II III I C. Johannesson I. The Nature of Solutions (p , ) Ch. 13 & 14 - Solutions

C. Johannesson A. Definitions  Solution -  Solution - homogeneous mixture Solvent Solvent - present in greater amount Solute Solute - substance being dissolved

C. Johannesson A. Definitions Solute Solute - KMnO 4 Solvent Solvent - H 2 O

C. Johannesson

Terms C. Johannesson Suspensions – a mixture when particles in a solvent are so large that they settle out unless constantly stirred or agitated. (Ex: muddy water) Colloids – when particles that are intermediate in size between those in solutions and suspensions form mixtures

C. Johannesson B. Solvation  Solvation –  Solvation – the process of dissolving solute particles are separated and pulled into solution solute particles are surrounded by solvent particles

C. Johannesson B. Solvation Strong Electrolyte Non- Electrolyte solute exists as ions only - + salt - + sugar solute exists as molecules only - + acetic acid Weak Electrolyte solute exists as ions and molecules DISSOCIATIONIONIZATION View animation online.animation

C. Johannesson B. Solvation  Dissociation separation of an ionic solid into aqueous ions NaCl(s)  Na + (aq) + Cl – (aq)

C. Johannesson B. Solvation  Ionization breaking apart of some polar molecules into aqueous ions HNO 3 (aq) + H 2 O(l)  H 3 O + (aq) + NO 3 – (aq)

C. Johannesson B. Solvation  Molecular Solvation molecules stay intact C 6 H 12 O 6 (s)  C 6 H 12 O 6 (aq)

C. Johannesson B. Solvation NONPOLAR POLAR “Like Dissolves Like”

C. Johannesson B. Solvation  Soap/Detergent polar “head” with long nonpolar “tail” dissolves nonpolar grease in polar water

C. Johannesson C. Solubility SATURATED SOLUTION no more solute dissolves UNSATURATED SOLUTION more solute dissolves SUPERSATURATED SOLUTION becomes unstable, crystals form concentration

C. Johannesson C. Solubility  Solubility maximum grams of solute that will dissolve in 100 g of solvent at a given temperature varies with temp based on a saturated soln

C. Johannesson C. Solubility  Solubility Curve shows the dependence of solubility on temperature

C. Johannesson C. Solubility  Solids are more soluble at... high temperatures.  Gases are more soluble at... low temperatures & high pressures (Henry’s Law). EX: nitrogen narcosis, the “bends,” soda

II III I C. Johannesson II. Concentration (p ) Ch. 13 & 14 - Solutions

C. Johannesson A. Concentration  The amount of solute in a solution.  Describing Concentration % by mass - medicated creams % by volume - rubbing alcohol ppm, ppb - water contaminants molarity - used by chemists molality - used by chemists

C. Johannesson A. Concentration SAWS Water Quality Report - June 2000

C. Johannesson B. Molality mass of solvent only 1 kg water = 1 L water

C. Johannesson B. Molality  Find the molality of a solution containing 75 g of MgCl 2 in 250 mL of water. 75 g MgCl 2 1 mol MgCl g MgCl 2 = 3.2 m MgCl kg water

C. Johannesson B. Molality  How many grams of NaCl are req’d to make a 1.54m solution using kg of water? kg water1.54 mol NaCl 1 kg water = 45.0 g NaCl g NaCl 1 mol NaCl

C. Johannesson B. Molarity L of solution

Practice  You have 0.8 L of a 0.5 M HCl solution. How many moles of HCl does this solution contain? C. Johannesson

Practice  What volume of 3.00 M NaCl is needed for a reaction that requires 146.3g of NaCl? C. Johannesson

C. Dilution  Preparation of a desired solution by adding water to a concentrate.  Moles of solute remain the same.

C. Johannesson C. Dilution  What volume of 15.8M HNO 3 is required to make 250 mL of a 6.0M solution? GIVEN: M 1 = 15.8M V 1 = ? M 2 = 6.0M V 2 = 250 mL WORK: M 1 V 1 = M 2 V 2 (15.8M) V 1 = (6.0M)(250mL) V 1 = 95 mL of 15.8M HNO 3

C. Johannesson D. Preparing Solutions  500 mL of 1.54M NaCl 500 mL water 45.0 g NaCl mass 45.0 g of NaCl add water until total volume is 500 mL mass 45.0 g of NaCl add kg of water 500 mL mark 500 mL volumetric flask  1.54m NaCl in kg of water

C. Johannesson D. Preparing Solutions Copyright © NT Curriculum Project, UW-Madison (above: “Filling the volumetric flask”)

C. Johannesson D. Preparing Solutions Copyright © NT Curriculum Project, UW-Madison (above: “Using your hand as a stopper”)

C. Johannesson D. Preparing Solutions  250 mL of 6.0M HNO 3 by dilution measure 95 mL of 15.8M HNO 3 95 mL of 15.8M HNO 3 water for safety 250 mL mark combine with water until total volume is 250 mL Safety: “Do as you oughtta, add the acid to the watta!”

C. Johannesson Solution Preparation Lab  Turn in one paper per team.  Complete the following steps: A) Show the necessary calculations. B) Write out directions for preparing the solution. C) Prepare the solution.  For each of the following solutions: 1) 1 L of 0.21M NaCl 2) 0.65m NaCl in mL of water 3) mL of 3.0M HCl from 12.1M concentrate.

II III I C. Johannesson III. Colligative Properties (p ) Ch. 13 & 14 - Solutions

C. Johannesson A. Definition  Colligative Property property that depends on the concentration of solute particles, not their identity

C. Johannesson B. Types  Freezing Point Depression  Freezing Point Depression (  t f ) f.p. of a solution is lower than f.p. of the pure solvent  Boiling Point Elevation  Boiling Point Elevation (  t b ) b.p. of a solution is higher than b.p. of the pure solvent

C. Johannesson B. Types View Flash animation.Flash animation Freezing Point Depression

C. Johannesson B. Types Solute particles weaken IMF in the solvent. Boiling Point Elevation

C. Johannesson B. Types  Applications salting icy roads making ice cream antifreeze cars (-64°C to 136°C) fish & insects

C. Johannesson C. Calculations  t :change in temperature (° C ) k :constant based on the solvent (° C·kg/mol ) m :molality ( m ) n :# of particles  t = k · m · n

C. Johannesson C. Calculations  # of Particles Nonelectrolytes (covalent) remain intact when dissolved 1 particle Electrolytes (ionic) dissociate into ions when dissolved 2 or more particles

C. Johannesson C. Calculations  At what temperature will a solution that is composed of 0.73 moles of glucose in 225 g of phenol boil? m = 3.2m n = 1  t b = k b · m · n WORK: m = 0.73mol ÷ 0.225kg GIVEN: b.p. = ?  t b = ? k b = 3.60°C·kg/mol  t b = (3.60°C·kg/mol)(3.2m)(1)  t b = 12°C b.p. = 181.8°C + 12°C b.p. = 194°C

C. Johannesson C. Calculations  Find the freezing point of a saturated solution of NaCl containing 28 g NaCl in 100. mL water. m = 4.8m n = 2  t f = k f · m · n WORK: m = 0.48mol ÷ 0.100kg GIVEN: f.p. = ?  t f = ? k f = 1.86°C·kg/mol  t f = (1.86°C·kg/mol)(4.8m)(2)  t f = 18°C f.p. = 0.00°C - 18°C f.p. = -18°C

II III I C. Johannesson I. Introduction to Acids & Bases (p ) Ch. 15 & 16 - Acids & Bases

C. Johannesson A. Properties  electrolytes  turn litmus red  sour taste  react with metals to form H 2 gas  slippery feel  turn litmus blue  bitter taste ChemASAP  vinegar, milk, soda, apples, citrus fruits  ammonia, lye, antacid, baking soda

C. Johannesson B. Definitions  Arrhenius - In aqueous solution… HCl + H 2 O  H 3 O + + Cl – AcidsAcids form hydronium ions (H 3 O + ) H HHHH H Cl OO – + acid

C. Johannesson B. Definitions  Arrhenius - In aqueous solution… BasesBases form hydroxide ions (OH - ) NH 3 + H 2 O  NH OH - H H H H H H N NO O – + H H H H base

C. Johannesson B. Definitions  Brønsted-Lowry HCl + H 2 O  Cl – + H 3 O + AcidsAcids are proton (H + ) donors. BasesBases are proton (H + ) acceptors. conjugate acid conjugate base baseacid

C. Johannesson B. Definitions H 2 O + HNO 3  H 3 O + + NO 3 – CBCAAB

C. Johannesson B. Definitions - can be an acid or a base.  Amphoteric - can be an acid or a base. NH 3 + H 2 O  NH OH - CACBBA

C. Johannesson B. Definitions F - H 2 PO 4 - H2OH2O HF H 3 PO 4 H 3 O +  Give the conjugate base for each of the following: - an acid with more than one H +  Polyprotic - an acid with more than one H +

C. Johannesson B. Definitions Br - HSO 4 - CO 3 2- HBr H 2 SO 4 HCO 3 -  Give the conjugate acid for each of the following:

C. Johannesson B. Definitions  Lewis AcidsAcids are electron pair acceptors. BasesBases are electron pair donors. Lewis base Lewis acid

C. Johannesson C. Strength  Strong Acid/Base 100% ionized in water strong electrolyte - + HCl HNO 3 H 2 SO 4 HBr HI HClO 4 NaOH KOH Ca(OH) 2 Ba(OH) 2

C. Johannesson C. Strength  Weak Acid/Base does not ionize completely weak electrolyte - + HF CH 3 COOH H 3 PO 4 H 2 CO 3 HCN NH 3

II III I C. Johannesson Ch. 15 & 16 - Acids & Bases II. pH (p )

C. Johannesson A. Ionization of Water H 2 O + H 2 O H 3 O + + OH - K w = [H 3 O + ][OH - ] = 1.0 

C. Johannesson A. Ionization of Water  Find the hydroxide ion concentration of 3.0  M HCl. [H 3 O + ][OH - ] = 1.0  [3.0  ][OH - ] = 1.0  [OH - ] = 3.3  M Acidic or basic? Acidic

C. Johannesson pH = -log[H 3 O + ] B. pH Scale 0 7 INCREASING ACIDITY NEUTRAL INCREASING BASICITY 14 pouvoir hydrogène (Fr.) “hydrogen power”

C. Johannesson B. pH Scale pH of Common Substances

C. Johannesson B. pH Scale pH = -log[H 3 O + ] pOH = -log[OH - ] pH + pOH = 14

C. Johannesson B. pH Scale  What is the pH of M nitric acid (HNO 3) ? pH = -log[H 3 O + ] pH = -log[0.050] pH = 1.3 Acidic or basic? Acidic

C. Johannesson B. pH Scale  What is the molarity of hydrobromic acid (HBr) in a solution that has a pOH of 9.6? pH + pOH = 14 pH = 14 pH = 4.4 Acidic pH = -log[H 3 O + ] 4.4 = -log[H 3 O + ] = [H 3 O + ] [H 3 O + ] = 4.0  M HBr

II III I C. Johannesson III. Titration (p ) Ch. 15 & 16 - Acids & Bases

C. Johannesson A. Neutralization  Chemical reaction between an acid and a base.  Products are a salt (ionic compound) and water.

C. Johannesson A. Neutralization ACID + BASE  SALT + WATER HCl + NaOH  NaCl + H 2 O HC 2 H 3 O 2 + NaOH  NaC 2 H 3 O 2 + H 2 O Salts can be neutral, acidic, or basic. Neutralization does not mean pH = 7. weak strong neutral basic

C. Johannesson B. Titration  Titration Analytical method in which a standard solution is used to determine the concentration of an unknown solution. standard solution unknown solution

C. Johannesson  Equivalence point (endpoint) Point at which equal amounts of H 3 O + and OH - have been added. Determined by… indicator color change B. Titration dramatic change in pH

C. Johannesson B. Titration moles H 3 O + = moles OH - M  V  n = M  V  n M:Molarity V:volume n:# of H + ions in the acid or OH - ions in the base

C. Johannesson B. Titration  42.5 mL of 1.3M KOH are required to neutralize 50.0 mL of H 2 SO 4. Find the molarity of H 2 SO 4. H3O+H3O+ M = ? V = 50.0 mL n = 2 OH - M = 1.3M V = 42.5 mL n = 1 MV# = MV# M(50.0mL)(2) =(1.3M)(42.5mL)(1) M = 0.55M H 2 SO 4