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When aqueous sodium carbonate, \(\mathrm{Na}_{2} \mathrm{CO}_{3}\), is treated with dilute hydrochloric acid, HCl, the products are sodium chloride, water, and carbon dioxide gas. What is the net ionic equation for this reaction?

Short Answer

Expert verified
The net ionic equation for the reaction is: \(CO_3^{2-} + 2H^+ ➔ H_2O + CO_2\)

Step by step solution

01

Write the balanced molecular equation

First, start by writing down the balanced molecular equation for the reaction. Sodium Carbonate (\(Na_2CO_3\)) reacts with Hydrochloric acid (2\(HCl\)) to form Sodium Chloride (2\(NaCl\)), Water (\(H_2O\)) and Carbon dioxide (\(CO_2\)). So, the balanced molecular equation is: \(Na_2CO_3 + 2HCl ➔ 2NaCl + H_2O + CO_2\)
02

Break into ions

Next, dissociate all strong electrolytes into their ions. Sodium Carbonate (\(Na_2CO_3\)) breaks down into 2 Sodium ions (\(2Na^+\)) and one Carbonate ion (\(CO_3^{2-}\)). Hydrochloric acid (2\(HCl\)) breaks down into 2 Hydrogen ions (\(2H^+\)) and 2 Chloride ions (\(2Cl^-\)). Sodium Chloride (2\(NaCl\)) breaks down into 2 Sodium ions (\(2Na^+\)) and 2 Chloride ions (\(2Cl^-\)). Water and Carbon dioxide exist as molecules: \(H_2O\) and \(CO_2\). So, the complete ionic equation is:\(2Na^+ + CO_3^{2-} + 2H^+ + 2Cl^- ➔ 2Na^+ + 2Cl^- + H_2O + CO_2\)
03

Cross out spectator ions

Spectator ions appear on both sides of the equation and do not directly participate in the reaction. In this case, Sodium ions (\(Na^+\)) and Chloride ions (\(Cl^-\)) appear in their same form on both sides of the equation, so these are spectators and can be crossed out. What remains are the ions that reacted directly, forming the net ionic equation: \(CO_3^{2-} + 2H^+ ➔ H_2O + CO_2\)

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Key Concepts

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

Chemical Reactions
A chemical reaction occurs when substances known as reactants transform into different substances called products.
This process involves a reorganization of atoms, and it is often represented by a chemical equation that shows what happens during the reaction and in what proportions.
In the case of the reaction between aqueous sodium carbonate (\(\text{Na}_2\text{CO}_3\)) and hydrochloric acid (HCl), the chemical equation is:
  • \(\text{Na}_2\text{CO}_3 + 2\text{HCl} → 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2\)
Here, sodium carbonate and hydrochloric acid combine to form sodium chloride, water, and carbon dioxide gas.
This chemical equation is balanced, meaning it shows that the number of each type of atom on the reactant side matches the number on the product side.
Balancing ensures the law of conservation of mass is upheld, indicating no atoms are lost or gained, merely rearranged.
Aqueous Solutions
In chemistry, an aqueous solution refers to a solution where water is the solvent.
In these solutions, the dissolved substances (solutes) are usually ions or molecules that are dispersed evenly throughout the water.
Aqueous solutions enable chemical reactions to occur more readily, particularly because ions in the solution can move freely and interact.
For instance, when sodium carbonate (\(\text{Na}_2\text{CO}_3\)) is mixed with hydrochloric acid (HCl) in water, they dissociate into ions:
  • \(\text{Na}_2\text{CO}_3\) dissociates to \(2\text{Na}^+\) and \(\text{CO}_3^{2-}\)
  • HCl dissociates to \(2\text{H}^+\) and \(2\text{Cl}^-\)
These ions are then free to interact, leading to the formation of new products.
The aqueous medium is crucial for this process, providing a platform for the ions to collide and react effectively.
Spectator Ions
Spectator ions are ions present in a chemical reaction that do not actively participate in it.
They appear unchanged on both sides of the chemical equation when considering the ionic form.
In the reaction between sodium carbonate and hydrochloric acid, spectator ions can be identified in the complete ionic equation:
  • \(2\text{Na}^+ + \text{CO}_3^{2-} + 2\text{H}^+ + 2\text{Cl}^- → 2\text{Na}^+ + 2\text{Cl}^- + \text{H}_2\text{O} + \text{CO}_2\)
Here, the sodium ions (\(\text{Na}^+\)) and the chloride ions (\(\text{Cl}^-\)) remain unchanged during the reaction.
They do not directly participate and are thus known as spectator ions.
Spectator ions can be removed to simplify the equation, focusing only on the ions that do change, which leads to the net ionic equation:
  • \(\text{CO}_3^{2-} + 2\text{H}^+ → \text{H}_2\text{O} + \text{CO}_2\)
The net ionic equation shows the essence of the chemical reaction, highlighting the components that actually change.

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Most popular questions from this chapter

Balance these equations for redox reactions occurring in basic solution. (a) \(\mathrm{CrO}_{4}^{2-}+\mathrm{S}_{2} \mathrm{O}_{4}^{2-} \longrightarrow \mathrm{Cr}(\mathrm{OH})_{3}(\mathrm{s})+\mathrm{SO}_{3}^{2-}\) (b) \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}+\mathrm{N}_{2} \mathrm{H}_{4} \longrightarrow\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}+\mathrm{N}_{2}(\mathrm{g})\) (c) \(\operatorname{Fe}(\mathrm{OH})_{2}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{Fe}(\mathrm{OH})_{3}(\mathrm{s})\) (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}+\mathrm{MnO}_{4}^{-} \longrightarrow\) \(\mathrm{CH}_{3} \mathrm{COO}^{-}+\mathrm{MnO}_{2}(\mathrm{s})\)

Select the (a) best and (b) poorest electrical conductors from the following solutions, and explain the reason for your choices: \(0.10 \mathrm{M} \mathrm{NH}_{3} ; 0.10 \mathrm{M} \mathrm{NaCl} ; 0.10 \mathrm{M}\) \(\mathrm{CH}_{3} \mathrm{COOH}\) (acetic acid); \(0.10 \mathrm{M} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) (ethanol).

A sample of battery acid is to be analyzed for its sulfuric acid content. A \(1.00 \mathrm{mL}\) sample weighs \(1.239 \mathrm{g}\). This \(1.00 \mathrm{mL}\) sample is diluted to \(250.0 \mathrm{mL}\), and \(10.00 \mathrm{mL}\) of this diluted acid requires \(32.44 \mathrm{mL}\) of \(0.00498 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) for its titration. What is the mass percent of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) in the battery acid? (Assume that complete ionization and neutralization of the \(\mathrm{H}_{2} \mathrm{SO}_{4}\) occurs.)

In this chapter, we described an acid as a substance capable of producing \(\mathrm{H}^{+}\) and a salt as the ionic compound formed by the neutralization of an acid by a base. Write ionic equations to show that sodium hydrogen sulfate has the characteristics of both a salt and an acid (sometimes called an acid salt).

Thiosulfate ion, \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\), is a reducing agent that can be oxidized to different products, depending on the strength of the oxidizing agent and other conditions. By adding \(\mathrm{H}^{+}, \mathrm{H}_{2} \mathrm{O},\) and/or \(\mathrm{OH}^{-}\) as necessary, write redox equations to show the oxidation of \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) to (a) \(\mathrm{S}_{4} \mathrm{O}_{6}^{2-}\) by \(\mathrm{I}_{2}\) (iodide ion is another product) (b) \(\mathrm{HSO}_{4}^{-}\) by \(\mathrm{Cl}_{2}\) (chloride ion is another product) (c) \(\mathrm{SO}_{4}^{2-}\) by \(\mathrm{OCl}^{-}\) in basic solution (chloride ion is another product)

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