SolutionsThe composition of the solvent and the solute determine whether a substance will dissolve. Stirring, temperature, and surface area of dissolving particles determine how fast the substance will dissolve.

Heterogeneous = not the same throughoutSuspensions = look uniform when stirred;

however, separate when left alone (Italian dressing, OJ w/ pulp)

Colloids = mixture of very tiny particles dispersed in another substance but do not settle out (gelatin, whipped cream, smoke)

Emulsion = immiscible liquids that are spread throughout one another (mayo, cream, butter)

Homogeneous = same throughoutSolution = 2 or more substances

uniformly spread throughout a single phase

Solute = substance that dissolves in a solution

Solvent = substance that does the dissolving

Hot Water Sugar Sugar Water

SolubilityThe quality or condition of being soluble.

The amount of solute that can be dissolved in a given amount of solvent at a certain temperature to produce a saturated solution

Often expressed in grams of solute per 100g of solvent

Miscible – when two liquids completely dissolve in each other forming a homogeneous solutionWater & ethanol

Immiscible – liquids that are insoluble in one anotherWater & oil

Dissolving ProcessSurface Area = faster dissolving

Loose sugar dissolves faster than sugar cube

Stir/Shake = faster dissolvingAllows molecules to interact with one

another heat = faster dissolving

Solutes dissolve faster in warmer liquidsheat speed collisions and more transfer of

energy

Unsaturated solution = soln that is able to dissolve more solute

Saturated solution = soln that can not dissolve any more solute

Factors affecting solubility

Temperature – as Temp solubilitySuper-saturated solution = soln holding more

dissolved solute than is specified by its solubility at a given temperature

Pressure – strongly influences solubility of gases. Solubility as the partial pressure of the gas above the solution Carbonated beverages are bottled under

pressure of CO2 when opened pressure above liquid and bubbles are released

16.1 Concentrations of solutions

A measure of the amount of solute that is dissolved in a given quantity of solventDilute – one that contains a small amount of

soluteConcentrated – one that contains a large

amount of solute

Molarity – the number of moles of solute dissolved in one liter of solutionM = Moles solute/liters of solutionUnits = moles/L

Calculating molarityA saline solution contains 0.900g NaCl in

exactly 100.0ml of solution what is the molarity of the solution?

Known0.90g NaCl (molar mass 58.453g/mol)

0.90g X (1mol/58.453g) = 0.0154 mol100 mL = 0.1L

Use equation M = moles/liter0.0154mol/0.1L = 0.154M

Calculation 1A solution has a volume of 250mL and

contains 0.70mol NaCl. What is its molarity?

Known 0.70mol NaCl250mL = 0.250L

Use equation M = moles/literM = 0.70mol/0.250L = 2.8M

Calculation 2How many moles of ammonium nitrate are

in 335mL of 0.425M NH4NO3?

Known M = 0.425M335mL = 0.335L

Manipulate equation to solve for moles Moles = Molarity X Liters

Moles = 0.425M X 0.335L = 0.142 moles

Dilutions

Diluting a solution reduces the number of moles of solute per unit volume, but the total number of moles of solute in solution does not changeMoles solute before = moles solute afterM1V1 = M2V2

Preparing a dilute solutionHow many milliliters of 2.00M MgSO4

solution must be diluted with water to prepare 100.00mL of 0.400M MgSO4?

Known M1 = 2.00M

M2 = 0.4M

V2 = 100mL

Manipulate equation to solve for V1 V1 = (M2V2)/M1

V1 = (0.40M X 100.0mL)/2.00M = 20.0mL

Calculation 1How many milliliters of 4.00M KI are needed

to prepare 250mL of 0.760M KI?Known

M1 = 4.00M

M2 = 0.760M

V2 = 250mL

Manipulate equation to solve for V1 V1 = (M2V2)/M1

V1 = (0.760M X 250mL) / 4.00M = 47.5 mL

Calculation 2How many milliliters of 0.20M NaCl are

formed when you dilute a 50mL solution of 1.0M NaCl using water?

KnownV1 = 50mLM2 = 0.20MM1 = 1.0M

Manipulate equation to solve for V2 V2 = (M1V1)/M2

V2 = (1.0M X 50mL) / 0.20M = 250mL

Percent solutions

The concentration of a solution in percent can be expressed in two waysRatio of volume of solute to the volume of the

solution (% (v/v)% volume = volume solute/volume soln X

100%Ratio of the mass of the volume to the mass of the

solution (% (m/m)% mass = mass of solute/mass soln X 100%

Calculating percent solution (v/v)What is the percent by volume of ethanol

(C2H6O) in the final solution when 85mL of ethanol is diluted to a volume of 250mL with water?

Known Volume ethanol 85mLVolume of solution 250mL

% v/v = Volume solute/volume solution X 100%

% v/v = (85mL/250mL) X 100% = 34%

Calculation 1If 10mL of acetone (C3H6O) is diluted with

water to a total solution volume of 200mL, what is the percent by volume of the solution?

Known Volume of acetone 10mLVolume of solution 200mL

% v/v = Volume solute/volume solution X 100%

% v/v = (10mL/200mL) X 100% = 5%

Calculation 2 (m/m)How many grams of K2SO4 would you

need to prepare 1500g of 5% K2SO4?

KnownMass of solution 1500g% by mass 5%

Manipulate equation to solve for mass solute mass solute = mass solution X % by mass

Mass solute = 1500g X 5% = 75g