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According to Bronsted-Lowry theory, in a neutralization reaction (1) A salt is formed (2) Two salts are formed(3) Two conjugate acid-base pairs are formed (4) One conjugate acid-base pair is formed

Short Answer

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One conjugate acid-base pair is formed.

Step by step solution

01

Understand the Neutralization Reaction

In a neutralization reaction according to Bronsted-Lowry theory, an acid reacts with a base to produce a salt and water. For example, hydrochloric acid (HCl) reacts with sodium hydroxide (NaOH) to form sodium chloride (NaCl) and water (H2O).
02

Identify Acid and Base

In the reaction, identify the acid and the base. The acid is a proton (H+) donor and the base is a proton (H+) acceptor. For example, in HCl + NaOH → NaCl + H2O, HCl is the acid because it donates a proton and NaOH is the base because it accepts a proton.
03

Form Conjugate Acid-Base Pairs

Determine the conjugate acid-base pairs formed in the reaction. For each acid that donates a proton, its conjugate base is formed. For each base that accepts a proton, its conjugate acid is formed. In our example, HCl (acid) forms Cl^- (conjugate base) and OH^- (base) forms H2O (conjugate acid).
04

Final Answer

From the steps above, it is clear that one conjugate acid-base pair is formed in the reaction. Specifically, the conjugate acid of the base and the conjugate base of the acid.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Neutralization Reaction
A neutralization reaction is a chemical process where an acid and a base react together to produce salt and water. According to the Bronsted-Lowry theory, this reaction involves the transfer of a proton (H+) from the acid to the base. For example, when hydrochloric acid (HCl) reacts with sodium hydroxide (NaOH), we get sodium chloride (NaCl) and water (H2O).

The primary goal of a neutralization reaction is to balance out the effects of acids and bases, ultimately bringing the solution closer to a neutral pH.
  • This is why it is called a 'neutralization' reaction
  • It usually ends with the production of salt and water.
  • It can be vital in many practical applications, such as treating acid indigestion or neutralizing acidic soil for agriculture.
Conjugate Acid-Base Pairs
In the context of the Bronsted-Lowry theory, acids and bases always come in pairs known as conjugate acid-base pairs. When an acid donates a proton, it forms its conjugate base. For instance, in the reaction between HCl and NaOH, hydrochloric acid (HCl) donates a proton to produce chloride ion (Cl^-), its conjugate base.

Likewise, when a base accepts a proton, it forms its conjugate acid. In the same reaction, the hydroxide ion (OH^-) from sodium hydroxide (NaOH) accepts a proton to form water (H2O), its conjugate acid.
  • This relationship is crucial for understanding the direction of acid-base reactions.
  • It helps predict the strength of acids and bases.
  • It also provides insight into the reaction's mechanism and outcome.
Proton Donor and Acceptor
Key to the Bronsted-Lowry theory is the idea of proton donors and acceptors. Acids are defined as proton donors because they give up a proton (H+) during the reaction. For instance, in the reaction HCl + NaOH → NaCl + H2O, hydrochloric acid (HCl) donates a proton to the hydroxide ion (OH^-) from sodium hydroxide (NaOH).

Bases, on the other hand, are proton acceptors. This means they accept a proton during the reaction. In the example given, the hydroxide ion (OH^-) accepts a proton to become water (H2O).
  • This concept helps distinguish between different acids and bases.
  • Allows us to categorize substances based on their behavior in chemical reactions.
  • Provides a simple yet powerful framework for understanding acid-base chemistry.

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Most popular questions from this chapter

Thc pair of salts that does not hydrolyze (1) \(\mathrm{FcCl}_{3}, \mathrm{Sn} \mathrm{Cl}_{4}\) (2) \(\mathrm{CaCl}_{2}, \mathrm{~K}_{2} \mathrm{SO}_{4}\) (3) \(\mathrm{CuSO}_{4}, \Lambda 1 \mathrm{Cl}_{3}\) (4) \(\mathrm{NII}_{4} \mathrm{Cl}, \mathrm{Na}_{2} \mathrm{CO}_{3}\)

For which reaction high pressure and high temperature is helpful in obtaining a high cquilibrium yicld? (1) \(2 \mathrm{NF}_{3}(\mathrm{~g}) \rightleftharpoons \mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{~F}_{2}(\mathrm{~g}) \quad 54.40 \mathrm{kcal}\) (2) \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{II}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NI}_{3}(\mathrm{~g})+22.08 \mathrm{kcal}\) (3) \(\mathrm{Cl}_{2}(\mathrm{~g})+2 \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{ClO}_{3}(\mathrm{~g}) \quad 49.4 \mathrm{kcal}\) (4) \(2 \mathrm{Cl}_{2} \mathrm{O}_{7}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{Cl}_{2}(\mathrm{~g}) \times 7 \mathrm{O}_{2}(\mathrm{~g})+126.8 \mathrm{kcal}\)

The \(\mathrm{p} K_{\mathrm{a}}\) of certain weak acid is \(4.0 .\) What should be the salt to acid ratio if we have to prepare a buffer with \(\mathrm{pH}=5\) using the acid and of its salts? (1) \(4: 5\) (2) \(5: 4\) (3) \(10: 1\) (4) \(1: 10\)

Sodium carbonate cannot be used in place of ammonium carbonate for the identification of the fifth group radicals. This is because the (1) sodium ions will interfere in the detection of the fifth group radicals (2) concentration of carbonate ions is very low (3) sodium will react with acidic radicals (4) magnesium will be precipitated

The most important buffer in blood consists of (1) \(\mathrm{HCl}\) and \(\mathrm{Cl}\) (2) \(\mathrm{H}_{2} \mathrm{CO}_{3}\) and \(\mathrm{HCO}_{3}\) (3) \(\mathrm{H}_{2} \mathrm{CO}_{3}\) and \(\mathrm{Cl}\) (4) \(\mathrm{HCl}\) and \(\mathrm{HCO}_{3}\)

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