Enter the net ionic equation representing solid chromium (iii) hydroxide reacting with nitrous acid. express your answer as a ba

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Ronch

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Enter the net ionic equation representing solid chromium (iii) hydroxide reacting with nitrous acid. express your answer as a ba

lanced net ionic equation. identify all of the phases in your answer.

Chemistry

2 answers:

Anarel [2.9K] · Answer 1 · 2026-02-17T08:09:19+03:00

The net ionic equation for the reaction between chromium (III) hydroxide and nitrous acid is:

Cr(OH)₃ (s) + 3H⁺ (aq) ⇒ Cr³⁺ (aq) + 3H₂O (l)

Additional Details

An electrolyte dissociates into ions in solution.

Chemical equations can also be represented with ionic species.

Strong electrolytes (fully ionized) are written as separate ions, whereas weak electrolytes (partially ionized) remain as intact molecules.

In ionic equations, spectator ions are those unchanged by the chemical process—they are present both before and after the reaction.

Removing these spectators results in the net ionic equation.

Gases, solids, and water (H₂O) are written as molecules, without ionization.

Therefore, only dissolved compounds are represented by their ions (aq).

The problem involves chromium (III) hydroxide reacting with nitrous acid.

The reaction occurring is:

Cr(OH)₃ (s) + 3HNO₂ (aq) ⇒ Cr(NO₂)₃ (aq) + 3H₂O (l)

Chromium (III) hydroxide is a solid and remains un-ionized, as does water.

Thus, the ionic equation is:

Cr(OH)₃ (s) + 3H⁺ (aq) + 3NO₂⁻ (aq) ⇒ Cr³⁺ (aq) + 3NO₂⁻ (aq) + 3H₂O (l)

The ion 3NO₂⁻ is a spectator ion; removing it yields the net ionic equation:

Cr(OH)₃ (s) + 3H⁺ (aq) ⇒ Cr³⁺ (aq) + 3H₂O (l)

alisha [2.9K] · Answer 2 · 2026-02-17T08:07:36+03:00

The net ionic equation is $\boxed{{\text{Cr}}{{\left( {{\text{OH}}} \right)}_3}\left( s \right)+3{{\text{H}}^+}\left( {aq} \right)\to{\text{C}}{{\text{r}}^{3+}}\left( {aq}\right)+3{{\text{H}}_{\text{2}}}{\text{O}}\left(l\right)}$ .

Among the substances, ${\mathbf{Cr}}{\left( {{\mathbf{OH}}} \right)_{\mathbf{3}}}$ is in the solid state, ${{\mathbf{H}}_{\mathbf{2}}}{\mathbf{O}}$ is liquid, while ${{\mathbf{H}}^{\mathbf{ + }}}$ and ${\mathbf{C}}{{\mathbf{r}}^{{\mathbf{3 + }}}}$ exist in an aqueous solution.

Explanation:

Chemical reactions can be portrayed by three equations:

1. Molecular equation

2. Total ionic equation

3. Net ionic equation

The molecular equation shows reactants and products as unchanged compounds. The total ionic equation displays all dissociated ions present in the reaction mixture. The net ionic equation includes only the ions actively involved in the reaction.

Steps to derive the net ionic equation:

Step 1: Write the balanced molecular equation, indicating phases.

Here, ${\text{Cr}}{\left( {{\text{OH}}} \right)_3}$ reacts with ${\text{HN}}{{\text{O}}_2}$ producing ${\text{Cr}}{\left( {{\text{N}}{{\text{O}}_{\text{2}}}} \right)_3}$ and ${{\text{H}}_2}{\text{O}}$ . The balanced molecular equation is:

${\text{Cr}}{\left( {{\text{OH}}} \right)_3}\left( s \right) + {\text{3HN}}{{\text{O}}_2}\left( {aq} \right) \to {\text{Cr}}{\left( {{\text{N}}{{\text{O}}_2}} \right)_3}\left( {aq} \right) + {\text{3}}{{\text{H}}_2}{\text{O}}\left( l \right)$

Step 2: Break apart all aqueous compounds into ions to form the total ionic equation. Compounds in liquids or solids remain unchanged. Example total ionic equation:

${\text{Cr}}{\left( {{\text{OH}}} \right)_3}\left( s \right) + {\text{3}}{{\text{H}}^ + }\left( {aq} \right) + {\text{3NO}}_2^ - \left( {aq} \right) \to 3{\text{C}}{{\text{r}}^{3 + }}\left( {aq} \right) + {\text{3NO}}_2^ - \left( {aq} \right) + {\text{3}}{{\text{H}}_2}{\text{O}}\left( l \right)$

Step 3: Cancel out ions that appear unchanged on both sides to obtain the net ionic equation:

${\text{Cr}}{\left( {{\text{OH}}} \right)_3}\left( s \right) + {\text{3}}{{\text{H}}^ + }\left( {aq} \right) + \boxed{{\text{3NO}}_2^ - \left( {aq} \right)} \to 3{\text{C}}{{\text{r}}^{3 + }}\left( {aq} \right) + \boxed{{\text{3NO}}_2^ - \left( {aq} \right)} + {\text{3}}{{\text{H}}_2}{\text{O}}\left( l \right)$

Thus, the net ionic equation is:

${\text{Cr}}{\left( {{\text{OH}}} \right)_3}\left( s \right) + {\text{3}}{{\text{H}}^ + }\left( {aq} \right) \to 3{\text{C}}{{\text{r}}^{3 + }}\left( {aq} \right) + {\text{3}}{{\text{H}}_2}{\text{O}}\left( l \right)$

Note: ${\mathbf{Cr}}{\left( {{\mathbf{OH}}} \right)_{\mathbf{3}}}$ is solid, ${{\mathbf{H}}_{\mathbf{2}}}{\mathbf{O}}$ is liquid, and ${{\mathbf{H}}^{\mathbf{ + }}}$ and ${\mathbf{C}}{{\mathbf{r}}^{{\mathbf{3 + }}}}$ are aqueous.

Enter the net ionic equation representing solid chromium (iii) hydroxide reacting with nitrous acid. express your answer as a ba

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