acid base

Chemistry: An Introduction to General, Organic, and Biological Chemistry

Acid-Base and Gas-Evolution Reactions

Learning Goal

Identify and write equations for acid-base and gas-evolution reactions.
Observation: Many chemical reactions result in the formation of gas bubbles.

Gas-Formation Reactions

Overview

In certain double-displacement reactions, an insoluble gas is produced, which drives the reaction to completion.

Example Reactions

Metal Carbonates Reaction with Acids:
 - Reaction: $CaCO_3(s) + 2HCl(aq) ightarrow CaCl_2(aq) + H_2CO_3(aq)$
 - Further decomposes to:
 - $CaCO_3(s) + 2HCl(aq) ightarrow CaCl_2(aq) + H_2O(l) + CO_2(g)$
 - Metal carbonates (or bicarbonates) react with acids to form carbonic acid, which subsequently decomposes into carbon dioxide gas and water.
Metal Sulfites Reaction with Acids:
 - Reaction: $MgSO_3(s) + 2HCl(aq) ightarrow MgCl_2(aq) + H_2SO_3(aq)$
 - Further decomposes to:
 - $MgSO_3(s) + 2HCl(aq) ightarrow MgCl_2(aq) + H_2O(l) + SO_2(g)$
 - Metal sulfites (or bisulfites) react with acids to yield sulfurous acid, which decomposes into sulfur dioxide gas and water.
Ammonium Salt Reaction with Bases:
 - Reaction: $NH_4Cl(aq) + NaOH(aq) ightarrow NaCl(aq) + NH_4OH(aq)$
 - Further decomposes to:
 - $NH_4Cl(aq) + NaOH(aq) ightarrow NaCl(aq) + NH_3(g) + H_2O(l)$
 - Ammonium salts react with bases to yield ammonium hydroxide, which decomposes into ammonia gas and water.

Summary Table of Gas Evolution Reactions

Table 7.4: Types of Compounds That Undergo Gas Evolution Reactions
  - Reactant Type: Sulfides
    - Intermediate product: None
    - Gas Evolved: H₂S
  - Reactant Type: Carbonates and Bicarbonates
    - Intermediate product: H₂CO₃
    - Gas Evolved: CO₂
  - Reactant Type: Sulfites and Bisulfites
    - Intermediate product: H₂SO₃
    - Gas Evolved: SO₂
  - Reactant Type: Ammonium
    - Intermediate product: NH₄OH
    - Gas Evolved: NH₃

Acid-Base Neutralization Reactions

Definition

A neutralization reaction is categorized as a double-displacement reaction between an acid and a base.

Characteristics

A salt is defined as an ionic compound that does not feature a hydrogen ion as its cation or a hydroxide ion as its anion and is generally soluble in water.
During the neutralization process, an acid reacts with a base to yield an ionic compound (the salt) and water.

Driving Force

The formation of water $H_2O$ , a molecular compound, acts as the primary driving force behind acid-base neutralization reactions.
The reaction between an $H^+$ ion from the acid and an $OH^-$ ion from the base results in the formation of water:
$H^+(aq) + OH^-(aq) ightarrow H_2O(l)$

Learning Check

Example Reaction

Copper (II) Hydroxide and Sulfuric Acid Reaction:
 - Write the balanced molecular equation:
 - $H_2SO_4(aq) + Cu(OH)_2(aq) ightarrow CuSO_4(aq) + H_2O(l)$
 - Balanced Equation:
 - $H_2SO_4(aq) + Cu(OH)_2(aq) ightarrow CuSO_4(aq) + 2H_2O(l)$

Ionic Representation

Starting components: $H^+(aq), SO_4^{2-}(aq), Cu^{2+}(aq), OH^-(aq)$
Products contain: $SO_4^{2-}(aq), Cu^{2+}(aq)$
Note that water remains as is in the complete ionic equation because it is a liquid component.

Complete Ionic Equation

$2H^+(aq) + SO_4^{2-}(aq) + Cu^{2+}(aq) + 2OH^-(aq) ightarrow Cu^{2+}(aq) + SO_4^{2-}(aq) + 2H_2O(l)$

Net Ionic Equation

After canceling the spectator ions, the net ionic equation simplifies to:
$2H^+(aq) + 2OH^-(aq) ightarrow 2H_2O(l)$

Additional Learning Check

Write the balanced molecular, complete ionic, and net ionic equations for:
1. $HClO_4(aq) + Ba(OH)_2(aq) ightarrow$
2. $HC_2H_3O_2(aq) + NaOH(aq) ightarrow$

Acid-Base Titration

Definition

Titration is a laboratory practice used to ascertain the unknown concentration of an acid by neutralizing it with a base such as NaOH.
The process utilizes a few drops of an indicator, such as phenolphthalein, to determine the endpoint of the titration.

Procedure

In this titration example:
- Base: Sodium hydroxide (NaOH)
- Acid: Unknown concentration of acidic solution

Indicators

The phenolphthalein indicator is employed:
- It changes to pink when the solution reaches neutralization.

Endpoint of Titration

At the endpoint:
  - The moles of the base match the moles of the acid present in the solution.
  - The concentration of the base is known.
  - The volume of the base used to achieve the endpoint is documented.
  - The molarity of the acid is computed utilizing the neutralization equation for the reaction.

Acid-Base Titration Calculations

Problem 1: Molarity of HCl Solution

Given: 18.5 mL of 0.225 M NaOH neutralizes 0.0100 L of HCl.
Neutralization Reaction:
$HCl(aq) + NaOH(aq) ightarrow NaCl(aq) + H_2O(l)$

Step 1: Analyze Given and Needed Quantities

Needed: Molarity of HCl solution.
Given: 18.5 mL of 0.225 M NaOH and 0.0100 L of HCl.

Step 2: Plan for Calculation

Convert measurements and utilize a mole-mole factor between NAOH and HCl for molarity determination.

Step 3: State Equalities and Conversion Factors

Concentration relationships are established through stoichiometric equivalencies.

Step 4: Calculation Setup

Utilize the gathered data to compute the desired molarity of the HCl solution.

Problem 2: Learning Check

What is the molarity of HCl if 25.5 mL of 0.438 M NaOH is required to neutralize 0.0250 L of HCl?
Neutralization Reaction:
$HCl(aq) + NaOH(aq) ightarrow NaCl(aq) + H_2O(l)$

Solution Steps

Repeat the analysis, planning, equalities, and setup as in the previous problem for calculation of molarity.