Solutions, Acids, and Bases: Key Concepts and Calculations

Chapter 13: Solutions

Solute and Solvent Identification

In chemistry, a solution is a homogeneous mixture composed of two or more substances. The solute is the substance dissolved, and the solvent is the substance doing the dissolving.

• Vinegar: Solute: Acetic acid; Solvent: Water

• Soda water: Solute: Carbon dioxide; Solvent: Water

• Air: Solute: Various gases (e.g., oxygen, carbon dioxide); Solvent: Nitrogen

• Vodka (40% ethanol): Solute: Ethanol; Solvent: Water

Example: In saltwater, salt is the solute and water is the solvent.

Solubility and Dissolving Agents

Solubility depends on the nature of both solute and solvent. "Like dissolves like"—polar solvents dissolve polar solutes, and non-polar solvents dissolve non-polar solutes.

• Motor oil: Dissolves best in non-polar solvents (e.g., lard)

• KF (potassium fluoride): Dissolves best in polar solvents (e.g., water)

• Lard: Dissolves best in non-polar solvents

• Ethanol: Dissolves in both polar and non-polar solvents due to its amphipathic nature

Common Solvents

• Polar solvents: Water, ethanol, methanol

• Non-polar solvents: Hexane, benzene, carbon tetrachloride

Mass Percent Calculations

Mass percent is the mass of solute divided by the total mass of solution, multiplied by 100%.

• Formula: $\text{Mass percent} = \frac{\text{mass of solute}}{\text{mass of solution}} \times 100\%$

• Example: If 8.72 g ethanol is dissolved in 76.1 g water, total mass = 8.72 + 76.1 = 84.82 g. Mass percent = $\frac{8.72}{84.82} \times 100\% = 10.3\%$

Concentration and Solution Calculations

• Mass of solute in solution: For a 2.45% NaCl solution, grams of NaCl in 6.71 g solution = $6.71 \times \frac{2.45}{100} = 0.164$ g

• Mass of solution from % (m/m): For 12.0 g sucrose in a 4.1% solution: $\text{Mass of solution} = \frac{12.0}{0.041} = 293$ g

Mole and Molarity Calculations

• Moles of solute: $\text{Moles} = \text{Molarity} \times \text{Volume (L)}$

• Volume from moles and molarity: $\text{Volume (L)} = \frac{\text{Moles}}{\text{Molarity}}$

• Example: 3.4 L of 0.101 M NaI: $3.4 \times 0.101 = 0.343$ mol NaI

• To find volume for 0.325 mol NaI at 0.982 M: $\frac{0.325}{0.982} = 0.331$ L

Solution Table Completion

The table compares solute mass, moles, volume, and molarity.

Solute	Solute Mass	Mole solute	Volume of Solution	Molarity (M)
MgSO4	0.588	Additional info: Calculate using molar mass	25.0 mL	Additional info: Calculate using moles and volume
NaOH	Additional info: Calculate using molarity and volume	Additional info: Calculate using molar mass	100.0 mL	1.75 M

Dilution Calculations

• Formula: $M_1V_1 = M_2V_2$

• Example: Diluting 25.0 mL of 5.5 M KBr to 100.0 mL: $M_2 = \frac{M_1V_1}{V_2} = \frac{5.5 \times 25.0}{100.0} = 1.375$ M

Preparation of Solutions from Stock

• To prepare 500.0 mL of 0.200 M NaOH from 1.5 M stock: $V_1 = \frac{M_2V_2}{M_1} = \frac{0.200 \times 500.0}{1.5} = 66.7$ mL stock, dilute to 500.0 mL

Stoichiometry in Solution Reactions

• Use balanced equations to determine moles required for reactions

• Example: $K_2S(aq) + Co(NO_3)_2(aq) \rightarrow 2 KNO_3(aq) + CoS(s)$

• Calculate moles and volumes using molarity and stoichiometry

Freezing Point Depression

• Formula: $\Delta T_f = m \times K_f$

• Example: For 1.75 m solution, $\Delta T_f = 1.75 \times 1.86 = 3.26 ^\circ C$

Chapter 14: Acids and Bases

Properties of Acids and Bases

Acids are substances that donate protons (H+), while bases accept protons.

• Properties of acids: Sour taste, turn litmus red, react with metals

• Properties of bases: Bitter taste, slippery feel, turn litmus blue

Definitions of Acids and Bases

• Arrhenius definition: Acids produce H+ in water; bases produce OH-

• Bronsted-Lowry definition: Acids are proton donors; bases are proton acceptors

Conjugate Acid-Base Pairs

In acid-base reactions, the acid forms its conjugate base after donating a proton, and the base forms its conjugate acid after accepting a proton.

• Example: $NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-$

• Acid: $NH_4^+$, Base: $NH_3$, Conjugate acid: $NH_4^+$, Conjugate base: $OH^-$

Conjugate Bases of Acids

• Conjugate base is formed by removing a proton from the acid

• Example: $H_2O$ → $OH^-$; $HBr$ → $Br^-$

Strengths of Acids and Bases

Acids and bases are classified as strong or weak based on their degree of ionization in water.

Acid	Strength
HF	Weak
H2SO4	Strong
HNO3	Strong
HCl	Strong
HC2H3O2	Weak

Base	Strength
LiOH	Strong
NH4OH	Weak
NaOH	Strong
NH3	Weak
Ca(OH)2	Strong

Neutralization Reactions

• Acid and base react to form water and a salt

• Example: $H_3PO_4(aq) + 3NaOH(aq) \rightarrow Na_3PO_4(aq) + 3H_2O(l)$

• Use stoichiometry to calculate volumes required for neutralization

Additional info:

• Some calculations require knowledge of molar mass and solution preparation techniques

• Tables above are partially inferred based on standard chemistry curriculum