Outline:Outline: 2/28/072/28/07 Hand in Seminar Reports – to me Pick up Quiz #6 – from me Pick up CAPA 14 - outside 5 more lectures until Exam 2…Today: End Chapter 17

Start Chapter 18

Polyprotic acids

Buffers

Factors that affect acid strength:

1. Polarity of HX bond

2. Bond strength of HX bond

3. Stability of conjugate base X

Acid-Base Behavior and Acid-Base Behavior and Chemical StructureChemical Structure

Which is a stronger acid?

Acid-Base Behavior and Acid-Base Behavior and Chemical StructureChemical Structure

HCl or HF

HClO or HBrO or HIO

HBrO2 or HBrO3

more electroneg

weaker bond

more O’s = electroneg

Carboxylic Acids

• Carboxylic acids all contain COOH.

• All carboxylic acids are weak acids.

• When the carboxylic acid loses a proton, it generate the carboxylate anion, COO.

Acid-Base Behavior and Acid-Base Behavior and Chemical StructureChemical Structure

RC

OH

O

Carboxylic AcidsCarboxylic Acids

RC

OH

O

H

RC

OH

O

CH3

RC

OH

O

RC

OH

O

CH3

RC

OH

O

RC

OH

O

H

Formic Acid

Formate Ion

Acetic Acid

Benzoic Acid

Acetate Ion

Benzoate Ion

Representative Polyprotic Acids

Polyprotic Acid Problem Determine the concentration of all ions

in 0.050 M H2CO3 Step #1: Write down all reactions H2CO3 + H2O CO3

- + H3O+ Ka1=4.5 x 10-7

HCO3- + H2O CO3

2- + H3O+ Ka2=4.7 x10-11

H2O + H2O H3O+ + OHKw=1.0 x10-14

Most of the H3O+ will come from the first Dissociation step. Treat as a simple weak acid problem.

H2CO3 + H2O CO3- + H3O+ Ka1=4.5 x 10-7

0.050 M 0 0 init (0.050x)M x M x M equil Assume x is small and we obtain: x= (0.050 4.5x10-7)1/2 = 1.5 10-4 M Hence: [CO3

- ]= [H3O+] = 1.5 10-4 M H2CO3 = 0.050 M 1.5 10-4 = 0.0498 M

Step 2. Set up ICE Table

What is the concentration of CO32-?

1.5 10-4 0 1.5 10-4 init x +x +x change(1.5104x) x 1.5 10-4 + x equil

x(1.5 10-4 +x)/(1.5 10-4 –x) = 4.7 10-11

Notice Ka2 . Assume x is small.

x = [CO32- ] = 4.7 10-11

HCO3- + H2O CO3

2- + H3O+ Ka2=4.7 x10-11

Chapter 17 : No Common Ions A typical weak acid dissociation:

HOAc H+ + OAc

0.25 0.0 0.0

Ka

1.8 × 105

Ka =x2

0.25x

x = 2.1 × 103

pH = 2.67

0.25 M

Chapter 18 : Common Ions Add ions to both sides of equation:

HOAc H+ + OAc

0.25 0.0 0.25

Ka

1.8 × 105

Ka =x(0.25+x)

0.25x

x = 1.8 × 105

pH = 4.74 vs. 2.67LeChâtelier

again

Acid Base reactions:

Add 0.10 mol NaOH to 1L H2O :

Strong Base = complete dissociation

pOH = log(OH) = log(0.1) = 1.0

pH = 13.0

No common ions: pH goes from 7.0 to 13.0

Acid Base reactions: Add 0.10 mol NaOH to 1L HOAc:

HOAc + OH H2O + OAc

What is the correct Keq to use?

HOAc H+ + OAc Ka

H+ + OH H2O 1/Kw

HOAc + OH H2O + OAc Ka/Kw

Ka/Kw= 1.8 × 105 /1.0 × 10= 1.8 × 10

Acid + Base reaction

0.25M buffer

Reaction with Common Ions Add 0.10 moles of NaOH:

HOAc + OH H2O + OAc

0.25 0.10 0.25

Ka 1.8 × 10

-0.10 -0.10 +0.10

0.15 0.0 0.35+x +x -x

Keq =(0.35x)

(0.15x)xx = 1.3 × 109 pOH = 8.89

pH = 5.11

Acid Base reactions: Add 0.10 mol NaOH to ….

No common ions: pH goes from 7.00 to 13.00

With common ions: pH goes from 4.74 to 5.11

change = +0.37

change = +6.00

A Buffer!

Chapter 18 : Buffers Can add lots of HA or A and pH

doesn’t change much mathematically.Easier to use Henderson-Hasselbach

equation (p. 811)….

pH = pKa + log [base][acid]

pH = 4.74 + log [OAc]/[HOAc]

pH = pKa + log ([conj base]/[acid])

If 50.0 g of sodium acetate is added to 100 mL of 0.100 M solution of acetic acid, what

will be the pH of the resultant buffer?

50 g CH3COONa = 0.61 mol/0.1L = 6.1 M

pH = 4.74 + log (6.1/0.1)

= 4.74 + 1.78 = 6.53

Where does this come from?

Ka = [H+][A]/[HA]

pKa = pH + p[A]/[HA]

pKa = pH log[A]/[HA]

pH = pKa + log ([conj base]/[acid])

Buffer calculations….

pH = pKa + log ([conj base]/[acid])

This one also exists:pOH = pKb + log ([conj acid]/[base])

Practice!