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

Short Answer

Expert verified
\(\mathrm{Cd^{2+} (aq) + 2 OH^- (aq) \rightarrow Cd(OH)_2 (s)}\)

Step by step solution

01

Identify the Compounds and Assign States

Before balancing the equation, identify the compounds involved and assign their states based on solubility rules and common sense. Here, CdCl\(_2\) and NaOH are aqueous (aq) because they are soluble in water. Cd(OH)\(_2\) is a precipitate and therefore solid (s). NaCl is also soluble, so it is aqueous (aq). The equation now looks like this: \[\mathrm{CdCl_2 (aq) + NaOH (aq) \rightarrow Cd(OH)_2 (s) + NaCl (aq)}\].
02

Balance the Chemical Equation

To balance the equation, start by balancing the number of atoms for each element on the reactant and product sides. Currently, the equation has: 1 Cd, 2 Cl, 1 Na, and 1 OH on the left and 1 Cd, 1 Cl, 2 OH, and 1 Na on the right. Adjust to balance: \[\mathrm{CdCl_2 (aq) + 2 NaOH (aq) \rightarrow Cd(OH)_2 (s) + 2 NaCl (aq)}\].
03

Write the Complete Ionic Equation

Break down all aqueous compounds into their ions, since they dissociate in water. The complete ionic equation is: \[\mathrm{Cd^{2+} (aq) + 2 Cl^- (aq) + 2 Na^+ (aq) + 2 OH^- (aq) \rightarrow Cd(OH)_2 (s) + 2 Na^+ (aq) + 2 Cl^- (aq)}\].
04

Write the Net Ionic Equation

Identify and remove the spectator ions, which appear on both sides of the equation without undergoing change. In this case, Na\(^+\) and Cl\(^-\) are spectator ions. Removing them, the net ionic equation is: \[\mathrm{Cd^{2+} (aq) + 2 OH^- (aq) \rightarrow Cd(OH)_2 (s)}\].

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Key Concepts

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

Precipitation Reaction
A precipitation reaction is a type of chemical reaction where two soluble salts in aqueous solution combine and form an insoluble solid, known as a precipitate. In this reaction, the solid forms because the resulting compound has very low solubility in water. For example, when solutions of cadmium chloride (CdCl\(_2\)) and sodium hydroxide (NaOH) are mixed, cadmium hydroxide (Cd(OH)\(_2\)) forms as a precipitate. This reaction can be generally represented as:
  • Soluble salt A + Soluble salt B → Precipitate + Soluble salt C
Recognizing a precipitation reaction can often be done using solubility rules, which predict the formation of a solid from the mixing of various aqueous solutions.
Net Ionic Equation
A net ionic equation simplifies a chemical equation by showing only the species that actually change during the reaction. It removes the spectator ions, which do not participate in the actual chemical change and remain unchanged in solution. This is useful for focusing on the core chemical changes without unnecessary details. In our example of the reaction between CdCl\(_2\) and NaOH, the net ionic equation eliminates sodium (Na\(^+\)) and chloride (Cl\(^-\)) ions, which are spectator ions. Thus, the net ionic equation for the formation of the precipitate Cd(OH)\(_2\) is:
  • Cd\(^{2+}\) (aq) + 2 OH\(^-\) (aq) → Cd(OH)\(_2\) (s)
This highlights the formation of the solid product by the direct combination of cadmium ions and hydroxide ions.
Balancing Chemical Equations
Balancing chemical equations is crucial for representing the law of conservation of mass. This law states that matter cannot be created or destroyed in a chemical reaction. As such, each side of the equation must have the same number of each type of atom. The balancing process involves:
  • Listing the different elements involved in the reaction.
  • Counting the number of atoms of each element in the reactants and products.
  • Adjusting coefficients (the numbers before molecules) to balance the atoms on both sides.
For the reaction between CdCl\(_2\) and NaOH, we balanced the equation by ensuring each element appeared in equal amounts on both sides: \[\mathrm{CdCl_2 (aq) + 2 NaOH (aq) \rightarrow Cd(OH)_2 (s) + 2 NaCl (aq)}\] Each step ensures that the chemically equivalent ingredients carry through the reaction without discrepancies.
Solubility Rules
Solubility rules are guidelines that help predict whether a compound will dissolve in water. These rules are essential for identifying whether a reaction involves a precipitate. Some key solubility rules include:
  • Most nitrate (NO\(_3\)^-) salts are soluble.
  • Sodium, potassium, and ammonium salts are often soluble.
  • Chlorides are generally soluble, except those formed with silver, lead, and mercury.
  • Hydroxides are generally insoluble, except those of alkali metals and barium.
These rules help in determining the states of compounds in a chemical reaction, like in our exercise where we identified cadmium hydroxide (Cd(OH)\(_2\)) as the insoluble precipitate based on its very low solubility in water.

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Most popular questions from this chapter

Balance equations for these reactions that occur in aqueous solution, and then classify each as a precipitation, acid-base, or gas-forming reaction. Show states for the products \((s, \ell, g, a q),\) give their names, and write the net ionic equation. (a) \(\mathrm{K}_{2} \mathrm{CO}_{3}+\mathrm{HClO}_{4} \rightarrow \mathrm{KClO}_{4}+\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{FeCl}_{2}+\left(\mathrm{NH}_{4}\right)_{2} \mathrm{S} \rightarrow \mathrm{FeS}+\mathrm{NH}_{4} \mathrm{Cl}\) (c) \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{2}+\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{aq}) \rightarrow \mathrm{FeCO}_{3}+\mathrm{NaNO}_{3}\) (d) \(\mathrm{NaOH}+\mathrm{FeCl}_{3} \rightarrow \mathrm{NaCl}+\mathrm{Fe}(\mathrm{OH})_{3}\)

Write a balanced equation for the ionization of nitric acid in water.

Write an equation that describes the equilibrium that exists when the weak acid benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}\right)\) dissolves in water. Identify each of the four species in solution as either Bronsted acids or Bronsted bases. Does the equilibrium favor the products or the reactants? (In acting as an acid, the \(\left.-\mathrm{CO}_{2} \mathrm{H} \text { group supplies } \mathrm{H}^{+} \text {to form } \mathrm{H}_{3} \mathrm{O}^{+} .\right)\)

Determine the oxidation number of each element in the following ions or compounds. (a) \(\mathrm{PF}_{6}\) (b) \(\mathrm{H}_{2} \mathrm{AsO}_{4}^{-}\) (c) \(\mathrm{UO}^{2+}\) (d) \(\mathrm{N}_{2} \mathrm{O}_{5}\) (e) \(\mathrm{POCl}_{3}\) (f) \(\mathrm{XeO}_{4}^{2-}\)

Gas evolution was observed when a solution of \(\mathrm{Na}_{2} \mathrm{S}\) was treated with acid. The gas was bubbled into a solution containing \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2},\) and a black precipitate formed. Write net ionic equations for the two reactions.

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