Solutions and Their Properties

16.1 Properties of Solutions

• Groundwater dissolves rock, forming limestone caves; sinkholes occur when cave roofs collapse.

• Solution Formation: Factors affecting the dissolving rate:

  • Sugar type (granulated vs. cube).

  • Temperature.

  • Stirring (agitation).

• Solutions are homogeneous mixtures (solid, liquid, or gas).

• Dissolving Factors:

  • Solute/solvent compositions determine if dissolving occurs.

  • Stirring, temperature, and particle surface area influence the dissolving rate.

• Key Concepts:

  • Dissolving rate factors.

  • Solubility expression.

  • Solute amount in a solvent.

• Vocabulary: saturated, solubility, unsaturated, miscible, immiscible, supersaturated, Henry's law.

Stirring and Solution Formation

• Stirring granulated sugar in tea increases the rate at which it dissolves.

• Stirring brings fresh solvent into contact with the solute's surface, speeding up dissolving at the surface of sugar crystals.

• Agitation affects the dissolving RATE, not the AMOUNT of solute that dissolves.

• Insoluble substances remain undissolved with agitation.

Temperature and Solution Formation

• Higher temperatures cause sugar to dissolve more rapidly.

• Increased kinetic energy of water molecules at higher temperatures increases the frequency/force of collisions with sugar crystals.

Particle Size and Solution Formation

• Smaller solute particles (granulated sugar) dissolve faster due to greater surface area exposure.

• Dissolving is a surface phenomenon.

Solubility

• Example: 36.0 g NaCl in 100 g water at 25°C dissolves completely.

• Adding more salt results in only 0.2 g dissolving.

• Water molecules continuously bombard excess solid, solvating ions (kinetic theory).

• Exchange process: solvation of new particles and crystallization of dissolved particles occur simultaneously.

• Dynamic Equilibrium: In a saturated solution, solvation and crystallization rates are equal; undissolved crystal mass remains constant.

• Saturated Solution: Contains the maximum solute amount for a given solvent quantity at constant temperature/pressure.

  • Example: 36.2 g NaCl in 100 g water at 25°C.

• Solubility: The Solute amount dissolving in a solvent quantity at a specific temperature/pressure to produce a saturated solution.

  • Expressed as grams of solute per 100 g solvent, or g/L for gases.

• Unsaturated Solution: At a given temperature/pressure, it contains less solute than a saturated solution; additional solute will dissolve.

• Miscibility:

  • Miscible Liquids: Infinitely soluble (e.g., water and ethanol).

  • Partially Miscible Liquids: Slightly soluble (e.g., water and diethyl ether).

  • Immiscible Liquids: Insoluble (e.g., oil and water).

• Saturated Solution: Contains the maximum amount of solute for a given quantity of solvent at a constant temperature and pressure.

Factors Affecting Solubility

• Solubility: Mass of solute dissolving in a solvent mass at a specified temperature.

• Temperature: Affects solid, liquid, and gaseous solutes; pressure affects gaseous solutes.

Temperature

• Solid Solubility: Usually increases with temperature increase.

• Mineral Deposits: Occur around hot springs due to cooling of saturated mineral solutions.

• Ytterbium Sulfate (Yb2(SO4)3): Solubility decreases with temperature increase (44.2 g/100g water at 0°C to 5.8 g/100g water at 90°C).

• Supersaturated Solution: Contains more solute than theoretically possible at a given temperature; crystallization starts with a seed crystal.

• Warm a saturated solution to dissolve more solute, then cool it carefully.

• Crystallization can be initiated by adding a seed crystal or scratching the container.

• Rock Candy: Produced by sugar crystallization from a supersaturated solution.

• Gases in Liquids: Solubility decreases as temperature rises (opposite of solids).

• Oxygen: Less soluble in water at higher temperatures, impacting aquatic life.

• Thermal Pollution: Industrial plants increase lake temperature, lowering dissolved oxygen concentration.

Pressure

• Solids/Liquids: Pressure has little impact on solubility.

• Gases: Solubility increases as gas's partial pressure above solution increases.

• Carbonated Beverages: High CO2 pressure forces gas into solution; opening decreases pressure, releasing CO2 bubbles.

• Henry's Law: At a given temperature, gas solubility (S) is directly proportional to gas pressure (P) above the liquid.

  

• $S1$: Solubility at pressure $P1$; $S2$: Solubility at pressure $P2$.

Sample Problem 16.1: Calculating the Solubility of a Gas

• Given: Gas solubility is 0.77 g/L at 3.5 atm. Find: Solubility at 1.0 atm (constant temperature).

• Solution:

  

• Solubility at 1.0 atm is 0.22 g/L.

A Solution for Kidney Failure

• Kidneys filter blood, excreting toxins in urine. Failure leads to hemodialysis.

• Nephrons filter blood; collecting ducts carry toxic waste to the ureter.

• Hemodialysis: Cleanses blood outside the body. Blood from vein -> dialysis machine -> dialyzing bath -> back to vein.

• Semi-permeable membrane: Allows small particles (waste) to flow through, while retaining large particles (blood cells).

• Maintaining concentration gradient: Fresh dialyzing fluid is added while removing waste-filled solution.

16.2 Concentrations of Solutions

• Clean Water: Governments set limits on contaminants (metals, pesticides, bacteria).

• Concentration of a solution: Amount of solute in a solvent quantity.

• Dilute Solution: Small solute amount; Concentrated Solution: Large solute amount.

• Qualitative terms: "concentrated" and "dilute". Quantitative expression: Molarity.

Molarity

• Molarity (M): Moles of solute per liter of solution.
  $M = \frac{\text{moles of solute}}{\text{liters of solution}}$

• Volume: Total solution volume(solvent + solute).

• 3M $NaCl$ is “three molar sodium chloride solution”.

Sample Problem 16.2: Calculating the Molarity of a Solution

• IV saline solution: 0.90 g $NaCl$ in 100 mL solution. What is the molarity?

• Solution:

  • Convert g/100 mL to mol/L.

  • Use molar mass of $NaCl$ (58.5 g/mol).
    $\frac{0.90 g \ NaCl}{100 mL} \times \frac{1 mol \ NaCl}{58.5 g \ NaCl} \times \frac{1000 mL}{1 L} = 0.15 M$

Sample Problem 16.3: Finding the Moles of Solute in a Solution

• Laundry bleach: dilute $NaCIO$ solution. How many moles of solute in 1.5 L of the 0.70M solution?

• Solution:
  $0.70 \frac{mol \ NaCIO}{L} \times 1.5 L = 1.1 mol \ NaCIO$

• If the molarity and volume are known, the moles of solute is $M \times V$.

Making Dilutions

• Dilution: Reducing solute's moles per unit volume (total solute moles remain constant).

• Moles of solute before dilution = moles of solute after dilution

• $M<em>1 \times V</em>1 = M<em>2 \times V</em>2$ i.e (Molarity Initial x Volume Initial = Molarity Final x Volume Final)

• Using the same volume units throughout is essential.

Sample Problem 16.4: Preparing a Dilute Solution

• Dilute $2.00M \, MgSO4$ to prepare $100.0 \, mL$ of $0.400 M\, MgSO4$. Volume of $2.00 M$ needed?

• Solution:
  $V<em>1 = \frac{M</em>2 \times V<em>2}{M</em>1} = \frac{0.400 M \times 100.0 mL}{2.00 M} = 20.0 mL$

Percent Solutions

• Expressing concentration as solute percentage.

  • Ratio of solute volume to solution volume (% (v/v)).

  • Ratio of solute mass to solution mass (% (m/m)).

Concentration in Percent (Volume/Volume)

• Ratio of solute volume to solution volume when both are liquid
  $\% (v/v) = \frac{volume \, of \, solute}{volume \, of \, solution} \times 100\%$

Sample Problem 16.5: Calculating Percent (Volume/Volume)

• What is the percent by volume of ethanol (C2H6O) when the 85 mL of ethanol is diluted to 250 mL with water?

• Solution:
  $\frac{85 mL}{250 mL} \times 100 \% = 34 \%$

Concentration in Percent (Mass/Mass)

• Solution concentration as grams of solute per 100 g of solution, particularly with solids solute.
  $\% (m/m) = \frac{mass \, of \, solute}{mass \, of \, solution} \times 100 \%$

16.3 Colligative Properties of Solutions

• Colligative Property: Depends on solute particle number, not identity. Includes vapor-pressure lowering, boiling-point elevation, freezing-point depression.

Vapor-Pressure Lowering

• Nonvolatile solute lowers vapor pressure.

• Solvation reduces solvent molecules escaping as vapor.

• Ionic solutes (e.g., NaCl, CaCl2) lower vapor pressure more than non-dissociating solutes (e.g., glucose).

• Vapor pressure decrease is proportional to solute particles.

Freezing-Point Depression

• Solute disrupts solid formation; more kinetic energy needed to solidify solution compared to pure solvent.

• Freezing point of the solution is lower than the original solvent.

• Also a colligative property: The number of solute particles determines the magnitude of the depression.

• Salting icy surfaces lowers freezing point.

• Ethylene glycol in car cooling systems prevents freezing.

Boiling-Point Elevation

• Adding nonvolatile solute decreases solvent's vapor pressure.

• More kinetic energy increases liquid's vapor pressure & initiates boiling.

• Solution's boiling point is higher than the pure solvent's.

• It depends on particle concentration rather than identity.

16.4 Calculations Involving Colligative Properties

Molality and Mole Fraction

• Molality (m): Moles of solute per kilogram of solvent.

• For water: 1 kg (1000 g) equals 1 L (1000 mL).

Sample Problem 16.6: Using Solution Molality

• How many grams of potassium iodide are needed in $500.0\, g$ of water to produce a $0.060 \, m$ KI solution?
  Solution:
  $0.5000 \, kg \, H<em>2O \times \frac{0.060 \, mol \, KI}{1 \, kg \, H</em>2O} \times \frac{166.0 \, g \, KI}{1 \, mol \, KI} = 5.0 \, g \, KI$

Mole Fraction

• Ratio of one substance's moles to total solution moles.

• If solute A is $nA$ and the solvent B is $nB$, then the mol fractions are these equations
  $X<em>A = \frac{n</em>A}{n<em>A + n</em>B}$
  $X<em>B = \frac{n</em>B}{n<em>A + n</em>B}$

Sample Problem 16.7: Calculating Mole Fractions

• What is the mole fraction of each compound in a solution containing 1.25mol ethylene glycol(EG) & 4.00 mol of water?
  *Solution:

• EG mole fraction calculation

$X_{EG}=\frac{1.25 mol}{1.25 mol+4.00 mol}=0.238$

• Water mole fraction calculation
  $X_{H20}=\frac{4.00 mol}{1.25 mol+4.00 mol}=0.762$

Freezing-Point Depression and Boiling-Point Elevation

• Nonvolatile solute addition lowers solvent's freezing point and increases boiling point.

• Change in freezing temperature ($\Delta T_f$) is the difference between solution and pure solvent's freezing point

• Boiling point elevation related to molality of solute and equation is
  ${\Delta}T<em>f = K</em>f \times m$

• Where m is equal to molality

• If the solute it boiling, the equation is
  ${\Delta}T<em>b = K</em>b \times m$

Sample Problem 16.8: Calculating the Freezing-Point Depression of a Solution

• Calculate freezing-point depression and freezing point for solution with 100g ethylene glycol in 0.500 kg water

1. Solve for the moles
   $moles C<em>2H</em>6O_2 = 100 g \times \frac{1 mol}{62.0 g}=1.61 mol$

2. Molality from previous calculation
   $Molality=\frac{1.61 mol}{0.500 kg}=3.22m$

3. Solve for freezing point using the equation
   ${\Delta}T_f = 1.86 \frac{C}{m} \times 3.22m=5.99 C$

4. Answer

• Freezing point of the solution: $0.00{\degree}C - 5.99{\degree}C = -5.99{\degree}C$

Sample Problem 16.9: Calculating the Boiling Point of a Solution

• What is the boiling point of a 1.50m $NaCl$ solution?$\Delta T_b = 0.512 \frac{\degree C}{m} \times 3.00m = 1.54 \degree C$

• The solution will boil at $100 \degree C + 1.54 \degree C = 101.54 \degree C$