unit 18 acid-base equilibria: buffers & hydrolysis

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

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

AcidsandBases


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)

AcidsandBases


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?

AcidsandBases

AcidsandBases

[ ]

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?

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-

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.

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 = −log (1.4 10−4) + log(0.10)(0.12)

pH = 3.85 + (−0.08)

pH = 3.77

Problem:

[H3O+] [A−][HA]

Ka =

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.


pKa

3.

3.

4.

4.

7.

9.

9.

pH = pKa + log[1M][1M]

pH = pKa

AcidsandBases

[H3O+] [A−][HA]

Ka =


[ ] [ ]

AcidsandBases

pH= pKa + log [base][acid]

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-)

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.


[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:

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*


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


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

AcidsandBases

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

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

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.

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.

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

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

AcidsandBases

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}

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

AcidsandBases

2007 (B)

AcidsandBases

Making Solution Molarity

AcidsandBases

2007B Q5

Carboxylic acid

Organic Acids: carboxylic acid functional group

-COOH -COO- + H+

AcidsandBases

2005A Q1

AcidsandBases

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

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

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.

AcidsandBases

conjugate

The Effect of Cations

unit 18 acid-base equilibria: buffers & hydrolysis

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