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Unit 3

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Unit 3. Acids and Bases. Hydrogen ions and pH. Ion product constant of water (K w ) H 2 O  H + + OH - In pure water : [H + ] = [OH - ] * [ ] are used to indicate concentration in mol/L K w = [H + ] [OH - ] = 1.0x10 -14 - PowerPoint PPT Presentation

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Page 1: Unit 3
Page 2: Unit 3

Hydrogen ions and pHIon product constant of water (Kw)

H2O H+ + OH-

In pure water : [H+ ] = [OH- ] * [ ] are used to indicate concentration in mol/L

Kw = [H+ ] [OH- ] = 1.0x10 -14 If [H+ ] > 1.0 x10-7 the solution is ACIDICIf [OH- ] > 1.0 x10-7 the solution is BASICIf [H+ ] or [OH- ] = 1.0 x10-7 the solution is

NEUTRAL

Page 3: Unit 3

pH and pOHValues for concentrations of H+ and OH- are

small so scientist use a scale based on common logarithms.

pH= - log [H+] * [H+] : concentration hydrogen ions (related to acids)

[H+] = 10-pH

pOH= -log [OH-] * [OH-] : concentration hydroxide ions (related to bases)

[OH-] = 10-pOH

pH +pOH=14

Page 4: Unit 3

A pH scale from 0 to 14 is used. (It has no units)

If pH< 7: solution is acidic ; if pH=7: solution is neutral; if pH>7: solution is basic

If pOH<7: solution is basic; if pOH=7: solution is neutral; if pOH>7 : solution is acidic

pH and pOH

Page 5: Unit 3

pH of Common Substances

Timberlake, Chemistry 7th Edition, page 335

1.0 MHCl0

gastricjuice1.6

vinegar2.8

carbonated beverage3.0

orange3.5

apple juice3.8

tomato4.2

lemonjuice2.2 coffee

5.0

bread5.5

soil5.5

potato5.8

urine6.0

milk6.4

water (pure)7.0

drinking water7.2

blood7.4

detergents8.0 - 9.0

bile8.0

seawater8.5

milk of magnesia10.5

ammonia11.0

bleach12.0

1.0 MNaOH(lye)14.0

8 9 10 11 12 14133 4 5 621 70

acidic neutral basic[H+] = [OH-]

Page 6: Unit 3

Example 1What is the pH of a solution with [H+]=1.00x10-

4 M? Is it acidic, basic, or neutral?

Given : [H+]=1.00x10-4 MUnknown: pH

pH = -log[H+]

pH = -log[1.00x10-4]

pH = 4.00 (acidic)

Page 7: Unit 3

Example 2What is the pH and pOH of a solution with

[ H+]= 0.00350mol /L?

Given : [ H+]= 0.00350mol /L?Unknown: pH and pOH

pH = -log[H+]

pH = -log[0.00350]

pH = 2.46

pH + pOH = 14

pOH = 14 – pH= 14 - 2.46 = 11.54

Page 8: Unit 3

Example 3What is the [H+ ] and [OH- ] of a solution if

pOH=4.40Given: pOH= 4.40 Unknown: [H+ ] and [OH- ]

[OH-] = 10-pOH

[OH-] = 10-4.40

[OH-] = 3.98x10-5 M

[H+] [OH-] = 1 x10-14

[H+] = 1 x10-14 = 1 x10-14 = 2.51x10-10M

[OH-] 3.98x10-5

Page 9: Unit 3

classworkRead p650-655Do problems p651#22 p653 #24-26 p655

#30-31

Page 10: Unit 3

Strengths of acids and basesStrong acids and bases : ionize completely in

aqueous solution.Examples: HCl, HNO3, H2SO4, KOH, NaOH

Weak acids and bases: ionize slightly in aqueous solution.

Examples: HClO, H3PO4 , NH3

Page 11: Unit 3

Reactions between acids and basesWhen and acid and a base react with each other it is called neutralization reaction.

ACID + BASE → SALT + WATER

HHCl + NaOHOH → NaCl + H2O

H-OHSalt: ionic compound made up of cation of base and anion from acid.

Page 12: Unit 3

Ex. Write the neutralization reaction between hydroiodic acid and potassium hydroxide.

Page 13: Unit 3

TitrationTitration: adding a known amount of a

solution of known concentration to determine the concentration of another solution.

Equivalence point: when number of moles of hydrogen ions equals the number of moles of hydroxide ions.

Standard solution: solution of known concentration

End point: point at which the indicator changes color

Point of neutralization is the end point of the titration.

M1V1= M2V2

Page 14: Unit 3

Ex. 1 How many milliliters of 0.45M HCl will neutralize 25.0 mL of 1.00M KOH?

mL HCl= 25.0mL x 1.00M KOH 0.45 M HCl

mL HCl= 55.6mL HCl

M1V1= M2V2

Page 15: Unit 3

Ex. 2 What is the molarity of a NaOH solution if 20.0 mL of the solution is neutralized by 28.0mL of 1.00 M HCl?

M NaOH= 1.00 M HCl x 28.0 mL 20.0 mL

M NaOH = 1.40 M

Classwork: p 989 #22-25

M1V1= M2V2