Systems with Two or More Components: Solutions, Distillation, and Colligative Properties

Classification of Mixtures and Solutions

A mixture, or miscela, is a system composed of two or more components that can exist in a gaseous, liquid, or solid state. These systems are categorized based on the uniformity of their properties. A homogeneous mixture is characterized by chemical-physical properties that are identical at any point within the system. In contrast, a heterogeneous mixture possesses chemical-physical properties that vary depending on the specific point being considered.

Solutions are defined specifically as homogeneous mixtures of two or more components, which may be solid, gaseous, or liquid. In any solution, the component present in the largest quantity is designated as the solvent, which also determines the physical state of the solution. The species present in a smaller quantity is referred to as the solute, or solutes if multiple substances are dissolved.

Several examples illustrate the various physical states of solutions. Air is a gaseous solution consisting of oxygen (O2O_2) and other gases dissolved in nitrogen (N2N_2). Sea water is a liquid solution containing sodium chloride (NaClNaCl) and other salts in water. Wine and beer are liquid solutions of ethanol in water. Seltzer water consists of carbon dioxide (CO2CO_2) in water, also a liquid solution. Solid solutions include brass, which is solid zinc (ZnZn) in solid copper (CuCu), and catalysts like gaseous hydrogen (H2H_2) in solid palladium (PdPd).

Methods of Expressing Solution Concentration

The concentration of a solution is an intensive property that measures the quantity of solute present in a specific amount of solvent or solution. While multiple methods exist to express this value, they all follow the general conceptual formula:

Solute Concentration=Quantity of soluteQuantity of solvent or solution\text{Solute Concentration} = \frac{\text{Quantity of solute}}{\text{Quantity of solvent or solution}}

Concentrations are classified based on the types of units used for the solute and the solution/solvent. Weight/Weight (p/pp/p) expressions include weight percentage (%p/p\%p/p), molality (mm), and mole fraction (xx). Weight/Volume (p/Vp/V) expressions include weight-volume percentage (%p/V\%p/V), molarity (MM), normality (NN), and parts per million (ppmppm). Finally, Volume/Volume (V/VV/V) expressions include volume percentage (%V/V\%V/V).

Molarity and Ion Concentration

Molarity (MM) represents the number of moles of solute dissolved in one liter of solution. The formula is expressed as:

Molarity(M)=moles soluteL of solutionMolarity(M) = \frac{\text{moles solute}}{\text{L of solution}}

For example, if 10.00g10.00\,g of sodium hydroxide (NaOHNaOH) are dissolved in water to form 0.5L0.5\,L of solution, the molarity is calculated by first finding the moles of NaOHNaOH using its molar mass (MM=40.00g/molMM = 40.00\,g/mol):

n=10.00g40.00g/mol=0.25moln = \frac{10.00\,g}{40.00\,g/mol} = 0.25\,mol

M=0.25mol0.5L=0.5MM = \frac{0.25\,mol}{0.5\,L} = 0.5\,M

Conversely, to prepare 300mL300\,mL of a 2M2\,M sodium hydroxide solution, the required moles are found by:

n=2M×0.3L=0.6molesn = 2\,M \times 0.3\,L = 0.6\,moles

The mass required is then:

m=0.6mol×40.00g/mol=24gm = 0.6\,mol \times 40.00\,g/mol = 24\,g

In the case of ionic compounds, the concentration of individual ions must be considered. For 10.00g10.00\,g of calcium chloride (CaCl2CaCl_2) in 1.6L1.6\,L of solution (MM=110.98g/molMM = 110.98\,g/mol):

n=10.00g110.98g/mol=0.09moln = \frac{10.00\,g}{110.98\,g/mol} = 0.09\,mol

M=0.09mol1.6L=0.06MM = \frac{0.09\,mol}{1.6\,L} = 0.06\,M

Upon dissolution according to CaCl2(s)+H2O(l)Ca2+(aq)+2Cl(aq)CaCl_2(s) + H_2O(l) \rightarrow Ca^{2+}(aq) + 2Cl^{-}(aq), the concentration of calcium ions is equal to the molarity of the salt (0.06M0.06\,M), while the concentration of chloride ions is doubled (0.12M0.12\,M).

Percentage and Parts Per Million Concentrations

Weight/weight percentage (%p/p\%p/p) represents the grams of solute dissolved in 100g100\,g of solution. The formula is:

%p/p=g soluteg solution×100\%p/p = \frac{\text{g solute}}{\text{g solution}} \times 100

If 450.0g450.0\,g of an aqueous solution contains 55.0g55.0\,g of potassium hydroxide (KOHKOH), the percentage is:

%p/p=55.0g450.0g×100=12.2%\%p/p = \frac{55.0\,g}{450.0\,g} \times 100 = 12.2\%

A common error occurs when the mass of the solvent is used instead of the total mass of the solution. If 55.0g55.0\,g of KOHKOH is added to 450.0g450.0\,g of water, the total solution mass is 505.0g505.0\,g, making the concentration 10.9%10.9\%, not 12.2%12.2\%.

Volume/volume percentage (%V/V\%V/V) represents the milliliters of solute in 100mL100\,mL of solution. In a 1.5L1.5\,L bottle of wine with 12.5%V/V12.5\%V/V ethyl alcohol, the volume of alcohol is:

V=12.5100×1500mL=187.5mLV = \frac{12.5}{100} \times 1500\,mL = 187.5\,mL

Weight/volume percentage (%p/V\%p/V) represents grams of solute in 100mL100\,mL of solution. A solution with 15g15\,g of salts in 0.8L0.8\,L (800mL800\,mL) of solution has a concentration of 1.875%p/V1.875\%p/V. For Gatorade, which is a 12%p/V12\%p/V saline solution, drinking 1.5L1.5\,L results in the ingestion of 180g180\,g of salts (12.0g:100mL=x:1500mL12.0\,g : 100\,mL = x : 1500\,mL).

Parts per million (ppmppm) represents milligrams of solute in 1L1\,L of solution. For very dilute solutions where the density is approximately 1g/mL1\,g/mL, 1ppm1\,ppm is equivalent to 0.001g0.001\,g of solute in 1000g1000\,g of solution, or 1g1\,g of solute in 1,000,000g1,000,000\,g of solution.

Molality and Mole Fraction

Molality (mm) represents the moles of solute dissolved in 1kg1\,kg of pure solvent. This unit is temperature-independent because it is based on mass. The formula is:

m=moles solutekg solventm = \frac{\text{moles solute}}{\text{kg solvent}}

If 30.0g30.0\,g of sulfuric acid (H2SO4H_2SO_4, MM=98.08g/molMM = 98.08\,g/mol) are dissolved in 300g300\,g of water, the molality is:

n=30.098.08=0.306moln = \frac{30.0}{98.08} = 0.306\,mol

m=0.306mol0.300kg=1.02mm = \frac{0.306\,mol}{0.300\,kg} = 1.02\,m

To convert molarity to molality, the solution density must be known. For a 3.4M3.4\,M solution of sulfuric acid with a density (dd) of 1.200g/mL1.200\,g/mL: the mass of 1L1\,L of solution is 1200g1200\,g. The mass of the solute in that Liter is 3.4mol×98.08g/mol=333.5g3.4\,mol \times 98.08\,g/mol = 333.5\,g. The mass of the solvent is 1200g333.5g=866.5g1200\,g - 333.5\,g = 866.5\,g or 0.8665kg0.8665\,kg. Thus, the molality is 3.4/0.8665=3.924m3.4 / 0.8665 = 3.924\,m.

Mole fraction (xx) is the ratio of the moles of one component to the total moles in the solution. For XsoluteX_{\text{solute}} and XsolventX_{\text{solvent}}, the sum always equals 1. In a solution of 15g15\,g of sodium nitrate (NaNO3NaNO_3, MM=84.99g/molMM = 84.99\,g/mol) in 650g650\,g of water (MM=18.01g/molMM = 18.01\,g/mol), the moles of water are 36.0836.08 and the moles of solute are 0.180.18. The mole fraction of water is 36.08/36.26=0.99536.08 / 36.26 = 0.995, and for the solute it is 0.18/36.26=0.0050.18 / 36.26 = 0.005.

Solution Dilution and Dissolution Processes

Dilute solutions can be prepared from concentrated ones by adding solvent. During dilution, the number of moles of solute remains constant (molesinitial=molesfinalmoles_{initial} = moles_{final}). This leads to the dilution formula:

Mi×Vi=Mf×VfM_i \times V_i = M_f \times V_f

The principle of "like dissolves like" explains chemical solubility. Substances with similar molecular structures exhibit similar intermolecular forces and are soluble in each other. Polar solvents dissolve polar compounds, while non-polar solvents dissolve non-polar compounds. Highly polar and apolar solvents do not mix.

Dissolution is viewed as a reaction: AA+BB2ABAA + BB \rightarrow 2AB, where AAAA is the solvent, BBBB is the solute, and ABAB is the solution. The enthalpy change associated with this is ΔHsol\Delta H_{sol}. In ideal solutions, where there are no specific interactions between solute and solvent, ΔHsol=0\Delta H_{sol} = 0. In real solutions, ΔHsol0\Delta H_{sol} \neq 0. Spontaneity is governed by Gibbs free energy:

ΔGsol=ΔHsolTΔSsol\Delta G_{sol} = \Delta H_{sol} - T\Delta S_{sol}

Because the entropic factor (ΔSsol\Delta S_{sol}) is always positive in dissolution (e.g., NaClNa++ClNaCl \rightarrow Na^+ + Cl^- increases disorder), a process with ΔHsol=0\Delta H_{sol} = 0 is always favorable. If ΔHsol0\Delta H_{sol} \neq 0, spontaneity depends on whether the entropic term outweighs the enthalpy.

Energetics and Hydration

The formation of a solution occurs in three stages: the separation of solute particles (ΔHsolute>0\Delta H_{solute} > 0), the separation of solvent molecules (ΔHsolvent>0\Delta H_{solvent} > 0), and the mixing of solute with solvent (ΔHmix<0\Delta H_{mix} < 0). The total enthalpy is:

ΔHsol=ΔHsolute+ΔHsolvent+ΔHmix\Delta H_{sol} = \Delta H_{solute} + \Delta H_{solvent} + \Delta H_{mix}

When ionic crystals dissolve in water, the water dipoles orient around surface ions. Negative dipole ends point toward positive ions, and vice versa. Intense ion-dipole interactions lead to the formation of hydrated ions and the disintegration of the ionic crystal. A hydrated ion is surrounded by a hydration shell. This process is always exothermic. The hydration heat depends on charge density (charge/surface area ratio). Higher charges and smaller ionic radii lead to stronger attraction and higher hydration energy.

Most ionic compound solubilization is endotermic, yet remains spontaneous because the TΔSsol-T\Delta S_{sol} term is larger than ΔHsol\Delta H_{sol}. In apolar solvents like hexane, ionic compounds do not dissolve because ion-solvent interactions are too weak compared to the intense bonds within the ionic solid.

Solubility and Environmental Factors

Solubility is the maximum quantity of solute that can dissolve in a given volume of solvent. A saturated solution contains this maximum amount at a given temperature in the presence of undissolved solute (the bottom body), establishing a dynamic equilibrium where the rate of dissolution equals the rate of deposition. An unsaturated solution contains less than the equilibrium amount.

For most ionic compounds, solubility increases with temperature. Recrystallization is a purification technique that exploits this: a near-saturated solution is created at high temperatures and then cooled, causing the solid to precipitate and reform as pure crystals. Exceptions include compounds containing SO32SO_3^{2-}, SO42SO_4^{2-}, PO43PO_4^{3-}, and AsO43AsO_4^{3-}.

Gas solubility in water generally decreases as temperature increases. Consequently, carbonated drinks are more "fizzy" when refrigerated, and warm lakes contain less dissolved O2O_2 than cold lakes. However, in organic solvents, gases often become more soluble at higher temperatures. Gas solubility in water increases with molecular weight due to increased polarizability and is generally low for noble gases.

Henry's Law, formulated in 1803, describes gas solubility relative to pressure: the solubility of a gas is directly proportional to the pressure of the gas above the liquid:

C=k×PgasC = k \times P_{gas}

Increasing pressure increases the frequency of molecular collisions with the surface, forcing more gas into the solution. Opening a carbonated drink causes a sudden drop in pressure, resulting in the rapid release of CO2CO_2.

Colligative Properties and Raoult's Law

Colligative properties are physical properties of solutions that depend exclusively on the number of solute particles dissolved, regardless of their chemical nature. These include vapor pressure lowering, boiling point elevation, freezing point depression, and osmotic pressure.

Raoult's Law states that dissolved solutes lower the vapor pressure of the solution compared to the pure solvent. This applies to ideal solutions (ΔHsol=0\Delta H_{sol} = 0) and non-ideal solutions with non-volatile solutes. For a two-component system:

p=pA0xA+pB0xBp = p_A^0 \cdot x_A + p_B^0 \cdot x_B

For non-volatile solutes, where pBp_B is negligible:

p=pA0xAp = p_A^0 \cdot x_A

Since xA+xB=1x_A + x_B = 1, rearranging gives the vapor pressure lowering:

Δpsolution=xsolute×psolvent0\Delta p_{solution} = x_{solute} \times p_{solvent}^0

This reduction in vapor pressure shifts the state diagram of the solvent: it expands the liquid range while reducing the solid and vapor ranges. Consequently, the boiling point (TebT_{eb}) increases and the melting point (TfusT_{fus} or TcrT_{cr}) decreases.

Boiling and Freezing Point Changes

Boiling point elevation (ΔTeb\Delta T_{eb}) occurs because a non-volatile solute reduces the number of solvent particles on the surface available to vaporize. Higher temperatures are required to establish the equilibrium where vapor pressure equals atmospheric pressure:

ΔTeb=TfinalTinitial=Keb×m\Delta T_{eb} = T_{final} - T_{initial} = K_{eb} \times m

Freezing point depression (ΔTcr\Delta T_{cr}) occurs because the solute particles interfere with the ability of solvent molecules to adhere to the solid surface. Lower kinetic energy (lower temperature) is needed to capture molecules into the solid phase:

ΔTcr=TfinalTinitial=Kcr×m\Delta T_{cr} = T_{final} - T_{initial} = -K_{cr} \times m

KebK_{eb} and KcrK_{cr} are proportionality constants specific to the solvent. Molality (mm) is used because it does not change with temperature as volume does. In pure water, temperature stays constant during phase changes, but in solutions, the temperature varies continuously as the solvent is removed (via boiling or freezing), thereby increasing the solute concentration.

Distillation Techniques and Azeotropes

Distillation separates substances based on differences in volatility (boiling points). Simple distillation is used when the components' boiling points differ by at least 70C70^{^\circ}C. The more volatile component boils off, passes through a condenser, and is collected as distillate.

Fractional distillation is required when the temperature difference is less than 25C25^{^\circ}C. It utilizes a rectification column to perform multiple vaporization-condensation cycles. According to Raoult’s and Dalton’s laws, the vapor phase is always richer in the more volatile component. For a mixture with xA=0.7x_A = 0.7 (pA0=500.0torrp_A^0 = 500.0\,torr) and xB=0.3x_B = 0.3 (pB0=250.0torrp_B^0 = 250.0\,torr), the initial vapor mole fraction is xAv=0.82x_A^v = 0.82. Condensing and re-boiling this enriched mixture leads to progressively higher concentrations until pure component A is obtained.

Non-ideal solutions show deviations from Raoult’s Law. Positive deviations occur in endothermic dissolutions (ΔHsol>0\Delta H_{sol} > 0), where ABA-B interactions are weaker than AAA-A and BBB-B; this results in higher vapor pressure and a minimum boiling point azeotrope. Negative deviations occur in exothermic dissolutions (ΔHsol<0\Delta H_{sol} < 0), where ABA-B interactions are stronger; this leads to lower vapor pressure and a maximum boiling point azeotrope. Azeotropic mixtures cannot be separated by distillation because the liquid and vapor phases have the same composition at point C.

Eutectic Mixtures and Osmosis

An eutectic mixture consists of two or more substances that melt at a temperature (TeutT_{eut}) lower than the melting points of the individual components. Cooling a solution (H2O+BH_2O + B) results in the separation of pure ice, enriching the remaining liquid in BB until the eutectic point (EE) is reached, where both components solidify together. For H2OH_2O and NaClNaCl, the eutectic point occurs at 21.3C-21.3^{^\circ}C with 23.3%p/p23.3\%p/p of salt. Practical applications include road de-icing with NaClNaCl or CaCl2CaCl_2, using ethylene glycol as aircraft anti-freeze, and the protection of fruit juice from freezing due to its solute content.

Osmosis is the spontaneous movement of solvent through a semipermeable membrane from a less concentrated solution to a more concentrated one. The membrane allows solvent but not large solute particles or socialized ions through. Osmotic pressure (π\pi) is the pressure required to stop this flow. For dilute solutions of non-electrolytes:

πV=nRTπ=MRT\pi V = nRT \rightarrow \pi = MRT

Solutions with equal osmotic pressure are isotonic. A hypertonic solution has higher pressure (causing solvent to flow out of a cell), while a hypotonic solution has lower pressure (causing solvent to flow into a cell). Reverse osmosis occurs if external pressure exceeding π\pi is applied, reversing the flow to produce pure water; this is used in desalinization.

Electrolytes and the van’t Hoff Factor

Electrolytes are substances (acids, bases, salts) that dissociate into ions in water, enabling the conduction of electricity. Strong electrolytes (e.g., HNO3HNO_3, NaClNaCl) dissociate completely (α=1\alpha = 1), while weak electrolytes (e.g., CH3COOHCH_3COOH, NH3NH_3) dissociate partially. Non-electrolytes (e.g., glucose, ethanol) do not dissociate.

Svante Arrhenius developed the theory of electrolytic dissociation to explain why colligative properties in electrolyte solutions deviate from theoretical values. For example, a 0.100m0.100\,m NaClNaCl solution shows nearly double the freezing point depression (0.361C-0.361^{^\circ}C) compared to a 0.100m0.100\,m glucose solution (0.186C-0.186^{^\circ}C).

Van’t Hoff introduced the factor ii to measure this deviation, defined as the ratio of the measured property to the expected property if the solute were a non-electrolyte:

i=moles of particles in solutionmoles of formula units dissolvedi = \frac{\text{moles of particles in solution}}{\text{moles of formula units dissolved}}

For weak electrolytes, the degree of dissociation α\alpha is used, where ν\nu is the number of ions produced per formula unit. The total moles of particles is ntot=n[1+α(ν1)]n_{tot} = n[1 + \alpha(\nu - 1)], meaning:

i=1+α(ν1)i = 1 + \alpha(\nu - 1)

All colligative property formulas incorporate this factor for electrolytes:

π=i×M×RT\pi = i \times M \times RT

Δp=i×xsolute×psolvent0\Delta p = i \times x_{solute} \times p_{solvent}^0

ΔTeb=i×Keb×m\Delta T_{eb} = i \times K_{eb} \times m

ΔTcr=i×Kcr×m\Delta T_{cr} = -i \times K_{cr} \times m