General Chemistry: Key Concepts and Problem-Solving Practice

Stoichiometry and Solution Chemistry

Electrolytes and Battery Chemistry

Electrolytes are substances that dissociate into ions in solution, enabling the conduction of electricity. In automotive batteries, sulfuric acid (H2SO4) is commonly used as an electrolyte. Calculating the amount of acid required involves stoichiometry and molarity concepts.

• Molarity (M): The concentration of a solution, defined as moles of solute per liter of solution.

• Stoichiometry: The quantitative relationship between reactants and products in a chemical reaction.

• Example: To find the volume of 3.0 M H2SO4 needed to supply 2.5 mol of battery acid:

  • Use $M = \frac{\text{mol}}{\text{L}}$ to solve for volume.

  • $\text{Volume} = \frac{2.5\ \text{mol}}{3.0\ \text{mol/L}} = 0.83\ \text{L}$

Spectator Ions in Chemical Reactions

In aqueous reactions, spectator ions are ions that do not participate in the actual chemical change but remain in solution unchanged.

• Example: In the reaction $\text{Pb(NO}_3)_2 (aq) + 2\text{NaI}(aq) \rightarrow 2\text{NaNO}_3(aq) + \text{PbI}_2(s)$, the spectator ions are Na+ and NO3-.

Calculating Molarity from Mass and Volume

Molarity is calculated by dividing the number of moles of solute by the volume of solution in liters.

• Formula: $M = \frac{\text{mol solute}}{\text{L solution}}$

• Example: Dissolving 16.65 g of NaCl (molar mass = 58.44 g/mol) in 0.925 L of water:

  • Moles NaCl: $\frac{16.65\ \text{g}}{58.44\ \text{g/mol}} = 0.285\ \text{mol}$

  • Molarity: $\frac{0.285\ \text{mol}}{0.925\ \text{L}} = 0.308\ \text{M}$

Physiological Saline Solution Preparation

Physiological saline is a 0.150 M NaCl solution, similar to the concentration of sodium chloride in human blood.

• Preparation: To prepare 275 mL (0.275 L) of 0.150 M NaCl:

  • Moles NaCl: $0.150\ \text{mol/L} \times 0.275\ \text{L} = 0.04125\ \text{mol}$

  • Mass NaCl: $0.04125\ \text{mol} \times 58.44\ \text{g/mol} = 2.41\ \text{g}$

Molecular Formula Determination

The molecular formula gives the actual number of atoms of each element in a molecule. It can be determined from the empirical formula and molar mass.

• Example: Terephthalic acid (empirical formula C4H3O2, molar mass between 150 and 167 g/mol):

  • Calculate empirical formula mass: $4 \times 12.01 + 3 \times 1.01 + 2 \times 16.00 = 83.06\ \text{g/mol}$

  • Divide molar mass by empirical formula mass: $\frac{166}{83.06} \approx 2$

  • Molecular formula: C8H6O4

Acid-Base Chemistry

pH and [H+] Calculations

pH is a measure of the hydrogen ion concentration in a solution. It is calculated using the formula:

• $\text{pH} = -\log[\text{H}^+]$

• To find [H+] from pH: $[\text{H}^+] = 10^{-\text{pH}}$

• Example: For pH = 5.63:

  • $[\text{H}^+] = 10^{-5.63} = 2.34 \times 10^{-6}\ \text{M}$

Acid-Base Reaction Classification

Chemical reactions can be classified as precipitation, acid-base, combination, or redox reactions.

• Acid-Base Reaction: Involves the transfer of protons (H+) between reactants.

• Example: $2\text{NaClO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{Cl}_2 + 2\text{H}_2O$ is an acid-base reaction.

Conjugate Base Identification

The conjugate base of an acid is formed when the acid donates a proton (H+).

• Example: The conjugate base of HSO4- is SO42-.

Precipitation and Ionic Equations

Precipitate Formation

A precipitate is an insoluble solid formed when two solutions are mixed and a product is insoluble in water.

• Example: Mixing Na2CO3 and BaCl2 forms BaCO3 as a precipitate.

Net Ionic Equations

Net ionic equations show only the species that actually participate in the reaction, omitting spectator ions.

• Example: For LiOH(aq) + HBr(aq):

  • Complete ionic: Li+(aq) + OH-(aq) + H+(aq) + Br-(aq) → Li+(aq) + Br-(aq) + H2O(l)

  • Net ionic: OH-(aq) + H+(aq) → H2O(l)

Stoichiometry of Chemical Reactions

Limiting Reactant and Mass Calculations

The limiting reactant is the reactant that is completely consumed first, limiting the amount of product formed.

• Example: For the reaction of Al and O2 to form Al2O3:

  • Calculate moles of each reactant.

  • Determine which reactant is limiting.

  • Calculate mass of product using stoichiometry.

Gas Reactions and Mole Calculations

Stoichiometry can be used to determine the amount of product formed from a given amount of reactant.

• Example: 17,400 mol of Cl2 reacts with excess SO2 to form SOCl2 and Cl2O:

  • Use balanced equation to find mole ratio.

  • Calculate moles of Cl2O formed.

Summary Table: Key Concepts and Examples

Concept	Definition	Example/Application
Molarity	Concentration of solute in solution (mol/L)	$M = \frac{\text{mol}}{\text{L}}$
Spectator Ion	Ions not involved in chemical change	Na+, NO3- in precipitation reactions
Precipitate	Insoluble solid formed in reaction	BaCO3 from Na2CO3 + BaCl2
pH	Measure of [H+] in solution	$\text{pH} = -\log[\text{H}^+]$
Limiting Reactant	Reactant consumed first in reaction	Al in Al + O2 → Al2O3
Conjugate Base	Species formed after acid donates H+	SO42- from HSO4-

Additional info: Some questions required inference of context and standard chemical procedures. All calculations and definitions are based on standard General Chemistry curriculum.