PRINCIPLES OF CHEMISTRY I

CHEM 1211

CHAPTER 4

DR. AUGUSTINE OFORI AGYEMANAssistant professor of chemistryDepartment of natural sciences

Clayton state university

CHAPTER 4

CHEMICAL REACTIONS IN SOLUTION

- A homogeneous mixture of two or more substances

Solvent - The substance present in the greatest quantity

Solute- The other substance(s) dissolved in the solvent

SOLUTION

- Solutions can exist in any of the physical states

Solid Solution- dental fillings, metal alloys (steel), polymers

Liquid Solution- sugar in water, salt in water, wine, vinegar

Gas Solution- air (O2, Ar, etc. in N2),

- NOx, SO2, CO2 in the atmosphere

SOLUTION

- A solution in which water (H2O) is the solventNaCl solution: solvent is H2O and solute is NaCl

Hydrophilic- Substances that dissolve in water

- Water loving (NaCl)

Hydrophobic- Substances that do not dissolve well in water

- Water fearing (hydrocarbons)

AQUEOUS SOLUTION

- Ions make aqueous solutions good conductors of electricity

- Solution conductivity indicates the presence of ions

AQUEOUS SOLUTION

Ionic Compounds

- Form ions in aqueous solution (dissociate into component ions)

Example- NaCl solution contains Na+ and Cl- ions

NaCl(aq) → Na+(aq) + Cl-(aq)- Each ion is surrounded by water molecules

- Good conductor of electricity

AQUEOUS SOLUTION

Solvation Process

- Ions in aqueous solution are surrounded by the H2O molecules

- The O atom in each H2O molecule has partial negative charge (δ-) - Attract positive ions

- The H atoms have partial positive charge (δ+)- Attract negative ions

- Cations and anions are prevented from recombining

- Ions disperse uniformly throughout the solution (homogeneous)

AQUEOUS SOLUTION

Molecular Compounds

- Most molecular compounds do not form ions in aqueous solution- The molecules disperse throughout the solution

- Molecules are surrounded by H2O molecules

Example- Sucrose solution contains neutral sucrose molecules

- Each molecule is surrounded by water molecules - Poor conductor of electricity

- A few molecular compounds form ions in aqueous solution- HCl dissociates into H+(aq) and Cl-(aq)

- HNO3 dissociates into H+(aq) and NO3-(aq)

AQUEOUS SOLUTION

- A solution in which another substance other than water is the solvent

ExamplesAlcohol

petroleum etherPentane

Carbon tetrachloride

NONAQUEOUS SOLUTION

The rate at which solutes dissolve can be increased by

- Grinding or crushing solute particles (size reduction)

- Heating

- Stirring or agitation

RATE OF DISSOLUTION

- Substances whose aqueous solutions contain ions NaCl(aq) → Na+(aq) + Cl-(aq)

- Two categories: strong and weak electrolytes

Strong Electrolytes- Solutes that completely or nearly completely

ionize when dissolved in water

Salts: NaCl, NH4Cl, KBr, NaNO3

Strong acids: HCl, HNO3, H2SO4

Strong Bases: NaOH, KOH, Ca(OH)2

ELECTROLYTES

- Substances whose aqueous solutions contain ionsNaCl(aq) → Na+(aq) + Cl-(aq)

- Two categories: strong and weak electrolytes

Weak Electrolytes- Only a small fraction of solutes ionize when

dissolved in water (exhibit a small degree of ionization)

Weak acids: acetic acid (HC2H3O2), citric acid (C6H8O7)Weak bases: ammonia (NH3) methylamine, cocaine, morphine

ELECTROLYTES

- Single arrow is used to represent ionization of strong electrolytesH2SO4(aq) → H+(aq) + HSO4

-(aq)- Ions have no tendency of recombining to form H2SO4

- Double arrow is used to represent ionization of weak electrolytesHC2H3O2(aq) ↔ H+(aq) + C2H3O2

-(aq)- This implies reaction occurs in both directions

- Chemical equilibrium is when there is a balance in both directions

ELECTROLYTES

NONELECTROLYTES

- Substances whose aqueous solutions do not contain ions

ExamplesMany molecular compounds

Sucrose (C12H22O11) ethanol (C2H5OH)

- A measure of how much of a solute can be dissolved in a solventat a given temperature

- Units: grams/100 mL

ExampleSolubility of sugar in water at 20 oC is 204 g/100 mL H2O

Three factors that affect solubility- Temperature

- Pressure- Polarity

SOLUBILITY

Unsaturated Solution- More solute can still be dissolved at a given temperature

Saturated Solution- No more solute can be dissolved at a given temperature

Supersaturated Solution- Too much solute has temporarily been dissolved

(more than solute solubility)

Precipitate- Solute (solid) that falls out of solution

SOLUBILITY

The best way to determine the solubility of a substance is by experiment

- Most nitrates (NO3-) are soluble

- Most salts of alkali metals (Group 1A), ammonium (NH4+),

acetates (C2H3O2-), and perchlorates (ClO4

-) are soluble

- Most salts containing Cl-, Br-, and I- are soluble Exceptions: salts of Ag+, Hg2

2+, Pb2+

SOLUBILITY RULES

The best way to determine the solubility of a substance is by experiment

- Most sulfates (SO42-) are soluble

Exceptions: BaSO4, PbSO4, Hg2SO4, SrSO4

- Most hydroxides (OH-) are slightly solubleHydroxides of Ba2+, Sr2+, and Ca2+ are marginally soluble

- Most salts containing S2-, CO32-, PO4

3-, CrO42- are insoluble

Exceptions: salts of alkali metals and NH4+

SOLUBILITY RULES

- Reactions that result in the formation of an insoluble product

- The insoluble product (solid) is known as the precipitate

- These products have very low solubility in water

- Attraction between the oppositely charged ions is so strong that water molecules cannot separate them

- A solute is insoluble if less than 0.01 mol of the solute dissolves in 1 L of solvent

PRECIPITATION REACTIONS

To predict solubility

- Examine the reactants

- Identify the ions present

- Predict the products

- Identify which are soluble and which are insoluble

PRECIPITATION REACTIONS

PRECIPITATION REACTIONSExample

AgNO3(aq) + KCl(aq) → white precipitate

- Ions present: Ag+, NO3-, K+, Cl-

- Possible combinations: AgNO3, AgCl, KCl, KNO3

- Predict products: AgCl and KNO3

- KNO3 is soluble and AgCl is not

AgNO3(aq) + KCl(aq) → AgCl(s) + KNO3(aq)

- When all soluble strong electrolytes are shown as ions

- Chemical equation is balanced

- Soluble compounds (aq) are separated into ions (only strong electrolytes)

- Insoluble compounds (s), liquids (l), and gases (g) are NOT separated into ions

IONIC EQUATIONS

Complete ionic equation

- When all ions in both reactants and products are shown

AgNO3(aq) + KCl(aq) → AgCl(s) + KNO3(aq)

Ag+(aq) + NO3-(aq) + K+(aq) + Cl-(aq)

→

AgCl(s) + K+(aq) + NO3-(aq)

IONIC EQUATIONS

Net Ionic Equation- When spectator ions are cancelled from the complete ionic equation- Net charge on reactant side must equal net charge on product side

Ag+(aq) + NO3-(aq) + K+(aq) + Cl-(aq)

→ AgCl(s) + K+(aq) + NO3

-(aq)

Ag+(aq) + Cl-(aq) → AgCl(s)

- Some ions appear on both reactant and product sides- These ions play no direct role in the reaction

- These ions are called spectator ions

IONIC EQUATIONS

Neutralization Reaction

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

Complete Ionic EquationH+(aq) + Cl-(aq) + Na+(aq) + OH-(aq)

→ Na+(aq) + Cl-(aq) + H2O(l)

Net Ionic EquationH+(aq) + OH-(aq) → H2O(l)

IONIC EQUATIONS

CONCENTRATION OF SOLUTIONS

- The amount of solute dissolved in a given quantity of solution

MOLARITY (M)

- The number of moles of solute per liter of solution

Lsolutionofvolume

solutemolesMolarity

- A solution of 1.00 M (read as 1.00 molar) contains 1.00 mol of solute per liter of solution

CONCENTRATION OF SOLUTIONS

Calculate the molarity of a solution made by dissolving 2.56 g ofNaCl in enough water to make 2.00 L of solution

- Calculate moles of NaCl using grams and molar mass- Convert volume of solution to liters

- Calculate molarity using moles and liters

NaClmol0.0438NaClg58.44

NaClmol1xNaClg2.56

mol/L)(orM0.0219solutionL2.00

NaClmol0.0438Molarity

CONCENTRATION OF SOLUTIONS

After dissolving 1.56 g of NaOH in a certain volume of water, the resulting solution had a concentration of 1.60 M. Calculate the

volume of the resulting NaOH solution

- Convert grams NaOH to moles using molar mass- Calculate volume (L) using moles and molarity

NaOHmol0.0380NaOHg41.00

NaOHmol1xNaOHg1.56

solutionL0.0237NaOHmol1.60

solutionLxNaOHmol0.0380solution Volume

CONCENTRATION OF IONS

Consider:1.00 M NaCl: 1.00 M Na+ and 1.00 M Cl-

1.00 M ZnCl2: 1.00 M Zn2+ and 2.00 M Cl-

1.00 M Na2SO4: 2.00 Na+ and 1.00 M SO42-

Square brackets are commonly used to represent concentration

The concentrations of Na+ and Cl- abovemay be represented as

[Na+] = 1.00 M and [Cl-] = 1.00 M

CONCENTRATION OF IONS

Calculate the number of moles of Na+ and SO42- ions in 1.50 L

of 0.0150 M Na2SO4 solution

0.0150 M Na2SO4 solution contains:2 x 0.0150 M Na+ ions and 0.0150 M SO4

2- ions

moles Na+ = 2 x 0.0150 M x 1.50 L = 0.0450 mol Na+

moles SO42- = 0.0150 M x 1.50 L = 0.0225 mol SO4

2-

DILUTION

Consider a stock solution of concentration M1 and volume V1

If water is added to dilute to a new concentration M2 and volume V2

moles before dilution = moles after dilution

M1V1 = M2V2

Calculate the volume of 3.50 M HCl needed to prepare 500.0 mL of 0.100 M HCl

(3.50 M)(V1) = (0.100 M)(500.0 mL)

V1 = 14.3 mL

CHEMICAL ANALYSIS (TITRATIONS)

Volumetric Analysis- Analysis by volume- Acid-base titrations

Gravimetric Analysis- Analysis by mass

- Determination of halides by addition of silver nitrateCl- + AgNO3 → AgCl (white ppt) + NO3

-

- Determination of sulfates by addition of barium chlorideBaCl2 + SO4

2- → BaSO4 (white solid) + 2Cl-

CHEMICAL ANALYSIS (TITRATIONS)

Calculate the concentration of NaOH solution if 24.50 mL of this base is needed to neutralize 12.00 mL of 0.225 M HCl solution

- Write balanced equation and determine mole ratio

- Calculate moles of HCl (convert mL to L)

- Determine moles of NaOH

-Calculate molarity of NaOH

CHEMICAL ANALYSIS (TITRATIONS)

NaOH + HCl → NaCl + H2O

1 mol NaOH : 1 mol HCl

Volume HCl = 12.00 mL = 0.01200 L

mol HCl = 0.225 M x 0.01200 mL = 0.00270 mol = mol NaOH

NaOHM0.110NaOHL1

NaOHmL1000x

NaOHmL24.50

NaOHmol0.00270NaOHMolarity

CHEMICAL ANALYSIS (TITRATIONS)

How many grams of KOH are needed to neutralize 25.00 mL of 0.250 M H2SO4 solution

- Write balanced equation and determine mole ratio

- Calculate moles of H2SO4

- Determine moles of KOH

- Calculate grams of KOH using molar mass

CHEMICAL ANALYSIS (TITRATIONS)

2KOH + H2SO4 → K2SO4 + 2H2O

2 mol KOH : 1 mol H2SO4

mol H2SO4 = 0.250 M x 0.02500 L = 0.00625 mol

mol KOH = 2 x 0.00625 mol = 0.0125 mol

g0.701KOHmol1

KOHg56.11xKOHmol0.0125KOHmass