Problem 2 - Chemistry: Principles and Reactions

Question

Write a net ionic equation for the reaction between aqueous solutions of (a) sodium acetate \(\left(\mathrm{NaC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)\) and nitric acid. (b) hydrobromic acid and strontium hydroxide. (c) hypochlorous acid and sodium cyanide. (d) sodium hydroxide and nitrous acid.

Step-by-Step Solution

Answer

Question: Write the net ionic equations for the following reactions between aqueous solutions: (a) Sodium acetate and nitric acid (b) Hydrobromic acid and strontium hydroxide (c) Hypochlorous acid and sodium cyanide (d) Sodium hydroxide and nitrous acid Answer: (a) \(\mathrm{H}^{+} \mathrm{(aq)} + \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-} \mathrm{(aq)} \rightarrow \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2} \mathrm{(aq)}\) (b) \(2 \mathrm{H}^{+} \mathrm{(aq)} + 2 \mathrm{OH}^{-} \mathrm{(aq)} \rightarrow 2 \mathrm{H}_{2}\mathrm{O} \mathrm{(l)}\) (c) \(\mathrm{HOCl} \mathrm{(aq)} + \mathrm{CN}^{-} \mathrm{(aq)} \rightarrow \mathrm{HCN} \mathrm{(aq)} + \mathrm{OCl}^{-} \mathrm{(aq)}\) (d) \(\mathrm{OH}^{-} \mathrm{(aq)} + \mathrm{HNO}_{2} \mathrm{(aq)} \rightarrow \mathrm{NO}_{2}^{-} \mathrm{(aq)} + \mathrm{H}_{2}\mathrm{O} \mathrm{(l)}\)

Step 1: Write the balanced molecular equation

The balanced molecular equation for sodium acetate and nitric acid is: \(\mathrm{NaC}_{2}\mathrm{H}_{3}\mathrm{O}_{2} \mathrm{(aq)} + \mathrm{HNO}_{3} \mathrm{(aq)} \rightarrow \mathrm{NaNO}_{3} \mathrm{(aq)} + \mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2} \mathrm{(aq)}\)

Step 2: Write the total ionic equation

Both nitric acid and sodium acetate will dissociate completely in water, so: \(\mathrm{Na}^{+} \mathrm{(aq)} + \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-} \mathrm{(aq)} + \mathrm{H}^{+} \mathrm{(aq)} + \mathrm{NO}_{3}^{-} \mathrm{(aq)} \rightarrow \mathrm{Na}^{+} \mathrm{(aq)} + \mathrm{NO}_{3}^{-} \mathrm{(aq)} + \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2} \mathrm{(aq)}\)

Step 3: Eliminate the spectator ions to find the net ionic equation

Sodium and nitrate ions are spectator ions, so the net ionic equation is: \(\mathrm{H}^{+} \mathrm{(aq)} + \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-} \mathrm{(aq)} \rightarrow \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2} \mathrm{(aq)}\) (b) Hydrobromic acid and strontium hydroxide

Step 1: Write the balanced molecular equation

The balanced molecular equation for hydrobromic acid and strontium hydroxide is: \(2 \mathrm{HBr} \mathrm{(aq)} + \mathrm{Sr}(\mathrm{OH})_{2} \mathrm{(aq)} \rightarrow \mathrm{SrBr}_{2} \mathrm{(aq)} + 2 \mathrm{H}_{2}\mathrm{O} \mathrm{(l)}\)

Step 2: Write the total ionic equation

Both hydrobromic acid and strontium hydroxide will dissociate completely in water: \(2 \mathrm{H}^{+} \mathrm{(aq)} + 2 \mathrm{Br}^{-} \mathrm{(aq)} + \mathrm{Sr}^{2+} \mathrm{(aq)} + 2 \mathrm{OH}^{-} \mathrm{(aq)} \rightarrow \mathrm{Sr}^{2+} \mathrm{(aq)} + 2 \mathrm{Br}^{-} \mathrm{(aq)} + 2 \mathrm{H}_{2}\mathrm{O} \mathrm{(l)}\)

Step 3: Eliminate the spectator ions to find the net ionic equation

The bromide ion is a spectator ion, and the net ionic equation is: \(2 \mathrm{H}^{+} \mathrm{(aq)} + 2 \mathrm{OH}^{-} \mathrm{(aq)} \rightarrow 2 \mathrm{H}_{2}\mathrm{O} \mathrm{(l)}\) (c) Hypochlorous acid and sodium cyanide

Step 1: Write the balanced molecular equation

The balanced molecular equation for hypochlorous acid and sodium cyanide is: \(\mathrm{HOCl} \mathrm{(aq)} + \mathrm{NaCN} \mathrm{(aq)} \rightarrow \mathrm{HCN} \mathrm{(aq)} + \mathrm{NaOCl} \mathrm{(aq)}\)

Step 2: Write the total ionic equation

Hypochlorous acid and hydrogen cyanide don't dissociate in water, but sodium cyanide does: \(\mathrm{HOCl} \mathrm{(aq)} + \mathrm{Na}^{+} \mathrm{(aq)} + \mathrm{CN}^{-} \mathrm{(aq)} \rightarrow \mathrm{HCN} \mathrm{(aq)} + \mathrm{Na}^{+} \mathrm{(aq)} + \mathrm{OCl}^{-} \mathrm{(aq)}\)

Step 3: Eliminate the spectator ions to find the net ionic equation

The sodium ion is a spectator ion, so the net ionic equation is: \(\mathrm{HOCl} \mathrm{(aq)} + \mathrm{CN}^{-} \mathrm{(aq)} \rightarrow \mathrm{HCN} \mathrm{(aq)} + \mathrm{OCl}^{-} \mathrm{(aq)}\) (d) Sodium hydroxide and nitrous acid

Step 1: Write the balanced molecular equation

The balanced molecular equation for sodium hydroxide and nitrous acid is: \(\mathrm{NaOH} \mathrm{(aq)} + \mathrm{HNO}_{2} \mathrm{(aq)} \rightarrow \mathrm{NaNO}_{2} \mathrm{(aq)} + \mathrm{H}_{2}\mathrm{O} \mathrm{(l)}\)

Step 2: Write the total ionic equation

Sodium hydroxide will dissociate completely in water, but nitrous acid won't: \(\mathrm{Na}^{+} \mathrm{(aq)} + \mathrm{OH}^{-} \mathrm{(aq)} + \mathrm{HNO}_{2} \mathrm{(aq)} \rightarrow \mathrm{Na}^{+} \mathrm{(aq)} + \mathrm{NO}_{2}^{-} \mathrm{(aq)} + \mathrm{H}_{2}\mathrm{O} \mathrm{(l)}\)

Step 3: Eliminate the spectator ions to find the net ionic equation

The sodium ion is a spectator ion, so the net ionic equation is: \(\mathrm{OH}^{-} \mathrm{(aq)} + \mathrm{HNO}_{2} \mathrm{(aq)} \rightarrow \mathrm{NO}_{2}^{-} \mathrm{(aq)} + \mathrm{H}_{2}\mathrm{O} \mathrm{(l)}\)

Key Concepts

Molecular Equation

A molecular equation offers a simple overview of a chemical reaction by showing all reactants and products as if they were intact compounds. This is the first step in understanding a reaction before breaking it down further. For example, when we look at the reaction between sodium acetate (\( \mathrm{NaC}_{2}\mathrm{H}_{3}\mathrm{O}_{2} \)) and nitric acid (\( \mathrm{HNO}_{3} \)), the molecular equation is:
\[\mathrm{NaC}_{2}\mathrm{H}_{3}\mathrm{O}_{2} \mathrm{(aq)} + \mathrm{HNO}_{3} \mathrm{(aq)} \rightarrow \mathrm{NaNO}_{3} \mathrm{(aq)} + \mathrm{HC}_{2}\mathrm{H}_{3} \mathrm{O}_{2} \mathrm{(aq)}\]This depicts all reactants and products but does not inform us about which ions participate in the reaction. To gain deeper insight, we need to proceed to the ionic equations.

Spectator Ions

Spectator ions are ions in a chemical reaction that do not take part in the formation of the reaction product. Their main "role" is to balance the charge in the reactant and product sides of a chemical equation. In the reaction of nitric acid and sodium acetate, sodium (\( \mathrm{Na}^{+} \)) and nitrate (\( \mathrm{NO}_{3}^{-} \)) ions are considered spectator ions. They appear on both side of the reaction without directly participating in the chemical change, thus allowing us to exclude them when writing the net ionic equation. This helps focus on the actual chemical change occurring.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products through various processes. In a typical reaction, bonds break and new bonds form, resulting in different substances. Understanding chemical reactions involves:

• Writing balanced molecular equations to see the overall change.
• Determining which ions or molecules contribute to the reaction.
• Identifying spectator ions to clarify which particles are actively involved in bonding changes.
  This structure helps in moving from a superficial understanding, represented by the molecular equation, to a granular view that excludes irrelevant ions.

Acid-Base Reactions

Acid-base reactions are a fundamental type of chemical reaction characterized by the exchange of protons (\(\mathrm{H}^{+}\) ions). When an acid reacts with a base, typically water and a salt result. A common example involves: hydrobromic acid (\( \mathrm{HBr} \)) reacting with strontium hydroxide (\( \mathrm{Sr(OH)}_{2} \)).
The net effect is:\[2 \mathrm{H}^{+} \mathrm{(aq)} + 2 \mathrm{OH}^{-} \mathrm{(aq)} \rightarrow 2 \mathrm{H}_{2}\mathrm{O} \mathrm{(l)}\]This net ionic equation shows the essence of the water formation process from acids and bases. Understanding this allows for predicting the outcome of numerous acid-base interactions, vital for grasping concepts like pH balance and buffer solutions.