unit four

Unit FourAcids-Bases Reactions

Acids & Bases• What causes acid rain? And

how can we prevent the damage?

• Why do Perrier drinking chickens give better eggs than chickens who drink regular water?

These are the types of questions we will be able to answer after this unit.

First Proposed Theory:Arrhenius Theory of Acids & BasesAn acid is any substance which releases

___________ in water- it is any ionic species that start with “____” - it tastes sour, & turns blue litmus paper redeg. HCl, HNO3, H2SO4

An base is any substance which releases ___________ in water- it is any ionic species that ends with “____”- it tastes bitter, feels slippery, & turns red litmus paper blue eg. NaOH, KOH, Ca(OH)2

A ___________ results from the neutralization rxn. of an acid & base

Neutralization rxn:Acid + Base → salt + H2O

Net Ionic Equation:________________________

Eg. HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

Q: Predict the products & balance the following Neutralization Reactions

H2SO4(aq) + NaOH(aq) →

H3PO4(aq) + KOH(aq) →

H4P2O7(aq) + NaOH(aq) →

Second Proposed Theory:Bronsted-Lowry Theory of Acids & Bases- more general & includes equilibrium rxns

An acid is a substance that DONATES a ___________ (___)

An base is a substance that Accepts a ___________ (___)

Eg. NH3 + H2O NH4+ + OH-

NH3 gained a proton → NH4+

Thus, NH3 = ___________

H2O lost a proton → OH- Thus, H2O = ___________

Your Turn!

CH3COOH + H2O CH3COO- + H3O+

(lost/donated H+) (gained H+)

ACID BASE Which reactants acts as an acid & which as

a base?

1. HNO3 + H2O NO3- + H3O+

2. HCO3- + SO3

2- CO32- + HSO3

-

3. HS- + H2PO4- H2S + HPO4

2-

4. H3PO4 + CH3COO- H2PO4- + CH3COOH

5. CO32- + HF HCO3

- + F-

Classifying Acids

A ___________ protic acid can donate 1 proton.

A ___________ protic acid can donate up to 2 proton.

A ___________ protic acid can donate up to 3 proton.

All acids that can donate more than 1 proton are known as ___________ protic acids.

Amphiprotic SubstancesAmphiprotic substances can act

as an ___________ or a ___________.

eg. Water NH3 + H2O NH4

+ + OH-

Base acid

CH3COOH + H2O CH3COO- + H3O+

Acid base

Apart from water, amphiprotic substances start with ____ and have a (_______) charge.

eg. H2PO4- , HS-, HCO3

-

Base acid

H3PO4 H2PO4- HPO4

2-

Accepts H+ Donates H+

NOTE: In every Bronsted-Lowry rxn. there is an acid and a base on both sides of the eqn.

CH3COOH + H2O CH3COO- + H3O+

acid base base acid

Q: Which do you think exhibit amphiprotic behavior?

Se2-

HSe-

H2Se

H3PO4

HPO42-

HSO3-

Q: Identify each species as an acid or base.

1. HF + SO32- F- + HSO3

-

2. H2O + HCO3- H3O+ + CO3

2-

3. NO2- + H2O OH- + HNO2

4. H2PO- + S2- HS- + HPO42-

5. N2H5+ + SO4

2- N2H4 + HSO4-

Conjugate Acid & BasesA “conjugate” acid-base pair is a pair

of chemical species that only differ by ONE proton.

Eg. NH4+ + H2O NH3 + H3O+

* Acid has the extra proton (ability to donate it)

Conjugate Pair

Conjugate Acid

Conjugate Base

NH4+ , NH3

H2O, H3O+

Assignment:

Acid & Base Handout Questions 1 4

Buffer Solutions

Buffers are solutions with the ability to resist the addition of strong acids or strong bases, within limits. eg. CH3COOH and its salt NaCH3COO

Most buffer solutions are made up using a WEAK ACID and its sodium salt!

When a strong base such as NaOH is added to the buffer CH3COOH …

• CH3COOH reacts with and consumes the excess OH-

• OH- reacts with the H3O+ ion from the acid in the following reaction:

H2O + CH3COOH <---> H3O+ + CH3COO-

H3O+ + OH- <--------> H2O  

When a strong acid such as HCl is added to the buffer CH3COOH …

• H3O+ reacts with the CH3COOH- ions of the salt and form more undissociated CH3COOH

H3O+ + CH3COO- <-------> H2O + CH3COOH

There is a limit to the quantity of H+ or OH- that a buffer can absorb without undergoing a significant change in pH.

Buffer Components

A buffer has two components. HA NaA ---> Na+ + A-

(weak acid) (a soluble salt of the weak acid)

Therefore any extra H3O+ will be neutralized by the A- in the buffer

H3O+ + A- <-------> HA + H2O

And any extra OH- that is added will be neutralized by the acid

    HA + OH- <------> A- + H2O

Sample Problem:CH3COOH + H2O CH3COO + H3O+

1.0 M 1.0M

Original Ratio: Acid = 1.0 = 1.0 Base 1.0

New Ratio: 1.1 = 1.22 0.9

ACID

BASE

1.1

0.9

Add 0.1 mol H3O+

Using Ka

Original[H3O+] = (1.8 x 10-5)(1.0)

= 1.8 x 10-5

pH = 4.74After

[H3O+] = (1.8 x 10-5)(1.22) = 2.5 x 10-5

pH = 4.66pH change of 0.08

With no buffer present…

Adding 0.1 mol of acid to H2O would change the pH by 6.00!

pH water = 7pH water with 0.1M H3O+ = 1

Thus a difference of 6.