Unit 1: Solutions - Practice Flashcards

Introduction to Solutions and Mixtures

  • Definition of Solutions: Solutions are homogeneous mixtures of two or more components. A homogeneous mixture is characterized by a uniform composition and uniform properties throughout its entire mass.
  • Components of a Solution:
    • Solvent: The component present in the largest quantity. It determines the physical state (solid, liquid, or gas) in which the solution exists.
    • Solutes: One or more components present in the solution other than the solvent.
  • Binary Solutions: These are solutions consisting of exactly two components. This unit focuses primarily on liquid binary solutions.
  • Importance of Composition: The utility of mixtures often depends on the specific ratio of their components. Examples include:
    • Brass: A mixture of Copper (CuCu) and Zinc (ZnZn).
    • German Silver: A mixture of Copper (CuCu), Zinc (ZnZn), and Nickel (NiNi).
    • Bronze: A mixture of Copper (CuCu) and Tin (SnSn).
    • Fluoride Ion (FF^-) Concentration in Water:
      • 1ppm1\,ppm prevents tooth decay.
      • 1.5ppm1.5\,ppm causes teeth to become mottled.
      • High concentrations are poisonous; for example, Sodium Fluoride (NaFNaF) is used as rat poison.
    • Medical Applications: Intravenous injections must be dissolved in water containing salts at specific ionic concentrations that match blood plasma concentrations.

Types of Solutions

Solutions are categorized based on the physical state of the solute and the solvent:

  • Gaseous Solutions:
    • Gas in Gas: Mixture of Oxygen (O2O_2) and Nitrogen (N2N_2) gases.
    • Liquid in Gas: Chloroform (CHCl3CHCl_3) mixed with Nitrogen (N2N_2) gas.
    • Solid in Gas: Camphor in Nitrogen (N2N_2) gas.
  • Liquid Solutions:
    • Gas in Liquid: Oxygen (O2O_2) dissolved in water.
    • Liquid in Liquid: Ethanol dissolved in water.
    • Solid in Liquid: Glucose dissolved in water.
  • Solid Solutions:
    • Gas in Solid: Solution of Hydrogen (H2H_2) in Palladium (PdPd).
    • Liquid in Solid: Amalgam of Mercury (HgHg) with Sodium (NaNa).
    • Solid in Solid: Copper (CuCu) dissolved in Gold (AuAu).

Expressing Concentration of Solutions

Concentration can be described qualitatively (dilute or concentrated) or quantitatively. Quantitative methods include:

  • Mass Percentage (w/w): Defined as the mass of the component per 100g100\,g of solution.     \text{Mass % of a component} = \frac{\text{Mass of the component in the solution}}{\text{Total mass of the solution}} \times 100
    • Example: 10%w/w10\%\,w/w glucose means 10g10\,g glucose in 90g90\,g water (100g100\,g total solution).
    • Industrial Application: Commercial bleaching solution contains 3.62%3.62\% mass percentage of Sodium Hypochlorite (NaClONaClO).
  • Volume Percentage (V/V): Defined as the volume of the component per 100mL100\,mL of solution.     \text{Volume % of a component} = \frac{\text{Volume of the component}}{\text{Total volume of solution}} \times 100
    • Example: 35%v/v35\%\,v/v ethylene glycol (antifreeze) lowers the freezing point of water to 255.4K255.4\,K (17.6C-17.6\,^{\circ}C).
  • Mass by Volume Percentage (w/V): Mass of solute dissolved in 100mL100\,mL of solution. Commonly used in medicine and pharmacy.
  • Parts per Million (ppm): Used for solutes present in trace quantities.     ppm=Number of parts of the componentTotal number of parts of all components×106\text{ppm} = \frac{\text{Number of parts of the component}}{\text{Total number of parts of all components}} \times 10^6
    • Example: Sea water contains 6×103g6 \times 10^{-3}\,g of dissolved O2O_2 per litre (1030g1030\,g), which is 5.8ppm5.8\,ppm.
  • Mole Fraction (xx): Ratio of the moles of a component to the total moles in the solution.     xA=nAnA+nBx_A = \frac{n_A}{n_A + n_B}
    • The sum of all mole fractions in a solution is always unity: x1+x2++xi=1x_1 + x_2 + \dots + x_i = 1.
  • Molarity (M): Number of moles of solute dissolved in one litre (or 1dm31\,dm^3) of solution.     M=Moles of soluteVolume of solution in litreM = \frac{\text{Moles of solute}}{\text{Volume of solution in litre}}
    • Example: 0.25molL1NaOH0.25\,mol\,L^{-1}\,NaOH means 0.25mol0.25\,mol of NaOHNaOH in 1L1\,L solution.
  • Molality (m): Number of moles of solute per kilogram (kgkg) of solvent.     m=Moles of soluteMass of solvent in kgm = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}}
    • Note: Mass %, ppm, mole fraction, and molality are independent of temperature. Molarity depends on temperature because volume changes with temperature.

Solubility of Solids in Liquids

  • Intermolecular Interactions: Solvent and solute must have similar interactions ("Like dissolves like"). Polar solutes (e.g., NaClNaCl, sugar) dissolve in polar solvents (water); non-polar solutes (naphthalene, anthracene) dissolve in non-polar solvents (benzene).
  • Processes:
    • Dissolution: Solute dissolves, increasing concentration in the solution.
    • Crystallisation: Solute particles collide with solid solute and separate from solution.
  • Saturated Solution: A solution where no more solute can dissolve at a given temperature and pressure. Dissolution and crystallisation are in dynamic equilibrium.
  • Unsaturated Solution: A solution containing less than the maximum amount of solute.
  • Effect of Temperature:
    • If dissolution is endothermic (ΔsolH>0\Delta_{sol} H > 0), solubility increases with temperature.
    • If dissolution is exothermic (ΔsolH<0\Delta_{sol} H < 0), solubility decreases with temperature.
  • Effect of Pressure: Pressure has no significant effect on the solubility of solids in liquids as they are highly incompressible.

Solubility of Gases in Liquids

  • General Rule: Solubility of gases in liquids increases with an increase in pressure.
  • Henry’s Law: At constant temperature, the solubility of a gas is directly proportional to its partial pressure above the surface of the liquid.
  • Mathematical Form: The partial pressure of the gas in the vapour phase (pp) is proportional to the mole fraction of the gas in the solution (xx).     p=KH×xp = K_H \times x
    • KHK_H is the Henry’s law constant. Higher KHK_H at a given pressure indicates lower solubility.
    • KHK_H increases with temperature, meaning gas solubility decreases as temperature rises. This is why aquatic species prefer cold water (higher dissolved O2O_2).
  • Applications of Henry's Law:
    1. Soft Drinks: Bottles are sealed under high pressure to increase CO2CO_2 solubility.
    2. Scuba Diving: High underwater pressure increases Nitrogen (N2N_2) solubility in blood. Rapid ascent causes bubbles (bends). Dive tanks are filled with air diluted with Helium (11.7%He11.7\%\,He, 56.2%N256.2\%\,N_2, 32.1%O232.1\%\,O_2) to prevent this.
    3. Anoxia: At high altitudes, low partial pressure of O2O_2 leads to low blood oxygen, causing weakness and mental confusion.

Vapour Pressure of Liquid Solutions

  • Raoult’s Law for Volatile Liquids: For a solution of volatile liquids, the partial vapour pressure of each component is directly proportional to its mole fraction in the solution.     p1=x1×p10p_1 = x_1 \times p_1^0
    • p10p_1^0 is the vapour pressure of the pure component.
  • Total Pressure (Dalton’s Law):     ptotal=p1+p2=x1p10+x2p20p_{total} = p_1 + p_2 = x_1 p_1^0 + x_2 p_2^0ptotal=p10+(p20p10)x2p_{total} = p_1^0 + (p_2^0 - p_1^0)x_2
  • Vapour Phase Composition: If y1y_1 and y2y_2 are mole fractions in the vapour phase:     pi=yi×ptotalp_i = y_i \times p_{total}
  • Raoult’s Law as a Special Case of Henry’s Law: Both laws state partial pressure is proportional to mole fraction. In Raoult's Law, the constant is pi0p_i^0; in Henry's Law, it is KHK_H.

Ideal and Non-Ideal Solutions

  • Ideal Solutions: Obey Raoult’s law at all concentrations.
    • Properties: ΔmixH=0\Delta_{mix} H = 0 and ΔmixV=0\Delta_{mix} V = 0.
    • Molecular Level: Interactions AABBABA-A \approx B-B \approx A-B.
    • Examples: n-hexane and n-heptane; bromoethane and chloroethane; benzene and toluene.
  • Non-Ideal Solutions: Do not obey Raoult’s law.
    • Positive Deviation: ABA-B interactions are weaker than individual component interactions. Molecules escape more easily. Vapour pressure is higher than predicted.
      • Example: Ethanol and Acetone (acetone breaks ethanol hydrogen bonds).
    • Negative Deviation: ABA-B interactions are stronger (e.g., hydrogen bonding). Molecules find it harder to escape. Vapour pressure is lower than predicted.
      • Example: Phenol and Aniline; Chloroform and Acetone.
  • Azeotropes: Constant boiling mixtures where liquid and vapour phases have the same composition.
    • Minimum Boiling Azeotrope: Formed by solutions with large positive deviations (e.g., 95%v/v95\%\,v/v ethanol in water).
    • Maximum Boiling Azeotrope: Formed by solutions with large negative deviations (e.g., 68%HNO368\%\,HNO_3 and 32%H2O32\%\,H_2O by mass, boiling at 393.5K393.5\,K).

Colligative Properties

Colligative properties depend solely on the number of solute particles, not their identity.

1. Relative Lowering of Vapour Pressure

Adding a non-volatile solute reduces the surface area available for solvent molecules to evaporate, lowering vapour pressure. p10p1p10=x2=n2n1+n2\frac{p_1^0 - p_1}{p_1^0} = x_2 = \frac{n_2}{n_1 + n_2} For dilute solutions: p10p1p10=w2×M1M2×w1\frac{p_1^0 - p_1}{p_1^0} = \frac{w_2 \times M_1}{M_2 \times w_1}

2. Elevation of Boiling Point (ΔTb\Delta T_b)

A solution's boiling point is always higher than the pure solvent's because of reduced vapour pressure. ΔTb=TbTb0=Kb×m\Delta T_b = T_b - T_b^0 = K_b \times m

  • KbK_b: Molal Elevation Constant (Ebullioscopic Constant) in Kkgmol1K\,kg\,mol^{-1}.
  • M2=Kb×w2×1000ΔTb×w1M_2 = \frac{K_b \times w_2 \times 1000}{\Delta T_b \times w_1}
3. Depression of Freezing Point (ΔTf\Delta T_f)

A solution freezes at a lower temperature than the pure solvent. ΔTf=Tf0Tf=Kf×m\Delta T_f = T_f^0 - T_f = K_f \times m

  • KfK_f: Molal Depression Constant (Cryoscopic Constant).
  • Formulas for KfK_f and KbK_b involving Enthalpy of Fusion (ΔfusH\Delta_{fus} H) and Vapourisation (ΔvapH\Delta_{vap} H):     Kf=R×M1×(Tf)21000×ΔfusHK_f = \frac{R \times M_1 \times (T_f)^2}{1000 \times \Delta_{fus} H}Kb=R×M1×(Tb)21000×ΔvapHK_b = \frac{R \times M_1 \times (T_b)^2}{1000 \times \Delta_{vap} H}
4. Osmosis and Osmotic Pressure (Π\Pi)
  • Osmosis: Flow of solvent through a semipermeable membrane (SPM) from pure solvent to solution (or dilute to concentrated).
  • Osmotic Pressure: The excess pressure applied to the solution side to stop osmosis.     Π=CRT=n2VRT\Pi = C R T = \frac{n_2}{V} R T
  • Advantages of Osmotic Pressure Method: Measured at room temperature; uses molarity; magnitude is large even for dilute solutions; suitable for sensitive biomolecules.
  • Isotonic Solutions: Same osmotic pressure. Blood cells are isotonic with 0.9%w/VNaCl0.9\%\,w/V\,NaCl (normal saline).
    • Hypertonic: Concentration >0.9%> 0.9\%, cells shrink.
    • Hypotonic: Concentration <0.9%< 0.9\%, cells swell.
  • Reverse Osmosis: Applied pressure greater than Π\Pi forces solvent to move from solution to pure solvent. Used in desalination of sea water using cellulose acetate membranes.

Abnormal Molar Masses and van’t Hoff Factor (ii)

  • Abnormal Molar Mass: Molar mass values that are lower (due to dissociation) or higher (due to association) than expected.
  • Association: e.g., Ethanoic acid dimerises in benzene: 2CH3COOH(CH3COOH)22CH_3COOH \rightleftharpoons (CH_3COOH)_2.
  • van’t Hoff Factor (ii):     i=Normal molar massAbnormal molar mass=Observed colligative propertyCalculated colligative propertyi = \frac{\text{Normal molar mass}}{\text{Abnormal molar mass}} = \frac{\text{Observed colligative property}}{\text{Calculated colligative property}}i=Number of moles of particles after association/dissociationNumber of moles of particles before association/dissociationi = \frac{\text{Number of moles of particles after association/dissociation}}{\text{Number of moles of particles before association/dissociation}}
  • Values of i:
    • i>1i > 1 for dissociation (e.g., KCli2;K2SO4i3KCl \rightarrow i \approx 2; K_2SO_4 \rightarrow i \approx 3).
    • i<1i < 1 for association (e.g., ethanoic acid in benzene i0.5i \approx 0.5).
  • Modified Equations:
    • Relative lowering of V.P.: p10p1p10=in2n1\frac{p_1^0 - p_1}{p_1^0} = i \frac{n_2}{n_1}
    • Boiling Point: ΔTb=iKbm\Delta T_b = i K_b m
    • Freezing Point: ΔTf=iKfm\Delta T_f = i K_f m
    • Osmotic Pressure: Π=iCRT\Pi = i C R T