In the neutralization process of ${{H}_{3}}P{{O}_{4}}$ and NaOH, the number of buffers formed will be __________:
[A] 3
[B] 1
[C] 2
[D] 4
Answer
284.1k+ views
Hint: A buffer solution is a solution which maintains the pH of a solution. Here, upon neutralisation the acid will lose its protons. The correct answer here is the number of protons in the acid i.e. the number of protons it has that is the number of buffers that will be formed here upon neutralisation.
Complete Step by Step Solution: We know that generally we can define a buffer solution as a solution which is resistant to change in pH upon addition of a small amount of base or acid. Buffer solution usually consists of a weak acid and salt of its conjugate base or of a weak base and a salt of its conjugate acid.
Here, we have phosphoric acid with sodium hydroxide. Let us write down its neutralization reaction. Here as we can see phosphoric acid has three removable protons thus for neutralisation three protons must be replaced.
We can write the reaction as-
\[{{H}_{3}}P{{O}_{4}}+NaOH\to Na{{H}_{2}}P{{O}_{4}}+{{H}_{2}}O\]
When the first proton is replaced, we get monobasic sodium phosphate.
Similarly, we can write the reaction of the neutralisation of the second proton as-
\[Na{{H}_{2}}P{{O}_{4}}+NaOH\to N{{a}_{2}}HP{{O}_{4}}+{{H}_{2}}O\]
Here, we get dibasic sodium phosphate.
And the third proton neutralisation reaction will be-
\[N{{a}_{2}}HP{{O}_{4}}+NaOH\to N{{a}_{3}}P{{O}_{4}}+{{H}_{2}}O\]
Here, we get tribasic sodium phosphate.
Here we can see that phosphoric acid is a weak acid and it is neutralised thus, the buffers we obtained are acidic buffers.
Here, we obtained three buffers upon neutralisation of each proton as we can see form the above reactions.
Therefore, the correct answer is option [A] 3.
Note: Buffer solutions are of two types- Acidic buffer and Basic buffer. Acidic buffer consists of a weak acid and its salt, along with a strong base. Basic buffer is prepared by mixing a weak base with its conjugate salt along with a strong acid.
A basic buffer can be formed from a mixture of ammonia and ammonium chloride in water.
$N{{H}_{3}}+HCl\to N{{H}_{4}}Cl$
An acidic buffer can be made from the mixture of sodium acetate and acetic acid in water.
$C{{H}_{3}}COOH-{{H}^{+}}\to C{{H}_{3}}CO{{O}^{-}}$
Complete Step by Step Solution: We know that generally we can define a buffer solution as a solution which is resistant to change in pH upon addition of a small amount of base or acid. Buffer solution usually consists of a weak acid and salt of its conjugate base or of a weak base and a salt of its conjugate acid.
Here, we have phosphoric acid with sodium hydroxide. Let us write down its neutralization reaction. Here as we can see phosphoric acid has three removable protons thus for neutralisation three protons must be replaced.
We can write the reaction as-
\[{{H}_{3}}P{{O}_{4}}+NaOH\to Na{{H}_{2}}P{{O}_{4}}+{{H}_{2}}O\]
When the first proton is replaced, we get monobasic sodium phosphate.
Similarly, we can write the reaction of the neutralisation of the second proton as-
\[Na{{H}_{2}}P{{O}_{4}}+NaOH\to N{{a}_{2}}HP{{O}_{4}}+{{H}_{2}}O\]
Here, we get dibasic sodium phosphate.
And the third proton neutralisation reaction will be-
\[N{{a}_{2}}HP{{O}_{4}}+NaOH\to N{{a}_{3}}P{{O}_{4}}+{{H}_{2}}O\]
Here, we get tribasic sodium phosphate.
Here we can see that phosphoric acid is a weak acid and it is neutralised thus, the buffers we obtained are acidic buffers.
Here, we obtained three buffers upon neutralisation of each proton as we can see form the above reactions.
Therefore, the correct answer is option [A] 3.
Note: Buffer solutions are of two types- Acidic buffer and Basic buffer. Acidic buffer consists of a weak acid and its salt, along with a strong base. Basic buffer is prepared by mixing a weak base with its conjugate salt along with a strong acid.
A basic buffer can be formed from a mixture of ammonia and ammonium chloride in water.
$N{{H}_{3}}+HCl\to N{{H}_{4}}Cl$
An acidic buffer can be made from the mixture of sodium acetate and acetic acid in water.
$C{{H}_{3}}COOH-{{H}^{+}}\to C{{H}_{3}}CO{{O}^{-}}$
Recently Updated Pages
Disproportionation Reaction: Definition, Example & JEE Guide

JEE Main 2025-26 Mock Test: Binomial Theorem & Quick Practice

Properties of Solids and Liquids Mock Test 2025

JEE Main Mock Test 2025-26: Principles Related To Practical

JEE Main Mock Test 2025-26: Dual Nature of Matter & Radiation

JEE Main 2025-26 Work, Energy and Power Mock Test – Free Practice Online

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Understanding the Electric Field of a Uniformly Charged Ring

Understanding Atomic Structure for Beginners

Derivation of Equation of Trajectory Explained for Students

Electron Gain Enthalpy and Electron Affinity Explained

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

JEE Advanced 2026 Notification Out with Exam Date, Registration (Extended), Syllabus and More

NCERT Solutions For Class 11 Chemistry In Hindi Chapter 1 Some Basic Concepts Of Chemistry - 2026-27

Understanding Electromagnetic Waves and Their Importance

Understanding Instantaneous Velocity

NCERT Solutions For Class 11 Chemistry Chapter 8 Organic Chemistry - Some Basic Principles And Techniques - 2026-27

