Chem 14A

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Chem 14A

• Friday July 27, 2007

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Lecture Outline

• Strength of acids and bases
• Weak acid equilibria
• Weak base equilibria
• Neutralization Reactions
• Titrations

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Strong and Weak Acids/Bases
The strength of an acid (or base) is determined
by the amount of IONIZATION.
HNO3, HCl, H2SO4 and HClO4 are among the only
known strong acids.
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Strong and Weak Acids/Bases

• STRONG ACID
• HNO3 (aq) H2O (l) ---gt H3O (aq)
  NO3- (aq)
• HNO3 is about 100 dissociated in water.

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Strong and Weak Acids/Bases

• Weak acids are much less than 100 ionized in
  water.
• One of the best known is acetic acid CH3CO2H

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percent ionization
For a monoprotic acid HA
HA0 initial concentration
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Acids
Conjugate Bases
Increase strength
Increase strength
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Strong and Weak Acids/Bases

• Strong Base 100 dissociated in water.
• NaOH (aq) ---gt Na (aq) OH- (aq)

Other common strong bases include KOH and
Ca(OH)2. CaO (lime) H2O --gt Ca(OH)2
(slaked lime)
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Weak Bases
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Strong and Weak Acids/Bases

• Weak base less than 100 ionized in water
• One of the best known weak bases is ammonia
• NH3 (aq) H2O (l) ? NH4 (aq) OH- (aq)

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Molecular Structure and Acid Strength
The weaker the acid
HF ltlt HCl lt HBr lt HI
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Molecular Structure and Acid Strength

• The O-H bond will be more polar and easier to
  break if
• Z is very electronegative or
• Z is in a high oxidation state

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Molecular Structure and Acid Strength
1. Oxoacids having different central atoms (Z)
that are from the same group and that have the
same oxidation number
Acid strength increases with increasing
electronegativity of Z
Cl is more electronegative than Br
HClO3 gt HBrO3
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Molecular Structure and Acid Strength
2. Oxoacids having the same central atom (Z) but
different numbers of attached groups.
Acid strength increases as the oxidation number
of Z increases.
HClO4 gt HClO3 gt HClO2 gt HClO
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Equilibria Involving Weak Acids and Bases

• Consider acetic acid, HC2H3O2 (HOAc)
• HC2H3O2 H2O ? H3O C2H3O2 -
• Acid Conj. base

(K is designated Ka for ACID) K gives the ratio
of ions to molecules
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Equilibrium Constants for Weak Acids
Weak acid has Ka lt 1 Leads to small H3O and a
pH of 2 - 7
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Equilibrium Constants for Weak Bases
Weak base has Kb lt 1 Leads to small OH- and a
pH of 12 - 7
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Relation of Ka, Kb, H3O and pH
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Calculations with Ka or Kb

• Ka or Kb can be used to calculate
• equilibrium concentrations
• pH
• percent ionization

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Ionization Constants of Conjugate Acid-Base Pairs
Ka
Kb
Kw
KaKb Kw
Weak Acid and Its Conjugate Base
15.7
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Weak Acid and Base Ionizations

• How do we calculate the H3O and pH of weak
  acid or base solutions if these do not ionize
  completely?
• Consider the specific weak acid or base
  dissociation equilibrium involved.
• Perform equilibrium calculations to determine
  concentrations of species in the reaction mixture
  at equilibrium.

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Weak Acid Ionization

• Consider the ionization of a weak acid
• HA H2O ltgt H3O A-
• The equilibrium constant for this reaction is the
  
• acid dissociation constant, Ka
• Ka H3OA-/HA
• (Note that these are all equilibrium
  concentrations.)

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Ka values for some Weak Acids
Increasing acid strength
H ion that ionizes shown in red
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Weak Acid Ionization

• Some acids have more than one ionizable proton.
• Called polyprotic acids
• Ionize in steps
• Consider the ionization of a weak diprotic acid
• H2A H2O ltgt H3O HA- Ka1
  H3OHA-/H2A
• HA- H2O ltgt H3O A2- Ka2
  H3OA2-/HA-
• The 1st ionization effectively controls solution
  pH for most polyprotic acids
• Ka1 gtgt Ka2 gt Ka3, etc.
• When calculating pH, ignore any ionization after
  the first ionization if Ka1 gtgt Ka2.

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• Solving weak acid ionization problems
• Identify the major species that can affect the
  pH.
• In most cases, you can ignore the autoionization
  of water.
• Ignore OH- because it is determined by H.
• Use ICE to express the equilibrium concentrations
  in terms of single unknown x.
• Write Ka in terms of equilibrium concentrations.
  Solve for x by the approximation method. If
  approximation is not valid, solve for x exactly.
• Calculate concentrations of all species and/or pH
  of the solution.

15.5
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Equilibria Involving A Weak Acid

• You have 1.00 M HOAc
• Calculate the pH and equilibrium conc of HOAc,
  H3O, and OAc-
• Step 1. Define equilibrium conc in ICE box
• HOAc H3O OAc-
• initial
• change
• equilib

1.00 0 0
-x x x
1.00-x x x
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Equilibria Involving A Weak Acid
You have 1.00 M HOAc. Calculate the equilibrium
conc. of HOAc, H3O, OAc-, and the pH.

• Step 2. Write Ka expression with terms from the
  ICE box

This is a quadratic. Solve using quadratic
formula.
Or you can make an approximation if x is very
small (x should be within 5 of initial conc)!
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Equilibria Involving A Weak Acid
You have 1.00 M HOAc. Calculate the equilibrium
conc of HOAc, H3O, OAc-, and the pH.

• Step 3. Solve Ka expression

First assume x is very small because Ka is so
small.
Now we can more easily solve this approximate
expression.
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Equilibria Involving A Weak Acid
You have 1.00 M HOAc. Calculate the equilibrium
conc. of HOAc, H3O, OAc-, and the pH.

• Step 3. Solve Ka approximate expression

x H3O OAc- 4.2 x 10-3 M pH - log
H3O -log (4.2 x 10-3) 2.37
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What is the pH of a 0.122 M monoprotic acid whose
Ka is 5.7 x 10-4?
Initial (M)
0.122
0.00
0.00
Change (M)
-x
x
x
Equilibrium (M)
0.122 - x
x
x
0.122 x ? 0.122
Ka ltlt 1
x2 6.95 x 10-5
x 0.0083 M
More than 5 Approximation not ok.
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x2 0.00057x 6.95 x 10-5 0
ax2 bx c 0
x 0.0081
x - 0.0081
H x 0.0081 M
pH -logH 2.09
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Quick Exercise I.

• What is the pH of a 0.5 M HF solution
• (at 250C)?

Ka 7.1 x 10-4
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What is the pH of a 0.5 M HF solution (at 250C)?
Initial (M)
0.50
0.00
0.00
Change (M)
-x
x
x
Equilibrium (M)
0.50 - x
x
x
0.50 x ? 0.50
Ka ltlt 1
x2 3.55 x 10-4
x 0.019 M
H F- 0.019 M
pH -log H 1.72
HF 0.50 x 0.48 M
15.5
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When can I use the approximation?
0.50 x ? 0.50
Ka ltlt 1
When x is less than 5 of the value from which it
is subtracted.
Less than 5 Approximation ok.
x 0.019
What is the pH of a 0.05 M HF solution (at 250C)?
x 0.006 M
More than 5 Approximation not ok.
Must solve for x exactly using quadratic equation
or method of successive approximation.
15.5
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Example of when not to use approximation

• Calculate the pH of a 0.0010 M solution of formic
  acid, HCO2H.
• HCO2H H2O ? HCO2- H3O
• Ka 1.8 x 10-4
• Approximate solution
• H3O 4.2 x 10-4 M, pH 3.37
• Exact Solution
• H3O HCO2- 3.4 x 10-4 M
• HCO2H 0.0010 - 3.4 x 10-4 0.0007 M
• pH 3.47

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Weak Bases and Base Ionization Constants
Kb is the base ionization constant
15.6
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15.6
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Weak Base Ionization

• Consider the ionization of a weak base
• B H2O ltgt BH OH-
• The equilibrium constant for this reaction is the
  
• base ionization constant
• Kb BHOH-/B
• (Note that these are all equilibrium
  concentrations)
• The same principles that apply to weak acids
  apply to weak bases

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Equilibria Involving A Weak Base

• You have 0.010 M NH3 Calculate the pH
• NH3 H2O ? NH4 OH-
• Kb 1.8 x 10-5
• Step 1. Define equilibrium concs. in ICE table
• NH3 NH4 OH-
• initial
• change
• equilib

0.010 0 0
-x x x
0.010 - x x x
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Equilibria Involving A Weak Base

• You have 0.010 M NH3. Calc. the pH.
• NH3 H2O ? NH4 OH-
• Kb 1.8 x 10-5
• Step 1. Define equilibrium concs. in ICE table
• NH3 NH4 OH-
• initial
• change
• equilib

0.010 0 0
-x x x
0.010 - x x x
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Equilibria Involving A Weak Base

• You have 0.010 M NH3. Calc. the pH.
• NH3 H2O ? NH4 OH-
• Kb 1.8 x 10-5
• Step 2. Solve the equilibrium expression

Assume x is small, so x OH- NH4
4.2 x 10-4 M and NH3 0.010 - 4.2 x 10-4
0.010 M The approximation is valid !
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Equilibria Involving A Weak Base

• You have 0.010 M NH3 Calculate the pH
• NH3 H2O ? NH4 OH-
• Kb 1.8 x 10-5
• Step 3. Calculate pH
• OH- 4.2 x 10-4 M
• so pOH - log OH- 3.37
• Because pH pOH 14,
• pH 10.63

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Types of Acid/Base Reactions Summary
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Acid-Base Properties of Salts

• NH4 H2O ltgt NH3 H3O
• Acids and bases can be either neutral molecules
  (CH3COOH, NH3) or ions (NH4, F-).
• Ionic acids are typically conjugate acids of weak
  molecular bases
• Ionic bases are conjugate bases of weak molecular
  acids
• Examples of ionic acid/base hydrolysis reaction
• weak acid
• F- H2O ltgt HF OH-
• weak base

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Acid-Base Properties of Salts

• What is the relationship between the strength of
  an acid and its conjugate base?
• The stronger the acid, the weaker the conjugate
  base.
• We can quantify this mathematically in terms of
  Ka and Kb.
• Kw KaKb
• For ionic acids and bases, Ka and Kb must often
  be calculated using this equation.

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pH testing

• There are several ways to test pH
• Blue litmus paper (red acid)
• Red litmus paper (blue basic)
• pH paper (multi-colored)
• pH meter (7 is neutral, lt7 acid, gt7 base)
• Universal indicator (multi-colored)
• Indicators like phenolphthalein
• Natural indicators like red cabbage, radishes

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Paper testing

• Paper tests litmus paper and pH paper
• Place a drop of the solution onto a piece of
  litmus paper
• Read and record the color change. Note what the
  color indicates.
• You should only use a small portion of the paper.
  You can use one piece of paper for several tests.

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pH paper
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pH meter

• Tests the voltage of the electrolyte
• Converts the voltage to pH
• Very cheap, accurate
• Must be calibrated with a buffer solution

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pH indicators

• Indicators are dyes that can be added that will
  change color in the presence of an acid or base
• Some indicators only work in a specific range of
  pH
• Some dyes are natural, like radish skin or red
  cabbage

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ACID-BASE REACTIONSTitrations

• H2C2O4(aq) 2 NaOH(aq) ---gt
• acid base
• Na2C2O4(aq) 2 H2O(liq)
• Carry out this reaction using a TITRATION.

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Setup for titrating an acid with a base
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Titration

• 1. Add solution from the buret
• 2. Reagent reacts with compound in solution in
  the flask
• Indicator should change color near equivalence
  point (end point color change)
• Equivalence Point
• (eq Acid eq Base)
• This is called NEUTRALIZATION

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Dilution/Titration Equation

• Remember
• M1 V1 M2 V2

Use this equation to determine titration
quantities
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Quick Exercise II.

• 35.62 mL of NaOH is neutralized with 25.2 mL
  of 0.0998 M HCl by titration to an equivalence
  point. What is the concentration of the NaOH?

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Answer to Quick Exercise II.

• M1 V1 M2 V2
• M1 (.03562L) .0998M (.0252L)
• M1 0.0706M