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

Short Answer

Expert verified
(a) Balanced. (b) Balanced. Net: \(\text{H}_2\text{CO}_3 + 2\text{OH}^- \rightarrow 2\text{H}_2\text{O} + \text{CO}_3^{2-}\). (c) Balanced. Net: \(2\text{H}^+ + 2\text{OH}^- \rightarrow 2\text{H}_2\text{O}\).

Step by step solution

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01

Balance Equation (a)

The reaction involves acetic acid (\(\text{HC}_2\text{H}_3\text{O}_2\)) and potassium hydroxide (\(\text{KOH}\)). This is a simple acid-base neutralization reaction resulting in water and salt.\[ \text{HC}_2\text{H}_3\text{O}_2(aq) + \text{KOH}(aq) \rightarrow \text{H}_2\text{O}(l) + \text{KC}_2\text{H}_3\text{O}_2(aq) \] The equation is already balanced.
02

Write Ionic and Net Ionic Equations for (a)

In aqueous solutions, strong electrolytes dissociate into ions.Ionic Equation:\[ \text{HC}_2\text{H}_3\text{O}_2(aq) + \text{K}^+(aq) + \text{OH}^-(aq) \rightarrow \text{H}_2\text{O}(l) + \text{K}^+(aq) + \text{C}_2\text{H}_3\text{O}_2^-(aq) \]Net Ionic Equation:\[ \text{HC}_2\text{H}_3\text{O}_2(aq) + \text{OH}^-(aq) \rightarrow \text{H}_2\text{O}(l) + \text{C}_2\text{H}_3\text{O}_2^-(aq) \]
03

Balance Equation (b)

The reaction involves carbonic acid (\(\text{H}_2\text{CO}_3\)) and sodium hydroxide (\(\text{NaOH}\)). \[ \text{H}_2\text{CO}_3(aq) + 2\text{NaOH}(aq) \rightarrow 2\text{H}_2\text{O}(l) + \text{Na}_2\text{CO}_3(aq) \] The balanced equation consumes two moles of \(\text{NaOH}\) for every mole of \(\text{H}_2\text{CO}_3\).
04

Write Ionic and Net Ionic Equations for (b)

Ionic Equation:\[ \text{H}_2\text{CO}_3(aq) + 2\text{Na}^+(aq) + 2\text{OH}^-(aq) \rightarrow 2\text{H}_2\text{O}(l) + 2\text{Na}^+(aq) + \text{CO}_3^{2-}(aq) \] Net Ionic Equation:\[ \text{H}_2\text{CO}_3(aq) + 2\text{OH}^-(aq) \rightarrow 2\text{H}_2\text{O}(l) + \text{CO}_3^{2-}(aq) \]
05

Balance Equation (c)

The reaction involves nitric acid (\(\text{HNO}_3\)) and barium hydroxide (\(\text{Ba(OH)}_2\)).\[ 2\text{HNO}_3(aq) + \text{Ba(OH)}_2(aq) \rightarrow 2\text{H}_2\text{O}(l) + \text{Ba(NO}_3\text{)}_2(aq) \] Balancing requires two moles of \(\text{HNO}_3\) to neutralize one mole of \(\text{Ba(OH)}_2\).
06

Write Ionic and Net Ionic Equations for (c)

Ionic Equation:\[ 2\text{H}^+(aq) + 2\text{NO}_3^-(aq) + \text{Ba}^{2+}(aq) + 2\text{OH}^-(aq) \rightarrow 2\text{H}_2\text{O}(l) + \text{Ba}^{2+}(aq) + 2\text{NO}_3^-(aq) \]Net Ionic Equation:\[ 2\text{H}^+(aq) + 2\text{OH}^-(aq) \rightarrow 2\text{H}_2\text{O}(l) \]This net ionic equation showcases the neutralization.

Key Concepts

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

Acid-Base Reactions
Acid-base reactions are fundamental chemical reactions where an acid and a base react to form water and a salt. These reactions are crucial in chemistry as they help us understand everything from the acidity of household products to the functioning of biological systems.

An acid is a substance that donates protons (H+ ions), while a base is a substance that accepts protons or provides hydroxide ions (OH-) in solution. When they react:
  • The H+ ion from the acid meets the OH- ion from the base to form water (H2O).
  • The remaining parts of the acid and base form a salt, which is usually dissolved in the solution.
For example, when acetic acid (HC2H3O2) reacts with potassium hydroxide (KOH), water and potassium acetate, a salt, are formed. Understanding these reactions helps us analyze more complex systems and chemical behavior.
Ionic Equations
Ionic equations are a step forward in understanding reactions that occur in aqueous (water) solutions. Unlike molecular equations, which show reactants and products as compounds, ionic equations break down soluble ionic compounds into their individual ions.

Writing ionic equations involves:
  • Dissociating all strong electrolytes into their respective ions. Strong electrolytes include strong acids, strong bases, and salts.
  • Including only those species that are in ionic form in the aqueous solution. For example, KOH dissociates into K+ and OH- ions in water.
This method allows us to observe more clearly what is happening during the reaction, specifically which ions are involved. This is particularly helpful when identifying spectator ions, which are ions that do not participate in the overall chemistry of the reaction.
Net Ionic Equations
Net ionic equations present the essence of chemical reactions by including only the species directly involved in the reaction, excluding the spectator ions. Spectator ions are present in the reactants and products but do not affect the outcome.

To write a net ionic equation, you:
  • First write the full ionic equation, showing all ions present in the reaction.
  • Then, remove the ions that appear unchanged on both sides of the equation.
For instance, in the reaction between nitric acid and barium hydroxide, the spectator ions are NO3- and Ba2+. These are omitted in the net ionic equation, leaving only: \( 2\text{H}^+(aq) + 2\text{OH}^-(aq) \rightarrow 2\text{H}_2\text{O}(l) \). This stripped-down equation highlights the neutralization process that is central to the reaction.
Balancing Chemical Equations
Balancing chemical equations is a crucial skill in chemistry that ensures the law of conservation of mass is respected, meaning atoms are neither created nor destroyed during a chemical reaction. In a balanced equation, the number of atoms for each element is equal on both sides of the equation.

To balance a chemical equation:
  • Start by counting all atoms for each element in the reactants and the products.
  • Adjust coefficients (numbers in front of molecules) to make sure the number of atoms for each element is the same on both sides of the equation.
  • Keep coefficients as the lowest possible whole numbers.
For example, in the reaction between HNO3 and Ba(OH)2, by balancing one molecule of water formed from each reaction of an H+ and OH- pair, you ensure the equation follows this pattern: \( 2\text{HNO}_3(aq) + \text{Ba(OH)}_2(aq) \rightarrow 2\text{H}_2\text{O}(l) + \text{Ba(NO}_3)_2(aq) \). Mastering this skill is essential for accurately predicting the results of chemical reactions.

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Most popular questions from this chapter

Give the oxidation number for the following species: \(\mathrm{H}_{2}, \mathrm{Se}_{\mathrm{g}}, \mathrm{P}_{4}, \mathrm{O}, \mathrm{U}, \mathrm{As}_{4}, \mathrm{~B}_{12}\)

You are given a colorless liquid. Describe three chemical tests you would perform on the liquid to show it is water.

A 22.02-mL solution containing \(1.615 \mathrm{~g} \mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) is mixed with a 28.64-mL solution containing \(1.073 \mathrm{~g}\) \(\mathrm{NaOH}\). Calculate the concentrations of the ions remaining in solution after the reaction is complete. Assume volumes are additive.

For each of the following pairs of combinations, indicate which one will produce the greater mass of solid product: a) \(105.5 \mathrm{~mL} 1.508 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) and \(250.0 \mathrm{~mL}\) \(1.2075 \mathrm{M} \mathrm{KCl}\) or \(138.5 \mathrm{~mL} 1.469 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) and \(100.0 \mathrm{~mL} 2.115 \mathrm{M} \mathrm{KCl}\) b) \(32.25 \mathrm{~mL} 0.9475 \mathrm{M} \mathrm{Na}_{3} \mathrm{PO}_{4}\) and \(92.75 \mathrm{~mL} 0.7750 \mathrm{M}\) \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) or \(52.50 \mathrm{~mL} 0.6810 \mathrm{M} \mathrm{Na}_{3} \mathrm{PO}_{4}\) and \(39.50 \mathrm{~mL} 1.555 \mathrm{M}\) \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) c) \(29.75 \mathrm{~mL} 1.575 \mathrm{M} \mathrm{AgNO}_{3}\) and \(25.00 \mathrm{~mL} 2.010 \mathrm{M}\) \(\mathrm{BaCl}_{2}\) or \(52.80 \mathrm{~mL} 2.010 \mathrm{M} \mathrm{AgNO}_{3}\) and \(73.50 \mathrm{~mL} 0.7500 \mathrm{M}\) \(\mathrm{BaCl}_{2}\)

Calculate the volume in milliliters of a solution required to provide the following: (a) \(2.14 \mathrm{~g}\) of sodium chloride from a \(0.270-M\) solution, (b) \(4.30 \mathrm{~g}\) of ethanol from a \(1.50-M\) solution, (c) \(0.85 \mathrm{~g}\) of acetic acid \(\left(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)\) from a \(0.30-M\) solution.

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