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Chapter 18 Acids and Bases Rainbow Connection #2

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Chapter 18. Acids and Bases Rainbow Connection #2. Ch. 18- Acids and Bases. Acids and bases have a central role in chemistry They affect our daily life Uses: manufacturing processes, environmental issues, functioning of our bodies Acid/ Base Video. Properties of Acids. Have pH  0-6 - PowerPoint PPT Presentation

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Page 1: Chapter 18

Chapter 18

Acids and Bases

Rainbow Connection#2

Page 2: Chapter 18

Ch. 18- Acids and Bases• Acids and bases have a central role in

chemistry• They affect our daily life• Uses: manufacturing processes,

environmental issues, functioning of our bodies

• Acid/ Base Video

Page 3: Chapter 18
Page 4: Chapter 18
Page 5: Chapter 18

Properties of Acids• Have pH 0-6• Tart or sour taste• Will conduct electricity• Cause indicators to change color (turns

blue litmus red)• Reacts w/ metals (Mg,Zn) to form H2 gas• Neutralize w/ a base forms a salt and H20• Ex. Citrus foods, tomatoes, vinegar

Page 6: Chapter 18

Acid Formulas (memorize)• HCl ( Hydrochloric Acid)• HNO3 (nitric acid)• H2SO4 (sulfuric acid)• H2CO3 (carbonic acid)• HC2H3O2 (acetic acid)• H3PO4 (phosphoric acid)

Page 7: Chapter 18

Properties of Bases• Also known as alkaline• pH 8-14• Have a bitter taste, slippery feel• Causes indicators to change color (turns

red litmus blue)• Conducts electricity• Reacts w/ acid to neutralize form a

salt and H2O• Ex. Cleaning products, soap, baking soda

Page 8: Chapter 18

Acid/ Base Theories• Definitions have changed over the years

as new information has been found• Arrhenius Theory video

Page 9: Chapter 18

Arrhenius Theory (1887)• Applies to a H2O solution• Svante Arrhenius (Swedish) saw that not

only do acids/ bases conduct electricity, they ionize (or release charged particles) when dissolved in water

• Theory:1. Acids- substance that ionize & produced (H+)

hydrogen ions in H2O2. Bases- ionize to produce (OH-) hydroxide

ions in H2O

Page 10: Chapter 18

Ex. HCl H+(aq) + Cl-(aq)

ACID

NaOH Na+(aq) + OH-

(aq) BASE

Page 11: Chapter 18

Brønsted- Lowry Theory (1923) (video)

• Working independently of each other, Johannes Brønsted (Danish) and Thomas Lowry (English) defined a theory that can be used w/ all solvents not just H2O (they found that substance lost or gained protons)

• Acid- in a chemical reaction, this is the substance that loses or donates a proton (H+ ion)

• Base- substance that accepts or gains a proton (H+ ion = proton)

Page 12: Chapter 18

• HCl + H2O H3O+ + Cl-

Accepts proton (base)

Donates (loses) proton – (acid)

H3O+ hydronium ion (formed when H2O gains H+ ion)

Page 13: Chapter 18

• NH3 + H2O NH4+ + OH-

Accepts proton (base)

Donates (loses) proton – (acid)

Amphoteric – (H2O) acts as an acid or a base (depending on the situation)

Page 14: Chapter 18

Conjugates• The particles that are formed as

products can react again (reversible reaction), they behave like acids and bases (we call these conjugates)

• Conjugate video

Page 15: Chapter 18

• Conjugate Acid - Particle that forms after the base accepts a proton (H+) from the acid

• Conjugate Base – particle that remains after a proton (H+) has been released by the acid

Page 16: Chapter 18

Ex. HNO3 + NaOH H2O + NaNO3

(H-OH)Acid

Conjugate BaseBaseConjugate acid

Page 17: Chapter 18

Disappearing ink

Page 18: Chapter 18

Ex. KOH + HBr KBr + H2O

Base Conjugate AcidAcid

Conjugate Base

Page 19: Chapter 18
Page 20: Chapter 18

Neutralization Reaction• Occurs when an acid and a base react

and there is a complete removal of all of the H+ and OH- ions

• Water will be formed w/ a salt in this double displacement reaction

• The solution will be neutral in pH• Important in: neutralization of soil,

antacids

Page 21: Chapter 18

Salt• Crystalline compound composed of the

negative (non-metal) ion of the acid and the positive (metallic ) ion of the base

• Salt examples: CaSO4 (plaster board), NaCl, KCl, (NH4)2SO4 (fertilizer)

Page 22: Chapter 18

Examples of Neutralization Reactions

• Sodium hydroxide + hydrochloric acid sodium chloride + water

• NaOH + HCl NaCl + H2O (H-OH)

Page 23: Chapter 18

• Potassium hydroxide + sulfuric acid water + potassium sulfate

• KOH + H2SO4 H- OH + K2SO4

2 KOH + H2SO4 2 H-OH + K2SO4

Page 24: Chapter 18

• Aluminum hydroxide + acetic acid

Page 25: Chapter 18

Titration

Page 26: Chapter 18

Titration• Used for a convenient method to

determine the concentration or molarity of an acid or base

• Uses 2 burets (long glass tubes used to measure volume)- Buret + .05 ml

Page 27: Chapter 18

Definition• Analytical method in which a standard

solution is used to determine the concentration of another solution

• Standard solution- one in which the concentration is known

Page 28: Chapter 18

Process• Using a given amount of acid in a flask, add

phenolphthalein and titrate to the end point with the base (making sure to mark down all the volumes)

• The whole reaction is a neutralization reaction

• Use an indicator to see the endpoint in which complete neutralization occurs (wait for phenolphthalein stays light pink for 30 sec)

Page 29: Chapter 18
Page 30: Chapter 18

Endpoint Past endpoint

Page 31: Chapter 18

Titration Process

Page 32: Chapter 18

Calculations• Reminder: M= moles• liters• So: Base (titrated soln)

– Moles (base) = volume (base) x Molarity (base)

Acid (standard soln) Moles (acid) = volume (acid) x Molarity (acid)

Page 33: Chapter 18

• Look at balanced equation and find the moles of the standard and the moles of the titrated unknown (look at coefficients)

• Most times it is a 1:1 ratio• 1 NaOH + 1 HCl NaCl + H2O (H-OH)• So: 1 mole (base) = 1 mole (acid)• Therefore: MA VA = MBVB

Page 34: Chapter 18

Problem• A titration of 15.00 ml of HCl, required

38.57 ml of a 0.152 M base NaOH. Calculate the molarity of the HCl (acid).

• MA = ?

Page 35: Chapter 18

• MA VA = MBVB

• MA = M BVB

• VA

• MA = (.152M) (38.57 ml)• (15.00ml)• MA = .391 M

Page 36: Chapter 18

Titration Calculation

Page 37: Chapter 18

ID- A, B, CA, CB

• CaCO3 + HCl CaCl2 + H2CO3

• KOH + H3PO4 K3PO4 + H2O

Page 38: Chapter 18

• phosphoric acid + calcium hydroxide calcium phosphate + water

• HBr + Al(OH)3

• Sulfuric acid + potassium hydroxide

Page 39: Chapter 18

Indicators• Used to find out if things are acidic or

basic• Def: weak organic acids or bases whose

colors differ from the colors of their conjugate acids or bases

acidbase

Page 40: Chapter 18
Page 41: Chapter 18

Needs of Indicators• Solution being tested needs to be

colorless• You need to be able to distinguish the

color change• Need several indicators to cover entire

pH range (0-14)• Liquid solution

Page 42: Chapter 18

Indicator Examples• Litmus red turns blue = base, blue

turns red = acid• Phenolphthalein hot pink > pH 10• Bromothymol blue blue= base (8),

Green= neutral, yellow = acid (6)• Universal Indicator Rainbow (all pH’s 4-

10)

• (bromothymol blues)

Page 43: Chapter 18

How Chemists use Indicators

Page 44: Chapter 18

Ionization of H2O

• Pure water can self ionize, it also acts as an acid or a base (amphoteric)

• So: H2O(aq) H+(aq) + OH- (aq)

• Experiments have shown that the concentration of [H+] = 1 X 10 –7M and [OH-] = 1 x 10 –7 M in pure water

• [ ] means concentration

Page 45: Chapter 18

Ion Product Constant for water• Equilibrium expression from the

multiplication of the concentrations of the products

• [H+] [OH-] =?

• (1 x 10 –7)(1 x 10 –7) = 1.0 x 10 -14 1. [H+] [OH-] = 1.0 x 10 -14

Page 46: Chapter 18

• If [H+] = 1.5 x 10 -6 M, what is the [OH-] = ?

• [H+] [OH-] = 1.0 x 10 -14

• (1.5 x 10 -6) [OH-] = 1.0 x 10 -14

• [OH-] = 1.0 x 10 -14

• 1.5 x 10 -6

• [OH- ] = 6.7 x 10-9 M

Page 47: Chapter 18

pH Concept (video)• Acidity scale developed by Soren

Sorenson base on the “power of the hydrogen”

• pH – measure of the hydrogen ion concentration of the solution– Equal to the negative logarithm of the

hydrogen ion concentration2. pH = - log [H+]

Page 48: Chapter 18

Ex. [H+] = 1.5 x 10-8

pH= ?

Graphing Calc. (-) log 1.5 (2nd) EE -8

Regular calc. 1.5 EE -8 log +/- pH= 7.8

Page 49: Chapter 18

To go from pH to [H+]concentration

3. [H+] = antilog (-pH)

Antilog = 10x key

Page 50: Chapter 18

• pH= 3.5• [H+] = ?[H+] = antilog (-pH)[H+] = antilog (-3.5)Graphing = 2nd log -3.5Regular 3.5 +/- 2nd log (to put in sci.not. Use 2nd #5)[H+] = 3.2 x 10 -4 M

Page 51: Chapter 18

pOH (hydroxide power)4. pH + pOH =14

• pH= 8 • pOH = ?8 + ? = 14 pOH= 6

5. pOH = - log [OH-]6. [OH-] = antilog (-pOH)

Page 52: Chapter 18

• [H+] = 3.5 x 10 -7 M, [OH-] = ?

• pH = 8.95 , [H+] = ?

• [OH-] = 5.65 x 10 -2 M, pOH = ?

• pOH= 11.9, [H+] = ??

• Formula 1

• Formula 3• Formula 5

• Formula 4, then 3 or• Formula 6, then 1

Page 53: Chapter 18

Rosengarten acid and base video