Problem 17 - Chemistry and Chemical Reactivity

Question

Balance the equation for the following precipitation reaction, and then write the net ionic equation. Indicate the state of each species \((\mathrm{s}, \ell, \mathrm{aq}, \text { or } \mathrm{g})\). $$\mathrm{CdCl}_{2}+\mathrm{NaOH} \rightarrow \mathrm{Cd}(\mathrm{OH})_{2}+\mathrm{NaCl}$$

Step-by-Step Solution

Answer

The balanced equation is \(\mathrm{CdCl}_{2} (aq) + 2\mathrm{NaOH} (aq) \rightarrow \mathrm{Cd(OH)}_{2} (s) + 2\mathrm{NaCl} (aq)\). The net ionic equation is \(\mathrm{Cd^{2+}} (aq) + 2\mathrm{OH^{-}} (aq) \rightarrow \mathrm{Cd(OH)}_{2} (s)\).

Step 1: Write the Unbalanced Equation

Start by writing down the unbalanced chemical equation with each compound in its appropriate state: \[\mathrm{CdCl}_{2} (aq) + \mathrm{NaOH} (aq) \rightarrow \mathrm{Cd(OH)}_{2} (s) + \mathrm{NaCl} (aq)\] Here, \(\mathrm{CdCl}_{2}\) and \(\mathrm{NaOH}\) are aqueous, \(\mathrm{Cd(OH)}_{2}\) is solid, and \(\mathrm{NaCl}\) is aqueous.

Step 2: Balance the Equation

To balance the equation, ensure that the number of atoms of each element on the reactant side is equal to the number on the product side:Initial unbalanced equation:\[\mathrm{CdCl}_{2} (aq) + \mathrm{NaOH} (aq) \rightarrow \mathrm{Cd(OH)}_{2} (s) + \mathrm{NaCl} (aq)\]- There are 2 chlorine (Cl) atoms in \(\mathrm{CdCl}_{2}\), so you need 2 molecules of \(\mathrm{NaCl}\) on the product side. - Balance sodium (Na) by noting that you now need 2 \(\mathrm{NaOH}\) molecules.Balanced chemical equation:\[\mathrm{CdCl}_{2} (aq) + 2\mathrm{NaOH} (aq) \rightarrow \mathrm{Cd(OH)}_{2} (s) + 2\mathrm{NaCl} (aq)\]

Step 3: Write the Total Ionic Equation

Dissociate all strong electrolytes into their ions:\[\mathrm{Cd^{2+}} (aq) + 2\mathrm{Cl^{-}} (aq) + 2\mathrm{Na^{+}} (aq) + 2\mathrm{OH^{-}} (aq) \rightarrow \mathrm{Cd(OH)}_{2} (s) + 2\mathrm{Na^{+}} (aq) + 2\mathrm{Cl^{-}} (aq)\]

Step 4: Identify the Spectator Ions

Spectator ions appear on both the reactant and product sides without undergoing any change. These are:- \(\mathrm{Cl^{-}} (aq)\)- \(\mathrm{Na^{+}} (aq)\)Spectator ions are thus excluded from the net ionic equation.

Step 5: Write the Net Ionic Equation

Remove the spectator ions (\(\mathrm{Cl^{-}}\) and \(\mathrm{Na^{+}}\)) from the total ionic equation to write the net ionic equation:\[\mathrm{Cd^{2+}} (aq) + 2\mathrm{OH^{-}} (aq) \rightarrow \mathrm{Cd(OH)}_{2} (s)\] This equation shows the formation of the solid product, \(\mathrm{Cd(OH)}_{2}\).

Key Concepts

Net Ionic Equation

The net ionic equation distills a chemical reaction down to its essentials by removing the spectator ions, leaving only the ions and molecules directly involved in the formation of the product. This approach highlights the core chemical change occurring in the reaction.
To write a net ionic equation, follow these steps:

• First, depict all strong electrolytes and soluble ionic compounds in their dissociated ionic forms in the equation.
• Next, identify and remove the spectator ions, which are present on both sides of the equation without undergoing any transformation.
• What's left is the net ionic equation, representing the actual chemical process taking place.

In the example given about cadmium chloride and sodium hydroxide, the net ionic equation showcases the key interaction: \[\mathrm{Cd^{2+}} (aq) + 2\mathrm{OH^{-}} (aq) \rightarrow \mathrm{Cd(OH)}_{2} (s)\] This equation focuses on the formation of cadmium hydroxide as a solid product, filtering out the uninvolved ions.

Spectator Ions

Spectator ions are ions that exist in the same form on both the reactant and product sides of a chemical equation. Although they don't participate directly in the chemical change, they stabilize the overall reaction environment.
Understanding spectator ions is crucial because they are essentially unchanged during the reaction. Here's how to identify them:

• During precipitation reactions, like in the given example, spectator ions do not form any part of the new solid product.
• They appear as free ions in both the reactants and products.
• In our balanced precipitation reaction, \(\mathrm{Cl^{-}}\) and \(\mathrm{Na^{+}}\) were spectator ions.

These ions were present as free ions before and after the reaction, thus not represented in the net ionic equation. They underscore the importance of simplifying chemical equations to focus on the substances that undergo actual change.

Precipitation Reaction

A precipitation reaction is a type of chemical reaction in which two aqueous solutions react to form an insoluble solid known as a precipitate. This happens when the product formed is not soluble in water and thus forms a distinct solid. Here is what happens during a precipitation reaction:

• Two ionic compounds in solution swap components to form a new insoluble compound.
• The new compound falls out of the solution as a precipitate due to its low solubility.

In the cadmium chloride and sodium hydroxide reaction, the insoluble cadmium hydroxide, \(\mathrm{Cd(OH)}_{2}\), is the precipitate.
Understanding precipitation reactions is important for predicting the formation of new compounds, which can be essential in fields like water treatment, chemistry synthesis, and various industrial processes. Recognizing these reactions involves using solubility rules to predict whether a precipitate will form when two aqueous solutions are mixed.