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Baking powder contains baking soda \(\left(\mathrm{NaH} \mathrm{CO}_{3}\right)\) and an acidic substance such as sodium alum, \(\operatorname{NaAl}\left(\mathrm{SO}_{4}\right)_{2} \cdot 12 \mathrm{H}_{2} \mathrm{O} .\) These components react in an aqueous medium to produce \(\mathrm{CO}_{2}\) gas, which "raises" the dough. Write a balanced net ionic equation for the reaction.

Short Answer

Expert verified
\( \text{HCO}_3^-(aq) + \text{H}^+(aq) \rightarrow \text{CO}_2(g) + \text{H}_2\text{O}(l) \)

Step by step solution

01

Understand the Reactants

Baking soda, \( \text{NaHCO}_3 \), acts as the bicarbonate source, and sodium alum, \( \text{NaAl(SO}_4)_2 \cdot 12 \text{H}_2 \text{O} \), provides the acidic component. Together, these will react in water to generate carbon dioxide, water, and possibly other ions or molecules.
02

Write Molecular Equations

In water, baking soda dissociates into \( \text{Na}^+ \) and \( \text{HCO}_3^- \). Sodium alum dissociates into \( \text{Na}^+ \), \( \text{Al}^{3+} \), and \( \text{SO}_4^{2-} \). The equation for this reaction in aqueous solution can initially be considered as:\[\text{NaHCO}_3(aq) + \text{NaAl(SO}_4)_2(aq) \rightarrow \text{CO}_2(g) + \text{H}_2\text{O}(l) + \text{other products}\]
03

Identify the Actual Reaction Components

The reaction primarily involves the bicarbonate ion \( \text{HCO}_3^- \) reacting with the acidic components. The acid will likely donate protons, leading to the formation of carbon dioxide \( \text{CO}_2 \) and water \( \text{H}_2\text{O} \). The key ionic components are \( \text{HCO}_3^- \) and possibly \( \text{H}^+ \) from the acid dissociation of \( \text{Al}^{3+} \) with water.
04

Write the Net Ionic Equation

The bicarbonate reacts in the following way:\[ \text{HCO}_3^-(aq) + \text{H}^+(aq) \rightarrow \text{CO}_2(g) + \text{H}_2\text{O}(l) \]This net ionic equation represents the main chemical process that produces carbon dioxide, which causes the dough to rise.
05

Balance the Equation

Verify that the equation is balanced. Each side of the equation has one \( \text{HCO}_3^- \), one \( \text{H}^+ \), and produces one \( \text{CO}_2 \) and one \( \text{H}_2\text{O} \). Thus, the equation is already balanced.

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Key Concepts

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

Baking Soda Reaction
The reaction of baking soda, also known as sodium bicarbonate (\(\mathrm{NaHCO}_3\)), forms the cornerstone of baking and many chemical reactions. Baking soda is a basic compound that reacts with acidic substances to produce carbon dioxide gas. This is the gas that causes baked goods to rise and become fluffy.
In the context of a reaction, baking soda dissociates in water to form bicarbonate ions \((\mathrm{HCO}_3^-\)) and sodium ions \((\mathrm{Na^+}\)). The bicarbonate ions are the real players in the reaction, as they interact directly with acid to produce carbon dioxide \((\mathrm{CO}_2)\).
  • This reaction is essential in baking because it leads to the production of carbon dioxide gas.
  • Carbon dioxide is crucial for creating the airy texture in baked goods as it expands when heated, making the dough rise.
  • Beyond baking, this reaction principle is used in chemistry demonstrations and experiments due to the visible effervescence.
Understanding this reaction is key in various culinary practices and experimental chemistry settings.
Acid-Base Reaction
An acid-base reaction is a fundamental chemical process involving the transfer of protons \((\mathrm{H}^+)\) between reactants. Baking soda reactions often involve these types of chemical interactions.
In our example, baking soda reacts with an acidic compound, sodium alum \((\mathrm{NaAl(SO}_4)_2 \cdot 12 \mathrm{H}_2 \mathrm{O})\). The acidic component dissociates in water, allowing it to donate protons.
When the bicarbonate ion from baking soda \((\mathrm{HCO}_3^-)\) meets the protons from an acid, the bicarbonate ion picks up a proton to form carbonic acid \((\mathrm{H}_2\mathrm{CO}_3)\). This then quickly decomposes into carbon dioxide \((\mathrm{CO}_2)\) and water \((\mathrm{H}_2\mathrm{O})\).
  • Proton transfer is a hallmark of acid-base reactions.
  • Formation of \(\mathrm{CO}_2\) is typical when bicarbonate acts as the base in these reactions.
  • The occurrence of such reactions can be identified by the production of bubbles or gas in a system.
Appreciating acid-base reactions provides a deeper insight into both everyday chemical happenings and complex industrial processes.
Chemical Equation Balancing
Balancing chemical equations is a critical skill for understanding reactions. It involves ensuring that the same number of atoms of each element are present on both sides of the equation. This satisfies the Law of Conservation of Mass.
For our baking soda and sodium alum reaction, the net ionic equation is:
\[ \mathrm{HCO}_3^-(aq) + \mathrm{H}^+(aq) \rightarrow \mathrm{CO}_2(g) + \mathrm{H}_2\mathrm{O}(l) \]
This equation is already balanced:
  • One bicarbonate ion \((\mathrm{HCO}_3^-))\) reacts with one proton \((\mathrm{H}^+\)).
  • They produce one carbon dioxide molecule \((\mathrm{CO}_2)\) and one water molecule \((\mathrm{H}_2\mathrm{O})\).
Balancing steps ensure clarity and correctness in chemical descriptions. Even complex reactions can be untangled by focusing on the basic principles of balancing as they apply to each component. Methodical balancing is essential for predicting yields, understanding reaction dynamics, and validating experimental results.

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Most popular questions from this chapter

Locate sulfur, selenium, chlorine, and bromine in the periodic table: (a) Which binary acid \(\left(\mathrm{H}_{2} \mathrm{~S}, \mathrm{H}_{2} \mathrm{Se}, \mathrm{HCl}\right.\), or \(\left.\mathrm{HBr}\right)\) is the strongest? Which is the weakest? Explain. (b) Which oxoacid \(\left(\mathrm{H}_{2} \mathrm{SO}_{3}, \mathrm{H}_{2} \mathrm{SeO}_{3}, \mathrm{HClO}_{3}\right.\), or \(\left.\mathrm{HBrO}_{3}\right)\) is the strongest? Which is the weakest? Explain.

For each of the following solutions, calculate [OH ] from \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\), or \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) from \(\left[\mathrm{OH}^{-}\right]\). Classify each solution as acidic, basic, or neutral. (a) \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=2.5 \times 10^{-4} \mathrm{M}\) (b) \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=2.0 \mathrm{M}\) (c) \(\left[\mathrm{OH}^{-}\right]=5.6 \times 10^{-9} \mathrm{M}\) (d) \([\mathrm{OH}]=1.5 \times 10^{-3} \mathrm{M}\) (e) \([\mathrm{OH}]=1.0 \times 10^{-7} \mathrm{M}\)

The reaction of \(\mathrm{PCl}_{4}^{+}\) with \(\mathrm{Cl}^{-}\) is a Lewis acid- base reaction. Draw dectron-dot structures for the reactants and products, and use the curved arrow notation (Section 15.15) to represent the donation of a lone pair of electrons from the Lewis base to the Lewis acid.

You may have been told not to mix bleach and ammonia. The reason is that bleach (sodium hypochlorite) reacts with ammonia to produce toxic chloramines, such as \(\mathrm{NH}_{2} \mathrm{Cl}\). For example, in basic solution: \(\mathrm{OCl}^{-}(a q)+\mathrm{NH}_{3}(a q) \longrightarrow \mathrm{OH}^{-}(a q)+\mathrm{NH}_{2} \mathrm{Cl}(a q)\) (a) The following initial rate data for this reaction were obtained in basic solution at \(25^{\circ} \mathrm{C}:\) $$ \begin{array}{lccc} \hline \mathrm{pH} & \text { Initial }\left[\mathrm{OCl}^{-}\right] & \text {Initial }\left[\mathrm{NH}_{3}\right] & \text { Initial Rate }(\mathrm{M} / \mathrm{s}) \\ \hline 12 & 0.001 & 0.01 & 0.017 \\ 12 & 0.002 & 0.01 & 0.033 \\ 12 & 0.002 & 0.03 & 0.100 \\ 13 & 0.002 & 0.03 & 0.010 \\ \hline \end{array} $$ What is the rate law for the reaction? What is the numerical value of the rate constant \(k\), including the correct units? (b) The following mechanism has been proposed for this reaction in basic solution: \(\mathrm{H}_{2} \mathrm{O}+\mathrm{OCl} \rightleftharpoons \mathrm{HOCl}+\mathrm{OH}^{-} \quad\) Fast, equilibrium constant \(K_{1}\) \(\mathrm{HOCl}+\mathrm{NH}_{3} \longrightarrow \mathrm{H}_{2} \mathrm{O}+\mathrm{NH}_{2} \mathrm{Cl}\) Slow, rate constant \(\mathrm{k}_{2}\) Assuming that the first step is in equilibrium and the second step is rate- determining, calculate the value of the rate constant \(k_{2}\) for the second step. \(K_{a}\) for \(\mathrm{HOCl}\) is \(3.5 \times 10^{-8} .\)

Give the formula for the conjugate base of each of the following Bronsted- Lowry acids: (a) \(\mathrm{HSO}_{4}^{-}\) (b) \(\mathrm{H}_{2} \mathrm{SO}_{3}\) (c) \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) (d) \(\mathrm{NH}_{4}^{+}\) (e) \(\mathrm{H}_{2} \mathrm{O}\) (f) \(\mathrm{NH}_{3}\)

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