Equilibrium and acids and bases

1
Equilibrium and acids and bases

• Chapter 14 and 15 in text

2
Chemical Equilibrium

• Concentrations of all reactants and products are
  constant with time.
• That does not mean reactant product
• Rather the rate of decomposition of react and
  formation of prod are constant.

3
Equilibrium is a Dynamic Situation
Demonstration
B
A
4

• When a car from island A moves to island B there
  is a stress placed on the equilibrium and the
  reaction (the members of the groups) must shift
  either left or right to relive the stress.
• If a stress is placed on Island A then the
  equilibrium will shift to the right to
  compensate.

5
What Just Happened?

•  When a chemical reaction is at equilibrium, any
  disturbance of the system, such as a change in
  temperature, or addition or removal of one of the
  reaction components, will "shift" the composition
  of the reaction to a new equilibrium state.

6
A Chemical Example of a Shift in Equilibrium

• Co(H2O)62 4Cl CoCl42 6H20
• Pink Blue

7
Demonstration of Equilibrium
8
Question Which direction would the equilibrium
shift if HCl was added to the reaction?

• Co(H2O)62 4Cl CoCl42 6H20
• Pink Blue

HCl
9
Answer

• The addition of HCl would cause an increase in
  Cl- and thus the reaction would react by shifting
  the reaction to the right (blue side) in order to
  consume the excess Cl- and return to a state
  of equilibrium.
• Co(H2O)62 4Cl CoCl42 6H20
• Pink Blue

10
The Law of Mass Action

• For any reversible reaction
• jA kB lC mD
• The law of mass action is represented by the
  following equilibrium expression. (Prod /React)
• K Cl Dm
• Aj Bk
• K equilibrium constant
• Solids and liquids are not written in equilibrium
  expressions

11

• K gt 1 Favors products
• Klt1 Favors reactants

12
Graph of Equilibrium
13
Question

• Write the equilibrium expression for the
  following equations
• PCl5 (g) PCl3 (g) Cl2 (g)
• Cl2O7(g) 8H2(g) 2HCl(g) 7H20(g)
• NOTE Make sure the RXN is BALANCED!!!

14
Answer

• K PCl3 Cl2
• PCl5

K H2O7 HCl2 Cl2O7 H28
15
Calculating Equilibrium

• Example
• Calculate the equilibrium constant K, for the
  following reaction at 25C,
• H2 (g) I2 (g) 2HI (g)
• if the equilibrium concentrations are
• H2 0.106M I2 0.022M HI 1.29M

16
Answer

• Equilibrium Expression
• K HI2
• H2 I2
• Equilibrium Constant
• K (1.29)2 7.1 x 102
• (0.106) (0.022)

Note the units cancel out! Thats one less thing
to remember!
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Homework

• Pg 633 1,8,9,14,16
• Equilibrium wks

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How do we predict which direction equilibrium
will shift to?

• Le Chateliers Principle
• If we stress a reaction out the reaction will
  shift (either towards R or P) to reduce the
  stress.
• Stress a change in Concentration
• Pressure
• Volume
• Temperature

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Concentration

• Ask yourself
• Is what is being added a solid or liq? (No
  effect)
• a gas or aqueous reactant or a product?
  (effect!)
• If its a reactant the reaction will shift toward
  the products to consume the extra reactant.
• If its a product the reaction will shift toward
  the reactants to consume the extra product.

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Example

• N2 g 3H2 g 2NH3 g
• Add NH3
• Remove N2
• Add H2

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Changes in V or P

• P changes in L or aq have little effect
• P changes of gases have huge effects (PVnRT)
  (PVPV)
• Concentration is effected by pressure.

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• An increase in Pressure (due to a decrease in V)
  will shift to decreases the total number of
  moles of gas.
• If the number of moles on R and P side are the
  same then a change in P will have no effect on
  equilibrium.

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Example

• Lets do Pg 626 example 14.12 in your text
• ?P shift to side of reaction with greatest number
  of moles of gas
• ?P shift to side of reaction with less moles of
  gas

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Temperature

• ONLY a change in TEMPERATURE can change the value
  of K (equilibrium constant).
• You must look at the H of the reaction to see how
  T will effect it
• Temperature is our stress so the reaction will
  move in the direction that removes the stress.

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• ? temperature the rxn moves in the endothermic
  direction (?H)
• ENDO ? Right ( K ? )
• ? temperature the rxn moves in the exothermic
  direction (-?H)
• EXO ? Left ( K ? )
• K only depends on temperature, catalysts have
  NO EFFECT on K
• Kgtgt1 favors products
• Kltlt 1 favors reactants

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Think about heat like a R or P

• Endothermic ?H
• Heat A ? B H 400 kJ
• Favors ? in T
• therefore K ? when it is heated and K ? when it
  is cooled.
• Exothermic -?H
• A ? Heat B H - 400 kJ
• Favors ? in T therefore K ? when it is cooled
  and K ? when it is heated

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Example

• Examples 14.13 on pg 628-629 in text

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Homework

• Pg 636 pg 49,51,53-55,62

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Nature of Acids and Bases

• Acids Sour Taste
• If a solution has a high
• H acidic
• Base Bitter Taste / Slippery feel
• If a solution has a high
• OH- base

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About the scale

• P quantity
• So
• pH quantity of H ion
• pOH quantity of OH ion
• pH pOH 14

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Concentrations

• H OH- Neutral
• H gt OH- Acid
• H lt OH- Base

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Arrhenius Concept

• Focuses on what ions were formed when acids and
  bases dissolved in water.
• Acids dissociate in water give hydrogen ions (H
  or H3O hydronium ion)
• Bases dissociate in water give hydroxide ions
  (OH- hydroxide ion ) .

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• Arrhenius acid - Any substance that ionizes when
  it dissolves in water to give the H ion. 
• e.g.   
• Arrhenius base - Any substance that ionizes when
  it dissolves in water to give the OH- ion. 
• e.g.    

34

• The theory can only classify substances when they
  are dissolved in water since the definitions are
  based upon the dissociation of compounds in
  water.
• It does not explain why some compounds containing
  hydrogen such as HCl dissolve in water to give
  acidic solutions and why others such as CH4 do
  not.
• The theory can only classify substances as bases
  if they contain the OH- ion and cannot explain
  why some compounds that don't contain the OH-
  such as Na2CO3 have base-like characteristics.

35
Bronsted Lowery Acid Base Concept

• Acid substance that can donate a proton ()
• Base substance that accepts a proton () (aka
  they have a lone pair of e- to accept a proton)
• Unlike Arrhenius concept this is applicable in
  both aqueous and non-aqueous states.

36
Equilibrium

• Reaction produces reactants and products at the
  same rate, but not necessarily in the same
  amounts
• Bathroom theory

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Example

• NH3 (aq) H20 (l) ? NH4 (aq) OH- (aq)
• Equilibrium will favor the formation of the
  weaker acid and the weaker base.
• In this rxn the NH4 and OH- will be low
  because they are the stronger acid and base.

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• In the above reaction, the H from HCl is donated
  to H2O which accepts the H to form H3O, leaving
  a Cl- ion. 

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Conjugate Acid and Base Pairs

• The part of the acid remaining when an acid
  donates a H ion is called the conjugate base. 
• The acid formed when a base accepts a H ion is
  called the conjugate acid. 

40
For the generic acid HA
Formed when a proton Is transferred to the base
Everything that is left after a proton to the
base.
NOTE
Strong acids have weak conjugate bases. Strong
bases have weak conjugate acids.
41
Question

• If H20 is an acid what would its conjugate base
  be?
• What is the conjugate acid of HPO42- ?
• What is the conjugate base of HS-

42
Answer

• H20 take away a proton OH-
• HPO42- add an H H2PO4-
• ( we added a proton and that needs to be
  reflected in the molecular charge)
• HS- S2-

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Amphoteric

• The ability of a substance to act as an acid or a
  base.
• Ex H2PO4- and H2O can act as both acids and
  bases.

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Back to Old Faithful
In this equation the stronger base will win the
competition for H. If H2O is a stronger base
than A-, then it will have a greater affinity for
the protons and the equilibrium will lie to the
right favoring the formation of H3O. If A- is
stronger then equilibrium will fall to the left
and acid in the form HA will form.
45
Strong Acids Conj. Bases

• HCl
• HBr
• HI
• HNO3
• HClO4
• HClO3
• H2SO4

• Cl-
• Br -
• I-
• NO3-
• ClO4-
• ClO3-
• HSO4-

You must memorize all of
these
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Common Strong Bases
Formula Name
NaOH sodium hydroxide
LiOH lithium hydroxide
KOH Mg(OH)2 potassium hydroxide Magnesium Hydroxide
Ca(OH)2 Ba(OH)2 Sr(OH)2 Calcium hydroxide Barium Hydroxide Strontium Hydroxide
47
Example

• Identify the CA and CB for each reaction
• HNO3 H2O ?
• NH3 H2O ?

48
Acid-dissociation equilibrium constant (Ka)

• The relative strength of an acid is described as
  an acid-dissociation equilibrium constant.
• The acid-dissociation equilibrium constant is the
  mathematical product of the equilibrium
  concentrations of the products of this reaction
  divided by the equilibrium concentration of the
  original acid

49
Think Products over reactants
50
Question

• Write an ionization equation for the following
  and then write the acid dissociation constant for
  both. (all occur in water)
• Hydrochloric acid
• Acetic acid HC2H3O2

51
Answer

• HCl ? H Cl-
• Ka H Cl- / HCl
• HC2H3O2 ? H C2H3O2-
• Ka H C2H3O2- / HC2H3O2

52
Homework

• Pg 686-687
• s 1,5 then write equilibrium expressions for
  number 5.

53

• Ch 16 homework
• Read pages 618-619
• 3, 5, 6,8, 9, 10

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Temperature increases as Kw increases
56
Question

• Calculate the H in aqueous household ammonia if
  the OH- concentration is 0.0025 M. Then tell me
  if the solution is an acid or a base.
• This solution is in water so we can use Kw

57
Answer

• Kw H OH 1.0 x 10 -14
• H Kw /OH 1.0 x 10 -14
• H Kw /0.0025M 1.0 x 10 -14
• H 4.0 x 10-12 M
• 0.0025M OH gt 4.0 x 10-12 M H solution is a base

58
Question

• Calculate the values of H OH- in a neutral
  solution at 25ºC
• If we know Kw HOH and H OH in a
  neutral solution
• Kw 1.0 E-14
• then 1.0 E-14 xx
• 1.0 E-14 X2 (QUADRATIC TIME!!!!)
• X 1.0E-7 M H and OH

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Example

• Calculate the concentrations using
• Kw H OH- 10-14
• For the following and state if it is an acid,
  base, or neutral
• OH- 10-5 M
• H 10.0 M

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Answer

• Kw 10-14 H 10-5 1.0 E-9 M
• H lt OH- Base
• Kw 10-1410 OH- 1.0E-15 M
• H gtOH- Acid

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pH and pOH

• Tell us the concentration of H and OH in
  solution.
• pH -log H pOH -log OH-
• H 10-pH OH
  10-pOH
• pH pOH 14
• (really strong acids may have a negative pH)

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Find pH

• pH -log H
• Calculate the pH of a neutral solution that has a
  H of 1.0 x 10-7 M
• pH - log 1.0 x 10-7 M
• pH 7.00

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pOH

• pOH -log OH-
• or
• pH pOH 14
• So we can use the pH (7) from the last problem to
  find the pOH of that solution.
• 14-7 7 or we could have been given the OH-

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Given pH find H

• pH -log H
• pH 4 find H
• 4 -log X
• 10 -4 X

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• homework pg 687
• 14,15,17, 18, 19,23

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Strong Acids

• Strong acids dissociate almost completely in
  water and therefore have relatively large Ka
  values.
• Equilibrium lies far to the right
• HA dissociates almost completely
• Yield weak conjugate bases
• Strong electrolytes (100 conductivity)
• Ka Large gt 1 

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Weak Acids

• Weak acids dissociate only slightly in water and
  therefore have relatively small Ka values.
• Equilibrium lies far to the left (HA does not
  dissociate)
• Yields strong conjugate base
• Ka small lt 1

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Solving Strong Acid Equations(and Bases too)

• 1. Strong acids dissociate 100 and are the main
  source of H in a solution.
• 2. Therefore H equals the original
  concentration of the acid
• ex 0.2 M HCl H 0.2 M Cl 0.2 M

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Calculating pH for a strong acid example

• What is the pH of a 0.04 M solution of HClO4?
• HClO4 strong acid equilibrium lies to the right
  and completely dissociates.
• H ClO4- 0.04 M
• pH -log 0.04 1.40

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Calculating Ka

• Write the ionization equation for the reaction
• Write the equilibrium expression for the reaction
  (Ka)
• Calculate the H using a given pH
• ICE box (initial, change, equilibrium to
  determine the values for concentration that you
  will substitute into your equilibrium equation.
• Insert concentration into equilibrium expression
  and solve for Ka

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Solving for Ka example

• A student prepared a 0.1M solution of formic acid
  (HCHO2) and measured its pH to be 2.38. Calculate
  Ka.
• How will you attack this problem?

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Answer

• Write the ionization rxn.
• HCHO2 ?H CHO2-
• Write Ka equation
• Ka H CHO2-
• HCHO2
• Solve for H using pH given
• pH -log H
• 2.38 -log H
• H 10-2.38 0.004168 4.2 X 10-3

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Not done yet

• ICE Box
• HCHO2 ? H
  CHO2-
• Initial 0 .10 0
  0
• Change -4.2 x10-3 4.2 x10-3 4.2 x10-3
  
• Equilibrium 0.1- 4.2 x10-3 M 4.2 x10-3 M 4.2
  x10-3M

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Still not yet

• Plug values into Ka equation
• HCHO2 ? H
  CHO2-
• Equilibrium 0.1- 4.2 x10-3 4.2 x10-3 4.2
  x10-3
• Ka HCHO2 4.2 x10-3 4.2 x10-3
• HCHO2 0.1- 4.2 x10-3
• Solve for Ka 1.8 x 10-4

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Lets do another one

• A student calculates the pH of HOCl to be 3.5
  find Ka. The initial concentration of the acid is
  0.01M
• Write equation for reaction
• Write Ka equation
• Use pH to solve for H
• ICE Box
• Plug values into Ka equation

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Using Ka and acid to calculate pH

• Write the dissociation equation for the reaction.
  
• Write the equation for Ka
• ICE Box but now we know our initial concentration
  of our acid so we can use it to solve for the
  H
• Use found H to calculate pH

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Weak acid example

• Calculate the pH of a 0.1 M aqueous solution of
  HOCl (Ka 3.4x10-8 weak acid).
• HOCl has a higher Ka value than water and will
  dissociate to produce the most H ions, so we use
  it to write our equation for Ka

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• Write the rxn HOCl H
  OCl-
• Write the equation for Ka 3.4x10-8
  HOCl-/HOCL
• Plug into ICE Box using X for unknown
  concentrations
• HOCl H OCl-
• I 0.1 0 0
• C -X X X
• E 0.1-X X X
• Substitute E values into Ka equation
• 3.4x10-8 (x) (x)
• 0.1-X

X the amount of HOCl that dissociates

We can ignore this x since it is so small since
stronger acid
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• Plug into Ka and solve for X H
• 3.4x10-8 X2
• 0.1
• X 5.9 x 10-5 M
• Use pH -log H to solve for pH
• pH -log 5.9 x 10-5
• pH 4.23

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Homework

• Ka Worksheet
• Todays
• 28, 33, 34, 41, 43, 47,

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Percent Dissociation (aka ionization)

• dissociation Hfinal X 100
• Hinitial
• This equation allows us to identify for the exact
  concentration of H that must dissociate for
  the equation to reach equilibrium.
• The stronger the acid the greater the ionization.
  

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Example

• Calculate the dissociation for 1.0 M HC2H3O2
• Ka 1.8 X 10-5
• HC2H3O2 H C2H3O2
• I 1.0 0 0
• C -X X X
• E 1.0-X X X

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• Ka 1.8 X 10-5 X2/ 1.0
• X 4.2 x 10-3
• diss H / HC2H3O2 100
• 4.2 x 10-3/ 1.0
• 0.42

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Bases

• When strong bases are dissolved in aqueous
  solutions they dissociate 100 in OH-, so we can
  treat strong base equations like we treat strong
  acid equations.
• Strong Base Large Kb high pOH

85

• All hydroxides of group 1A and 2A are strong
  bases.
• Exception Be(OH)2
• Strong bases will have large Kb values.

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Weak Bases

• B (aq) H2O BH (aq) OH-(aq)
• Base Acid Conj. Acid Conj Base
• React with water to form conjugate acid of the
  base and OH- ion.
• Small Kb values (low pOH)

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Example

• Calculate the pH of 0.05 M solution of Base X, Kb
  1.7 x 10-9
• Kb is small and tells us Base x is a weak
  base

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Answer

• B (aq) H2O BH (aq) OH-(aq)
• Base Acid Conj. Acid Conj Base
• ( H2O is and acid so we can ignore it since we
  are given kb)
• I 0.05 0
  0
• C -x x
  x
• E 0.05-x x
  x
• Plug equilibrium values into Kb BHOH-
• 
  B

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• 1.7 x 10-9 x2
• 0.05 (Ignore the
  X)
• x OH- 9.2 x 10-6
• pOH - log (9.2 x 10-6) 5.04
• pH pOH 14
• pH 5.04 14
• pH 8.96

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Other Questions

• Which of the following compounds produce the
  highest pOH as a 0.05M solution. (look for the
  highest Kb)
• Ammonia (Kb 1.8 x 10-5)
• Methylene (Kb 4.4 x 10-4)
• Dimethylamine (Kb 5.4 x 10-4)
• Pyridine (Kb 1.7 x 10-9)

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Lewis Acids

• Lewis Acid a substance that accepts an electron
  pair
• Lewis Base a substance that donates and electron
  pair.

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Keeping it straight
Model Definition of Acid Definition of Base
Arrehnius H producer In water OH- Producer In water
Bronsted- Lowery H donor H acceptor
Lewis Electron pair acceptor Electron pair donor
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Lewis Acid Base Example

• Brønsted-Lowry acid-base reaction donation and
  acceptance of a proton
• Lewis base (OH -) hydroxide ion donates a pair
  of electrons for covalent bond formation, Lewis
  acid (H ) accepts the pair of electrons.

Lewis Base
Lewis Acid
Lewis complex
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Lewis Example

• For each rxn identify the Lewis acid and base.
• H H2O ? H3O

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Answer

• The proton (H) is the Lewis acid and the water
  (H2O) is the Lewis base.

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Note

• Note
• Every Bronsted-Lowery base is a Lewis base
  because they all have a lone pair of e- to accept
  the protons.
• BUT not every Lewis base is a Bronsted-Lowery
  base because not all LB can accept protons.

97

• Chemicals which have no hydrogen to donate (aka
  the Bronsted-Lowry scheme) can still be acids
  according to the lewis scheme.
• example, BF3 . If we determine Lewis structure of
  BF3 , we find that B is octet deficient and can
  accept a lone pair. Thus it can act as a Lewis
  acid. Thus, when reacting with ammonia, the
  reaction would look like

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Titration

• Demo
• https//www.youtube.com/watch?v3P8FPsbseqM
• Titration Curves
• http//www.ausetute.com.au/titrcurv.html

100
Label the Diagram
101
Selecting an Indicator
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Which indicator would you choose to use in the
lab?
A
B
D
C
104
Sample problem

• 30 mL of 0.10M NaOH neutralized 25.0mL of
  hydrochloric acid. Determine the concentration of
  the acid
• Write the balanced chemical equation for the
  reaction         
• Calculate moles NaOH(known volume and
  concentration)
• From the balanced chemical equation find the mole
  ratio 
• Find moles HCl   
• Calculate concentration of HCl M n V     

105

•       NaOH(aq) HCl(aq) ? NaCl(aq) H2O(l)
• Extract the relevant information from
• the question      NaOH      V 30mL , M
  0.10M   HCl      V 25.0mL, M ?
• Calculate moles NaOH
• NaOH      V 30mL , M 0.10M     
• 3 x 10-3 moles
• From the balanced chemical equation find the mole
  ratio     
• Find moles HCl      NaOH HCl is 11      So
  NaOH HCl 3 x 10-3 moles at the
  equivalence point
• Calculate concentration of HCl M n V      n
  3 x 10-3 mol,       V 25.0 x 10-3L      0.12
  M HCl