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AcidsandBases
Unit 18Acid-Base Equilibria: Buffers & Hydrolysis
Dr. Jorge L. AlonsoMiami-Dade College –
Kendall CampusMiami, FL
CHM 1046: General Chemistry and Qualitative Analysis
Textbook Reference:
•Chapter 19 (sec. 1-4)
•Modules # 7-8
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AcidsandBases
The Common-Ion Effect
“The extent of ionization of a weak electrolyte is decreased by adding to the solution a strong electrolyte that has an ion in common with the weak electrolyte.”
HC2H3O2 (aq) + H2O (l) H3O+ (aq) + C2H3O2
− (aq)
Add strong electrolyte
Consider a solution of acetic acid, a weak electrolyte:
What will Le Châtelier predict will happen to the equilibrium?
It will shift to the left.
A solution composed of two substances, each containing a same ion in common. One is a weak electrolyte (equilibrium) the other strong(not equil).
Examples: (1) HF(aq) + NaF(s)
(2) HC2H3O2(aq) + NaC2H3O2 (s)
(3) HF(aq) + HCl (aq) NaC2H3O2
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AcidsandBases
The Common-Ion Effect
Calculate the fluoride ion concentration and pH of a solution that is 0.20 M in HF and 0.10 M in HCl. Ka for HF is 6.8 10−4.
[H3O+] [F−][HF]
Ka = = 6.8 10-4
Problem:
Because HCl, a strong acid, is also present, the initial [H3O+] is not 0, but rather 0.10 M.
[HF], M [H3O+], M [F−], M
Initially 0.20 0.10 0
Change −x +x +x
At Equilibrium 0.20 − x 0.20 0.10 + x 0.10 x
HF(aq) + H2O(l) H3O+(aq) + F−(aq)
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AcidsandBases
The Common-Ion Effect
X =
x = 1.4 10−3
(0.10+x) (x)(0.20-x)= 6.810−4 =
(0.20) (6.8 10−4)(0.10)
• So, pH = −log (0.10)
pH = 1.00
[H3O+] [F−][HF]
Ka =
Therefore, [F−] = x = 1.4 10−3
[H3O+] = 0.10 + x = 0.10 + 1.4 10−3 = 0.10 M
What is [F−] ? And pH?
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AcidsandBases
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AcidsandBases
[ ]
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AcidsandBases
Buffers• Solutions that are particularly resistant to pH changes,
even when strong acid or base is added.
• Buffers are solutions of a weak acid mixed with a salt of its conjugate base (a weak conjugate acid-base pair solution).
i.e., acetic acid + sodium acetate
HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2
−(aq)[ ] [ ]
Which is less acidic, buffer (WA + Salt of c.base) or same conc. of weak acid alone?
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AcidsandBases
Buffers
If an acid is added, the F− reacts with H+ to form HF and water.
If a hydroxide is added to an equimolar solution of HF in NaF, for example, the HF reacts with the OH− to make F− and water.
[HF] + H2O H3O+ + [F- ]
Add acid (H3O+)Add base (OH-)
OH- + HF H2O + F- HF H3O+ + F-
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AcidsandBases
pH of Buffer Calculations: The Henderson–Hasselbalch Equation
Rearranging slightly, this becomes[A−][HA]
Ka = [H3O+]
Taking the negative log of both side, we get[A−][HA]
−log Ka = −log [H3O+] + −log
For the dissociation of a generic acid, HA:[H3O+] [A−]
[HA]Ka =HA + H2O H3O+ + A−
pKa = pH − log[base][acid]
Rearranging, this becomes
pH = pKa + log[base][acid]
Henderson–Hasselbalch Equation.
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AcidsandBases
Henderson–Hasselbalch Equation
What is the pH of a buffer that is 0.12 M in lactic acid, HC3H5O3, and 0.10 M in sodium lactate? Ka = 1.4 10−4.
pH = pKa + log[base][acid]
pH = −log (1.4 10−4) + log(0.10)(0.12)
pH = 3.85 + (−0.08)
pH = 3.77
Problem:
[H3O+] [A−][HA]
Ka =
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AcidsandBases
pH Range of Buffers Systems
• The pH range is the range of pH values over which a buffer system works effectively.
• It is best to choose an acid with a pKa close to the desired pH.
pH = pKa + log[base][acid]
pKa
3.
3.
4.
4.
7.
9.
9.
pH = pKa + log[1M][1M]
pH = pKa
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AcidsandBases
[H3O+] [A−][HA]
Ka =
pH = pKa + log[base][acid]
[ ] [ ]
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AcidsandBases
pH= pKa + log [base][acid]
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AcidsandBases
When Strong Acids or Bases Are Added to a Buffer…
You end up with two problems: (1) an acid base neutralization, and (2) an equilibrium problem
[HX] + H2O H3O+ + [X−]
Strong acid (H3O+)Strong base (OH-)
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AcidsandBases
Addition of Strong Acid or Base to a Buffer
(1) Neutralization is NOT an equilibrium reaction.(2) However, it affects the amounts of the weak acid [HA]
and its conjugate base [A-] left over, and a new equilibrium will be established.
pH = pKa + log[base][acid]
[HA] + H2O H3O+ + [A−]
Mole (η) HA A- H3O+ or OH-
Initially η η + η
Change (H3O+ or OH-) + η + η - η
Add acid (H3O+)Add base (OH-)
End (after)reaction η η 0 η
(1) Use: Mole ICEnd Table: *used for neutralization
Rx*
Moles (η) Molarity (η/L)
Add some
OH-
Add some
H3O+
(HA + OH- A- + H2O) (H3O+ + A- HA + H2O)
What are the new
equilibrium conc. after
acid or base is added?
(2)For a buffer:
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AcidsandBases
Calculating pH Changes in Buffers A buffer is made by adding 0.300 mol HC2H3O2 and 0.300 mol
NaC2H3O2 to enough water to make 1.00 L of solution. The pH of the buffer is 4.74. Calculate the pH of this solution after 0.020 mol of NaOH is added.
Before the reaction, since
mol HC2H3O2 = mol C2H3O2−
pH = pKa = −log (1.8 10−5) = 4.74
The 0.020 mol NaOH will react with 0.020 mol of the acetic acid:
Mole HC2H3O2 C2H3O2− OH−
Initially 0.300 η 0.300 η 0.020 η
Change - 0.020 η + 0.020 η - 0.020 η
HC2H3O2(aq) + OH−(aq) C2H3O2−(aq) + H2O(l)
Problem:
End (after) reaction 0.280 η 0.320 η 0.000 η
Use: Mole ICEnd Table: *used for neutralization Rx*
pH = pKa + log[base][acid]
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AcidsandBases
Calculating pH Changes in Buffers
Now use the Henderson–Hasselbalch equation to calculate the new pH:
pH = 4.74 + log(0.320)(0.280)
pH = 4.74 + 0.058pH = 4.80
pH = pKa + log[base][acid]
Mole HC2H3O2 C2H3O2− OH−
Initially 0.300 η 0.300 η 0.020 η
Change - 0.020 η + 0.020 η - 0.020 ηEnd (After) reaction 0.280 η 0.320 η 0.000 η
yxy
xlogloglog
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AcidsandBases
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AcidsandBases
Hydrolysishydro = water lysis = breaking down
Organic/Biochemistry (enzymatic splitting of organic molecules by using water)
hydrolysis
dehydration synthesis
Electrochemistry (also called electrolysis or electrohydrolysis)
Acid Base Chemistry (splitting water by cations or anions to form acidic or basic solutions.
2 H2O 2 H2 + O2
elect.
Cation+ + HOH CationOH + H+
Anion- + HOH HAnion + OH-
Module 8
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AcidsandBases
The Acid-Base Properties of Salt Solutions and Hydrolysis
Cation+ + HOH CationOH + H+
Anion- + HOH HAnion + OH-
Salt (Ionic Compound) = Metal + Nonmetal = Cation+ + Anion-
Al 3+Al3+
Cations of Weak Bases
Anions of Weak Acids
Hydrolysis: splitting water by cations or anions to form acidic or basic solutions
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AcidsandBases
conjugate The Effect of Anions
• An anion that is the conjugate base of a weak acid will increase the pH (behave as bases).
• An anion that is the conjugate base of a strong acid will not affect the pH.
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AcidsandBases
Effect of Cations
• Cations of the strong bases will not affect the pH
SOLUBILITY RULES: for Ionic Compounds (Salts)All OH- are insoluble, except for IA metals, NH4
+, Ca2+, Ba2+ , and Sr2+ (heavy IIA).
• Cations of a weak bases will lower the pH (behave as acids)
• Greater charge and smaller size make a cation more acidic.
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AcidsandBases
Ka of Weak Acid and Kb of its Conjugate Base
The Ka of an acid and Kb of its conjugate base are related in this way: Ka Kb = Kw
Therefore, if you know one of them, you can calculate the other.Problem: Is NaC2H3O2 acidic or basic? Calculate its Ka or Kb.
Ka Kb = Kw
(1.8 x 10-5) Kb = (1.0 x 10-14) Kb =
(1.0 x 10-14)
(1.8 x 10-5) = 5.6 x 10-10
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AcidsandBases
Effect of Cations and Anions
When a solution contains both the conjugate base of a weak acid and the conjugate acid of a weak base, the affect on pH depends on the Ka and Kb values.
• Cations of the strong bases will not affect the pH
SOLUBILITY RULES: for Ionic Compounds (Salts)
All OH- are insoluble, except for IA metals, NH4
+, Ca2+, Ba2+ , and Sr2+ (heavy IIA).
• Cations of a weak bases will lower the pH (behave as acids)
CATION ANION
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AcidsandBases
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AcidsandBases
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AcidsandBases
Acidic and Basic SaltsAcid Salts: salts of weak polyprotic acids. Are not necessarily acidic, but do neutralize bases.
Basic Salts: salts of weak polyhydroxy bases. Are not necessarily basic, but do neutralize acids.
Examples:
Acid Salt Acid Salt(s)
HNO3 NaNO3 -----------
H2CO3 Na2CO3 NaHCO3
H3PO4 Na3PO4 Na2HPO4 NaH2PO4
Solubility Rule:• All OH- are insoluble except for IA metals, NH4
+ & slightly soluble Ca 2+ Ba2+ & Sr2+
Mg(OH) 2(s) Fe(OH)3(s) Cr(OH) 3(s){Milk of Magnesia}
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AcidsandBases
Buffers can act as either…acids or bases…
...they are amphiprotic
.
HCO3−
(aq)
H2PO4−
(aq)
H2O(l)
↔ CO32- + H2OH2O + H2CO3 ↔
H2O + H3PO4 ↔
H2O + H3O+ ↔
↔ HPO42- + H2O
↔ OH- + H2O
accepts protons / donates protons
OH-H3O+
Add acid Add base
Not a buffer
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AcidsandBases
2007 (B)
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AcidsandBases
Making Solution Molarity
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AcidsandBases
2007B Q5
Carboxylic acid
Organic Acids: carboxylic acid functional group
-COOH -COO- + H+
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AcidsandBases
2005A Q1
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AcidsandBases
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AcidsandBases
Review of Strong Acid & BasesStrong acids have very weak conjugate bases
HX + H2O H3O+ + X-
acid base conj. a conj. b
Strong bases have very weak conjugate acids
MOH M+ + OH-
base conj. a conj. b
Salts of Strong acids & bases have very weak conjugate acids & bases
MX + H2O M+(aq)
+ X- (aq)
salt conj. a conj. b
Has no affinity for H+
Has no affinity for OH-
Have no affinity for H+ or OH-
Example: HCl
Example: NaOH
Example: NaCl
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AcidsandBases
Salts: (1) of strong base & weak acid:
MA M+(aq) + A-
(aq)
A- + HOH HA + OH- (hydrolysis)
(2) of a weak base & strong acid:
BHX BH+(aq) + X-
(aq)
BH+ + HOH B + H3O+ (hydrolysis)
(3) of a weak base & weak acid:
BHA ↔ BH+(aq) + A-
(aq)
Weak acids (HA) have very strong conjugate bases (A-)
HA + H2O ↔ H3O+ + A-
acid base conj. a conj. b
Review of Weak Acid & Bases
Weak bases (B) have very strong conjugate acids (BH+)
B + HOH ↔ BH+ + OH-
base acids conj. a conj. b
Has affinity for H+
Has affinity for OH-
Example: HF
Example: NH3
Example: NaFExample: NH4Cl
Example: NH4F
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AcidsandBases
Reactions of Cations with Water
• Most metal cations that are hydrated in solution also lower the pH of the solution, acting as Lewis acids: The attraction between nonbonding
electrons on waters’ oxygen and the metal cation causes a shift of the electron density in water.
This makes the O-H bond more polar and the water more acidic.
• Cations with acidic protons (like NH4
+) will lower the pH of a solution:
NH4+ + H2O H3O
+ + NH3
OH
H
O
H
H
Greater charge and smaller size make a cation more acidic.
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AcidsandBases
conjugate
The Effect of Cations