Previously in Chem104:

• more acid/base reactions:

• weak / weak• strong / strong• strong / weak

• calculations

• Polyprotic acids

Today in Chem104:

•Titrations

•Buffers

• calculations

Titrations: a summary

1) strong acid + strong base titrations

2) weak acid or base titrations (by strong base or acid)

• Have pH 7 at equivalence pt• have flat slopes at beginning and end

• have pH at equivalence pt determine by conjugateweak acid titrations have basic pH at eq. pt.weak base titrations have acidic pH at eq. pt.

• have more pronounced slope at beginning• have pH = pKa at ½ volume to equivalence point • have buffer region where [AH] ~ [A], i.e., where conjugate species have about the same concentrations

Buffers: a summary

1. Resist change in pH

2. Made from conjugates in ~equal concentrations

• Acid form [AH] reacts with added base• Base form [A] reacts with added acid

• pH = pKa + log [A] / [AH]But you don’t need to memorize this:

you can derive it! Fast!

4. Buffer pH determined from theHenderson-Hasselbalch equation

3. An acid or base may have multiple buffer regions

• Give me 2 examples

Buffers: one new point

Buffer capacity: how much acid or base can it “absorb”, or compensate for before pH changes

Consider these two buffer solutions and answer,“Which has higher buffer capacity?”

• 0.100 M Acetic acid + 0.100 M sodium acetate

• 0.001 M Acetic acid + 0.001 M sodium acetate

Buffers: how would you make one?

My research students have that very problem in research lab. Let’s do it and I can report to them that my Gen Chem students can help them out!

How would you make 1 L of a 0.100 M phosphate buffer at pH 7?”

1st: find the Ka’s for the acid/base system

2nd: determine the conjugate pair appropriate for the pH

3rd: use the HH equation (or derive it) or the Ka expression to find the relative proportions of conjugates

Phosphoric acid, H3PO4 …which conjugate pair to use at pH 7?

Step 1. H3PO4 + H2O

H2PO4 - + H3O+

Ka1 = 7.6 x 10-3

Step 2. H2PO4- +

H2OHPO4

2- + H3O+

Ka2 = 6.2 x 10-8

Step 3. HPO42-

+ H2O

PO4 3- + H3O+

Ka3 = 2.12 x 10-13

Let’s do it!

Step 2. H2PO4- +

H2OHPO4

2- + H3O+

Ka2 = 6.2 x 10-8

pH = pKa + log [A] / [AH]

7.00 = 7.21 + log [A] / [AH]

-0.21 = log [A] / [AH]

0.62 = [A] / [AH] Or 0.62 = mol A / mol AH

For 1L of 0.100 M: mol A + mol AH = 0.100mol

So: (0.62 mol AH) + mol AH = 0.100mol

So 0.62 mol AH = 1.00 mol A

1.62 mol AH = 0.100mol

mol AH = 0.100 / 1.62 mol = 0.0617 mol AH

mol A = 0.62 x 0.0617 mol AH = 0.0383 mol A

Let’s do it!

Step 2. H2PO4- +

H2OHPO4

2- + H3O+

Ka2 = 6.2 x 10-8

To make the buffer solution:

0.0617 mol AH = 0.0617 mol NaH2PO4 0.0617 mol NaH2PO4 x 119.98 g/mol = 7.40 g NaH2PO4

0.0383 mol A = 0.0383 mol Na2HPO4 0.0383 mol Na2HPO4 x 141.96 g/mol = 5.44 g Na2HPO4

Dissolved in 1 L water

All Definitions of Acid and Base use Donor /Acceptor

Bronsted Acid/Base: proton H+ donor/acceptor

Remember this reaction?

Lewis Acid/Base: electron pair donor/acceptor

CuCl2(H2O)2 (s) + 3H2O [CuCl(H2O)5]+ + Cl-

Cu

H2O

H2O OH2

OH2

Cl

OH2Cu2+:OH2

e- acceptor :e- donorLewis Acid :Lewis Base

All ionic solids dissolve using Lewis A/B interactions

NaCl(s) + 6H2O [Na(H2O)6]+ + Cl-

Na+:OH2

e- acceptor :e- donorLewis Acid :Lewis Base

Mn+

H2O

H2O OH2

OH2

H2O

OH2

Written simply:

This is typical expression for solubility equilibriumGiven by the Solubility Product Ksp

All ionic solids dissolve using Lewis A/B interactions

AgCl(s) + 2H2O [Ag(H2O)2]+ + Cl-

Ksp = 1.8 x10-10 Ksp = [Ag+][Cl-]1.8 x10-10 = [Ag+][Cl-]1.3 x10-5 M = [Ag+] = [Cl-]

AgCl(s) Ag+ + Cl-

Very low solubility due to weak Lewis A/B interactions which does not compensate for large lattice energy

1.3 x10-5 M = [Ag+] = [Cl-] This is the molar solubility of AgCl

Ionic solids which completely dissolve are highly soluble and cannot be described with a Ksp

NaCl(s) + 6H2O [Na(H2O)6]+ + Cl-

Mn+

H2O

H2O OH2

OH2

H2O

OH2

Solubility obeys

AgCl(s) Ag+ + Cl- + excess Cl-

Ksp = 1.8 x10-10 Solubility =1.3 x10-5 M = [Ag+] = [Cl-]

AgCl(s) Ag+ + Cl-

If more chloride is added the equilbirum shifts left,and Solubility Product Ksp requires that less AgCl dissolves

Ksp = 1.8 x10-10 Solubility, [Ag+] <1.3 x10-5 M

otherwise called The Common Ion Effect obeys

AgCl(s) Ag+ + Cl- + excess Cl-

Ksp = 1.8 x10-10 Solubility =1.3 x10-5 M = [Ag+] = [Cl-]

AgCl(s) Ag+ + Cl-

If more chloride is added the equilbirum shifts left,and Solubility Product Ksp requires less AgCl dissolves

Ksp = 1.8 x10-10 butSolubility, [Ag+] <1.3 x10-5 M because [Cl- ] >>1.3 x10-5 M

Cleanliness is next to Godliness

So controlling solubility can make you more holy?

Let’s see how…

The pH Effect obeys

Ksp = 3.7 x10-9

Ca(CO3)(s) Ca2+ + CO32-

If pH is lowered by adding acid,more CaCO3 dissolves

…. and cleans the dishwasher:

+ AH

HCO3-

H2CO3

+ AHH2O + CO2

+ AH

The Chelate Effect obeys

Ksp = 3.7 x10-9

Ca(CO3)(s) Ca2+ + CO32-

If Ca2+ is removed by adding a ligand,

more CaCO3 dissolves

…. and also cleans the dishwasher:

+ citric acid

Ca(citrate)

COOHHOC

COOHHOO

-OCCO-

-OC

OH

O

O

OCa2+

Making better (stronger) Lewis A/B interactionscan improve solubility and clean, too

We have seen:AgCl(s) + 2H2O [Ag(H2O)2]+ + Cl-

Ksp = 1.8 x10-10

AgCl can be completely dissolved!

Very low solubility due to weak Lewis A/B interactions which do not compensate for large lattice energy

But if ammonia is Lewis base:AgCl(s) + 2 NH3 [Ag(NH3)2]+ + Cl-