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Write a balanced equation for the neutralization of each of the following: a. \(\mathrm{H}_{2} \mathrm{SO}_{4}(a q)\) and \(\mathrm{NaOH}(a q)\) b. \(\mathrm{HCl}(a q)\) and \(\mathrm{Fe}(\mathrm{OH})_{3}(s)\) c. \(\mathrm{H}_{2} \mathrm{CO}_{3}(a q)\) and \(\mathrm{Mg}(\mathrm{OH})_{2}(s)\)

Short Answer

Expert verified
a. \(\mathrm{H}_{2} \mathrm{SO}_{4}(a q) + 2\mathrm{NaOH}(a q) \rightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(a q) + 2\mathrm{H}_{2} \mathrm{O}(l)\) b. \(3\mathrm{HCl}(a q) + \mathrm{Fe}(\mathrm{OH})_{3}(s) \rightarrow \mathrm{FeCl}_{3}(a q) + 3 \mathrm{H}_{2} \mathrm{O}(l)\) c. \(\mathrm{H}_{2} \mathrm{CO}_{3}(a q) + \mathrm{Mg}(\mathrm{OH})_{2}(s) \rightarrow \mathrm{MgCO}_{3}(a q) + 2\mathrm{H}_{2} \mathrm{O}(l)\)

Step by step solution

01

Understanding Neutralization

Neutralization is a chemical reaction between an acid and a base forming a salt and water.
02

Write the equation for \(\mathrm{H}_{2} \mathrm{SO}_{4}(a q)\) and \(\mathrm{NaOH}(a q)\)

Start by writing the reactants: \(\mathrm{H}_{2} \mathrm{SO}_{4} + 2 \mathrm{NaOH} \). The products will be a salt and water: \(\mathrm{Na}_{2} \mathrm{SO}_{4} + 2 \mathrm{H}_{2} \mathrm{O}\).
03

Balanced equation for \(\mathrm{H}_{2} \mathrm{SO}_{4}(a q)\) and \(\mathrm{NaOH}(a q)\)

The balanced equation is: \(\mathrm{H}_{2} \mathrm{SO}_{4}(a q) + 2\mathrm{NaOH}(a q) \rightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(a q) + 2\mathrm{H}_{2} \mathrm{O}(l)\).
04

Write the equation for \(\mathrm{HCl}(a q)\) and \(\mathrm{Fe}(\mathrm{OH})_{3}(s)\)

Start by writing the reactants: \(3 \mathrm{HCl} + \mathrm{Fe}(\mathrm{OH})_{3} \). The products will be a salt and water: \(\mathrm{FeCl}_{3} + 3 \mathrm{H}_{2} \mathrm{O}\).
05

Balanced equation for \(\mathrm{HCl}(a q)\) and \(\mathrm{Fe}(\mathrm{OH})_{3}(s)\)

The balanced equation is: \(3\mathrm{HCl}(a q) + \mathrm{Fe}(\mathrm{OH})_{3}(s) \rightarrow \mathrm{FeCl}_{3}(a q) + 3 \mathrm{H}_{2} \mathrm{O}(l)\).
06

Write the equation for \(\mathrm{H}_{2} \mathrm{CO}_{3}(a q)\) and \(\mathrm{Mg}(\mathrm{OH})_{2}(s)\)

Start by writing the reactants: \(\mathrm{H}_{2} \mathrm{CO}_{3} + \mathrm{Mg}(\mathrm{OH})_{2} \). The products will be a salt and water: \(\mathrm{MgCO}_{3} + 2 \mathrm{H}_{2} \mathrm{O}\).
07

Balanced equation for \(\mathrm{H}_{2} \mathrm{CO}_{3}(a q)\) and \(\mathrm{Mg}(\mathrm{OH})_{2}(s)\)

The balanced equation is: \(\mathrm{H}_{2} \mathrm{CO}_{3}(a q) + \mathrm{Mg}(\mathrm{OH})_{2}(s) \rightarrow \mathrm{MgCO}_{3}(a q) + 2\mathrm{H}_{2} \mathrm{O}(l)\).

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Key Concepts

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

Balanced Chemical Equations
Balanced chemical equations are essential to accurately represent reactions. For a reaction to be balanced, the number of atoms for each element must be the same on both sides of the equation. This follows the Law of Conservation of Mass, which states that matter cannot be created or destroyed in a chemical reaction.

Let's take a look at an example from the exercise:

For the reaction \(\mathrm{H}_2\mathrm{SO}_4(aq) + 2\mathrm{NaOH}(aq)\rightarrow \mathrm{Na}_2\mathrm{SO}_4(aq) + 2\mathrm{H}_2\mathrm{O}(l)\),

The number of hydrogen (H), sulfur (S), oxygen (O), and sodium (Na) atoms are balanced to ensure the same number of each type of atom exists on both sides of the equation. Such balancing is crucial in all chemical reactions to correctly portray the substances involved and the products formed.
Acid-Base Reactions
Acid-base reactions, also known as neutralization reactions, occur when an acid and a base react to form water and a salt. The acid provides hydrogen ions (H+) while the base provides hydroxide ions (OH-), which combine to form water (H2O).

For instance, in the reaction between hydrochloric acid (HCl) and iron(III) hydroxide (Fe(OH)3):
\(3\mathrm{HCl}(aq) + \mathrm{Fe}(\mathrm{OH})_3(s) \rightarrow \mathrm{FeCl}_3(aq) + 3\mathrm{H}_2\mathrm{O}(l)\),

The H+ ions from HCl react with the OH- ions from Fe(OH)3 to produce water molecules. At the same time, the remaining parts, Cl- and Fe3+ combine to form the salt, FeCl3. Understanding how acids and bases interact is vital in predicting the outcomes of chemical reactions.
Salt Formation
During neutralization reactions, one of the key products formed is a salt. A salt is a compound made from the neutralization of an acid by a base.

Let’s consider the reaction between sulfuric acid (H2SO4) and sodium hydroxide (NaOH):
\(\mathrm{H}_2\mathrm{SO}_4(aq) + 2\mathrm{NaOH}(aq) \rightarrow \mathrm{Na}_2\mathrm{SO}_4(aq) + 2\mathrm{H}_2\mathrm{O}(l)\),

Here, the sodium (Na+) from NaOH combines with the sulfate (SO4^2-) from H2SO4 to form the salt sodium sulfate (Na2SO4). Every neutralization reaction will result in the formation of a specific salt along with water.
Water Production
Water production is a crucial aspect of neutralization reactions. In most acid-base reactions, the hydrogen ions from the acid react with the hydroxide ions from the base to produce water molecules.

Take, for example, the reaction between carbonic acid (H2CO3) and magnesium hydroxide (Mg(OH)2):
\(\mathrm{H}_2\mathrm{CO}_3(aq) + \mathrm{Mg}(\mathrm{OH})_2(s) \rightarrow \mathrm{MgCO}_3(aq) + 2\mathrm{H}_2\mathrm{O}(l)\),

Here, each H+ ion from H2CO3 pairs with an OH- ion from Mg(OH)2 to form water. Thus, two water molecules are produced in this case. Understanding water production helps in balancing chemical equations and understanding the complete outcome of reactions.

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Most popular questions from this chapter

A solution of \(0.205 \mathrm{M} \mathrm{NaOH}\) is used to \(20.0 \mathrm{~mL}\) of \(\mathrm{a} \mathrm{H}_{2} \mathrm{SO}_{4}\) solution. If \(45.6 \mathrm{~mL}\) of the \(\mathrm{NaOH}\) solution is required to reach the endpoint, what is the molarity of the \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution? $$ \begin{aligned} &\text { (14.8) } \\ &\mathrm{H}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{NaOH}(a q) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Na}_{2} \mathrm{SO}_{4}(a q) \end{aligned} $$

Phosphoric acid dissociates to form hydronium ion and dihydrogen phosphate. Phosphoric acid has a \(K_{\mathrm{a}}\) of \(7.5 \times 10^{-3}\). Write the equation for the reaction and the acid dissociation expression for phosphoric acid.

Someone with kidney failure excretes urine with large amounts of \(\mathrm{HCO}_{3}^{-}\). How would this loss of \(\mathrm{HCO}_{3}^{-}\) affect the \(\mathrm{pH}\) of the blood plasma?

A buffer solution is made by dissolving \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) and \(\mathrm{NaC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) in water. (14.9) a. Write an equation that shows how this buffer neutralizes added acid. b. Write an equation that shows how this buffer neutralizes added base. c. Calculate the \(\mathrm{pH}\) of this buffer if it contains \(0.20 \mathrm{M} \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) and \(0.40 \mathrm{M} \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-}\). The \(K_{\mathrm{a}}\) for \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) is \(1.8 \times 10^{-5}\)

Why would the pH of your blood plasma increase if you breathe fast?

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