Pre-AP Chemistry Unit 4 Study Notes: Chemical Reactions and Stoichiometry

Definition of Hard Water: Hard water is characterized by containing high concentrations of specific aqueous ions, primarily calcium ( $Ca^{2+}$ ) and magnesium ( $Mg^{2+}$ ).
Precipitation in Pipes: As hard water flows through plumbing systems, compounds containing these ions can undergo a chemical reaction where they precipitate (form a solid) and build up on the internal surfaces of the pipes. This buildup is often referred to as scale.
Reaction Identification: A precipitation reaction in this context is identified by the formation of an insoluble solid from aqueous reactants. A typical reaction representing this process involves the combination of these metallic cations with anions like carbonate ( $CO_3^{2-}$ ) or phosphate ( $PO_4^{3-}$ ) to form a solid crust.

Molarity and Concentration: Concentration is often measured in molarity ( $M$ ), defined as the number of moles of solute per cubic decimeter ( $dm^3$ ) of solution ( $mol\,dm^{-3}$ ).
Particle Diagrams: Particle-level diagrams are used to model the relative number of ions or molecules in a solution.
- For a $1.0\,M$ solution of copper(II) chloride ( $CuCl_2$ ), the diagram represents a specific ratio of ions ( $Cu^{2+}$ and $Cl^-$ ) in a given volume.
- A $2.0\,M$ solution of the same substance, assuming equal volume, must be represented by exactly double the number of particles compared to the $1.0\,M$ solution to reflect the doubled concentration.

Composition of Vinegar: Vinegar is an aqueous solution of acetic acid ( $CH_3COOH$ ).
Quantitative Data for Vinegar:
- Concentration of acetic acid ( $CH_3COOH$ ): $0.80\,M$ ( $0.80\,mol\,dm^{-3}$ ).
- Concentration of hydrogen ions ( $H^+$ ): $4 \times 10^{-3}\,M$ ( $4 \times 10^{-3}\,mol\,dm^{-3}$ ).
- Sample Volume: $500\,cm^3$ ( $0.500\,dm^3$ ).
Calculation of Moles of Acetic Acid:
- Formula: $\text{moles} = \text{concentration} \times \text{volume}$
- Calculation: $0.80\,mol\,dm^{-3} \times 0.500\,dm^3 = 0.40\,mol$ of $CH_3COOH$ .

Experimental Scenario: A student combines $125\,cm^3$ of $1.0\,M$ sodium phosphate ( $Na_3PO_4$ ) with an excess of calcium chloride ( $CaCl_2$ ).
Reaction Product: The reaction goes to completion to form a precipitate of calcium phosphate ( $Ca_3(PO_4)_2$ ).
Quantitative Constants:
- Molar mass of calcium phosphate ( $Ca_3(PO_4)_2$ ): $310\,g\,mol^{-1}$ .
Process of Recovery: The resulting mixture is filtered to isolate the solid precipitate from the aqueous liquid, and the precipitate is subsequently dried to determine its mass.

Drain Cleaner Composition: Some drain cleaners contain sodium hydroxide ( $NaOH$ ) and small particles of metallic aluminum ( $Al$ ).
Chemical Reaction: The reaction can be examined to determine the transfer of electrons and changes in oxidation states.
Oxidation State Analysis:
- Sodium ( $Na$ ): In many reactions of this type, sodium acts as a spectator ion or remains in the $+1$ oxidation state.
- Oxygen ( $O$ ): Typically maintains an oxidation number of $-2$ in these compounds.
- Hydrogen ( $H$ ): Oxidation state changes if gas ( $H_2$ ) is produced.
- Aluminum ( $Al$ ): Metallic aluminum ( $Al^0$ ) is oxidized as it loses three electrons to form the aluminum ion ( $Al^{3+}$ ) in a complex or salt.
Magnesium Bromide Formation: The reaction between magnesium ( $Mg$ ) and bromine ( $Br_2$ ) to form magnesium bromide ( $MgBr_2$ ) serves as a fundamental model for redox:
- $Mg$ is oxidized because it loses two electrons to form $Mg^{2+}$ .
- Each bromine atom in $Br_2$ is reduced because it gains one electron to form $Br^-$ .

Experimental Observations: Samples of four metals with equal mass were placed in $1.0\,M$ hydrochloric acid ( $HCl$ ). The observations were as follows:
- Calcium ( $Ca$ ): Reacted in less than $10\,s$ ; vigorous gas production. (Most reactive).
- Magnesium ( $Mg$ ): Reacted in about $3\,min$ ; moderate gas production. (Highly reactive).
- Iron ( $Fe$ ): Reacted very slowly; slow gas production; did not totally react. (Low reactivity).
- Silver ( $Ag$ ): No visible reaction. (Least reactive).
Reactivity Ranking (Least to Most): Ag < Fe < Mg < Ca.

Strong Acid Models: A strong acid in solution is modeled as being completely dissociated into its constituent ions. In a particle diagram, this is shown by the absence of neutral molecules ( $HA$ ) and the presence of individual $H^+$ and $A^-$ ions.
pH Calculations: The pH of a solution is determined by the concentration of hydrogen ions ( $[H^+]$ ).
- Scenario: A stomach acid sample has an $H^+$ concentration of $1.0 \times 10^{-2}\,M$ .
- Formula: $pH = -\log[H^+]$
- Calculation: $pH = -\log(1.0 \times 10^{-2}) = 2$ .
Strong Acid-Strong Base Reactions: These reactions involve a strong acid (like $HCl$ ) reacting with a strong base (like $NaOH$ ) to produce water ( $H_2O$ ) and a salt.
- Example Net Ionic Equation: $H^+(aq) + OH^-(aq) \rightarrow H_2O(l)$ .

Solution Preparation: A student creates $125\,cm^3$ ( $0.125\,dm^3$ ) of a $1.0\,M$ aluminum chloride ( $AlCl_3$ ) solution.
Balanced Molecular Equation: The creation of a precipitate (aluminum hydroxide) from aluminum chloride and sodium hydroxide is represented by:
- $AlCl_3(aq) + 3NaOH(aq) \rightarrow Al(OH)_3(s) + 3NaCl(aq)$
Net Ionic Equation: Focuses only on the species forming the precipitate:
- $Al^{3+}(aq) + 3OH^-(aq) \rightarrow Al(OH)_3(s)$
Moles of Aluminum Ions:
- Calculation: $0.125\,dm^3 \times 1.0\,mol\,dm^{-3} = 0.125\,mol$ of $Al^{3+}$ .
Mass of Precipitate Calculation:
- Based on the stoichiometry ( $1:1$ ratio between $Al^{3+}$ and $Al(OH)_3$ ), $0.125\,mol$ of $Al(OH)_3$ is formed.
- Molar mass of $Al(OH)_3$ : $26.98 + 3(16.00 + 1.01) = 78.01\,g\,mol^{-1}$ .
- Maximum mass: $0.125\,mol \times 78.01\,g\,mol^{-1} = 9.75\,g$ .

Reaction 1: $CH_3COOH(aq) + NaOH(aq) \rightarrow NaCH_3COO(aq) + H_2O(l)$
- Reaction Type: Acid-Base reaction.
- Justification: This is a neutralization reaction where a proton ( $H^+$ ) is transferred from the acetic acid (the acid) to the hydroxide ion (the base) to form liquid water.
Reaction 2: $Fe(s) + CuCl_2(aq) \rightarrow Cu(s) + FeCl_2(aq)$
- Reaction Type: Oxidation-Reduction (Redox) reaction.
- Justification: This is a single replacement reaction where the oxidation state of iron changes from $0$ to $+2$ (oxidation) and the oxidation state of copper changes from $+2$ to $0$ (reduction) via the transfer of electrons.