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Acids and Bases Chapter 19 DHS Chemistry

Acids and Bases Chapter 19 DHS Chemistry. I. Definition and Properties

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Page 1: Acids and Bases Chapter 19 DHS Chemistry. I. Definition and Properties

Acids and BasesChapter 19

DHS Chemistry

Page 2: Acids and Bases Chapter 19 DHS Chemistry. I. Definition and Properties

I. Definition and Properties

Page 3: Acids and Bases Chapter 19 DHS Chemistry. I. Definition and Properties

A. Definitions of Acids & Bases

•There are many theories and definitions of acids and bases.

Page 4: Acids and Bases Chapter 19 DHS Chemistry. I. Definition and Properties

Bronsted-Lowry Definitions

•An acid is a molecule or ion that acts as a proton (hydrogen ion) donor.

HCl + H20 H30+ + Cl-

acid

Page 5: Acids and Bases Chapter 19 DHS Chemistry. I. Definition and Properties

Bronsted-Lowry Definitions

•A base is a molecule or ion that acts as a proton (hydrogen ion) acceptor.

HCl + NH3 NH4+ + Cl-

base

Page 6: Acids and Bases Chapter 19 DHS Chemistry. I. Definition and Properties

Arrhenius Definitions:

oAn acid produces hydronium (H30+) ions when dissolved in water.

Ex. HCl + H2O Cl- + H3O+

Page 7: Acids and Bases Chapter 19 DHS Chemistry. I. Definition and Properties

Arrhenius Definitions:

oA base produces hydroxide (OH- ) ions when dissolved in water.

Ex. NH3 + H2O NH4+ + OH-

Page 8: Acids and Bases Chapter 19 DHS Chemistry. I. Definition and Properties

Polyprotic acids• Acids that can contain multiple

hydrogens to donate.

• Match the terms with the probable acid

monoprotic H2SO4

diprotic H3PO4

polyprotic HCl

Page 9: Acids and Bases Chapter 19 DHS Chemistry. I. Definition and Properties

•Acids that contain 1 ionizable hydrogen are monoprotic acids.Ex. HCl, HBr

Page 10: Acids and Bases Chapter 19 DHS Chemistry. I. Definition and Properties

•Acids that contain 2 ionizable hydrogens are diprotic acids.Ex. H2SO4

Page 11: Acids and Bases Chapter 19 DHS Chemistry. I. Definition and Properties

•Acids that contain 3 ionizable hydrogens are triprotic/polyprotic acids.Ex. H3PO4

Page 12: Acids and Bases Chapter 19 DHS Chemistry. I. Definition and Properties

•Some substances can act as both an acid and a base – these substances are considered amphoteric.

EX: HCl + H20 H30+ + Cl- water as a base

acid base

NH3 + H20 NH4+ + OH- water as

an acid

base acid

H20 + H20 H30+ + OH- water as an

acid

pure water has H2O, H3O+, OH- & base

Page 13: Acids and Bases Chapter 19 DHS Chemistry. I. Definition and Properties

B. Properties(refer to your previous

notes!)

Page 14: Acids and Bases Chapter 19 DHS Chemistry. I. Definition and Properties

C. Common Acids and Bases

(refer to your previous notes! Make sure you have those

particular ones memorized, both their formula and their name!)

Page 15: Acids and Bases Chapter 19 DHS Chemistry. I. Definition and Properties

II. Strength of Acids and Bases

Page 16: Acids and Bases Chapter 19 DHS Chemistry. I. Definition and Properties

A. Ionization(refer to your previous notes!)

Page 17: Acids and Bases Chapter 19 DHS Chemistry. I. Definition and Properties

B. pH Scale

Page 18: Acids and Bases Chapter 19 DHS Chemistry. I. Definition and Properties

Hydronium vs Hydroxide

•Hydronium H3O+

– A hydrogen ion in water

– H+ + H2O H3O+

– H+ and H3O+ used interchangeably

– For acids

•Hydroxide OH-

– For bases

Page 19: Acids and Bases Chapter 19 DHS Chemistry. I. Definition and Properties

B. pH Scale•a measure of the number of hydronium or hydroxide ions is the pH scale

•it is based on the concentration of hydrogen and hydroxide ions in solution

•pH is defined as the negative logarithm of the hydrogen ion concentration

Page 20: Acids and Bases Chapter 19 DHS Chemistry. I. Definition and Properties

•[ ] = concentration (molarity)•notice pH is based on a log scale

•log(base 10) is the power to which 10 must be raised to equal that number.

pH = -log([H+]) or -log([H3O+])

pOH = -log ([OH-])

Page 21: Acids and Bases Chapter 19 DHS Chemistry. I. Definition and Properties

Logarithms

b n = x n = logb xLogarithm is an exponent

23 = 8

Then, 3 is called the logarithm of 8 with a base 2

3 = log28

Page 22: Acids and Bases Chapter 19 DHS Chemistry. I. Definition and Properties

Logarithms – Base 10

104 = 10,000

The logarithm of 10,000 with a base 10 is 4

4 = log1010,000

We’ll use this concept for pH

Page 23: Acids and Bases Chapter 19 DHS Chemistry. I. Definition and Properties

pH and pOH are related by the following:

14 = pH + pOH

Page 24: Acids and Bases Chapter 19 DHS Chemistry. I. Definition and Properties

For example, calculate the pH of the following:

0.001 M HCl

In calculator: -log(0.001)

(some calculators are a little different)

pH = 3

Page 25: Acids and Bases Chapter 19 DHS Chemistry. I. Definition and Properties

For example, calculate the pH of the following:

0.01 M NaOH

Because NaOH is a base, we’re calculating the pOH

pOH = -log(0.01)

pH = 12

pOH = 214 = pH + pOH2

Page 26: Acids and Bases Chapter 19 DHS Chemistry. I. Definition and Properties

For example, calculate the pH of the following:

3.2 x 10-9 M HNO3

pH = -log(3.2 x 10-9)

pH = 8.49

Page 27: Acids and Bases Chapter 19 DHS Chemistry. I. Definition and Properties

pH Scale

•the typical pH scale runs from 0 to 14 .

Page 28: Acids and Bases Chapter 19 DHS Chemistry. I. Definition and Properties

III. Titrations

Page 29: Acids and Bases Chapter 19 DHS Chemistry. I. Definition and Properties

A. Titrations•The concentration of an acid (or base) in solution can be determined by performing a neutralization reaction.

Recall neutralization…

Page 30: Acids and Bases Chapter 19 DHS Chemistry. I. Definition and Properties

C. Neutralization Reactions•The reaction of an acid with a base produces water and a class of compounds called salts.

HA + BOH B A + H2O

ex. HCl(aq) + NaOH(aq) NaCl (aq) +

H2O(l)

acid base salt water

Page 31: Acids and Bases Chapter 19 DHS Chemistry. I. Definition and Properties

•An indicator is used to show when neutralization has occurred.

•An indicator is a substance that forms different colors in different pH solutions. Phenolphthalein is a common indicator used in acid-base titrations. It will change from colorless in acidic environments to pink in basic environments.

Page 32: Acids and Bases Chapter 19 DHS Chemistry. I. Definition and Properties

Steps in a formal titration1. A measured volume

of a solution of unknown concentration (acid or base) is added to an Erlenmeyer flask.

2. A solution of known molarity (acid or base) is added to a buret.

Page 33: Acids and Bases Chapter 19 DHS Chemistry. I. Definition and Properties

3. Several drops of an indicator are added to the unknown solution

4. Measured volumes of a solution of known molarity (acid or base) are mixed into unknown solution until the indicator just barely changes to a different color.

Page 34: Acids and Bases Chapter 19 DHS Chemistry. I. Definition and Properties

• The solution of known concentration is called the standard solution. The standard solution is added using a buret.

• The process of adding a known amount of solution of known concentration to determine the concentration of another solution is called titration.

• The point at which the indicator changes color is the end point of the titration.

Page 35: Acids and Bases Chapter 19 DHS Chemistry. I. Definition and Properties

•This can also be done less “formally” using any volume measures. The results won’t be as accurate, but it gets you close.

Example: (drops, substitute for mL)

Page 36: Acids and Bases Chapter 19 DHS Chemistry. I. Definition and Properties
Page 37: Acids and Bases Chapter 19 DHS Chemistry. I. Definition and Properties

The end point

Page 38: Acids and Bases Chapter 19 DHS Chemistry. I. Definition and Properties

How to read a buret

Page 39: Acids and Bases Chapter 19 DHS Chemistry. I. Definition and Properties

B. Solving Titration Problems

Remember, in order for the solution to be neutral,

Moles H30+ = moles OH-

a = # of H+ ions in acidsb = # of OH- ions in basesM = Molarity (M)V = volume

aMAVA = bMBVB

(similar to dilutions)

Page 40: Acids and Bases Chapter 19 DHS Chemistry. I. Definition and Properties

Examples

1) Determine a and b for the following

a = __ for H2SO4 and

b = __ for Ca(OH)2

2

2

Page 41: Acids and Bases Chapter 19 DHS Chemistry. I. Definition and Properties

Example

2) It takes 26.23 mL of a 1.008 M NaOH solution to neutralize 35.28 mL of a monoprotic acid solution. What is its molarity?

1(MA)(35.28 mL) = 1(1.008M)(26.23 mL)

MA = 0.749 M

Page 42: Acids and Bases Chapter 19 DHS Chemistry. I. Definition and Properties

Example

3) If 15.50 mL of NaOH solution were neutralized with 23.40 mL of 0.533 M H2SO4, what is the concentration of the NaOH ?

2(0.533 M)(23.40 mL) = 1(MB)(15.50 mL)

MB = 1.61 M

Page 43: Acids and Bases Chapter 19 DHS Chemistry. I. Definition and Properties

Example

4) 25.00 mL of 0.720 M nitric acid is used to completely neutralize a 1.0 M NaOH solution. What volume of NaOH is present?

1(0.720 M)(25.00 mL) = 1(1.0 M)VB

VB = 18.00 mL

Page 44: Acids and Bases Chapter 19 DHS Chemistry. I. Definition and Properties

Ex: A 25 mL solution of nitric acid is completely neutralized by 18mL of 1.0M NaOH. What is the concentration of the acid solution?

Extra Example

Page 45: Acids and Bases Chapter 19 DHS Chemistry. I. Definition and Properties

Ex: A 25 mL solution of nitric acid is completely neutralized by 18 mL of 1.0M NaOH. What is the concentration of the acid solution?

MA = ?? MB = 1MVA = 25mLVB = 18mL

MAVA = MBVB rearrangedto solve for MA MA= MBVB

VA

(1M)(18mL) = 0.720M 25mL

Page 46: Acids and Bases Chapter 19 DHS Chemistry. I. Definition and Properties

Practice1. What is the molarity of

phosphoric acid if 15.0 mL of the solution is completely neutralized by 38.5 mL of 0.150 M Al(OH)3?

2. It takes 26.23 mL of a 1.32 x 10-3 M NaOH solution to neutralize a 45.56 mL of a triprotic acid solution. What is the concentration of the acid?

0.385 M H3PO4

MB = 2.53 x 104 M

Page 47: Acids and Bases Chapter 19 DHS Chemistry. I. Definition and Properties

Practice

3. What is the molarity of sodium hydroxide if 20.0 mL of the solution is neutralized by 28.0 mL of 0.60 M HCl?

4. How many mL of 0.45 M HCl must be added to 25.00 mL of 1.00 M Sr(OH)2 to make a neutral solution? VA = 111.00 mL

0.840 M NaOH

Page 48: Acids and Bases Chapter 19 DHS Chemistry. I. Definition and Properties

1. What is the molarity of phosphoric acid if 15.0 mL of the solution is completely neutralized by 38.5 mL of 0.150 M Al(OH)3?

H3PO4 Al(OH)3

MA = ?? MB = 0.150MVA = 15mL VB = 38.5mL

MAVA = MBVB rearranged to solve for MA MA= MBVB

VA

(0.150M)(38.5mL) = 0.385M H3PO4 15mL

Page 49: Acids and Bases Chapter 19 DHS Chemistry. I. Definition and Properties

3. What is the molarity of sodium hydroxide if 20.0 mL of the solution is neutralized by 28.0 mL of 0.60 M HCl?

HCl NaOHMA = 0.60M MB = ??VA = 28.0mL VB = 20.0mL

MAVA = MBVB rearranged to solve for MB MB= MAVA

VB

(0.60M)(28.0mL) = 0.840M NaOH 20.0mL