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Chapter 4: Problem 34

Balance the following equations and write the corresponding ionic and net ionic equations (if appropriate). (a) \(\mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{KOH}(a q) \longrightarrow\) (b) \(\mathrm{H}_{2} \mathrm{CO}_{3}(a q)+\mathrm{NaOH}(a q) \longrightarrow\) (c) \(\mathrm{HNO}_{3}(a q)+\mathrm{Ba}(\mathrm{OH})_{2}(a q) \longrightarrow\)

Short Answer

Expert verified
The balanced chemical equations are: (a) \(CH_{3}COOH_{(aq)} + KOH_{(aq)} \rightarrow KCH_{3}COO_{(aq)} + H_{2}O_{(l)}\) (b) \(H_{2}CO_{3}_{(aq)} + 2NaOH_{(aq)} \rightarrow Na_{2}CO_{3}_{(aq)} + 2H_{2}O_{(l)}\)(c) \(2HNO_{3}_{(aq)} + Ba(OH)_{2}_{(aq)} \rightarrow Ba(NO_{3})_{2}_{(aq)} + 2H_{2}O_{(l)}\). And the net ionic equations are: (a) \(CH_{3}COOH_{(aq)} + OH^-_{(aq)} \rightarrow CH_{3}COO^-_{(aq)} + H_{2}O_{(l)}\) (b) \(2H^{+}_{(aq)} + 2OH^-_{(aq)} \rightarrow 2H_{2}O_{(l)}\)(c) \(2H^{+}_{(aq)} + 2OH^-_{(aq)} \rightarrow 2H_{2}O_{(l)}\)

Step by step solution

01

Balance the equations and identify the products

First, identify the products of each reaction, which will always be water and a salt. Then balance the equations. (a) \(CH_{3}COOH_{(aq)} + KOH_{(aq)} \rightarrow KCH_{3}COO_{(aq)} + H_{2}O_{(l)}\) (b) \(H_{2}CO_{3}_{(aq)} + 2NaOH_{(aq)} \rightarrow Na_{2}CO_{3}_{(aq)} + 2H_{2}O_{(l)}\)(c) \(2HNO_{3}_{(aq)} + Ba(OH)_{2}_{(aq)} \rightarrow Ba(NO_{3})_{2}_{(aq)} + 2H_{2}O_{(l)}\)
02

Write the ionic equations

In an ionic equation, we break all the strong electrolytes into their individual ions.(a) \(CH_{3}COOH_{(aq)} + K^{+}_{(aq)} + OH^-_{(aq)} \rightarrow CH_{3}COO^-_{(aq)} + K^{+}_{(aq)} + H_{2}O_{(l)}\) (b) \(2H^{+}_{(aq)}+ CO_{3}^{2-}_{(aq)} + 2Na^{+}_{(aq)} + 2OH^-_{(aq)} \rightarrow 2Na^{+}_{(aq)}+ CO_{3}^{2-}_{(aq)} + 2H_{2}O_{(l)}\) (c) \(2H^{+}_{(aq)} + 2NO_{3}^-_{(aq)} + Ba^{2+}_{(aq)} + 2OH^-_{(aq)} \rightarrow Ba^{2+}_{(aq)} + 2NO_{3}^-_{(aq)} + 2H_{2}O_{(l)}\)
03

Write the net ionic equations

Finally, simplify to the net ionic equations by canceling out the species that appear on both sides of the equation (spectator ions)(a) \(CH_{3}COOH_{(aq)} + OH^-_{(aq)} \rightarrow CH_{3}COO^-_{(aq)} + H_{2}O_{(l)}\) (b) \(2H^{+}_{(aq)} + 2OH^-_{(aq)} \rightarrow 2H_{2}O_{(l)}\)(c) \(2H^{+}_{(aq)} + 2OH^-_{(aq)} \rightarrow 2H_{2}O_{(l)}\)

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Key Concepts

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

Balancing Equations
Balancing chemical equations is a fundamental skill in chemistry that ensures the number of atoms for each element is the same on both sides of the equation. This conservation satisfies the law of conservation of mass, meaning matter cannot be created or destroyed in a chemical reaction. To achieve this balance, follow these steps:
  • Identify each compound's chemical formula on both sides of the equation.
  • Count the number of atoms for each element in the reactants and products.
  • Adjust coefficients (the numbers before molecules) to equalize the number of atoms for each element on both sides.
  • Ensure all coefficients are the smallest set of integers possible.
For example, when balancing the equation for acetic acid and potassium hydroxide:\[CH_{3}COOH_{(aq)} + KOH_{(aq)} \rightarrow KCH_{3}COO_{(aq)} + H_{2}O_{(l)}\] By adjusting the coefficients, we maintain balance in terms of both mass and charge across the reaction.
Ionic Equations
Ionic equations break down strong electrolytes (substances that completely dissociate into ions in solution) into their respective ions. This definition applies to soluble salts and strong acids and bases. Writing ionic equations helps us understand the behavior of ions in aqueous reactions:
  • Identify strong electrolytes and their dissociation into ions.
  • Write the complete ionic equation by expressing aqueous compounds as individual ions.
  • Maintain charge balance and atom count on both sides of the equation.
For instance, acetic acid reacting with potassium hydroxide involves dissociation of potassium hydroxide:\[CH_{3}COOH_{(aq)} + K^{+}_{(aq)} + OH^-_{(aq)} \rightarrow CH_{3}COO^-_{(aq)} + K^{+}_{(aq)} + H_{2}O_{(l)}\] This showcases the ionic participants in the chemical process.
Net Ionic Equations
Net ionic equations focus on the chemical species that undergo change during the reaction, omitting spectator ions, which do not participate directly in the reaction. This simplification highlights the essence of the reaction, providing a clearer understanding of the chemical change:
  • Begin with the complete ionic equation.
  • Identify and cancel out the spectator ions present on both sides of the equation.
  • Rewrite the equation showing only the ions and molecules directly involved in the reaction.
For acetic acid and potassium hydroxide, the net ionic equation becomes:\[CH_{3}COOH_{(aq)} + OH^-_{(aq)} \rightarrow CH_{3}COO^-_{(aq)} + H_{2}O_{(l)}\] This step distills the chemical equation to its core, highlighting the actual reaction components.

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Most popular questions from this chapter

How does an acid-base indicator work?

Someone spilled concentrated sulfuric acid on the floor of a chemistry laboratory. To neutralize the acid, would it be preferable to pour concentrated sodium hydroxide solution or spray solid sodium bicarbonate over the acid? Explain your choice and the chemical basis for the action.

Nitric acid is a strong oxidizing agent. State which of the following species is least likely to be produced when nitric acid reacts with a strong reducing agent such as zinc metal, and explain why: \(\mathrm{N}_{2} \mathrm{O}\) \(\mathrm{NO}, \mathrm{NO}_{2}, \mathrm{~N}_{2} \mathrm{O}_{4}, \mathrm{~N}_{2} \mathrm{O}_{5}, \mathrm{NH}_{4}^{+}\).

A \(60.0-\mathrm{mL} 0.513 \mathrm{M}\) glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) solution is mixed with \(120.0 \mathrm{~mL}\) of \(2.33 \mathrm{M}\) glucose solution. What is the concentration of the final solution? Assume the volumes are additive.

Hydrogen halides (HF, HCl, HBr, HI) are highly reactive compounds that have many industrial and laboratory uses. (a) In the laboratory, HF and \(\mathrm{HCl}\) can be generated by reacting \(\mathrm{CaF}_{2}\) and \(\mathrm{NaCl}\) with concentrated sulfuric acid. Write appropriate equations for the reactions. (Hint: These are not redox reactions.) (b) Why is it that HBr and HI cannot be prepared similarly-that is, by reacting NaBr and NaI with concentrated sulfuric acid? (Hint: \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is a stronger oxidizing agent than both \(\mathrm{Br}_{2}\) and \(\mathrm{I}_{2} .\) ) (c) HBr can be prepared by reacting phosphorus tribromide \(\left(\mathrm{PBr}_{3}\right)\) with water. Write an equation for this reaction.
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