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What is a salt? Give two balanced chemical equations showing how a salt is formed when an acid reacts with a base.

Short Answer

Expert verified
A salt is an ionic compound formed when an acid reacts with a base, in which the hydrogen ions (H+) from the acid are replaced by metal ions (or other positively charged ions) from the base. For example, the reaction between Hydrochloric acid (HCl) and Sodium Hydroxide (NaOH) produces Sodium Chloride (NaCl) and water: \[ HCl(aq) + NaOH(aq) \rightarrow NaCl(aq) + H_2O(l) \] Another example is the reaction between Sulfuric acid (H2SO4) and Magnesium Hydroxide (Mg(OH)2) which forms Magnesium Sulfate (MgSO4) and water: \[ H2SO4(aq) + Mg(OH)2(s) \rightarrow MgSO4(aq) + 2H_2O(l) \]

Step by step solution

01

Definition of a Salt

A salt is an ionic compound formed when an acid reacts with a base, and the hydrogen ions (H+) from the acid are replaced by metal ions (or other positively charged ions) from the base. The general formula for a salt is M+A-, where M is a metal (or another positively charged ion) and A is an anion.
02

Example 1: Formation of Sodium Chloride (NaCl)

In this example, we will consider the reaction between Hydrochloric acid (HCl) and Sodium Hydroxide (NaOH): HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) This is a balanced chemical equation since there is an equal number of atoms of each element on both sides of the equation.
03

Example 2: Formation of Magnesium Sulfate (MgSO4)

In this example, we will consider the reaction between Sulfuric acid (H2SO4) and Magnesium Hydroxide (Mg(OH)2): H2SO4(aq) + Mg(OH)2(s) → MgSO4(aq) + 2H2O(l) This is a balanced chemical equation since there is an equal number of atoms of each element on both sides of the equation.

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Key Concepts

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

Ionic Compounds
Ionic compounds are formed through the chemical reaction between two ions: a positively charged ion known as a cation, and a negatively charged ion known as an anion.
In a salt formation, the cation is usually a metal, and the anion comes from the decomposed acid.
These opposite charges attract each other, which leads to the formation of a stable, neutral ionic compound.
  • Cations are often metals like Sodium (Na⁺) or Magnesium (Mg²⁺).
  • Anions might be non-metallic ions like Chloride (Cl⁻) or Sulfate (SO₄²⁻).
When combined, these ions form a solid structure known as a crystal lattice, where ions are arranged in a repeating pattern. This structure makes salts typically hard and brittle, and many are highly soluble in water.
When dissolved, salts disassociate into their individual ions, enabling them to conduct electricity.
Acid-Base Reaction
An acid-base reaction is a fundamental chemical process that occurs when an acid and a base interact and exchange ions.
The process results in the neutralization of the acid's hydrogen ions ( H⁺ ) by the base’s hydroxide ions ( OH⁻ ), forming water ( H₂O ), and usually a salt.
  • Acids are substances that can donate hydrogen ions ( H⁺ ) in a chemical reaction, like Hydrochloric acid (HCl) or Sulfuric acid ( H₂SO₄ ).
  • Bases are compounds that can accept hydrogen ions, often dissociating into ions in water, such as Sodium hydroxide (NaOH) and Magnesium hydroxide ( Mg(OH)₂ ).
This exchange is an essential chemical process not just in laboratory reactions, but also in food science, pharmaceuticals, and even our own biological systems. It leads to a balanced chemical equation, as witnessed in the formation of sodium chloride ( NaCl ) or magnesium sulfate ( MgSO₄ ).
Balanced Chemical Equations
Balanced chemical equations are central to understanding chemical reactions as they depict how matter is conserved.
Each balanced equation illustrates that the number of atoms for each element involved remains constant throughout the reaction.
This balancing follows from the Law of Conservation of Mass, which states that matter cannot be created or destroyed.
  • The left side of the equation (reactants) should have the same number of atoms of each type as the right side (products).
  • For instance, the balanced equation: HCl + NaOH → NaCl + H₂O shows equal amounts of each element.
To achieve a balance, coefficients are adjusted, but never the subscripts in compounds, as this changes the actual substances involved.
Balancing equations ensures accuracy in representing the chemical process and allows for proper determination of reactant-product ratios.

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Most popular questions from this chapter

Complete and balance each of the following molecular equations for strong acid/strong base reactions. Underline the formula of the salt produced in each reaction. a. \(\mathrm{HNO}_{3}(a q)+\mathrm{KOH}(a q) \rightarrow\) b. \(\mathrm{H}_{2} \mathrm{SO}_{4}(a q)+\mathrm{Ba}(\mathrm{OH})_{2}(a q) \rightarrow\) c. \(\mathrm{HClO}_{4}(a q)+\mathrm{NaOH}(a q) \rightarrow\) d. \(\mathrm{HCl}(a q)+\mathrm{Ca}(\mathrm{OH})_{2}(a q) \rightarrow\)

Distinguish between the molecular equation, the complete ionic equation, and the net ionic equation for a reaction in solution. Which type of equation most clearly shows the species that actually react with one another?

If atoms of the metal calcium were to react with molecules of the nonmetal fluorine, \(\mathrm{F}_{2}\), how many electrons would each calcium atom lose? How many electrons would each fluorine atom gain? How many calcium atoms would be needed to react with one fluorine molecule? What charges would the resulting calcium and fluoride ions have?

For the reaction \(2 \mathrm{Na}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{Na}_{2} \mathrm{O}_{2}(s),\) show how electrons are gained and lost by the atoms.

Balance each of the following equations that describe decomposition reactions. a. \(\mathrm{CaSO}_{4}(s) \rightarrow \mathrm{CaO}(s)+\mathrm{SO}_{3}(g)\) b. \(\mathrm{Li}_{2} \mathrm{CO}_{3}(s) \rightarrow \mathrm{Li}_{2} \mathrm{O}(s)+\mathrm{CO}_{2}(g)\) c. \(\mathrm{LiHCO}_{3}(s) \rightarrow \mathrm{Li}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}_{2}(g)\) d. \(\mathrm{C}_{6} \mathrm{H}_{6}(l) \rightarrow \mathrm{C}(s)+\mathrm{H}_{2}(g)\) e. \(\operatorname{PBr}_{3}(l) \rightarrow \mathrm{P}_{4}(s)+\mathrm{Br}_{2}(l)\)

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