Unit 5: Solutions, Concentration, and Colligative Properties Notes

Fundatmentals of Solutions

• Solution Components:
  • Solute: The substance that is dissolved in a solution. It is present in the smaller quantity.
  • Solvent: The dissolving medium, which is present in the larger quantity. Water is the most common solvent.
  • Rule of Dissolution: The solute dissolves IN the solvent to create a solution.
• Suspension: A mixture containing small particles that are not dissolved in the solution.
• Electrolyte: An aqueous solution capable of conducting electricity. This occurs when ionic compounds have been dissolved in a solvent.
• Dissolution Principle ("Like Dissolves Like"): For a solute to successfully dissolve in a solvent, they must be of the same chemical "type."
  • Polar solutes will dissolve in polar solvents.
  • Non-polar solutes will dissolve in non-polar solvents.

Solubility Dynamics and Classifications

• Factors to Increase Solubility (Dissolving More Solute):
  • Temperature ($T$): Increasing temperature increases kinetic energy ($KE$) and contact between particles, thereby increasing solubility.
  • Stirring (Agitation): Increasing mechanical movement increases the contact between the solute and solvent.
  • Solute Size (Surface Area): Decreasing the size of the solute particles (increasing surface area) increases the contact for dissolution.
• Levels of Saturation:
  • Saturated Solution: Contains the maximum amount of solute that can be dissolved in a given quantity of solvent at a specific temperature.
  • Supersaturated Solution: Contains more than the maximum amount of solute. This is typically achieved by heating the solution to dissolve more solute and then cooling it carefully.
  • Unsaturated Solution: Contains less than the maximum amount of solute possible for that solvent.
• Solubility and Pressure in Gases (Henry's Law):
  • The solubility of a gas is directly proportional to the pressure of the gas above the liquid.
  • Formula: $\frac{S_1}{P_1} = \frac{S_2}{P_2}$
  • Where $S$ represents solubility and $P$ represents the pressure of the gas.

Hydrates, Colloids, and Mixtures

• Hydrate: A chemical compound that contains water as part of its structure.
  • Example: Copper(II) sulfate pentahydrate ($CuSO_4 \times 5H_2O$).
• Anhydrate: A compound that does not contain water.
• Colloids: A heterogeneous mixture containing particles ranging in size from $1\text{ nm}$ to $1000\text{ nm}$. These particles are usually scattered throughout the solution but do not settle out.
• Tyndall Effect: The phenomenon where light is scattered as it passes through a colloid. The light beams become visible because they bounce off the suspended particles.

Concentration Measures: Molarity and Dilution

• Molarity ($M$): Measured in units of mole per liter ($\text{mol/L}$).
  • Formula: $M = \frac{\text{moles of solute}}{\text{L of solution}}$
  • Note: Solution volume includes both solute and solvent.
  • Unit Conversion: To convert milliliters ($mL$) to liters ($L$), divide by $1000$.
• Example Problem (Molarity): $45.0\text{ g}$ of $LiBrO_3$ is added to $125\text{ mL}$ of water. The final volume of the solution is $135\text{ mL}$. Calculate the molarity ($M$).
• Example Problem (Grams from Molarity): Calculate the number of grams of $BaS$ required for a $250.0\text{ mL}$ volume of a $1.45\text{ M}$ solution.
• Example Problem (Volume from Molarity): $35.0\text{ g}$ of $KOH$ is dissolved in water to form a $0.654\text{ M}$ solution. Determine the volume of the solution.
• Dilution: The process of reducing the concentration of a solution by adding more solvent.
  • The total number of moles of solute remains constant during dilution.
  • Formula: $M_1 V_1 = M_2 V_2$
  • Volume ($V$) can be expressed in $mL$ for this calculation.
• Example Problem (Dilution): How many $mL$ of a $1.45\text{ M}$ solution is needed to produce $300.0\text{ mL}$ of a $0.750\text{ M}$ solution?

Concentration Measures: Percentages and Fractions

• Percent by Volume ($V/V$):
  • Formula: $\%V/V = \frac{\text{volume of solute}}{\text{volume of solution}} \times 100$
  • Requirement: Both volumes must be in the same units (typically $mL$).
  • Example Problem: Calculate the volume percentage of a solution made by mixing $25.0\text{ mL}$ of concentrated solution with $350.0\text{ mL}$ of water.
• Percent by Mass/Volume ($\text{m/v}$):
  • Formula: $\% \text{by m/v} = \frac{\text{mass of solute (g)}}{\text{volume of solution (mL)}} \times 100$
  • Example Problem: Determine the mass by volume percentage in a solution made by mixing $25.0\text{ g}$ of $NaCl$ in $225\text{ g}$ of water, where the final solution volume is $230.0\text{ mL}$.
• Mass Percent (Mass $\%$):
  • Formula: $\text{Mass } \% \text{ of A} = \frac{\text{Mass of A}}{\text{Total mass}} \times 100$
  • Total Mass = $\text{mass of solute} + \text{mass of solvent}$.
  • Example Problem: $25.0\text{ g}$ of $LiOH$ is added to $100.0\text{ g}$ of water. Calculate the mass percent of the solvent.
• Mole Fraction ($X_a$): The ratio of the moles of one component to the total moles in the solution.
  • Formula: $X_a = \frac{\text{moles of component } a}{\text{total moles of solution}}$
  • Total moles calculated by adding moles of solute and moles of solvent.
  • Example Problem: Calculate the mole fraction of solute in a solution made of $15.0\text{ g}$ of $MgO$ mixed in $125\text{ g}$ of water.

Molality

• Molality ($m$): Measured in units of mole per kilogram ($\text{mol/kg}$).
  • Formula: $m = \frac{\text{moles of solute}}{\text{kg of solvent}}$
  • Unit Conversion: To convert grams ($g$) to kilograms ($kg$), divide by $1000$.
• Example Problem (Molality): Calculate the molality of a solution made by dissolving $45.0\text{ g}$ of $C_6H_{12}O_6$ in $126\text{ g}$ of hot water.

Comprehensive Calculation Practice

• Example 1: $55.0\text{ grams}$ of $Ni(NO_3)_2$ is added to $250\text{ grams}$ of water. Total volume is $255\text{ mL}$. Find:
  • Mole fraction of solute.
  • Mass percent of solvent.
  • Molarity ($M$) and Molality ($m$).
  • Mass/volume percentage.
• Example 2: $17.5\text{ g}$ of $Zn(OH)_2$ is added to $668\text{ g}$ of water. Final volume is $680.0\text{ mL}$. Find:
  • Mass percent of solute.
  • Mole fraction of solvent.
  • Mass/volume percentage.
  • Molarity ($M$).
  • Molality ($m$).

Colligative Properties

• Definition: Properties of a solution that depend solely on the amount (concentration) of solute particles, not the chemical identity of the solute.
• Vapor Pressure Lowering: Vapor pressure decreases with the addition of a solute (related to Henry's Law).
• Boiling Point Elevation ($\Delta T_b$):
  • The boiling point of a solvent increases when a solute is added.
  • Formula: $\Delta T_b = k_b m$
  • $\Delta = \text{change}$.
  • $k_b = \text{molal boiling point constant for the specific solvent}$.
  • New boiling point = \text{Normal boiling point} + \text{\Delta T_b}.
• Freezing Point Depression ($\Delta T_f$):
  • The freezing point of a solvent decreases when a solute is added.
  • Formula: $\Delta T_f = k_f m$
  • $k_f = \text{molal freezing point constant for the specific solvent}$.
  • New freezing point = \text{Normal freezing point} - \text{\Delta T_f}.

Solving for Colligative Properties

• Scenario 1: Find New Temperature ($T$) or $\Delta T$:
  1. Use solute and solvent data to calculate molality ($m = \frac{\text{mole solute}}{\text{kg solvent}}$).
  2. Use Molality ($m$) and the constant ($k$) to find $\Delta T = km$.
  3. Calculate the New Temperature.
• Scenario 2: Given New Temperature ($T$) or $\Delta T$:
  1. Use $\Delta T$ and the constant ($k$) to solve for molality ($m = \frac{\Delta T}{k}$).
  2. Use Molality to find the mass/moles of the solute or the mass of the solvent ($m = \frac{\text{mole solute}}{\text{kg solvent}}$).
• Example Problem (NaCl): $45.0\text{ g}$ of $NaCl$ added to $450\text{ g}$ water. Find new BP and FP. ($k_b = 0.520^\circ\text{C/m}$; $k_f = 1.86^\circ\text{C/m}$).
• Example Problem (Solvent Mass): Calculate grams of water needed with $55.0\text{ g}$ of $LiBr$ to raise the BP to $100.50^\circ\text{C}$. ($k_b = 0.520^\circ\text{C}$).
• Example Problem (Solute Moles): Calculate moles of solute added to $765\text{ g}$ water to lower the FP by $1.45^\circ\text{C}$. ($k_f = 1.86^\circ\text{C/m}$).
• Example Problem (Formula Mass): $32.5\text{ g}$ of unknown substance added to $455\text{ g}$ of ethanol ($C_2H_5OH$). Normal BP is $78.5^\circ\text{C}$; new BP is $85.6^\circ\text{C}$. Determine Formula Mass ($FM$). ($k_b \text{ for ethanol} = 1.19^\circ\text{C/m}$).
  • $FM = \frac{\text{grams}}{\text{moles}}$

Summary Review Material

• Chapter 13 Focus: Vocabulary, Phase Diagrams.
• Chapter 15 Focus: Vocabulary, Hydrates, Solubility Curves.
• Chapter 16 Focus: Vocabulary, Concentration Calculations ($M, m, X_a, \text{mass } \%, \text{mass/v } \%, \text{v/v } \%$), and Colligative Properties.