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: $\Delta H<em>{\text{soln}} = \Delta H</em>{\text{solvent-bond breaking}} + \Delta H<em>{\text{solute-bond breaking}} + \Delta H</em>{\text{new solute-solvent interactions}}$
• 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%): $\text{v/v \%} = \left(\frac{V<em>{\text{solute}}}{V</em>{\text{solution}}}\right) \times 100$
  • 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):
  $x<em>i = \frac{n</em>i}{\sum<em>j n</em>j}, \quad \sum<em>j x</em>j = 1$
• Molarity (M): $M = \frac{n<em>{\text{solute}}}{V</em>{\text{solution}}}$
  • Note: volume of solution, not necessarily equal to volume of solvent
• Molality (m): $m = \frac{n<em>{\text{solute}}}{m</em>{\text{solvent}}}$
  • For dilute aqueous solutions at 25°C, $m \approx M$ since density of water ≈ 1 g/mL
• Dilute concentration measures
  • Parts per million (ppm):
    $\text{ppm} = \frac{m<em>{\text{solute}}}{m</em>{\text{solution}}} \times 10^6$
  • Parts per billion (ppb):
    $\text{ppb} = \frac{m<em>{\text{solute}}}{m</em>{\text{solution}}} \times 10^9$
  • Parts per trillion (ppt):
    $\text{ppt} = \frac{m<em>{\text{solute}}}{m</em>{\text{solution}}} \times 10^{12}$
• 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):
  $g/L = \frac{m<em>{\text{solute}}}{V</em>{\text{solution}}}$
• Formality (F):
  • Formal concentration uses formula weight units per liter of solution
  • $F = \frac{n<em>{\text{solute}}}{V</em>{\text{solution}}} = \frac{m<em>{\text{solute}}}{MW</em>{\text{formula}} \cdot V_{\text{solution}}}$

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: $n<em>{\text{water}} = \frac{90}{18} = 5\text{ mol},\quad n</em>{\text{glycerol}} = \frac{92}{92} = 1\text{ mol}$
  • Total moles = 6; $x<em>{\text{water}} = \frac{5}{6} \approx 0.833,\quad x</em>{\text{glycerol}} = \frac{1}{6} \approx 0.167$
  • Check: $x<em>{water} + x</em>{glycerol} = 1.000$
• Molarity example: 11 g CaCl₂ (MW = 110) in 100 mL solution
  • $n<em>{\mathrm{CaCl</em>2}} = \frac{11}{110} = 0.10\text{ mol}$
  • 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)
  • $M<em>i V</em>i = M<em>f V</em>f$
  • Example: To prepare 0.300 L of 1.2 M NaOH from 5.5 M stock:
  • $V<em>i = \frac{M</em>f V<em>f}{M</em>i} = \frac{(1.2)(0.300)}{5.5} \approx 0.065\,\text{L} = 65\,\text{mL}$

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