Question

Add the physical state for each substance in the following aqueous reactions. Then write and balance a net ionic equation for each reaction. (a) \(\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}+\mathrm{KOH} \longrightarrow \mathrm{Cr}(\mathrm{OH})_{3}+\mathrm{K}_{2} \mathrm{SO}_{4}\) (b) \(\mathrm{K}_{2} \mathrm{CrO}_{4}+\mathrm{PbCl}_{2} \longrightarrow \mathrm{KCl}+\mathrm{PbCrO}_{4}\) (c) \(\mathrm{Na}_{2} \mathrm{SO}_{3}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}+\mathrm{H}_{2} \mathrm{O}+\mathrm{SO}_{2}\)

Short answer

Balanced net ionic equations: \[2\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3(aq)} + 6\mathrm{OH^{-}}_{(aq)} \rightarrow 2\mathrm{Cr}(\mathrm{OH})_{3(s)} + 3\mathrm{SO}^{-2}_{4(aq)} \] \[ \mathrm{CrO}_{4^{-}2}}_{(aq)} +\mathrm{Pb^{2+}_{(aq)}} \rightarrow \mathrm{PbCrO}_{4(s)} \] \[ 2\mathrm{H}_{2} \mathrm{SO}_{4(aq)} +\mathrm{SO}_{3^{-2}}_{(aq)} \rightarrow 2\mathrm{SO}^{-2}_{4(aq)} + 2\mathrm{H}_{2}\mathrm{O}_{l} + 2\mathrm{SO}_{2(g)} \]

Step-by-step solution

Identify states and soluble compounds

For reaction (a) all the compounds involved are soluble except for \(\mathrm{Cr}(\mathrm{OH})_{3}\) which is not soluble in water and will therefore exist as a solid in the reaction. Reaction (b), has \(\mathrm{PbCrO}_{4}\) as the only compound that is not soluble in water thus it will be a solid. For reaction (c), \(\mathrm{SO}_{2}\) is a gas, and everything else is in aqueous solution.

Write net ionic equations

By removing the spectator ions from the total ionic equations we get the net ionic equations: \[\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3(aq)}+ 6\mathrm{OH^{-}}_{(aq)}\rightarrow 2\mathrm{Cr}(\mathrm{OH})_{3(s)}+ 3\mathrm{SO}^{-2}_{4(aq)} \] \[\mathrm{CrO}_{4^{-}2}}_{(aq)}+\mathrm{Pb^{2+}_{(aq)}} \rightarrow\mathrm{PbCrO}_{4(s)} \] \[\mathrm{H}_{2} \mathrm{SO}_{4(aq)} +\mathrm{SO}_{3^{-2}}_{(aq)} \rightarrow \mathrm{SO}^{-2}_{4(aq)} + \mathrm{H}_{2}\mathrm{O}_{l} + \mathrm{SO}_{2(g)} \]

Balance the net ionic equations

Balance the net ionic equations by making sure there is an equal number of atoms of each element on both sides of the equation: \[2\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3(aq)} + 6\mathrm{OH^{-}}_{(aq)}\rightarrow 2\mathrm{Cr}(\mathrm{OH})_{3(s)} + 3\mathrm{SO}^{-2}_{4(aq)} \] \[\mathrm{CrO}_{4^{-}2}}_{(aq)} +\mathrm{Pb^{2+}_{(aq)}} \rightarrow\mathrm{PbCrO}_{4(s)} \] \[2\mathrm{H}_{2} \mathrm{SO}_{4(aq)} +\mathrm{SO}_{3^{-2}}_{(aq)} \rightarrow 2\mathrm{SO}^{-2}_{4(aq)} + 2\mathrm{H}_{2}\mathrm{O}_{l} + 2\mathrm{SO}_{2(g)} \]

Key concepts

Aqueous Reactions

Aqueous reactions are fascinating because they occur when substances dissolve in water, forming solutions where the reactants and products can move freely. In these reactions, water acts as a medium that facilitates the movement of ions, allowing them to collide and react with each other. This explains why reactions in aqueous solutions can occur rapidly.
When considering aqueous reactions, it's important to identify the physical states of the substances involved, as this can affect the reaction pathway. For example, in the given reactions:

• In reaction (a), both chromium(III) sulfate \(\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) and potassium hydroxide \(\mathrm{KOH}\) are soluble and present in aqueous form, except chromium(III) hydroxide \(\mathrm{Cr}(\mathrm{OH})_{3}\), which precipitates as a solid.
• In reaction (b), lead(II) chromate \(\mathrm{PbCrO}_{4}\) is the solid formed, indicating that it is not soluble in water.
• Reaction (c) involves both aqueous and gaseous products, with sulfur dioxide \(\mathrm{SO}_{2}\) being the only gas.

Recognizing these states helps us predict how the substances interact within the solution.

Precipitation Reactions

Precipitation reactions occur when two aqueous solutions are mixed, and an insoluble solid, called a precipitate, forms. This is a type of double displacement reaction, where ions exchange partners to form new compounds. The formation of a precipitate is often indicated by the appearance of a solid in the otherwise clear solution.
In the context of the provided reactions:

• In reaction (a), when \(\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) and \(\mathrm{KOH}\) are combined, chromium(III) hydroxide \(\mathrm{Cr}(\mathrm{OH})_{3}\) forms as a solid precipitate.
• In reaction (b), the mixing of \(\mathrm{K}_{2} \mathrm{CrO}_{4}\) and \(\mathrm{PbCl}_{2}\) results in the precipitation of lead(II) chromate \(\mathrm{PbCrO}_{4}\).

To determine if a precipitation reaction occurs, it's vital to know the solubility rules of the substances involved. By predicting which new combinations of ions are insoluble, you can foresee the formation of a precipitate.

Balancing Chemical Equations

Balancing chemical equations is crucial for correctly representing chemical reactions. It ensures that the law of conservation of mass is followed, meaning the same number of each type of atom is present on the reactant and product sides of the equation. Balancing equations helps us find the correct proportions and relationships between different substances involved.
Here's how you approach balancing:

• Begin by writing the unbalanced equation.
• Identify the number of atoms of each element present in both reactants and products.
• Adjust the coefficients (numbers placed in front of compounds) to ensure an equal number of atoms for each element.
• Check your work to confirm that both sides are balanced.

For example, in the balanced net ionic equations provided: - Each reaction ensures that the number of each type of ion and molecule is the same on both sides. - This includes complex ions like \(\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) and \(\mathrm{OH^{-}}\).Properly balancing these equations reveals the stoichiometry of the reaction, guiding how much of each reactant is needed and how much product will be produced.