Title: CHEMISTRY Chapter 15 Applications of Aqueous Equilibria
1CHEMISTRYChapter 15Applications of Aqueous
Equilibria
2- Two important points
- Reactions with strong acids or strong bases go to
completion. - Reactions with only weak acids and bases reach an
equilibrium.
3The pH scale
- The pH scale ranges from 0 to 14.
- Acids have a pH less than 7.
- A base has a pH greater than 7.
- Pure water has a pH equal to 7.
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6 Acids and bases in your body
- Many reactions, such as the ones that occur in
your body, work best at specific pH values.
7pH and blood
- The pH of your blood is normally within the range
of 7.37.5. - Holding your breath causes blood pH to drop.
- High blood pH can be caused by hyperventilating.
8Acids Bases
- Strong acids
- Know the names and formulas of the 7 common
strong acids - HCl (aq) hydrochloric acid
- HBr (aq) hydrobromic acid
- HI (aq) hydroiodic acid
- HClO3 chloric acid
- HClO4 perchloric acid
- HNO3 nitric acid
- H2SO4 sulfuric acid
9Acids Bases
- Examples of Weak Acids
- HF (aq) hydrofluoric acid
- H3PO4 phosphoric acid
- CH3COOH acetic acid
10Acids Bases
- Strong Bases
- Know the names and formulas of the strong bases
- Alkali metal (1A) hydroxides
- LiOH lithium hydroxide
- NaOH sodium hydroxide
- KOH potassium hydroxide
- RbOH rubidium hydroxide
- CsOH cesium hydroxide
11Acids Bases
- Strong bases to know (cont)
- Heavy alkaline earth metal (2A) hydroxides
- Ca(OH)2 calcium hydroxide
- Sr(OH)2 strontium hydroxide
- Ba(OH)2 barium hydroxide
12Acids Bases
- Examples of Weak Bases
- ammonia (NH3)
- sodium bicarbonate (NaHCO3)
- baking soda
- a component of Alka-Seltzer
13Acids Bases
- Generally, when solutions of an acid and a base
are combined, the products are a salt and water.
- CH3COOH (aq) NaOH (aq) ??CH3COONa (aq) H2O
(l) - HCl (aq) NaOH (aq) ?? NaCl (aq) H2O (l)
- All neutralization reactions are double
displacement reactions -
Neutralization Reaction
14Neutralization Reactions
- When a
- strong acid reacts with a strong base,
- the net ionic equation is
- HCl (aq) NaOH (aq) ?? NaCl (aq) H2O (l)
- H (aq) Cl- (aq) Na (aq) OH-(aq) ??
- Na (aq) Cl- (aq) H2O (l)
-
- H (aq) OH-(aq) ?? H2O (l)
15Strong Acids and Bases
16Weak Acids and Strong Bases
17Strong Acid Weak Base
18Weak Acid Weak Base
19COMMON-ION EFFECT
- A shift in equilibrium due to addition of same
ion salt to an aqueous weak acid or weak base is
the common-ion effect. - This is an example of
- Le Chatliers Principle.
20The Common Ion Effect
Weak acid HA H2O ? H3O A- Salt of conj. Base NaA ? Na(aq) A-(aq)
two sources of A- Common Ion! two sources of A- Common Ion! two sources of A- Common Ion!
- What affect does the addition of its conjugate
base have on the weak acid equilibrium? On the
pH? - Used in making buffered solutions
21The Common-Ion Effect
- Common Ion Two dissolved solutes that contain
the same ion (cation or anion). - The presence of a common ion suppresses the
ionization of a weak acid or a weak base. - Common-Ion Effect is the shift in equilibrium
caused by the addition of a compound having an
ion in common with the dissolved substance.
22The Common-Ion Effect
- Q Calculate the pH of a 0.20 M CH3COOH solution
with no salt added. - Q Calculate the pH of a solution containing 0.20
M CH3COOH and 0.30 M CH3COONa. - Q What is the pH of a solution containing 0.30 M
HCOOH, before and after adding 0.52 M HCOOK?
23Buffer Solutions
- A Buffer Solution is a solution of
- (1) a weak acid or a weak base and
- (2) its salt
- both components must be present.
- A buffer solution has the ability to resist
changes in pH upon the addition of small amounts
of either acid or base. - Buffers are very important to biological systems.
24Buffer Solutions
25How do buffers work?
26Adding strong acid or base to a buffer
- Adding acid H3O HA or A- ?
- Adding base OH- HA or A- ?
- Calculating pH
- Stoichiometry of added acid or base
- Equilibrium problem (H-H equation)
27HendersonHasselbalch equation
- To determine the pH, we apply I.C.E. and then the
HendersonHasselbalch equation. - When the concentration of HA and salt are high
(0.1 M) we can neglect the ionization of acid
and hydrolysis of salt.
28HendersonHasselbalch equation
Henderson-Hasselbalch equation
29Buffer Solutions
- Buffer solutions must contain relatively high
acid and base component concentrations, the
buffer capacity. - Acid and base component concentrations must not
react together. - The simplest buffer is prepared from equal
concentrations of acid and conjugate base.
30Buffer Solutions
- Buffer Preparation Use the HendersonHasselbalch
equation in reverse. - Choose weak acid with pKa close to required pH.
- Substitute into HendersonHasselbalch equation.
- Solve for the ratio of conjugate base/acid.
- This will give the mole ratio of conjugate base
to acid. The acid should always be 1.0.
31Buffer Solutions
- Q Describe how you would prepare a phosphate
buffer with a pH of about 7.40. - Q How would you prepare a liter of carbonate
buffer at a pH of 10.10? You are provided with
carbonic acid (H2CO3), sodium hydrogen carbonate
(NaHCO3), and sodium carbonate (Na2CO3).
32Buffer Solutions
- Q Calculate the pH of a buffer system containing
1.0 M CH3COOH and 1.0 M CH3COONa. What is the pH
of the system after the addition of 0.10 mole of
gaseous HCl to 1.0 L of solution? - Q Calculate the pH of 0.30 M NH3/0.36 NH4Cl
buffer system. What is the pH after the addition
of 20.0 mL of 0.050 M NaOH to 80.0 mL of the
buffer solution?
33Acid-Base Titrations
- Titration a reaction used to determine
concentration (acid-base, redox, precipitation) - Titrant solution in buret usually a strong
base or acid - Analyte solution being titrated often the
unknown - equivalence point (or stoichiometric point)
- mol acid mol base
- Found by titration with an indicator
- Solution not necessarily neutral
- pH dependent upon salt formed
- pH titration curve plot of pH vs. titrant volume
34Solubility Equilibria
- Solubility Product is the product of the molar
concentrations of constituent ions and provides a
measure of a compounds solubility. - MX2(s) ? M2(aq) 2 X(aq)
- Ksp M2X2
35Solubility Equilibria
- Al(OH)3 1.8 x 1033
- BaCO3 8.1 x 109
- BaF2 1.7 x 106
- BaSO4 1.1 x 1010
- Bi2S3 1.6 x 1072
- CdS 8.0 x 1028
- CaCO3 8.7 x 109
- CaF2 4.0 x 1011
- Ca(OH)2 8.0 x 106
- Ca3(PO4)2 1.2 x 1026
- Cr(OH)3 3.0 x 1029
- CoS 4.0 x 1021
- CuBr 4.2 x 108
CuI 5.1 x 1012 Cu(OH)2 2.2 x 1020 CuS 6.0 x
1037 Fe(OH)2 1.6 x 1014 Fe(OH)3 1.1 x
1036 FeS 6.0 x 1019 PbCO3 3.3 x 1014 PbCl2 2.4
x 104 PbCrO4 2.0 x 1014 PbF2 4.1 x
108 PbI2 1.4 x 108 PbS 3.4 x 1028 MgCO3 4.0 x
105 Mg(OH)2 1.2 x 1011
MnS 3.0 x 1014 Hg2Cl2 3.5 x 1018 HgS 4.0 x
1054 NiS 1.4 x 1024 AgBr 7.7 x 1013 Ag2CO3
8.1 x 1012 AgCl 1.6 x 1010 Ag2SO4 1.4 x
105 Ag2S 6.0 x 1051 SrCO3 1.6 x 109 SrSO4
3.8 x 107 SnS 1.0 x 1026 Zn(OH)2 1.8 x
1014 ZnS 3.0 x 1023
36Solubility Equilibria
- Q The solubility of calcium sulfate (CaSO4) is
found experimentally to be 0.67 g/L. Calculate
the value of Ksp for calcium sulfate. - Q The solubility of lead chromate (PbCrO4) is
4.5 x 105 g/L. Calculate the solubility
product of this compound. - Q Calculate the solubility of copper(II)
hydroxide, Cu(OH)2, in g/L.
37Solubility Equilibria
- Ion Product (Q) solubility equivalent of the
reaction quotient. It is used to determine
whether a precipitate will form. - Q lt Ksp Unsaturated Q Ksp Saturated Q gt
Ksp Supersaturated precipitate forms.
38 Factors that Affect Solubility
- Common-Ion Effect
- LeChateliers Principle revisited
- Addition of a product ion causes the solubility
of the solid to decrease, but the Ksp remains
constant. - pH
- LeChateliers Principle again!
- Basic salts are more soluble in acidic solution.
- Acidic salts are more soluble in basic solution.
- Environmental example CaCO3 limestone
- Stalactites and stalagmites form due to changing
pH in the water and thus solubility of the
limestone.
39Solubility Equilibria
- Q Exactly 200 mL of 0.0040 M BaCl2 are added to
exactly 600 mL of 0.0080 M K2SO4. Will a
precipitate form? - Q If 2.00 mL of 0.200 M NaOH are added to 1.00 L
of 0.100 M CaCl2, will precipitation occur?
40The Common-Ion Effect and Solubility
- The solubility product (Ksp) is an equilibrium
constant precipitation will occur when the ion
product exceeds the Ksp for a compound. - If AgNO3 is added to saturated AgCl, the increase
in Ag will cause AgCl to precipitate. - Q Ag0 Cl0 gt Ksp
41The Common-Ion Effect and Solubility
42The Common-Ion Effect and Solubility
43The Common-Ion Effect and Solubility
- Q Calculate the solubility of silver chloride
(in g/L) in a 6.5 x 103 M silver chloride
solution. - Q Calculate the solubility of AgBr (in g/L)
in(a) pure water(b) 0.0010 M NaBr
44Complex Ion Equilibria and Solubility
- A complex ion is an ion containing a central
metal cation bonded to one or more molecules or
ions. - Most metal cations are transition metals because
they have more than one oxidation state. - The formation constant (Kf) is the equilibrium
constant for the complex ion formation.
45Complex Ion Equilibria and Solubility
46Complex Ion Equilibria and Solubility
47Complex Ion Equilibria and Solubility
- ION Kf
- Ag(NH3)2 1.5 x 107
- Ag(CN)2 1.0 x 1021
- Cu(CN)42 1.0 x 1025
- Cu(NH3)42 5.0 x 1013
- Cd(CN)42 7.1 x 1016
- CdI42 2.0 x 106
ION Kf HgCl42 1.7 x 1016 HgI42 3.0 x
1030 Hg(CN)42 2.5 x 1041 Co(NH3)63 5.0 x
1031 Zn(NH3)42 2.9 x 109
48Complex Ion Equilibria and Solubility
- Q A 0.20 mole quantity of CuSO4 is added to a
liter of 1.20 M NH3 solution. What is the
concentration of Cu2 ions at equilibrium? - Q If 2.50 g of CuSO4 are dissolved in 9.0 x 102
mL of 0.30 M NH3, what are the concentrations of
Cu2, Cu(NH3)42, and NH3 at equilibrium?
49Complex Ion Equilibria and Solubility
- Q Calculate the molar solubility of AgCl in a
1.0 M NH3 solution. - Q Calculate the molar solubility of AgBr in a
1.0 M NH3 solution.
50Complex Ion Equilibria and Solubility
51Qualitative Analysis Scheme
52Qualitative Analysis