SOLUTIONSWe all want them – but few of us have them.

Solutions

A solution is a completely homogeneous mixture.

Solutions have one or more solutes (dissolvee) and a solvent (dissolver).

When a substance dissolves it is soluble

Liquids that mix are miscible, those that don't are immiscible

Characteristics of solutions

Fluid solutions are clear (transparent).

Dissolving is a physical change. Dissolved substances retain their chemical identities.

The physical attributes of the solvent and solute will be altered (physical state, mp, etc).

Types of solutions

Solutions can involve any state of matter.

Liquid/liquid Solid/solid Gas/gas Solid in liquid Gas in liquid

Aqueous solutions - ionic

Solvation of ionic substances

Aqueous solutions - ionic

Solvation involves dissociation of the solid ionic solute.

Aqueous solutions - ionic

The energy needed to break strong ionic bonds comes from the energy liberated when solvent/ solute interactions are formed.

Water, being polar, has partially positive and negative regions which are attracted to the positive and negative ions of the ionic solute.

Aqueous solutions - ionic

Ionic salts are insoluble in non-polar solvents

Aqueous solutions - molecular Solvation of molecular substances. Molecular substances do not have

strong interactions to overcome. Dissociation in solution does not occur. Neutral molecules are separated from each

other. Molecules with polar regions will be

soluble in water. Examples – ethanol (CH3CH2OH), sucrose

Aqueous solutions - molecular Sucrose

Aqueous solutions - molecular Polar substances are soluble in water

because the polar water molecule can attach to the polar portions of the solute molecule.

Nonpolar molecules are not soluble in water because water is more attracted to itself than the potential solute.

Nonpolar molecules usually do not contain oxygen – example methane (CH4).

Aqueous solutions - molecular Molecule polarity depends on bond polarity

and geometry. Bond polarity depends on electronegativity

differences. Miscibility – Liquid solutes

Liquids which mix in all proportions are said to be miscible.

The liquid present in larger amounts is considered the solvent.

When mixing unlike liquids, volumes do not necessarily add.

Solubility

Rules of Thumb Like dissolves like Polar dissolves polar, nonpolar dissolves

nonpolar Measuring solubilty

Accepted unit is g/100 g solvent. Temperature must be specified – may be

0º C, may be 20º C, may be 100ºC

Solubility graph

Factors affecting solubility

Temperature Increase in temperature usually means

increased solubility for a solid or liquid solute. Supersaturation

A solution carefully cooled below temperature where solute is soluble

Disturbing solution or adding seed crystal causes rapid crystallization with attendant evolution or absorption of heat

Gases become less soluble with increasing temperature.

Factors affecting solubility

Pressure Pressure does not affect the solubility of

solid or liquid solutes. Gases become more soluble with

increasing pressure Henry's Law: Gas solubility is directly

related to pressureP = kC

P = pressure of dissolved gas over solution k = constant characteristic of system C = concentration (solubility)

Henry’s Law problem

A soft drink is bottled so that a bottle at 25ºC contains CO2 at a pressure of 5.0 atm over the liquid. Assuming that the partial pressure of CO2 in the atmosphere is 4.0x10-4atm, calculate the equilibrium concentration of CO2 in the soda both before and after the bottle is opened. The Henry's Law constant for CO2 in aqueous solution is 32 L‧atm/mol at 25ºC.

Solution to Henry’s Law problem Before opening: P = kC; k = 32 L‧

atm/mol, and P = 5.0 atm. So CCO2 = P/C = 5.0atm/32 L‧atm/mol

= 0.16M After opening: P = kC; k = 32

L‧atm/mol, and P = 4.0x10-4 atm. So CCO2 = P/C = 4.0x10-4atm/32

L‧atm/mol= 1.3x10-5M

Factors affecting solubility

Nature of solute and solvent Relative polarities (“like dissolves like”)

Factors affecting rate of solution Temperature – higher temperature

means faster dissolving Particle size – smaller particles mean

faster solution Stirring – stirring increases solution rate

Concentration

Per cent concentration Weight % (w/w)

w/w% = (mass solute)/(mass solution) mass solution = mass solute + mass solvent

Volume % (v/v) v/v% = (volume solute)/(volume solution) volume solution ≠ volume solution + volume

solvent!!

% concentration problem

A solution is prepared by mixing 1.00 g ethanol (C2H5OH) with 100.0 g water to give a final volume of 101 mL. Calculate the w/w and v/v % concentrations. (The density of ethanol is 0.789g/mL)

Solution to concentration problem w/w: [1.00g eth/101g soln]x100% =

0.990%w/w v/v: volume of ethanol is 1.00g/0.789g/mL

= 1.27mL[1.27mL/101mL]x100% = 1.26%v/v

Molarity

Unit: moles/(liter solution) Symbol: M Making a solution: Solute is

measured, dissolved in a small amount of water and diluted to desired volume in a volumetric flask.

Molarity examples

What mass of KCl is necessary to make 300 mL of a 0.15 M solution?

Solution: 0.15 M x 0.300 L = 0.045 mol KCl

molar mass of KCl = 74.6 g/mol0.0450 mol x 74.6 g/mol = 3.4 g KCl

Molarity example

You have 458 mL of a 0.29 M solution of sodium hydroxide. What mass of sodium hydroxide is contained therein?

Solution: 0.458 L x 0.29 M = 0.13 mol NaOH0.13 mol x 40. g/mol = 5.2 g NaOH

Molality

Unit: (moles solute)/(kg solvent) Symbol: m Examples: How would you prepare a

0.17 m solution of sodium phosphate using 800. mL water?

Solution: 0.17 mol/kg x .800 kg = 0.14 mol

Molar mass of Na3PO4 =163.9 g/m

0.14 mol x 163.9 g/mol = 23g Na3PO4

Molality example

How would you make 1200 g of a 0.235 m calcium chloride solution?

Solution: CaCl2: 111 g/mol 1.200 kg = mass solvent + mass solute mass solute = x 0.235mol CaCl2 = x/111g/mol

1 kg solvent (1.200 kg – x) Multiply both sides by 111

(111) (111)

Molality example solution

0.0261kg CaCl2 = x_____

1 kg solvent (1.200 kg – x)(Note that the mass of the solute must be in

kg) .0261(1.200 – mass solute) = mass solute .03132–.0261(mass solute) = mass solute .03132/1.0261 = mass solute = 0.0305 kg CaCl2 (30.5g)

Mole Fraction

XA = nA/(nA + nB) No unit (all units cancel) Example. What is the mole fraction of HCl in

concentrated hydrochloric acid (37%w/w)?37% = 37g HCl/(37g HCl + 63g H2O)

Change all masses to moles(37/36.5)/(37/36.5 + 63/18)

XHCl = 0.22 (22% of the particles in concentrated HCl are hydrogen chloride, neglecting dissociation)

Reactions in solution

Precipitates and ionic equations Overall equation

Pb(NO3)2(aq)+2KI(aq)PbI2(s)+2KNO3(aq)

Ionic equationPb+2

(aq)+2NO3‑(aq)+2K++2I‑(aq)PbI2(s)+2K+

+2NO3‑(aq)

Net ionic equation – eliminate spectator ions

Pb+2(aq) + 2I‑(aq) PbI2(s)

Reactions in solution

Other reactions may be driven by the formation of a gas.

2HCl(aq) + Na2CO3(aq) CO2(g) + H2O(l) + 2NaCl(aq)

overall equation2H++2Cl‑(aq)+2Na++CO3

‑2(aq)CO2(g)+H2O(l)+2Na+ +2Cl‑(aq)

ionic equation2H+ + CO3

‑2(aq) CO2(g) + H2O(l)

net ionic equation

Reactions in solution

Write overall, ionic and net ionic equations for the following reaction.

zinc metal reacts with hydrochloric acid to give aqueous zinc chloride and

hydrogen gasZn + 2HCl(aq) ZnCl2 + H2

Zn + 2H+(aq) + 2Cl-(aq) Zn+2

(aq) + 2Cl-(aq) + H2(g)

Zn + 2H+(aq) Zn+2

(aq) + H2(g)

Reactions in solution

Reactions can also be driven by formation of a complex, usually brightly colored.Cu2+

(aq ) + 4 NH3(aq ) Cu(NH3)42+

(aq )

0.1M Cu(II)0.1M Cu(II) with ammonia added

Reactions in solution

Iron solutions form a complex with thiocyanate (SCN-)

Fe+3 + SCN- FeSCN+2

0.1M Fe+30.1M Fe+3 with thiocyanate added

Reactions in solution

Formation of a small covalent molecule will also drive a reaction.2KOH(aq) +H2SO4(aq) K2SO4(aq) +H2O

Use the solubility table (Appendix D) to decide how to write reaction equations for precipitation reactions.

K2S(aq) + Pb(NO3)2(aq) NaOH(aq) + CuSO4(aq)

PbS(s) + 2KNO3(aq)

Na2SO4(aq)+Cu(OH)2(s)

Colligative properties

From Latin “colligare”, to bind together Colligative properties depend on the

number of solute particles present Molecular substances give one mole of

particles per mole substance Salts give more than one mole particles per

mole substance because of dissociation Number of particles per formula unit = i

(van’t Hoff factor)

Colligative properties

Vapor pressure lowering Solute particles take the place of some

of the solvent particles at the surface Vapor pressure of liquid is lowered by

presence of a solute Extent of lowering depends on number

of solute particles present

Colligative properties

Boiling point elevation Solutions have higher boiling points than

pure solvents. This is true with solid solutes and heavier liquid solutes.

Other liquid solutes may form azeotropes, which are mixtures with lower boiling points than either solute or solvent – example 95% ethanol/water.)

Solutes raise the boiling point of liquids because they lower the vapor pressure. When Pv = Pa, vaporization (boiling) occurs.

Colligative properties

Boiling point elevation depends on the number of particles present. (van't Hoff factor, i)

Sodium chloride elevates the bp of water twice as much per mole as sucrose, for it makes two particles per mole. (i = 2 for dilute solutions)

NaCl Na+ + Cl-

Colligative properties

Freezing point depression Solutes lower the freezing point of liquid

solvents. Solutes interfere with solvents’ ability to associate as a solid and crystallize.

Freezing point depression also depends on number of particles.

Osmosis and osmotic pressure Solutions of different concentrations on

either side of a semi-permeable membrane exert a net pressure toward the more concentrated solution

Colligative properties

Solvent moves so as to dilute the more concentrated solution.

Osmotic pressure is the pressure exerted by the greater height of solution on the concentrated side.

Colligative properties

Reverse osmosis is used to purify water.

Colligative properties

Calculations Colligative property calculations use

molality as the unit of concentration, for it does not depend on volume.

Colligative properties vary directly with concentration.

BP elevation: Tb = ikbm, where kb = the boiling point

elevation constant for that liquid.

Colligative properties

kb for water is 0.51ºC/molal FP depression:

Tf = ikfm kf for water is 1.86ºC/molal

Compare the freezing points of 1.5 molal aqueous solutions of NaCl and CaCl2. NaCl: Tf = (2)1.86ºC/m (1.5m) = 5.6ºC. FP = -

5.6ºC CaCl2: Tf = (3)1.86ºC/m (1.5m) = 8.4ºC FP = -

8.4ºC

Colligative properties

Molecular weight calculations Solving for molality can yield the molecular

weight of a substance if the mass is known. Example. 1.235 g of a molecular

substance dissolved in 100.0 g benzene lowers the freezing point of benzene by 0.468ºC. What is the molecular weight of the substance? kf for benzene is 5.12 ºC/molal.

Colligative properties

Solution: Tf = ikfm

m = mol/Kg solvent; mol = mass/mm Tf = ___ikfmass___

(mm)(Kg solvent) mm = ___ikfmass___

(Tf)(Kg solvent)

mm = _(1)(5.12ºC/m)(1.235g)____ = 135g/mol

(0.100Kg benzene)(0.468ºC)

Nonhomogeneous mixtures

Suspensions Particles larger than 103 nm (1 m or 10-6 m) Particles will settle out on standing Often opaque Examples: paint, muddy water, orange juice

Colloids Particles between 1 nm and 1m Particles will not settle out on standing

Nonhomogeneous mixtures

May appear clear, but exhibits Tyndall effect

Examples: fog, blood, milk, soapy water