Outline:412005

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Outline 4/1/2005

• Announcements
• pick up CAPA sets
• Today Chapter 16 Acids Bases
• Introduction to Acids and Bases

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Acid/Base Reactions

• some covalent compounds have weakly bound H atoms
  and can lose them to water (acids)
• some compounds produce OH- in water solutions
  when they dissolve (bases)
• acid/base reaction are very important to
  biochemistry and environmental chemistry

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Acids Bases
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Biological Effects of acids in the environment
aquatic animals develop abnormally acidic bodies
of water can be sterilized
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Acid/Base Definition

• Bronsted-Lowry definition
• an acid is a proton (H) donor (HCN, HCl, etc.)
• a base is a proton (H) acceptor (CN-, OH-, etc.)
• Lewis definition
• an acid is an electron pair acceptor
• a base is an electron pair donor

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What is an Acid?

• an acid is any chemical that donates a H
• in water H3O is generated (since H2O has the
  highest conc. of all species)
• but H is shorthand
• H is called a proton, H3O is called hydronium
• example - the strong acid HCl
• hydrochloric acid or muriatic acid
• HCl(aq) H2O (l) ? H3O (aq) Cl- (aq)
• HCl (aq) ? H (aq) Cl- (aq) (short-hand)
• other strong acids HNO3 (nitric)
• H2SO4 (sulfuric)

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What is a Base?

• a base is any chemical that accepts a proton
• in aqeuous solns, the H usually comes from H2O
• OH- is produced
• example - the strong base NaOH (sodium hydroxide
  or lye - commercial name)
• NaOH (aq) ? Na (aq) OH- (aq)
• another strong base KOH (potassium hydroxide)
• weak base NH3
• NH3 (aq) H2O ? NH4 (aq) OH- (aq)

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Acid Base Nomenclature

• HNO3, H3O, H3PO4 , etc.
• NaOH, OH- , PO43- , etc.

• Monoprotic, diprotic, polyprotic acids
• Conjugate acids, conjugate bases.

• Strong vs. Weak

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Other Acids and Bases

• organic acids
• COOH (carboxylic acid groups)
• NH3 (protonated amine) groups
• organic bases
• NH2 (amine) groups (have a lone pair of
  electrons)
• COO- (carboxylate) groups
• soluble metal oxides
• Li2O H2O ? 2 Li 2 OH-

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Neutralization

• when an acid and a base are mixed, they will
  neutralize each other
• HCl(aq) NaOH(aq) ? H2O(l) NaCl(aq)
• CH3COOH(aq)NaHCO3 (aq)?NaCH3COO(aq)CO2(g)H2O(l)
  
• the acid and base react to form water and a salt
  (usually)

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Solubility Rules and Strong Acids/Bases

• Solubility rules
• Highly soluble anions
• ionic salts of Cl-, Br-, I-,
• ionic salts of NO3-, SO4-,
• ionic salts of ClO4-, etc.

make Strong Acids when combined with H
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16_10.jpg
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16_11.jpg
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Outline 4/4/2005

• Announcements
• Today Chapter 16 Acids Bases
• Equilibrium Calcs
• worksheet

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More Equilibrium!

• Aqueous equilibria ? acids bases

What exactly is in an HNO3 solution?
effectively no HNO3 is present (its a strong
acid)
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• Does this differ from a weak acid?

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Major vs. Minor Species

• Major species are those much larger than 10-5 M
• Minor species are those much smaller than 10-5 M
• of course this is all relative and arbitrary

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you need to memorize these !!
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The extent of dissociation Ka and Kb

• Define acid dissociation constant
• Ka HA-/ HA

• Define base dissociation constant
• Kb BHOH-/B

• OK, what exactly does water do?
• H2O(l) H2O(l) ? H3O(aq) OH-(aq)
• Keq H3OOH- Kw

1 ? 10 -14 (by definition at 25 C)

• this is the water dissociation constant

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Define water as neutral

• Kw 10-14 (arbitrary scale definition)
• then H3O OH- x
• or x2 10-14 or x 10-7 M

Define pH - log H

• Whats the pH of neutral water?
• - log10-7 7.0
• pHlt7 solution is acidic
• pHgt7 solution is basic

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Same thing for bases.

• Define pOH -logOH-

• Since Kw H3OOH- 10-14
• -logH3O - logOH- - log (10-14)
• or pH pOH 14

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A note about Equilibrium Constants

• it contains some Kas, Kbs, and Ksps
• the table presents them as pKa, etc.
• pKa -log(Ka) so
• Ka 10-pKa

Why?
Which is easier to read 6.84 or 1.45 ?
10-7
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Neutralization

• when an acid and base are combined, they
  neutralize each other
• the extent of neutralization depends on the
  strength of the acid/base
• strong acids strong bases completely neutralize
  each other
• examples

1) CH3COO- H2O ? CH3COOH OH- 2) CH3COO-
H3O ? CH3COOH H2O 3) OH- H3O ? 2 H2O
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Neutralization

• strong acids strong bases completely neutralize
  each other (very large Keq)
• when strong acids or strong bases are combined
  with weak bases or acids, the reaction has a very
  large Keq
• when weak acids and bases are combined, the Keq
  is small or very small

1) CH3COO- H2O ? CH3COOH OH- 2) CH3COO-
H3O ? CH3COOH H2O 3) OH- H3O ? 2 H2O
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Equilibrium Calculations

• Calculate the pH of a solution of 2.5?10-2 M
  HClO4. (perchloric acid)

• Step 1 identify the acid
• If strong acid complete dissociation
• HClO4 (aq) ? H (aq) ClO4- (aq)
• 2.5?10-2 M 0 0
  init
• 0 2.5?10-2 2.5?10-2
  equil

• Step 2 pH - log H
  - log (2.5?10-2) 1.60

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Equilibrium Calculations

• Calculate the pH of a solution of 2.5?10-2 M
  HClO? (hypochlorous acid)

• Step 1 identify the acid
• if weak acid look up Ka 3.5 ? 10-8
• HClO (aq) ? H (aq) ClO- (aq)
• 2.5?10-2 M 0 0
  init
• -x x x
  change
• (2.5?10-2-x) x x
  equil

• Step 2 find x, then pH - log H
  - log (3.0?10-5)
  4.53

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Weak Base Example

• Calculate the pH of a solution of 2.5?10-2 M
  trimethylamine (CH3)3N ?

• Step 1 identify it as a weak base (pKb4.19)
• (CH3)3N(aq) H2O ? OH-(aq)(CH3)3NH(aq)
• 2.5?10-2 M 0 0
  init
• -x x x
  change
• (2.5?10-2-x) x x
  equil

• Step 2 find x, pOH - log OH-
  - log (1.3?10-3) 2.89

• Step 3 14 - pOH pH 11.11

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A trick for acid problems
What do you get when you use the 5 rule?

• HCO2H(aq) H2O(l) ? H3O(aq) HCO2-(aq)

• initial 0.10
  0 0 (M)
• change -x x
  x (M)
• equil. 0.10 -x x
  x (M)

• Ka x2/(0.10-x)
• assume x ltlt 0.10
• Ka x2/(0.10)
• x H3O Ka(CHA) ½

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Worksheet

• Lets do some harder problems
• start problem 1
• Strong acid weak base
• write the reaction, eliminate spectator ions
• what is the Keq?

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Step 1 Write the relevant reactions for
H(aq) NH3(aq) ? NH4(aq) Keq ?

• H2O(l)NH3(aq)?NH4(aq) OH-(aq) Kb1.8?10-5
• is Kw in there somewhere?

H2O(l)?H(aq)OH-(aq) Kw1.0?10-14 what
about just Ka?
Keq Kb / Kw 1/Ka 1.8 ? 109
Step 2 Now find the equil. conc
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Worksheet problem 1 (cont.)
H(aq) NH3(aq) ? NH4 (aq) 0.0667
0. 0667 0 init (M)
0.0 0.0 0. 0667 new
(M) x x -x change
x x (0.0667-x)
eq.(M)
Keq 1.8 ? 109 (0.0667 - x) / x2
x 6.1 ? 10-6 mol pH -log H
-log (x) 5.2
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Worksheet

• Strong base weak acid
• Worksheet - problem 2
• do this one on your own

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Outline 4/6/2005

• Announcements
• Today Chapter 16 Acids Bases
• buffers

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Conjugate acids bases

• For each acid there is a conjugate base.
• HCl ? H Cl-
• For each base there is a conjugate acid.
• NH3 H2O ? OH- NH4

• For every acidbase reaction, there are both
  conjugate basesconjugate acids.

• For example, what is HCO3-

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• HCO3- HCl ? H2CO3 Cl-

• H acceptor H donor
• Bronsted Base Bronsted Acid

• HCO3- OH- ? CO3 2 - H2O

• H donor H acceptor
• Bronsted Acid Bronsted Base

• Both! (amphoteric)

• H2CO3 polyprotic acid

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Acid Base Nomenclature
The salt of a weak acid is a base

• HOAc H2O ? H3O OAc-
• NaOAc H2O ? OH- HOAc Na

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Outline 4/8/2005

• Announcements
• Today Chapter 16 Acids Bases
• buffers and titrations

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Indicators - the magic wand
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Buffers

• Buffering demo
• Watch pH of water acid/base
• Watch pH of buffer acid/base

• Definition
• A buffer is usually a mixture of conjugate
  acid-base pairs in solution. A buffer resists
  strong changes in pH.

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How do you make a buffer?
add a weak acid and its conjugate base to a
solution
A weak acid only partially breaks up CH3COOH
H2O ? CH3COO- H3O A weak base only pulls some
H off H2O CH3COO- H2O ? CH3COOH OH-
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How do you make a buffer?
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Buffers

• Buffer capacity refers to how much of the buffer
  is there and how much acid or base it can absorb
  before it no longer works
• The more buffer you add to the solution, the
  higher the buffer capacity
• I could add
• 1 mole of HOAc and 1 mole of NaOAc
• OR 10 mol of HOAc and 10 mol of NaOAc
• Which has the largest buffer capacity?
• 10 mol of HOAc and 10 mol of NaOAc (? 10 !)

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Outline 4/11/2005

• Announcements
• Today Chapter 17 Acids Bases
• buffers and titrations

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How a Buffer Works
from Principles of Chemsitry, Munowitz, 1st ed,
Norton.
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What happens when you add OH- instead?
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• Worksheet - problem 3

50 g CH3COONa is added to 100 mL of 0.1 M
CH3COOH. pH ?
50 g 0.61 mol/0.1L 6.1 M
Relevant equation?
CH3COOH ? CH3COO- H
Init 0.1 M 6.1 M
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• Should we worry about x?

CH3COOH ? CH3COO- H
Eq 0.1 M - x 6.1 M x x
Ka 10-4.75 (6.1 x) x /(0.1- x) assume
xltlt0.1 10-4.75 (6.1) x /(0.1) x
(10-4.75)(0.1)/(6.1) 2.9?10-7 M
the initial conc. are the eq. conc.
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• An easy way to handle these?

The buffer equation pH pKa log (conj
base/acid)

• Where does this come from?

Ka HA-/HA pKa pH p(A-/HA) pKa
pH - log(A-/HA)
pH pKa log (conj base/acid) this assumes
the shift in equilibrium is insignificant (x is
real small)
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• Worksheet - problem 3

pH pKa log (conj base/acid)
pH 4.75 log (6.1/0.1) 4.75 1.78
6.53
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Worksheet - problem 4
2x 2(2.9?10-7 M) 5.9?10-7 M where do we go
from here? pH 4.75 log (6.1 2x)/(0.1- 2x)
4.75 1.78 6.53
Same answer! Not a lot of change a buffer!!!
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Buffer calculations.
pH pKa log (conj base/acid)
This one also exists pOH pKb log (conj
acid/base)
Practice!
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A buffer example

• Human blood must be maintained at a pH of 7.40 ?
  0.05 for the rest of the body biochemistry to
  function. This is achieved (in part) with a
  buffer made from CO2
• CO2 (aq) H2O (l) ? H2CO3 (aq)

• If the carbonic acid concentration of a healthy
  human is 1.35?10-2 M, what is the normal human
  concentration of HCO3-?

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pH pKa log (conj base/acid)

• 7.40 6.37 log (HCO3-/ 1.35?10 -2)
• HCO3- 1.4465?10-1 M

• A rather twisted addition to the problem
• Inject 10 mL of 1.0 M HCl into a (previously
  healthy) human with a total blood volume of 8.0
  L. What is the new pH?

Buffer capacity problem
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How many moles H added?
0 .010 L ? 1.0 M 0.010 mol
How many moles H there already?

• H 10-7.40 3.18?10-7 M ? 8.0 L
• 2.54?10-6 mol
• negligible!

Initial H?

• H(0.010 mol2.54?10-6 mol)/ 8.0 L
• 1.25 ? 10-3 M

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Same old procedure.

• H2CO3 ? HCO3- H

1.35?10-2 1.4465?10-1 1.25?10-3 Init
Ch
1.25?10-3 -1.25?10-3 -1.25?10-3 1.48?10-2
1.4340?10-1 0
- x x x
Ch solve for x x 2.0591?10-8 M pH 7.38
Ka 4. 27?10-7 which direction is favored?
or use pH pKalog (conj base
/acid) 6.37log(1.43?10-1/1.48?10-2) 7.38
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About the blood and respiration...
Why are you supposed to breath into a paper bag
when you hyperventilate?

• Breathing too fast expels CO2 too fast
• H2CO3 ? CO2 H2O (which way will it shift?)
• Breathing into a bag increases the conc. of CO2
  in the air
• H2CO3 ? CO2 H2O (which way will it shift?)

? your blood pH will increase
(alkalosis) and you will faint
? your blood pH will decrease
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Outline 4/13/2005

• Announcements
• Today Chapter 16 Acids Bases
• titrations

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Acid/Base Titrations

• titration a way of determining the
  concentration of one solution by using another
  solution of known concentration

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Terms

• equivalence point (or stoichiometric point) ? the
  calculated point when enough titrant has been
  added to react with all of the unknown
• physical change color, turbidity, temperature,
  conductivity, voltage
• mid-point ? the point in a weak acid titration
  where pH pKa

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weak acid
Buffer region
Stoichiometric (or equivalence) point
of OH- added
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Strong acid strong base
H OH- ? H2O
pH at equivalence point 7.0

• Strong base weak acid

• pH at equivalence point depends on the Kb of the
  conjugate base

• Strong acid weak base

• pH at equivalence point depends on the Ka of the
  conjugate acid

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What type of Keq problem is it? Well, what is in
the solution?
weak acid, so ICE
weak acid weak base, so buffer
weak base, so ICE
strong base, so pH14-pOH
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What is being titrated by what?
Are you starting with an acid or base?
Are you titrating with an acid or base?
Are you starting with a weak or strong acid?
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What is being titrated by what?
Are you starting with an acid or base?
Are you titrating with an acid or base?
Are you starting with a weak or strong base?
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Buffers and Buffer Capacity

• What is the pH of a solution made by adding 0.11
  mol of acetic acid and 0.14 mol of sodium
  acetate, to enough water to make 1.0 L solution?
• Acetic acid and sodium acetate
  A buffer !!!

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Same old procedure.

• HC2H3O2 ? H C2H3O2-

0.11 0
0.14 I C
-x x
x 0.11 - x x 0.14
x E
solve for x x 1.4?10-5 M pH 4.84
Ka 1.8?10-5
or use pH pKalog (conj base /acid)
4.74log(0.14/0.11) 4.84
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Addition of a Strong Acid/Base to a Buffer

• What is the pH of the buffer solution after
  adding 0.40 mol of NaOH to the 1.00 L of
  solution?
• First Write the reaction.
• HC2H3O2 OH-? H2O C2H3O2-
• Second calculate concentration of species after
  the reaction

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Same old procedure.

• HC2H3O2 OH-? C2H3O2- H2O

0.11 0.40 0.14
I C
-0.11 -0.11 0.11
0 0.29 0.25
NI
The acetic acid is consumed!! Try to use the
buffer equation...
pH pKalog (conj base /acid)
4.74log(0.30/0) Does not work!
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pH is determined by excess amount of hydroxide

• pOH -log(0.29) 0.54
• pH 14 - 0.54 13.46
• Do not forget buffer capacity

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Three Regions in a Titration

• 1. Starting pH
• 2. Buffer region calc
• 3. Equivalence point
• 4. Post-equivalence point
• see AcidBaseTitrations worksheet

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Hesss law and acid/base problems

• Whats the Keq of
• HCOO- H3O ? HCOOH H2O

• Keq HCOOH / HCOO-H3O

• Does this relate to any Ka you know?

• Ka HCOO-H3O / HCOOH
• 1.8 x 10-4

• Keq 1 / Ka 5555

• Look for reactions written backwards!

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Acid - Base reactions

• Whats the Keq of
• HClO NH3 ? ClO- NH4
• (w. acid) (w. base) (conj base)
  (conj acid)

• Keq ClO-NH4/ HClONH3
• Does this relate to any Ks you know?

• Ka ClO-H / HClO 3.0 ? 10-8

• Kb NH4OH- / NH3 1.8 ? 10-5

• Kw HOH- 1.0 ? 10-14

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• Keq Ka Kb / Kw 54
• HClO NH3 ? ClO- NH4

Does Keq favor products or reactants?
products

• Four types of acid-base reactions
• weak acid weak base
• strong acid weak base
• weak acid strong base
• strong acid strong base

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Relationship Between Conjugates

• Conjugate acids and bases are related by the
  following equation
• Ka Kb Kw

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Challenge Worksheet

• the pH of rain

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Outline 4/15/2005

• Announcements
• Today Chapter 16 Solubility
• Ksp calculations

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Dissolved Ionic Compounds

• When ionic compounds dissolve in water, they
  break up into ions
• NaCl(s) ? Na(aq) Cl-(aq)

• the attraction that water has for Na and Cl- is
  greater than the attraction between Na and Cl-
• so the ionic bond between Na and Cl- breaks and
  new bonds between water and the ions form

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Dissolved Ionic Compounds

• some ionic compounds do not dissolve in water (at
  least not very much)
• PbSO4 (s) ? Pb2(aq) SO42-(aq)

• the attraction that water has for Pb2 and SO42-
  is not greater than the attraction between Pb2
  and SO42-
• so the ionic bond between Pb2 and SO42- does not
  break and PbSO4 does not dissolve

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Solubility

• It get complicated
• Pb2 and other ions create more order in the
  solvent
• this result is a negative ?S, reducing
  spontaneity
• also consider the ?Hsolv endo or exothermic

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Dissolved Ionic Compounds

• When NaCl dissolves it forms 2 ions in solution
• NaCl(s) ? Na(aq) Cl-(aq)

• how many ions does Pb(NO3)2 form when it
  dissolves?
• 1 Pb2 and 2 NO3- ions 3 ions

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Reaction Types - Precipitation

• when solutions of ions mix, sometimes an
  insoluble salt forms

Pb(NO3)2 (aq) 2 NaI (aq) ? PbI2(s) 2 NaNO3(aq)

• the net ionic reaction is
• Pb2(aq) 2 I- (aq) ? PbI2(s)
• spectator ions NO3- and Na

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Ksp reaction format BaF2 (s) ? Ba2(aq) 2
F-(aq)
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Barium Sulfate

• Ba2 is a mildly toxic heavy metal when it is
  soluble in water

• linked to elevated blood pressure
• ingestion can cause nausea, vomiting, diarrhea,
  and crampy abdominal pain within minutes of
  consuming the meal
• BaSO4 is so insoluble (pKsp 9.96) that it
  is ingested as a contrast agent in X-rays and CAT
  scans

Olmsted and Williams 3rd ed pg. 161
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Kidney Stones
Ebbing 7th ed pg. 769
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Reaction Types - Precipitation

• sometimes nothing happens

KNO3 (aq) NaI (aq) ? KI(aq) NaNO3(aq)

• since nothing really happened, there is not a
  reaction

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Why Use Net Ionic Equations?

• some of the elements in the equation can be
  changed without changing the result

Pb(NO3)2 (aq) 2 NaI (aq) ? PbI2(s) 2
NaNO3(aq) Pb2(aq) 2 I- (aq) ? PbI2(s)
Pb(CH3COO)2 (aq) 2 KI (aq) ? PbI2(s) 2
KCH3COO(aq) Pb2(aq) 2 I- (aq) ? PbI2(s)

• the result is the same!

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Solubility Rules for Ionic Compounds in Water
Soluble Ionic Compounds
1) All common compounds of alkali ions (Na,K,
etc.) and ammonium (NH4) are soluble. 2) All
common nitrates (NO3-), acetates (CH3COO-), and
most perchlorates (ClO4-) are soluble. 3) All
common chlorides (Cl-), bromides (Br -), and
iodides (I-) are soluble, except those of Ag,
Pb2, Cu, and Hg22. 4) All common sulfates
(SO42-) are soluble, except those of Ca2, Sr2,
Ba2, and Pb2.
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Solubility Rules for Ionic Compounds in Water
Insoluble Ionic Compounds
1) All common metal hydroxides are insoluble,
except those of alkali metals, NH4, and the
larger members of alkaline earth metals
(beginning with Ca2). 2) All common carbonates
(CO32-) and phosphates (PO43-) are insoluble,
except those of alkali metals and NH4. 3) All
common sulfides are insoluble, except those of
alkali metals, alkaline earth metals, and NH4.
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Ksp Example

• What mass of AgCl will dissolve in a 500 mL soln
  of 5.0?10-4 M NaCl?
• AgCl(s) ? Ag(aq) Cl-(aq)
  pKsp9.74
• initial(M) 0 5.0?10-4

• change(M) x x
• eq. (M) x 5.0?10-4 x

• 10-9.74 1.82 ?10-10 (x)(5.0?10-4 x)
• assume xltlt 5.0?10-4
• x 1.82 ?10-10 / 5.0?10-4 3.64?10-7M

• ?

• find the mass of AgCl (3.64?10-7M)(0.5L)(143g/mol
  )
• 2.6 ?10-5g

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Another Ksp problem
Mix 500 mL of 0.200 M magnesium chloride with 200
mL of 0.400 M NaOH, what mass of precipitate
Mg(OH)2 forms?
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Another Ksp problem (cont.)

• (0.500L)(0.200 mol/L) 0.100 mol MgCl2
• (0.200L)(0.400 mol/L) 0.0800 mol NaOH

• 0.500 L 0.200 L 0.700 L
• 0.100/0.700 0.142 M MgCl2
• 0.0800/0.700 0.114 M NaOH

• What is the reaction of interest?
• Mg2 Cl- Na OH-

• Mg(OH)2(s) ? Mg2 2OH-

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Another Ksp problem (cont.)

• Mg(OH)2(s) ? Mg2(aq) 2 OH-(aq)
  pKsp11.25
• I(M) 0 0.143 0.114
  Ksp says?

• Ksp 10-11.25 (2x)2(0.0859x)
• assume xltlt0.0859
• x(10-11.25)/(0.08594)½4.04?10-6 M

• check

Mg2 0.0859x 0.086 M OH- 2x
8.1E-6 M g Mg(OH)2 (0.057M)(.7L)(58.3g/mol)2.
3 g
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stop here
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Outline 3/21/2003

• Announcements
• Today Chapter 15 Solubility
• Ksp calculations
• surfactants

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Solubility Equilibrium

• There is no solubility equilibrium until a
  solution is saturated.
• A saturated solution contains as much solute as
  possible at a given temperature and pressure.

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What would happen if.

• Excess NH3 were added to the solution?
• AgCl(s) ? Ag Cl-
• Ag 2 NH3 ? Ag(NH3)2

• Ag would be used up...more Ag(NH3)2
• for AgCl(s) ? Ag Cl-
• fewer products
• more AgCl would have to dissolve
• WHY?

• To satisfy the Keq expression!
• (LeChâteliers Principle )

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What would happen if.

• Heat were added to the solution?

• Ag 2 NH3 ? Ag(NH3)2

• DH0rxn-120 kJ/mol

• Ag 2 NH3 ? Ag(NH3)2 120 kJ

• Ag would be produced...less Ag(NH3)2
• for AgCl(s) ? Ag Cl-
• more products
• more AgCl would have to ppt