various industrial processes. Such reactions can often be detected by changes in pH, which indicates whether a substance is acidic or basic based on its hydrogen ion concentration. Understanding these reactions provides a foundation for exploring neutralization and equivalent weights, which further describe the stoichiometry of the involved substances.
By mastering acid-base reactions, you gain insights into how these chemical processes work, enabling you to predict the products and properly balance equations.

Equivalent Weight

The concept of equivalent weight is crucial in stoichiometry as it relates to the quantitative relationships in chemical reactions. Equivalent weight measures how much of a substance reacts with or supplies one mole of hydrogen ions (\(\mathrm{H}^+\)) or electrons in a given reaction.
For acids like \(\mathrm{H}_3\mathrm{PO}_4\), the equivalent weight is calculated by dividing the molar mass by the number of protons donated. In our example reaction, \(\mathrm{H}_3\mathrm{PO}_4\) donates two protons, resulting in an equivalent weight of \(\frac{98}{2} = 49\, \text{g/equiv}\). However, to match the problem context donating three protons, a calculation adjustment is needed to \(\frac{98}{3}\).
Similarly, for bases like \(\mathrm{X}(\mathrm{OH})_2\), the equivalent weight is determined by dividing the molar mass by the number of hydroxide ions (\(\mathrm{OH}^-\)) it provides. This precise measurement is essential for accurately determining the amounts needed for reactions involving full neutralization.

Molar Mass

To understand chemical reactions fully, one must comprehend the significance of molar mass. Molar mass is the weight of one mole of a substance, typically expressed in grams per mole (g/mol). It directly impacts the stoichiometric calculations needed for predicting the outcomes of reactions.

• In our exercise, the molar mass of \(\mathrm{H}_3\mathrm{PO}_4\) is 98 g/mol.

This means that one mole of phosphoric acid weighs 98 grams.
Molar mass plays an integral role in calculating the equivalent weight and in completing balanced chemical equations. By knowing the molar mass, chemists can determine how many moles of a reactant are needed, establish the proportions in a reaction, and calculate yields.
When dealing with compounds composed of multiple elements, it's crucial to add the atomic masses of each element involved based on their abundance in the given compound. Mastering this concept allows for seamless conversions between grams and moles, aiding with precise and accurate stoichiometric calculations.

Neutralization

Neutralization is an essential chemical process encapsulating the reaction between an acid and a base to form a neutralized product, generally a salt and water. The concept is fundamental when examining the outcome of acid-base reactions.
During neutralization, the hydrogen ions (\(\mathrm{H}^+\)) from an acid combine with the hydroxide ions (\(\mathrm{OH}^-\)) of a base, resulting in water ():

• \(\mathrm{H}^+ + \mathrm{OH}^- → \mathrm{H}_2\mathrm{O}\)

In the example given, \(\mathrm{H}_3\mathrm{PO}_4\) reacting with \(\mathrm{X}(\mathrm{OH})_2\) shows partial neutralization, evidenced by the formation of \(\mathrm{XHPO}_4\). To achieve complete neutralization, one more mole of \(\mathrm{X}(\mathrm{OH})_2\) is required, transforming \(\mathrm{XHPO}_4\) to \(\mathrm{X}_2\mathrm{HPO}_4\).
Neutralization is not only a concept observed in laboratories but also a daily occurrence, influencing natural processes and industrial applications. It’s vital for pH balance in the human body and essential in wastewater treatment. Understanding neutralization allows individuals to comprehend how acid-base interactions work in reflexive environments.