Chapter 11: Solutions and Colloids

Fundamental Definitions of Solutions and Components

• Solution: A homogeneous mixture consisting of two or more substances.

• Solvent: The component of a solution present in the highest concentration.

• Solute: The component of a solution present in a lower concentration compared to the solvent.

• Aqueous Solution: A solution in which water acts as the solvent.

• Example (Hummingbird Food): Sugar water.

  • Solvent: Water ($H_2O$).

  • Solute: Sugar.

The Process of Dissolution and Molecular Interactions

• Molecular Interactions Involved:

  • Solute-Solute Interactions: Attractive forces existing between solute particles that must be overcome for dissolution to occur.

  • Solute-Solvent Interactions: Attractive forces between solute particles and solvent particles.

• Condition for Dissolving:

  • A solute will dissolve and mix with solvent particles if the attractions between the solute and solvent are strong enough.

  • Rule of Thumb: Dissolution occurs when \text{Solvent-Solute attraction} > \text{Solute-Solute attraction}.

Electrolytes and Classification of Solutes

• Strong Electrolyte:

  • Definition: A substance that dissociates completely ($100\%$) into ions when dissolved in water.

  • Properties: Conducts electricity effectively.

  • Examples: Ionic compounds, strong acids, and strong bases.

  • Dissociation Reaction Example: $NaCl(s) \rightarrow Na^+(aq) + Cl^-(aq)$.

• Weak Electrolyte:

  • Definition: A substance that only partially dissociates into ions. The solution contains a mixture of both ions and intact molecules.

  • Properties: Conducts electricity weakly.

  • Examples: Weak acids (e.g., $HF$) and weak bases.

  • Dissociation Reaction Example: $HF(aq) \rightleftharpoons H^+(aq) + F^-(aq)$.

• Nonelectrolyte:

  • Definition: A substance that does not dissociate into ions and remains as intact molecules in solution.

  • Properties: Does not conduct electricity.

  • Examples: Covalent compounds, sugar ($C_{12}H_{22}O_{11}$), and large proteins.

  • Dissociation Reaction Example: $C_{12}H_{22}O_{11}(s) \rightarrow C_{12}H_{22}O_{11}(aq)$.

Solubility and Saturation States

• Solubility: The maximum concentration of a solute that can be achieved in a specific solvent under given conditions when the dissolution process is at dynamic equilibrium.

• Dynamic Equilibrium: A state where the $\text{Rate of dissolution} = \text{Rate of recrystallization}$.

  • If \text{Rate of dissolution} > \text{Rate of recrystallization}, the process is in the "Dissolution only" phase.

• Types of Solutions:

  • Saturated Solution: Contains the maximum amount of solute that can dissolve in the solvent at a specific temperature. Any additional solute added will not dissolve.

  • Unsaturated Solution: A solution containing a concentration of solute that is below its solubility limit.

  • Supersaturated Solution: A solution containing a concentration of solute that exceeds its theoretical solubility limit.

• Solubility Scales:

  • High Solubility: Sodium Chloride ($NaCl$) has a solubility of $360\,g/L$ at $25\,^oC$.

  • Low Solubility: Calcium Carbonate ($CaCO_3$) has a solubility of $0.013\,g/L$ at $25\,^oC$.

Factors Influencing Solubility

Temperature and Solids

• General Rule: In most cases (though not always), raising the temperature increases the solubility of solid solutes.

• Exceptions: Some substances, such as $Ce_2(SO_4)_3$, show a decrease in solubility as temperature increases.

• Using Solubility Curves:

  • On the curve: Saturated solution.

  • Below the curve: Unsaturated solution.

  • Above the curve: Supersaturated solution.

Temperature and Gases

• General Rule: The solubility of a gas in water decreases as temperature increases.

• Mechanical Explanation: Higher temperatures provide higher average kinetic energy to dissolved gas particles, allowing them to overcome intermolecular forces and enter the gas phase more easily.

Pressure and Gases

• General Rule: Increasing the pressure of a gas over a liquid increases the gas's solubility in that liquid.

• Mechanical Explanation: Increased pressure causes more gas molecules to collide with the liquid surface, increasing the probability of dissolution.

• Example: Carbon dioxide ($CO_2$) being forced into soda under pressure; bubbles escape when the pressure is released.

Units of Concentration

• Molarity (M):     $M = \frac{\text{amount solute (mol)}}{\text{volume solution (L)}}$

• Molality (m):     $m = \frac{\text{amount solute (mol)}}{\text{mass solvent (kg)}}$

• Mole Fraction ($\chi$)) of Solute:     $\chi_{solute} = \frac{n_{solute}}{n_{solute} + n_{solvent}}$

Mathematical Examples of Concentration

Calculating Molarity, Molality, and Mole Fraction

• Scenario: $17.2\,g$ of ethylene glycol ($C_2H_6O_2$, molar mass $62.07\,g/mol$) dissolved in $500\,g$ of water ($H_2O$). Final solution volume is $515\,mL$.

  • Moles of Solute: $17.2\,g / 62.07\,g/mol = 0.277\,mol$.

  • Molarity: $0.277\,mol / 0.515\,L = 0.538\,M$.

  • Molality: $0.277\,mol / 0.500\,kg = 0.554\,m$.

  • Mole Fraction:

    1. Moles $H_2O = 500\,g / 18.016\,g/mol = 27.75\,mol$.

    2. $\chi = 0.277 / (0.277 + 27.75) = 0.00989$.

Converting Molality to Mole Fraction

• Scenario: A $3.0\,m$ solution of sodium chloride ($NaCl$).

  1. Step 1: Interpret molality as $3.0\,mol$ $NaCl$ in $1.0\,kg$ $H_2O$.

  2. Step 2: Convert solvent mass to moles. $1.0\,kg \times (1000\,g / 1\,kg) \times (1\,mol / 18.016\,g) = 55\,mol\,H_2O$.

  3. Step 3: $\chi_{NaCl} = \frac{3.0\,mol}{3.0\,mol + 55\,mol} = 0.052$.

Colligative Properties and the van 't Hoff Factor

• Colligative Properties: Physical properties of solutions that depend solely on the concentration of solute particles (molecules or ions), regardless of their chemical identity.

• van 't Hoff Factor (i): The ratio of moles of particles in solution to the moles of formula units dissolved.

  • Nonelectrolytes: $i = 1$ (does not dissociate).

  • Electrolytes: i > 1.

  • Example $NaCl$: Dissociates into $1\,Na^+$ and $1\,Cl^-$; therefore, $i = 2$.

  • Example $Ca(NO_3)_2$: Dissociates into $1\,Ca^{2+}$ and $2\,NO_3^-$; therefore, $i = 3$.

• Note on Polyatomic Ions: These ions do not split further (e.g., $SO_4^{2-}$, $NO_3^-$, $NH_4^+$, $PO_4^{3-}$ stay intact).

Vapor Pressure Lowering

• Vapor Pressure: The pressure exerted by the vapor phase over its liquid at a given temperature.

• Mechanism: The presence of nonvolatile solute particles at the liquid-gas interface obstructs solvent molecules from escaping into the gas phase, lowering the vapor pressure of the solution compared to the pure solvent.

• Raoult's Law:     $P_{solution} = \chi_{solvent} \times P^o_{solvent}$     Where $\chi_{solvent} = \frac{n_{solvent}}{(i \times n_{solute}) + n_{solvent}}$.

• Example Calculation:

  • Given: $18.3\,g$ $NaNO_3$, $27.75\,mol$ $H_2O$, $P^o_{H2O} = 233.7\,mm\,Hg$ at $70\,^oC$.

  • i: $2$ (for $NaNO_3$).

  • Moles solute: $18.3\,g / 85.0\,g/mol = 0.215\,mol$.

  • $\chi_{solvent}$: $27.75 / [ (2 \times 0.215) + 27.75 ] = 0.9847$.

  • $P_{solution}$: $0.9847 \times 233.7\,mm\,Hg = 230\,mm\,Hg$.

Boiling Point Elevation and Freezing Point Depression

• Boiling Point Elevation: The addition of a nonvolatile solute increases the boiling point of the solution.

  • Formula: $\Delta T_b = i \times m \times K_b$

  • Solution B.P.: $B.P._{pure\,solvent} + \Delta T_b$

• Freezing Point Depression: The addition of a solute decreases the freezing point of the liquid.

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

  • Solution F.P.: $F.P._{pure\,solvent} - \Delta T_f$

  • Constants: $K_b$ and $K_f$ are specific to the solvent used.

Example Problem: Glucose Solution

• Given: $55.8\,g$ glucose ($C_6H_{12}O_6$, molar mass $180.16\,g/mol$) in $455\,g$ water. $K_f = 1.86\,^oC/m$, $K_b = 0.512\,^oC/m$.

  1. Determine i: Glucose is molecular, so $i = 1$.

  2. Moles glucose: $0.309725\,mol$.

  3. Molality (m): $0.309725\,mol / 0.455\,kg = 0.680714\,m$.

  4. $\Delta T_f$: $1 \times 1.86 \times 0.680714 = 1.27\,^oC$.        New F.P.: $0\,^oC - 1.27\,^oC = -1.27\,^oC$.

  5. $\Delta T_b$: $1 \times 0.512 \times 0.680714 = 0.349\,^oC$.        New B.P.: $100\,^oC + 0.349\,^oC = 100.349\,^oC$.

Osmosis and Osmotic Pressure

• Osmosis: The diffusion of solvent molecules through a semipermeable membrane from a region of higher solvent concentration to a region of lower solvent concentration.

• Semipermeable Membrane: A barrier that allows small solvent molecules to pass but blocks larger solute particles.

• Osmotic Pressure ($\Pi$)): The pressure required to stop the movement of solvent through the membrane.

  • Formula: $\Pi = i \times M \times R \times T$

  • Constants: $R = 0.08206\,L \cdot atm / mol \cdot K$. temperature ($T$) must be in Kelvin.

Example Calculation (Osmotic Pressure)

• Given: $0.10\,M$ $Na_3PO_4$ at $25\,^oC$.

  1. i: $4$ ($3\,Na^+ + 1\,PO_4^{3-}$).

  2. Temp: $25 + 273.15 = 298.15\,K$.

  3. $\Pi$: $4 \times 0.10\,mol/L \times 0.08206\,L \cdot atm / mol \cdot K \times 298.15\,K = 9.8\,atm$.

Example Calculation (Molar Mass via Osmotic Pressure)

• Given: $10.0\,g$ protein in $0.500\,L$ solution. $\Pi = 5.9\,torr$ ($5.9/760\,atm$) at $22\,^oC$ ($295\,K$).

  1. Find Molarity (M): $M = \frac{\Pi}{iRT} = \frac{5.9/760}{1 \times 0.08206 \times 295} = 3.2 \times 10^{-4}\,mol/L$.

  2. Find Moles: $0.500\,L \times 3.2 \times 10^{-4}\,mol/L = 1.6 \times 10^{-4}\,mol$.

  3. Calculate Molar Mass: $10.0\,g / 1.6 \times 10^{-4}\,mol = 6.2 \times 10^4\,g/mol$.