Concentration in Solutions: Key Concepts and Formulas

Fundamentals

  • Solvent: component present in the greatest amount in a solution
  • Solute: substance(s) dissolved in the solvent (present in lesser amounts)
  • Solutions: homogeneous mixtures with components uniformly distributed on a microscopic scale
  • Solvation: process of solute particles being surrounded by solvent molecules
  • Hydration: solvation when the solvent is water
  • Solubility: solute must be able to dissolve in the solvent to form a homogeneous solution
  • Miscible vs. immiscible: miscible substances mix to form a single phase; immiscible form separate phases
  • Formation of a solution is a physical process, not a chemical one
  • Enthalpy of solution: \Delta H{\text{soln}} = \Delta H{\text{solvent-bond breaking}} + \Delta H{\text{solute-bond breaking}} + \Delta H{\text{new solute-solvent interactions}}
  • Exothermic solvation (\Delta H_{\text{soln}} < 0) favors solution formation
  • Entropy change (disorder) plays a role: increasing entropy favors dissolution

Solvation and Concentration Essentials

  • Solvation vs hydration highlights the interaction of solute with solvent (water as solvent => hydration)
  • The state of the solvent largely influences the state of the solution; solute state can differ
  • Solutions are often considered in terms of concentration units to quantify how much solute is present

Types of Solutions (by phase of solute/solvent)

  • Gas in gas: example air
  • Gas in liquid: example carbonated beverages (CO₂ in water)
  • Liquid in liquid: example gasoline (miscible liquids)
  • Liquid in solid: Example tea (solutes dissolve in water)
  • Gas in solid: example hydrogen in palladium (H₂ in Pd)
  • Solid in liquid: example mercury in silver (alloy formation)
  • Solid in solid: example metal alloys
  • Note: miscibility determines whether a single phase forms or multiple phases persist

Concentration Units Overview

  • Percent by mass (mass percent):
    \%1 h{ by mass} = \frac{m{\text{solute}}}{m{\text{solution}}} \times 100
  • Volume percent (v/v%): \text{v/v \%} = \left(\frac{V{\text{solute}}}{V{\text{solution}}}\right) \times 100
    • Volume percent is based on volumes of solute and solution; liquids and gases volumes are not always additive
  • Mass percent (same as percent by mass): see above
  • Mass/Volume percent (m/v %):
    \%1 h{ m/v} = \frac{m{\text{solute}}}{V{\text{solution}}} \times 100
  • Mole fraction (X):
    xi = \frac{ni}{\sumj nj}, \quad \sumj xj = 1
  • Molarity (M): M = \frac{n{\text{solute}}}{V{\text{solution}}}
    • Note: volume of solution, not necessarily equal to volume of solvent
  • Molality (m): m = \frac{n{\text{solute}}}{m{\text{solvent}}}
    • For dilute aqueous solutions at 25°C, $m \approx M$ since density of water ≈ 1 g/mL
  • Dilute concentration measures
    • Parts per million (ppm):
      \text{ppm} = \frac{m{\text{solute}}}{m{\text{solution}}} \times 10^6
    • Parts per billion (ppb):
      \text{ppb} = \frac{m{\text{solute}}}{m{\text{solution}}} \times 10^9
    • Parts per trillion (ppt):
      \text{ppt} = \frac{m{\text{solute}}}{m{\text{solution}}} \times 10^{12}
  • Normality (N)
    • Normality = equivalents per liter of solution; reaction dependent
    • Example: 1 M (\mathrm{H2SO4}) is 2 N for acid-base reactions (provides 2 (\mathrm{H^+})) but 1 N for sulfate precipitation (1 mole of (\mathrm{SO_4^{2-}}) reacts)
  • Grams per liter (g/L):
    g/L = \frac{m{\text{solute}}}{V{\text{solution}}}
  • Formality (F):
    • Formal concentration uses formula weight units per liter of solution
    • F = \frac{n{\text{solute}}}{V{\text{solution}}} = \frac{m{\text{solute}}}{MW{\text{formula}} \cdot V_{\text{solution}}}

Examples and Key Calculations

  • Percent by mass example: 20 g salt in 100 g solution → \text{Mass \% NaCl} = \frac{20}{100} \times 100 = 20\%
  • Mole fraction example: 92 g glycerol with 90 g water
    • Moles: n{\text{water}} = \frac{90}{18} = 5\text{ mol},\quad n{\text{glycerol}} = \frac{92}{92} = 1\text{ mol}
    • Total moles = 6; x{\text{water}} = \frac{5}{6} \approx 0.833,\quad x{\text{glycerol}} = \frac{1}{6} \approx 0.167
    • Check: x{water} + x{glycerol} = 1.000
  • Molarity example: 11 g CaCl₂ (MW = 110) in 100 mL solution
    • n{\mathrm{CaCl2}} = \frac{11}{110} = 0.10\text{ mol}
    • Volume = 0.100 L; M = \frac{0.10}{0.100} = 1.0\,\text{M}
  • Molality example: 10 g NaOH (MW = 40) in 500 g water
    • n_{\mathrm{NaOH}} = \frac{10}{40} = 0.25\text{ mol}
    • Mass of solvent = 500 g = 0.500 kg; m = \frac{0.25}{0.500} = 0.50\,\text{m}
  • Dilutions (MiVi = MfVf)
    • Mi Vi = Mf Vf
    • Example: To prepare 0.300 L of 1.2 M NaOH from 5.5 M stock:
    • Vi = \frac{Mf Vf}{Mi} = \frac{(1.2)(0.300)}{5.5} \approx 0.065\,\text{L} = 65\,\text{mL}

Quick Reference Notes

  • Solvent is the component in greatest amount; solute is what is dissolved
  • Solutions are homogeneous; solute may be in a different phase than the solvent
  • Solvation (hydration if water) is the key interaction in solution formation
  • Enthalpy and entropy govern whether dissolution is favored
  • Use the appropriate concentration unit for the given data (mass, volume, mole, or equivalents)
  • Dilutions follow MiVi = MfVf to maintain moles of solute