The Chemistry of Acids and Bases - PowerPoint PPT Presentation

1 / 36
About This Presentation
Title:

The Chemistry of Acids and Bases

Description:

M = mol/L 0.0290 M x 0.010 L = x g 192.14 g/mol = 0.0557 g in 10 mL Using the density of citric acid in lemonade: 10.0 mL x 0 ... – PowerPoint PPT presentation

Number of Views:193
Avg rating:3.0/5.0
Slides: 37
Provided by: Richa619
Category:

less

Transcript and Presenter's Notes

Title: The Chemistry of Acids and Bases


1
The Chemistry of Acids and Bases
2
Acids and Bases TitrationAt the conclusion of
our time together, you should be able to
  1. Do a titration of an acid with a base
  2. Calculate the equivalence point (neutralization
    point) for a titration problem
  3. Define normality

3
Familiar Saying
Failure to be present causes the vital organ to
become more enamored. Distance makes the heart
grow fonder!!
4
ACID-BASE REACTIONSTitrations
  • H2C2O4(aq) 2 NaOH(aq) ---gt
  • acid base
  • Na2C2O4(aq) 2 H2O(liq)
  • Carry out this reaction using TITRATION.

5
Setup For Titrating An Acid With A Base
6
Titration
  • 1. Add solution from the buret.
  • 2. Reagent (base) reacts with compound (acid) in
    solution in the flask.
  • Indicator shows when exact stoichiometric
    reaction has occurred. (Acid Base)
  • This is called the NEUTRALIZATION point.
  • The equivalence point has been reached

7
The Formula for Titration
  • Use
  • M1 E V1 M2 E V2
  • M molarity of acid and base
  • E number of H or OH- per mole of acid or base
  • M x E Normality
  • V volume

8
Various Protic Titration Page 561 Standardize
a solution of NaOH i.e., accurately determine
its concentration.
  • 32.7 mL of 0.100 M NaOH is neutralized with 25.0
    mL of HCl by titration to an equivalence point.
    What is the concentration of the acid?

9
How do you do this??
  • Use
  • M1 E V1 M2 E V2
  • 0.100 M x 1 x 32.7 mL
  • M x 1 x 25.0 mL
  • 0.131 M HCl
  • pH -log (0.131 x 1) 0.883

10
A.P. TEST IN ENGLISH
  • 26,000 riot-crazed aborigines are storming the
    test room. You are to write an essay that will
    calm them. You are then to deliver your essay to
    them using only the "Shakespearean English" with
    quotes from the Greek and Latin languages.

11
Page 56-2 You have 50.0 mL of 0.250 M NaOH
neutralized 20.0 mL of sulfuric acid. Determine
the M of the acid?
  • 0.250 M x 1 x 50.0 mL
  • M x 2 x 20.0 mL
  • 0.313 M H2SO4
  • For each mole of sulfuric acid there are 2
    moles of H, therefore,
  • pH -log (0.313 x 2) 0.203

12
Bill Gates' RulesHere is a list of 11 things
that many high school and college graduates did
not learn in school. In his book, Bill Gates
talks about how feel-good, politically-correct
teachings created a full generation of kids with
no concept of reality and how this concept has
set them up for failure in the real world.
  • RULE 5
  • Flipping burgers is not beneath your dignity.
    Your grandparents had a different word for burger
    flipping they called it opportunity.

13
Acids and Bases TitrationLets see if you can
  1. Do a titration of an acid with a base
  2. Calculate the equivalence point (neutralization
    point) for a titration problem
  3. Define normality

14
Page 563 Various Protic
  • Use
  • M1 E V1 M2 E V2
  • 0.0500 M x 2 x 29.6 mL
  • M x 1 x 40.0 mL
  • 0.0740 M HNO3
  • pOH -log (0.0500 x 2) 1.00

15
Page 581 Titration with Grams
  • Use
  • M1 E V1 M2 E V2
  • Moles of base (1)
  • .800 M x 1 x 0.1000 L
  • 0.0800 M HCl
  • grams (1 mol/MM) (E) 0.0800 mol
  • grams 56.11 (0.0800)
  • 4.49 g KOH

16
Page 594 Titration with Grams
  • Use
  • M1 E V1 M2 E V2
  • 3 x Moles of acid
  • 0.623 M x 1 x 0.074 L
  • 0.0461 M KOH
  • multiply both sides by MM and divide by MEV
  • 3 x 2.73 g
  • 0.0461 mol Molar Mass
  • 178 g/mol

17
Pass the Clicker!!!
18
Normality is defined as
  1. M V
  2. E V
  3. M E
  4. M E V
  5. Not listed

19
A method used to determine the concentration of
an unknown acid or base.
  1. Equivalence point
  2. Neutralization point
  3. Titration
  4. Titrant
  5. Normality
  6. Flask
  7. Buret
  8. Not listed

20
The titrant is usually added to a(n)
  1. Equivalence point
  2. Neutralization point
  3. Titration
  4. Titrant
  5. Normality
  6. Flask
  7. Buret
  8. Not listed

21
The indicator is usually added to the
  1. Equivalence point
  2. Neutralization point
  3. Titration
  4. Titrant
  5. Normality
  6. Flask
  7. Buret
  8. Not listed

22
Going Down!!
23
3 M sulfuric acid would have a normality of
  1. 3 N
  2. 2 N
  3. 5 N
  4. 6 N
  5. 9 N
  6. Not listed

24
100.0 mL of 3.0 M sulfuric acid neutralizes 50.0
mL of sodium hydroxide. What is the molarity of
the base?
  1. 12 M
  2. 6.0 M
  3. 3.0 M
  4. 1.5 M
  5. 0.33 M
  6. Not listed

25
100.0 mL of 3.000 M nitric acid neutralizes 3.000
M of aluminum hydroxide. How many mL of the base
did you use?
  1. 100.0 mL
  2. 50.00 mL
  3. 33.33 mL
  4. 16.67 mL
  5. 8.333 mL
  6. Not listed

26
Murphy's LawsofScience and Technology
Whenever a system becomes completely defined,
some fool discovers something which either
abolishes the system or expands it beyond
recognition.
27
(No Transcript)
28
The Chemistry of Acids and Bases
29
Acids and Bases TitrationAt the conclusion of
our time together, you should be able to
  1. Do a titration of an acid with a base
  2. Calculate the equivalence point (neutralization
    point) for a titration problem
  3. Define normality
  4. Determine the percent acid with a titration
    problem

30
You will pay, I guarantee it!!!
31
During a titration, the indicator goes in the
  1. Erlenmyer flask
  2. Beaker
  3. Buret

32
During a titration, the substance of unknown
concentration goes in the
  1. Erlenmyer flask
  2. Beaker
  3. Buret

33
Lemonade Titration Problem
  • MEV (base) MEV (acid)
  • 0.100 M x 1 x 8.70 mL
  • M x 3 x 10.0 mL
  • 0.0290 M citric acid

34
Lemonade Titration Problem
  • M mol/L
  • 0.0290 M x 0.010 L
  • x g
  • 192.14 g/mol
  • 0.0557 g in 10 mL

35
Lemonade Titration Problem
  • Using the density of citric acid in lemonade
  • 10.0 mL x 0.960 g/mL
  • 9.60 g lemonade in 10.0 mL of solution
  • 0.0557 g citric acid in 10.0 mL x 100
  • 9.60 g lemonade in 10.0 mL
  • 0.580

36
(No Transcript)
Write a Comment
User Comments (0)
About PowerShow.com