General properties ACIDS Taste sour Turn litmus React with active metals – Fe, Zn React with bases BASES Taste bitter Turn litmus Feel soapy or slippery (react with fats to make soap) React with acids blue to redred to blue

Acid Nomenclature Review Binary  Ternary An easy way to remember which goes with which… “In the cafeteria, you ATE something ICky”

Acid Nomenclature Flowchart

HBr (aq)HBr (aq) H 2 CO 3H 2 CO 3 H 2 SO 3H 2 SO 3  hydrobromic acid  carbonic acid  sulfurous acid Acid Nomenclature Review

Name ‘Em! HI (aq)HI (aq) HCl (aq)HCl (aq) H 2 SO 4H 2 SO 4 HNO 3HNO 3 H 3 PO 4H 3 PO 4

Definitions Acids – produce H + Bases - produce OH - Acids – donate H + Bases – accept H + Acids – accept e - pair Bases – donate e - pair Arrehenius Bronsted-Lowry Lewis only in water any solvent used in organic chemistry, wider range of substances

Examples Arrhenius Bronsted-Lowry Lewis HCl NaOH HClNH 3 :NH 3 BF 3 HCN The hydrogen ion in aqueous solution H + + H 2 O  H 3 O + (hydronium ion)

Neutralization In general: Acid + Base  Salt + Water All neutralization reactions are double replacement reactions. HCl + NaOH  NaCl + HOH HCl + Mg(OH) 2  H 2 SO 4 + NaHCO 3 

HA Let’s examine the behavior of an acid, HA, in aqueous solution. What happens to the HA molecules in solution?

HA H+H+ A-A- Strong Acid 100% dissociation of HA Would the solution be conductive?

HA H+H+ A-A- Weak Acid Partial dissociation of HA Would the solution be conductive?

HA H+H+ A-A- Weak Acid HA  H + + A - At any one time, only a fraction of the molecules are dissociated.

Strong and Weak Acids/Bases Strong acids/bases – 100% dissociation into ions HClNaOH HNO 3 KOH H 2 SO 4 Weak acids/bases – partial dissociation, both ions and molecules CH 3 COOHNH 3

pH o C (H + ) = (OH - ) distilled water acidic (H + ) > (OH - ) basic or alkaline (H + ) < (OH - ) natural waters pH = normal rain (CO 2 ) pH = 5.3 – 5.7 acid rain (NO x, SO x ) pH of in Washington DC area 0-14 scale for the chemists fish populations drop off pH < 6 and to zero pH < 5

You are here! pH of Rainwater across United States in 2001 Increasing acidity Why is the eastern US more acidic? air masses

What is acid rain? CO 2 (g) + H 2 O  H 2 CO 3  H + + HCO 3 - Dissolved carbon dioxide lowers the pH Atmospheric pollutants from combustion NO, NO 2 + H 2 O …  HNO 3 SO 2, SO 3 + H 2 O …  H 2 SO 4 both strong acids pH < 5.3

Effects of Acid Rain on Marble (calcium carbonate) George Washington: BEFORE George Washington: AFTER

pH The biological view in the human body gastric juice urine saliva cerebrospinal fluid blood pancreatic juice bile acidicbasic/alkaline 7 Tortora & Grabowski, Prin. of Anatomy & Physiology, 10 th ed., Wiley (2003)

Chapter 16.2 “Concentration of Solutions”

Concentration is... a measure of the amount of solute dissolved in a given quantity of solvent A concentrated solution has a large amount of solute A dilute solution has a small amount of solute –These are qualitative descriptions But, there are ways to express solution concentration quantitatively

Concentrated vs. Dilute

Molarity Molarity = moles of solute liters of solution Abbreviated with a capital M, such as 6.0 M This is the most widely used concentration unit used in chemistry.

- Page 481

Practice Problem A sample of mol iron(III) chloride, Fe Cl 3, was dissolved in water to give mL of solution. What is the molarity of the solution? Since molarity = Moles of FeCl 3 Liters of solution moles of FeCl Liters of solution Since molarity = = 1.36 M FeCl 3

Making solutions 1)Pour in a small amount of the solvent, maybe about one-half 2)Then add the pre-massed solute (and mix by swirling to dissolve it) 3)Carefully fill to final volume. –Fig. 16.8, page 481 Can also solve: moles = M x L Sample Problem 16.3, page 482

Dilution Adding water to a solution will reduce the number of moles of solute per unit volume but the overall number of moles remains the same! Think of taking an aspirin with a small glass of water vs. a large glass of water You still have one aspirin in your body, regardless of the amount of water you drank, but a larger amount of water makes it more diluted.

Dilution water (solvent)solute concentrated, M initial diluted, M final adding water lowers the solute concentration moles of solute remain constant V initial V final moles initial = moles final M final x V final = M initial x V initial

Dilution The number of moles of solute in solution doesn’t change if you add more solvent! The # moles before = the # moles after Formula for dilution: M 1 x V 1 = M 2 x V 2 M 1 and V 1 are the starting concentration and volume; M 2 and V 2 are the final concentration and volume. Stock solutions are pre-made solutions of known Molarity. Sample 16.4, p.484

Percent solutions can be expressed by a) volume or b) mass Percent means parts per 100, so Percent by volume: = Volume of solute x 100% Volume of solution indicated %(v/v) Sample Problem 16.5, page 485

Percent solutions Percent by mass: = Mass of solute(g) x 100% Volume of solution (mL) Indicated %(m/v) More commonly used 4.8 g of NaCl are dissolved in 82 mL of solution. What is the percent of the solution? How many grams of salt are there in 52 mL of a 6.3 % solution?

Titration Calculation HCl + NaOH  NaCl + HOH at equivalence point: mole HCl = mole NaOH moles = M x V L M acid x V initial acid = M base x V buret A way to analyze solutions! indicator PGCC CHM 101 Sinex If the mole to mole ratio = 1:1

Titration Calculation HCl + NaOH  NaCl + HOH at equivalence point: mole HCl = mole NaOH moles = M x V L A way to analyze solutions! indicator PGCC CHM 101 Sinex

M acid x V initial acid = M base x V buret If the mole to mole ratio = 1:1 If the mole to mole ratio is not 1:1 you need to add a mole factor to the equation above based on the coefficients in the balanced equation. Example: 2 HCl + Sr(OH) 2  SrCl 2 + 2H 2 O Mole to mole ratio = 2 acid : 1 base

The number of moles of acid is twice the number of moles of base, therefore it will take twice as much base to neutralize the acid. moles acid = 2(moles base ) The '2' in this formula is called the mole factor. Substituting into equation (1) we have: M A V A = 2 M B V B

Sample Problem: If 25.0 mL of a standard 0.05 M HCl solution is required to neutralize 20.0 mL of a solution of Sr(OH) 2, what is the concentration of the base? (Use the equation in the example above.) M A V A = 2 M B V B (0.05 M)(25.0 mL) = 2(X)(20.0 mL) X =

35.62 mL of NaOH is neutralized with 25.2 mL of M HCl by titration to an equivalence point. What is the concentration of the NaOH? LAB PROBLEM #1: Standardize a solution of NaOH — i.e., accurately determine its concentration.

PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? Add water to the 3.0 M solution to lower its concentration to 0.50 M Dilute the solution!