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Mixtures: Solutions and Colloids

Mixtures: Solutions and Colloids

: and Intermolecular Forces in – Solutions – homogeneous mixtures (a single phase) • All types of IFs in pure substances also occur in – Colloids – heterogeneous mixtures (two or more phases) solutions (in addition, ion-dipole forces are very common) 13.1 Types of Solutions and • Ion-dipole forces – present in solutions of ionic – – the substance that dissolves; usually compounds in polar such as H2O the most abundant component of the ; has – Hydration () the same physical state as the solution • The water dipoles pull the ions – Solute – the substance that is dissolved away from the ionic crystal and surround them (typical coord. – Solubility (S) – the maximum amount of solute numbers 4 or 6) that can be dissolved in a given amount of • There is a short range order solvent or solution around the ions consisting of H- – of solute – various units are used bonded shells of water molecules

• Dipole-dipole forces – present in solutions of •The “like dissolves like rule” – substances with similar types of IFs dissolve in each other well polar molecules in polar solvents such as H2O • H-bonding forces – present in solutions of O- – Strong solute-solvent IFs lead to better solubility and N-containing molecules (sugars, alcohols, (lower the total energy of the system) amino acids, …) in protic solvents such as H O – The solute-solvent IFs created during dissolution 2 must have comparable strength to the solute-solute • Ion-induced dipole forces – solutions of ionic and solvent-solvent IFs destroyed in this process compounds in less polar or non-polar solvents • Dipole-induced dipole forces – present in solutions of polar molecules in non-polar solvents or non-polar molecules in polar solvents • Dispersion forces – present in all solutions (most important for solutions of non-polar molecules in non-polar solvents)

Example: Example: ¾ Water dissolves well alcohols (ROH) with short What is a better solvent for diethyl ether hydrocarbon chains (R) → methanol, ethanol, … (CH CH -O-CH CH ), water or propanol ¾ Strong H-bonding IFs in the solute and the solvent are 3 2 2 3 replaced with strong solute-solvent H-bonding IFs (CH3CH2CH2OH)? ¾ Water does not dissolve well alcohols (ROH) with ¾ Both water and propanol interact with ether long hydrocarbon chains (R) → hexanol, … through H-bonding ¾ Strong H-bonding in water is replaced with weaker IFs ¾ Propanol and ether interact well through dispersion between the water dipole and the large non-polar forces (similar non-polar hydrocarbon chains) hydrocarbon chains (R) → (H-bonding with the OH part ¾ Water can’t interact well through dispersion forces of the alcohol is a smaller fraction of the total IFs) with the hydrocarbon portion of ether ⇒ Propanol is a better solvent for ether

1 • Soaps –Na+ salts of long chain carboxylic acids • Liquid solutions – the solvent is a liquid – Long, non-polar hydrocarbon “tail” → hydrophobic – Solid-liquid and liquid-liquid solutions – Small, polar-ionic carboxyl “head” → hydrophilic • Some salts and polar molecular compounds dissolve – The polar head dissolves in water; the non-polar tail well in water and short chain alcohols dissolves in grease → washing action • Less polar solids dissolve well in less polar solvents like acetone, chloroform, ether, … • Detergents – contain surfactants (surface-active • Non-polar substances dissolve best in non-polar compounds) → similar to soaps solvents like hexane, benzene, … Tail Head – Gas-liquid solutions • Non-polar gases have poor solubility in water • Some gases dissolve in water through chemical reactions (HCl, CO2, SO2, …) • Gaseous solutions – the solvent is a gas – Gas-gas solutions – gases mix in all proportions

• Solid solutions – the solvent is a solid Example: – Gas-solid solutions – gas molecules penetrate the Quick cleaning of laboratory glassware: crystal lattices of some metals (H2/Pd, O2/Cu, …) ¾ Cleaning a sample tube with a salt residue – Solid-solid solutions – homogeneous alloys, Wash with water → dissolves salt (ion-dipole forces) waxes, … Wash with ethanol → dissolves water (H-bonding) ¾Substitutional alloys – Wash with acetone → dissolves ethanol (dipole- the atoms of one element dipole, dispersion forces) take some of the positions Dry → acetone evaporates easily (low T ) in the lattice of another b element ¾ Cleaning a sample tube with an oily residue ¾Interstitial alloys – the Wash with hexane → dissolves oil (dispersion forces) atoms of one element fit Wash with acetone → dissolves hexane (dispersion into the gaps of the lattice and dipole-induced dipole forces) of another element Dry → acetone evaporates easily (low Tb)

13.2 Energy Changes in the Solution Solute (aggregated) + heat → Solute (separated) ∆Hsolute>0 Solvent (aggregated) + heat → Solvent (separated) ∆H >0 Process solvent Solute (separ.) + Solvent (separ.) → Solution + heat ∆Hmix<0 • Solution cycle – the solution process can be – Heat of solution (∆Hsoln) – the total enthalpy divided into three steps: change during the solution cycle (can be either exothermic or endothermic) 1) Separation of solute particles to make room for the solvent – endothermic (energy is needed to overcome the IFs of attraction) 2) Separation of solvent particles to make room for the solute particles – endothermic (energy is needed to overcome the IFs of attraction) 3) Mixing of solvent and solute particles – exothermic (solute-solvent IFs lower the energy) ∆Hsoln = ∆Hsolute + ∆Hsolvent + ∆Hmix

2 Solutions of Ionic Solids • Individual ionic heats of hydration – ∆H for • For dilute solutions of ionic solids, the solute the hydration of 1 mol of separated gaseous separation has ∆H equal to the lattice enthalpy cations (or anions) + - M+(g) → M+(aq) or X-(g) → X-(aq) MX(s) → M (g) + X (g) ∆Hsolute = ∆Hlattice> 0 •The hydration of the separated solute ions is a – ∆Hhydr of ions is always exothermic (∆Hhydr< 0) combination of steps 2 and 3 in the solution cycle because the ion-dipole forces that replace some of → (solvent separation + mixing) the H-bonds in water are stronger – The overall heat of hydration is a sum of the heats – Heat (enthalpy) of hydration (∆H ) – ∆H for hydr for the cations and the anions hydration of separated gaseous ions M+(g) + X-(g) → M+(aq) + X-(aq) ¾Trends in ∆Hhydr of ions (same as for ∆Hlattice) • ∆H is larger (more exothermic) for ions with ∆Hhydr = ∆Hsolvent + ∆Hmix hydr ⇒∆H = ∆H + ∆H greater charges and smaller sizes → ions with soln solute hydr higher charge ⇒∆Hsoln= ∆Hlattice + ∆Hhydr • The charge factor is more important

Some ionic heats of hydration in kJ/mol The Tendency Toward Disorder Li+(-558) Be2+(-2533) F-(-483) – ∆Hsoln< 0 favors the solution process since the Na+(-444) Mg2+(-2003) Al3+(-4704) Cl-(-340) total energy of the system is lowered; but many K+(-361) Ca2+(-1657) Br-(-309) ionic solids have ∆Hsoln> 0 and still dissolve ¾ The sign of ∆Hsoln is hard to predict, since both readily in water ∆Hlattice and ∆Hhydr depend on the ionic charge and ⇒There is a second factor that affects the solution size and tend to cancel each other’s effects process and acts in addition to the enthalpy factor Example: NaCl • Disorder – systems have a natural tendency to ∆H = 787 kJ/mol lattice become more disordered ∆Hhydr = (-444) + (-340) = -784 kJ/mol – Entropy – a measure of the disorder in the system ↑Entropy ↔↑Disorder ∆Hsoln = (787) + (-784) = +4 kJ/mol – Mixing leads to greater disorder and ↑ the entropy

• The solution process is governed by the 13.3 Solubility as an Equilibrium Process combination of the enthalpy and entropy factors • Typically solutes have limited solubility in a – Compounds with ∆Hsoln< 0 are typically soluble since both factors are favorable given solvent – The solution process is countered by – Compounds with ∆Hsoln> 0 are soluble only if ∆Hsoln is relatively small so the favorable entropy factor recrystallization of the dissolved solute dominates – Dynamic equilibrium – the rates of dissolution – Compounds with ∆Hsoln>> 0 are typically insoluble and recrystallization become equal since the unfavorable enthalpy factor dominates Solute (undissolved) ↔ Solute (dissolved) Example: Benzene is insoluble in water because – Saturated solution – no more solute can be the solute-solvent IFs are weak so the (-) ∆Hmix dissolved (can be produced by equilibrating the is very small compared to the (+) ∆Hsolut & ∆Hsolv solution with an excess of the solute; the excess ⇒ ∆Hsoln= ∆Hsolute + ∆Hsolvent + ∆Hmix >> 0 solute can be filtered out after saturation)

3 – Molar solubility (S) – the concentration of the Solute + Solvent + heat ↔ Saturated solution saturated solution in mol/L ¾Increasing T provides the heat needed for the – Unsaturated solution – more solute can be forward reaction to proceed (S↑) dissolved – Supersaturated solution – the concentration is •Some solids are more soluble at higher T higher than the solubility, S (unstable, non- despite their exothermic (-) ∆Hsoln (Why?) equilibrium state; disturbances or addition of seed ¾Tabulated ∆Hsoln values refer to the standard state crystal lead to crystallization) (1M solution), but saturated solutions can be Effect of Temperature on Solubility much more concentrated •Most solids are more soluble at higher T due ¾At very high the hydration process is hindered so ∆H becomes less negative; thus to their endothermic (+) ∆Hsoln hydr ∆Hsoln can change dramatically and even switch ¾(+) ∆Hsoln means that heat is absorbed during dissolution (the heat can be viewed as a reactant) signs from (-) to (+)

• Gases are less soluble in water at higher T due ¾Henry’s law - Sgas is directly proportional to Pgas to their exothermic (-) ∆H soln Sgas = kH×Pgas ¾ ∆H is (-) since ∆H ≈ 0 and H < 0 soln solute hydr – kH – Henry’s law constant (depends on the gas, Gas + Water ↔ Saturated solution + heat solvent and T) ¾Increasing T provides the heat needed for the Example: What is the concentration of CO2 in rain reverse reaction to proceed (S↓) drops at 20°C and 1.0 atm in air that has 1.0% by CO ? [k (CO , 20°C) = 2.3×10-2 mol/L.atm] Effect of on Solubility 2 H 2 1.0% by volume → χ = 0.010 • Pressure has little effect on the solubility of CO2

solids and liquids (low compressibility) PCO2 = χCO2 Ptot = 0.010 × 1.0 atm = 0.010 atm • The molar solubility of gases (Sgas) increases -2 SCO2 = kH×PCO2 = 2.3×10 mol/L.atm × 0.010 atm with increasing their partial (P ) gas = 2.3×10-4 mol/L over the solution (collisions with liquid surface ↑)

13.4 Expressing Solute Concentration ¾Molarity (M) – the number of moles of solute • Concentration – the ratio of the quantity of per 1 liter of solution solute to the quantity of solution (or solvent) M = (mol of solute)/(liters of solution) – M is affected by temperature (the solution volume changes with T, so M changes too) – The solution volume is not a sum of the solvent and solute volumes (it must be measured after mixing) ¾ (m) – the number of moles of solute per 1 kilogram of solvent m = (mol of solute)/(kilograms of solvent) – m is not affected by temperature (the amounts of solute and solvent don’t change with T) – The solution volume is not needed; m can be calculated from the masses of solute and solvent

4 • M and m are nearly the same for dilute aqueous • Parts of solute by parts of solution solutions since 1 L of water is about 1 kg, so (liters ¾Parts by mass of solution) ≈ (kg of solvent) – Mass % – grams of solute per 100 grams of Example: Calculate M and m for a solution solution → % (w/w) prepared by dissolving 2.2 g of NaOH in 55 g of mass of solute Mass% = ×100% water if the density of the solution is 1.1 g/mL. mass of solution 1 mol NaOH mol solute = 2.2 g NaOH× = 0.055 mol – ppm or ppb – grams of solute per 1 million or 1 40 g NaOH billion grams of solution (for trace components) 0.055 mol NaOH mol m = = 1.0 → 1.0 m ()molal 0.055 kg water kg ¾Parts by volume – Volume % – volume of solute per 100 volumes mass 2.2 g + 55 g Volume = = = 52 mL of solution → % (v/v) density 1.1 g/mL volume of solute 0.055 mol NaOH mol M = = 1.1 → 1.1 M ()% = ×100% 0.052 L solution L volume of solution

– ppmv or ppbv – volume of solute per 1 million • Converting units of concentration or 1 billion volumes of solution (used for trace -6 gases in air) Example: A sample of water is 1.1×10 M in chloroform (CH3Cl). Express the concentration ¾ fraction (X) – ratio of the # mol of of chloroform in ppb. (Assume density of 1.0 solute to the total # mol (solute + solvent) g/mL) mol of solute X = 1.1×10-6 M → 1.1×10-6 mol CH Cl per 1 L solution mol of solute + mol of solvent 3

Example: Calculate the X of NaOH in a solution -6 50.5 g CH3Cl −5 1.1×10 mol CH3Cl× = 5.6×10 g CH3Cl containing 2.2 g of NaOH in 55 g of water. 1 mol CH3Cl

1 mol 1 mol g 3 2.2 g = 0.055 mol NaOH 55 g = 3.1 mol H O 1 L → 1000 mL×1.0 = 1.0×10 g solution 40 g 18 g 2 mL −5 0.055 mol 5.6×10 g CH3Cl 9 X = = 0.018 3 ×10 ppb = 56 ppb 0.055 mol + 3.1 mol 1.0×10 g solution

Example: What is the molality of a solution of Example: What is the molality of a 1.83 M methanol in water, if the mole fraction of NaCl solution with density of 1.070 g/mL? methanol in it is 0.250? Assume 1 L (103 mL) of solution: Assume 1 mol of solution: → nNaCl = 1.83 mol → n = 1 mol × 0.250 = 0.250 mol meth 1.070 g → n = 1 - 0.250 = 0.750 mol mass of solution = 103 mL× = 1070 g water 1 mL 18.0 g H O 1 kg 58.44 g NaCl 0.750 mol H O× 2 × = 0.0135 kg H O mass of NaCl = 1.83 mol × = 107 g 2 3 2 1 mol 1 mol H2O 10 g 0.250 mol methanol mass of water = 1070 g − 107 g = 963 g = 0.963 kg m = = 18.5 m 0.0135 kg H O 1.83 mol NaCl 2 m = = 1.90 m 0.963 kg H2O

5 Example: What is the molarity of a 1.20 m KOH 13.5 of Solutions solution in water having density of 1.05 g/mL? • Colligative properties – depend on the

Assume 1 kg (1000 g) of solvent (H2O): concentration of solute particles but not on their chemical identity → nKOH = 1.20 mol – The concentration of solute particles depends on 56.1 g KOH mass of KOH = 1.20 mol × = 67.3 g the amount of dissolved solute as well as on its 1 mol ability to dissociate to ions in solution mass of solution = 1000 g + 67.3 g = 1067 g • Strong electrolytes – dissociate completely (soluble 1 mL salts, strong acids and bases) volume of solution = 1067 g× = 1016 mL = 1.02 L 1.05 g • Weak electrolytes – dissociate partially (weak acids 1.20 mol KOH and bases) M = = 1.18 M 1.02 L solution • Nonelectrolytes – do not dissociate (many organic compounds)

Nonvolatile Nonelectrolyte Solutions ¾Raoult’s Law – the vapor pressure of the solvent – No dissociation; no vapor pressure (glucose, sugar, …) over the solution is directly proportional to the • Vapor pressure lowering (∆P) – the vapor mole fraction of the solvent pressure of the solvent over the solution (P ) – Followed strictly at all concentrations only by ideal solv solutions is always lower than the vapor pressure over Psolv = XsolvP°solv the pure solvent (P°solv) at a given temperature X ≤ 1 ⇒ P ≤ P° Pure solvent disorder↑↑ solv solv solv Least disordered Vapor Xsolv = 1 – Xsolute ⇒ Psolv = (1 - Xsolute)P°solv Most disordered P = P° - X P° → P° - P = X P° Solution solv solv solute solv solv solv solute solv More disordered disorder↑ ∆P= XsoluteP°solv ⇒ The solution has a lesser tendency to vaporize so ⇒ The vapor pressure lowering is directly its vapor pressure is lower proportional to the mole fraction of the solute

Example: The vapor pressure of water over a • elevation (∆Tb) and freezing solution of a nonelectrolyte is 16.34 torr at 20°C. point depression (∆Tf) Determine the mole fraction of the solute, if the – The solution boils at a higher temperature equilibrium vapor pressure of water at 20°C is compared to the pure solvent (the solution has 17.54 torr. lower vapor P so it needs higher T to boil) → Psolv = 16.34 torr P°solv = 17.54 torr – The solution freezes at a lower temperature → Xsolute = ? compared to the pure solvent

disorder↑ Pure solvent ∆P ()17.54 -16.34 torr More disordered X solute = o = = 0.0684 Solid Psolv 17.54 torr Least disordered disorder↑↑ Solution – Most real (non-ideal) solutions behave as ideal at Most disordered low concentrations (less than 0.1 m for ⇒ The solution has a greater tendency to melt so its nonelectrolytes and less than 0.01 m for electrolytes) melting (freezing) point is lower

6 ∆Tb = Tb -T°b > 0 → ∆Tb = kb m ∆Tf = T°f -Tf > 0 → ∆Tf = kf m → kb – boiling point elevation constant → Depend on → kf – freezing point depression constant → the solvent → m – molality of solution Example: What is the boiling point of a solution prepared by dissolving 12 g of glucose (MW = 180 g/mol) in 55 g of water? (For water, kb=0.512°C/m and kf=1.86°C/m) 1 mol gluc 0.067 mol gluc 12 g gluc = 0.067 mol m = = 1.2 m 180 g gluc 0.055 kg water T T° T° T f f b b °C ∆T = 0.512 ×1.2 m = 0.62°C T = 100 + 0.62 = 100.62°C b m b

• (Π) – Π is the hydrostatic pressure necessary to stop the net flow of solvent caused by osmosis – Osmosis – the flow of solvent trough a semipermeable membrane from a less Π = MRT Π = (nsolute/Vsoln)RT concentrated into a more concentrated solution → M – molarity of solution – Semipermeable membrane – the solute particles → R – ; T – temperature in K can’t pass through – The equation is the equivalent of the law ¾The solvent (P = nRT/V) applied to solutions tends to flow • Π is the pressure the solute would exert if it were an ideal gas occupying alone the volume of the solution into the – Osmosis is essential for controlling the shape and solution where size of biological cells and purifying blood the disorder is through dialysis greater – Reverse osmosis – reversing the flow by applying external pressure (used to purify sea water)

Example: What is the minimum pressure that Using Colligative Properties to Find the must be applied in order to purify a 0.82 M Solute nonelectrolyte solution by reverse osmosis at • The molar mass of the solute can be obtained by 25°C? measuring one of the colligative properties of the → Calc. Π (the necessary pressure must be ≥Π) solution, the mass of the solute and the mass of the solvent (or the volume of the solution) mol L⋅atm Π = MRT = 0.82 × 0.08206 × 298 K – ∆P and ∆Tb are not very sensitive to changes in the L mol ⋅ K solute concentration (rarely used for molar mass Π = 20 atm determinations) – ∆Tf is more sensitive especially when solvents with → This pressure is equivalent to a 200 meters large kf constants are used (1/8 mile) tall water column! – Π is most sensitive and can be used for substances with low molar solubility such as large biomolecules

7 Example: The Tf of camphor is 179.80°C and it’s Example: A 3.0 g polymer sample is dissolved in kf is 39.7°C/m. When 200.0 mg of a compound (X) enough benzene to produce 150. mL of solution. If are added to 100.0 g of camphor, it’s freezing point the solution’s osmotic pressure is 0.0119 atm at drops to 179.29°C. What is the molar mass of X? 25°C, what is the average MW of the polymer? → Calc. the m of X: Calc. M of the solution:

∆Tf ()179.80 -179.29 °C mol Π 0.0119 atm −4 mol ∆T = k m → m = = = 0.013 M = = = 4.87 ×10 f f RT 0.08206 L.atm/mol. K × 298 K L k f 39.7°C/m kg → Calc. the number of moles of X: → Calc. the number of moles of the polymer: mol( X ) mol mol( pol ) mol m = → 0.013 × 0.1000 kg = 0.0013 mol M = → 4.87×10−4 × 0.150 L = 7.30×10−5 mol kg( camphor ) kg L( so ln) L → Calc. the molar mass of X: → Calc. the molar mass of the polymer: grams( X ) 0.2000 g grams( polym ) 3.0 g MW = → = 160 g/mol MW = → = 41000 g/mol mol( X ) 0.0013 mol mol( polym ) 7.30×10−5 mol

Volatile Nonelectrolyte Solutions – The mole fractions of the solute and solvent in the – No dissociation; both solvent and solute have vapor vapor phase above the solution can be calculated pressures (mixtures of volatile organic compounds) using Dalton’s law vap vap • Raoult’s law can be applied to the vapor Psolv = X solv Ptot Psolute = X solute Ptot pressures of both the solute and the solvent P P X vap = solv X vap = solute Psolv = XsolvP°solv Psolute = XsoluteP°solute solv solute Ptot Ptot ⇒The presence of each component lowers the vapor pressure of the other component Example: The equilibrium vapor pressures of pure benzene and toluene are 95.1 and 28.4 torr, – The total pressure over the solution is the sum of respectively at 25°C. Calculate the total the partial pressures of the solute and the solvent pressure, partial pressures and mole fractions of P = P + P = X P° + X P° benzene and toluene over a solution of 1.0 mol of tot solv solute solv solv solute solute benzene in 3.0 mol of toluene at 25°C.

1.0 3.0 Electrolyte Solutions X b = = 0.25 X t = = 0.75 1.0 + 3.0 1.0 + 3.0 – The solute dissociates to ions o Pb = X b Pb = 0.25× 95.1 torr = 24 torr – The number of particles in the solution is greater than the one implied by the solute concentration P = X P o = 0.75× 28.4 torr = 21 torr t t t • The formulas for the colligative properties Ptot = Pb + Pt = 24 + 21 = 45 torr must be modified P 24 P 21 – van’t Hoff factor (i) – accounts for the X vap = b = = 0.53 X vap = t = = 0.47 b t dissociation of the solute Ptot 45 Ptot 45 ⇒The mole fractions of the solute and solvent in the – i is determined experimentally as a ratio of the vapor phase are different from those in the liquid measured colligative property versus the one phase → The vapor phase is enriched with the expected for a nonelectrolyte more volatile component → Condensation of the ∆P= i(X P° ) ∆T = i(k m) vapors leads to a solution which is enriched with the solute solv b b more volatile component → Π = i(MRT) ∆Tf = i(kf m)

8 – For ideal solutions (very dilute solutions), the ideal value of i represents the number of moles of particles to which a mole of the solute dissociates – For real solutions, i is smaller than the ideal value – The deviation is due to clustering of cations and anions (ionic atmosphere) which reduces the “effective” concentration of particles → (i×Conc.) Example: Select the solution with the higher osmotic pressure: 0.1M NaCl or 0.08M CaCl2 → Calculate the effective concentrations: → 0.1M NaCl → i×M = 2×0.1 = 0.2

→ 0.08M CaCl2 → i×M = 3×0.08 = 0.24 (higher)

⇒ 0.08M CaCl2 has higher Π (Π = i×MRT)

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