Solution Formation: Intermolecular Forces, Solubility, Temperature, and Pressure

Solute-Solvent Interactions and Solution Formation

• General Principle: Solution formation is governed by intermolecular forces (IMFs) between solute and solvent particles. The principle of "like dissolves like" is foundational.

• Ionic Compound in Polar Solvent (e.g., Magnesium Chloride (MgCl$_2$) in Water)

  • MgCl$_2$: An ionic compound, dissociates into ions ($Mg^{2+}$ and $Cl^-$).

  • Water (H$_2$O): A polar molecule (dipolar).

  • Interaction Type: The most important solute-solvent interaction is ion-dipolar force, where the partially charged poles of water molecules attract the fully charged ions of magnesium chloride.

  • Lattice Energy: While lattice energy is important for understanding the stability of the ionic solid, the specific solute-solvent interaction in solution is ion-dipolar.

• Nonpolar Compound in Nonpolar Solvent (e.g., Hexane and Benzene)

  • Hexane and Benzene: Both are nonpolar organic compounds. They lack polar atoms like nitrogen or oxygen that would create significant dipole moments.

  • Interaction Type: The most important solute-solvent interaction is London Dispersion Forces, which are present between all molecules but are the primary force for nonpolar substances.

• Compound with Hydrogen Bonding in Polar Solvent (e.g., Sucrose in Water)

  • Sucrose (Sugar): A disaccharide composed of glucose and fructose units (both contain numerous -OH groups).

  • Water (H$_2$O): A polar molecule capable of hydrogen bonding.

  • Interaction Type: The most important solute-solvent interaction determining the magnitude of $\Delta H$ for solvent expansion is hydrogen bonding. Sucrose's many hydroxyl (-OH) groups can form strong hydrogen bonds with water molecules, facilitating dissolution.

Effect of Temperature on Solubility

• Solids (e.g., KNO$_3$, Sugar)

  • Relationship: As temperature increases, the solubility of most solid solutes in a liquid solvent typically increases.

  • Example (KNO$_3$): Looking at a solubility vs. temperature graph (e.g., grams of solute per $100$ grams of water):

    • At $20 \degree C$, approximately $38$ g of KNO$_3$ dissolves in $100$ g of water.

    • At $50 \degree C$, approximately $82$ g of KNO$_3$ dissolves in $100$ g of water.

  • Real-world Application (Glazed Doughnuts): To make glazed doughnuts, sugar is dissolved in water at high temperatures (e.g., boiling). More sugar can dissolve at higher temperatures, creating a supersaturated solution upon cooling. Dipping doughnuts into this cooled, concentrated sugar solution results in a thick glaze.

• Gases (e.g., Oxygen, Carbon Dioxide)

  • Relationship: As temperature increases, the solubility of gas solutes in a liquid solvent typically decreases.

  • Units: Gas solubility is often measured in molarity ($M$) (moles per liter) or millimolar (mM). (Distinction: Big M = Molarity; Small m = Molality (moles solute / kg solvent)).

  • Example (Oxygen in Water):

    • At $20 \degree C$, a certain amount of oxygen dissolves in water.

    • At $50 \degree C$, only about $1$ mM of oxygen dissolves. This significant decrease explains why boiled water tastes 'off' (due to less dissolved oxygen) and why fish cannot survive in boiled, cooled water (oxygen does not re-dissolve quickly).

  • Real-world Application (Soda): Warm soda loses its fizz because warmer temperatures reduce the solubility of carbon dioxide gas, causing it to escape the solution.

Effect of Pressure on Solubility

• Solids: Pressure has a negligible effect on the solubility of solids in liquids.

• Gases:

  • Relationship: As pressure increases, the solubility of gas solutes in a liquid solvent increases.

  • Explanation: Higher pressure above a liquid pushes more gas molecules into the liquid phase, increasing their concentration in the solution.

  • Real-world Application (Soda): Soda cans are pressurized to force a large amount of carbon dioxide gas into the beverage. When the can is opened, the pressure decreases, and the dissolved CO$_2$ escapes as bubbles.

Henry's Law: Quantifying Gas Solubility in Liquids

• Direct Proportionality: Henry's Law states that the solubility of a gas ($S<em>{gas}$) in a liquid is directly proportional to the partial pressure of that gas ($P</em>{gas}$) above the liquid.

  • $\text{Solubility of gas } \varpropto \text{ Partial pressure of gas}$ or $S<em>{gas} \varpropto P</em>{gas}$

• Henry's Law Equation:

  • $S<em>{gas} = K</em>H P_{gas}$ where:

    • $S_{gas}$ = Molar solubility of the gas (typically in mol/L or M).

    • $K_H$ = Henry's Law constant (specific to the gas, solvent, and temperature, typically in M/atm or mol/(L·atm)).

    • $P<em>{gas}$ = Partial pressure of the gas above the solution (typically in atm, but can be kPa, mmHg, etc. – unit must match $K</em>H$).

• Example Problem 1: Calculating Molar Concentration of a Gas

  • Problem: Calculate the molar concentration of carbon dioxide in water at $25 \degree C$ if the partial pressure of CO$<em>2$ is $0.56$ atm. Henry's Law constant for CO$</em>2$ in water at $25 \degree C$ is $3.1 \times 10^{-2}$ mol/(L·atm).

  • Given: $P<em>{CO</em>2} = 0.56 \text{ atm}$; $K_H = 3.1 \times 10^{-2} \text{ mol/(L·atm)}$

  • Solution:
    $S<em>{CO</em>2} = K<em>H P</em>{CO<em>2}$ $S</em>{CO<em>2} = (3.1 \times 10^{-2} \text{ mol/(L·atm)}) \times (0.56 \text{ atm})$ $S</em>{CO_2} = 1.74 \times 10^{-2} \text{ M}$

  • Key: Always ensure units cancel out correctly to yield the desired unit (e.g., M or mol/L). Be prepared for different pressure units (e.g., kilopascals) requiring conversion.

• Example Problem 2: Calculating Concentration Change with Pressure

  • Problem: Carbon dioxide in a drink under a pressure of $2.3$ atmospheres has a concentration of $0.071$ M. Calculate the concentration after the drink is opened and the partial pressure of CO$_2$ drops to $0.02$ atmospheres.

  • Conceptual Understanding: Since pressure decreases from $2.3$ atm to $0.02$ atm, the solubility (and thus concentration) of CO$_2$ is expected to decrease significantly.

  • Given:

    • Initial concentration ($S<em>1$) = $0.071 \text{ M}$ at initial pressure ($P</em>1$) = $2.3 \text{ atm}$.

    • Final pressure ($P_2$) = $0.02 \text{ atm}$.

  • Relationship: Since $S<em>{gas} = K</em>H P<em>{gas}$, then the ratio $S</em>{gas}/P<em>{gas}$ is constant ($= K</em>H$). Therefore, for two different conditions:
    $S<em>1 / P</em>1 = S<em>2 / P</em>2$
    Rearranging for $S<em>2$: $S</em>2 = S<em>1 \times (P</em>2 / P_1)$

  • Solution:
    $S<em>2 = (0.071 \text{ M}) \times (0.02 \text{ atm} / 2.3 \text{ atm})$ $S</em>2 \approx 0.0006 \text{ M}$

  • Key: Understanding the direct proportionality allows for ratio calculations when $K_H$ is not explicitly given, or when comparing two states of the same gas-liquid system.