ACIDS and BASES (unit 11) Notes start on slide 35 ***

ACIDS, BASES & SALTS

Unit 11

The Arrhenius Theory of Acids and Bases

Arrhenius Theory of Acids and Bases:

an acid contains hydrogen and ionizes in solutions to produce H+ ions:

HCl H+(aq) + Cl-(aq)

Arrhenius Theory of Acids and Bases:

a base contains an OH- group and ionizes in solutions to produce OH- ions:

NaOH Na+(aq) + OH-(aq)

Neutralization Neutralization: the combination

of H+ with OH- to form water.

H+(aq) + OH-(aq) H2O (l)

Hydrogen ions (H+) in solution form hydronium ions (H3O+)

In Reality…

H+ + H2O H3O+

Hydronium Ion

(Can be used interchangeably with H+)

Commentary on Arrhenius Theory…

One problem with the Arrhenius theory is that it’s not comprehensive enough. Some compounds act like acids and bases that don’t fit the standard definition.

Bronsted-Lowry Theory of Acids & Bases

Bronsted-Lowry Theory of Acids & Bases:

An acid is a proton (H+) donor

A base is a proton (H+) acceptor

for example…

HCl(aq) + H2O(l) H3O+(aq) + Cl-(aq)

Proton transfer

Acid

Base

NH3(aq) + H2O(l) NH4+ (aq) + OH-

(aq)

BASE

ACID

CONJUGATE ACID

CONJUGATE BASE

Ammonia is a proton acceptor, and thus a base

another example…

Water is a proton donor, and thus an

acid.

Conjugate acid-base pairs

Conjugate acid-base pairs differ by one proton (H+)

A conjugate acid is the particle formed when a base gains a proton.

A conjugate base is the particle that remains when an acid gives off a proton.

Examples: In the following reactions, label the conjugate acid-base pairs:

H3PO4 + NO2- HNO2 + H2PO4

-

CN- + HCO3- HCN + CO3

2-

HCN + SO32- HSO3

- + CN-

H2O + HF F- + H3O+

acid base c. acid c. base

acidbase c. acid c. base

acid base c. acid c. base

acidbase c. acidc. base

Amphoteric Substances

A substance that can act as both an acid and a base (depending on what it is reacting with) is termed amphoteric.

Water is a prime example.

Properties of Acids and Bases ACIDS

Have a sour taste Change the color

of many indicators Are corrosive

(react with metals)

Neutralize bases Conduct an

electric current

BASES Have a bitter taste Change the color

of many indicators Have a slippery

feeling Neutralize acids Conduct an

electric current

Strength of Acids and Bases

A strong acid dissociates completely in sol’n: HCl H+(aq) + Cl-(aq)

A weak acid dissociates only partly in sol’n: HNO2 H+(aq) + NO2

-(aq)

A strong base dissociates completely in sol’n: NaOH Na+(aq) + OH-(aq)

A weak base dissociates only partly in sol’n: NH3(aq) + H2O(l) NH4

+(aq) + OH-(aq)

The Lewis Theory of Acids and Bases

The Lewis Theory of Acids & Bases

Lewis acid: a substance that can accept an electron pair to form a covalent bond (electron pair acceptor).

Lewis base: a substance that can donate an electron pair to form a covalent bond (electron pair donor).

Neutralization (using Lewis)

Neutralization: the formation of a coordinate covalent bond in which both electrons originated on the same (donor) atom.

Example 1:

Ionization of NH3: NH3 + H2O NH4

+ + OH-

N NH

H

H O HH+ H

H

H

H H+ O..

.

.

.

...

.

.

+-

acidbase

Acid = electron pair acceptor, base = electron pair donor (to form the covalent bond)

Example 2:

Auto-ionization of water: H2O + H2O H3O+ + OH-

O OH H O HH+ H

HH H+ O.. .

.

.

...

.

.

+-

acidbase

.

.

.

.

Acid = electron pair acceptor, base = electron pair donor (to form the covalent bond)

Example 3: Reaction of NH3 with HBr (a Lewis

AND a Bronsted-Lowry acid-base reaction): NH3 + HBr NH4

+ + Br-

N NH

H

H BrH+ H

H

H

H + Br..

.

.

.

...

.

.

+-acid

base

SUMMARY OF ACID-BASE THEORIES

Theory Acid Definition Base Definition

Arrhenius Theory

Any substance which releases H+ ions in water solution.

Any substance which releases OH- ions in water solution

Brǿnsted-Lowry Theory

Any substance which donates a proton.

Any substance which accepts a proton.

Lewis Theory

Any substance which can accept an electron pair.

Any substance which can donate an electron pair.

Acid-Base Reactions Neutralization reactions: reactions

between acids and metal hydroxide bases which produce a salt and water.

H+ ions and OH- ions combine to form water molecules:

H+(aq) + OH-(aq) H2O(l)

Example 1: the reaction of HCl and NaOH (there are 3 ways to write the chemical equation):

Balanced formula unit equation: HCl(aq) + NaOH(aq) H2O(l) + NaCl(aq)

Total ionic equation: H+ + Cl- + Na+ + OH- H2O + Na+ + Cl-

Net ionic equation: H+(aq) + OH-(aq) H2O(l)

Example 2: Write the 3 types of equations for the reaction of hydrobromic acid, HBr, with potassium hydroxide, KOH.

Balanced formula unit equation: HBr(aq) + KOH(aq) H2O(l) + KBr(aq)

Total ionic equation: H+ + Br- + K+ + OH- H2O + K+ + Br-

Net ionic equation: H+(aq) + OH-(aq) H2O(l)

Example 3: Write the 3 types of equations for the reaction of nitric acid, HNO3, with calcium hydroxide, Ca(OH)2.

Balanced formula unit equation: 2HNO3(aq) + Ca(OH)2(aq) 2H2O(l) +

Ca(NO3)2(aq)

Total ionic equation: 2H+ + 2NO3

- + Ca2+ + 2OH- 2H2O + Ca2+ + 2NO3

-

Net ionic equation: H+(aq) + OH-(aq) H2O(l)

Demos…

DEMO: Sponge

How does that work?... The sponge is soaked in Congo red. Congo red is a dye, a biological stain,

and a pH indicator. It has been used as a direct fabric dye for cotton to produce a bright red color.

Scientists use Congo red as a pH indicator (a substance that will change color in the presence of different ion concentrations, [H+])

Variety of pH indicators… There are many different types of

pH indicators, such as universal indicator and litmus paper.

Litmus paper comes in red Litmus paper and blue Litmus paper.

Red litmus paper in an acids turns…

Blue litmus paper in a base turns …

BLUE

RED

Demo: tap water vs. dH2O

Both waters have Universal indicator in them (= pH indicator (changes color in the presence of ions), which is a type of weak acids)

The water will change pH, and therefore COLOR (which helps us determine if a solution is acidic or basic) with the addition of HCl (acid) and NaOH (base)

Universal Indicator Color Chart

pH scale 0 7 14

Acid Neutral Base

Why does it take more drops of acid or base to make the tap water change color than it does for the distilled water?

What is distilled water made of? What is tap water made of?

Buffered Solutions

A solution of a weak acid and a common ion is called a buffered solution.

Thus, the solution maintains it’s pH in spite of added acid or base.

pH and pOH

Ionization of water Experiments have shown that pure

water ionizes very slightly: 2H2O H3O+ + OH-

Measurements show that: [H3O+] = [OH-]=1 x 10-7 M

Pure water contains equal concentrations of H3O+ + OH-, so it

is neutral.

pH

pH is a measure of the concentration of hydronium ions in a solution.

pH = -log [H3O+] or

pH = -log [H+]

Example: What is the pH of a solution where [H3O+] = 1 x 10-7 M?

pH = -log [H3O+] pH = -log(1 x 10-7)pH = 7

Example: What is the pH of a solution where [H3O+] = 1 x 10-5 M?

pH = -log [H3O+] pH = -log(1 x 10-5)pH = 5

When acid is added to water, the [H3O+] increases, and the pH decreases.

Example: What is the pH of a solution where [H3O+] = 1 x 10-10 M?

pH = -log [H3O+] pH = -log(1 x 10-10)pH = 10

When base is added to water, the [H3O+] decreases, and the pH increases.

The pH Scale

Acid Neutral Base

0 7 14

pOH

pOH is a measure of the concentration of hydroxide ions in a solution.

pOH = -log [OH-]

Example: What is the pOH of a solution where [OH-] = 1 x 10-5 M?

pOH = -log [OH-] pOH = -log(1 x 10-5)pOH = 5

How are pH and pOH related?

At every pH, the following relationships hold true:

[H3O+] • [OH-] = 1 x 10-14 M

pH + pOH = 14

Example 1: What is the pH of a solution where [H+] = 3.4 x 10-5 M?

pH = -log [H+] pH = -log(3.4 x 10-5 M)pH = 4.5

Example 2: The pH of a solution is measured to be 8.86. What is the [H+] in this solution?

pH = -log [H+] 8.86 = -log [H+] -8.86 = log [H+] [H+] = antilog (-8.86) [H+] = 10-8.86

[H+] = 1.38 x 10-9 M***you may have to put your calculator into sci

mode to get the decimals

Example 3: What is the pH of a solution where [H+] = 5.4 x 10-6 M?

pH = -log [H+] pH = -log(5.4 x 10-6)pH = 5.3

Example 4: What is the [OH-] and pOH for the solution in example #3?

[H3O+][OH-]= 1 x 10-14

(5.4 x 10-6)[OH-] = 1 x 10-14

[OH-] = 1.9 x 10-9 M ***you may have to put your calculator into sci

mode to get the decimals

pH + pOH = 14 pOH = 14 – 5.3 = 8.7

Extra Practice Classify each solution as acidic,

basic, or neutral ***MUST SOLVE FOR pH (write

down this #) and use the pH scale

a. [H+] = 6.0 x 10-10 Mb. [OH-] = 3.0 x 10-2 Mc. [H+] = 2.0 x 10-7 Md. [OH-] = 1.0 x 10-7 M

pH = 9.2, basicpOH = 1.5, pH = 12.5,

basicpH = 6.7, acidicpOH = 7 = pH, neutral

Example #5 Which is the MOST basic from

question #4?

B, pH = 12.7

Acids and bases: Titrations The amount of acid or base in a

solution is determined by carrying out a neutralization reaction;

an appropriate acid-base indicator (changes color in specific pH range) must be used to show when the neutralization is completed.

Buret

Solution with Indicator

DEMO of lab…

Read the buret correctly = 2 decimal places

Acid Base Titration: the addition of a known amount of solution to determine the volume or concentration of another solution

A very accurate method to measure concentration.

Acid + Base Salt + Water

H+ + OH- H2O

Moles H+ = Moles OH-

3 Steps: (SHOW DEMO OF TITRATION)

1. Add a measured amount of an acid of unknown concentration to a flask.

2. Add an appropriate indicator to the flask

3. Add measured amounts of a base of known concentration using a buret. Continue until the indicator shows that neutralization has occurred. This is called the end point of the titration(*** look at page 620 fig. 19-20)

Example: A 25 mL solution of H2SO4 is

neutralized by 18 mL of 1.0 M NaOH using phenolphthalein as an indicator. What is the concentration of the H2SO4 solution?

Equation: H2SO4 + 2 NaOH Na2SO4 + 2 H2O

(steps)

Steps1) How many mol of NaOH are

needed for neutraliztion?2) How many moles of H2SO4 were

neutralized?3) Calculate the concentration of the

acid:

1) How many mol of NaOH are needed for neutraliztion?

Mol NaOH = 1.0 mol x 0.018 L = 0.018 mol

1 L

2) How many moles of H2SO4

were neutralized?

Mol H2SO4 = 0.018 mol NaOH x 1 mol H2SO4 =

2 mol NaOH

9.0 x 10-3 mol H2SO4

3) Calculate the concentration of the acid:

M = mol = 9.0 x 10-3 mol = 0.36 ML 0.025 L

Titration CurvePAGE 619 fig 19-17

A graph showing how the pH changes as a function of the amount of added titrant in a titration.

Data for the graph is obtained by titrating a solution and measuring the pH after EVERY drop of added titrant.

Equivalence point = The point on the curve where the

moles of acid equal the moles of base;

the midpoint of the steepest part of the curve is a good approximation of the equivalence point.

PAGE 619 fig 19-17

PAGE 620 fig. 19-19 Knowledge of the equivalence

point can then be used to choose a suitable indicator for a given titration;

the indicator must change color at a pH that corresponds to the equivalence point.

Calculations of Titrations

The mole method and molarity

Calculate the molarity of a sulfuric acid solution if 23.2 mL of it reacts with 0.212 g of Na2CO3.

Na2CO3 + H2SO4 Na2SO4 + CO2 + H2O

****use steps 1-3 that we just did. answer

ANSWER

Molarity = of a sulfuric acid solution if 23.2 mL of it reacts with 0.212 g of Na2CO3.

Na2CO3 + H2SO4 Na2SO4 + CO2 + H2O0.212 2 3 1000 1 2 3 1 2 40.0862 2 4

23.2 1 105.99 2 3 1 2 3

mol g Na CO mL mol Na CO mol H SOM x x x M H SO

L mL L g Na CO mol Na CO

Normality The Normality (N) of a solution is

defined as the molarity x the total positive oxidation number of the solute.

EXAMPLESHCl H+ + Cl-H2SO4 2 H+ + OH-NaOH Na+ + OH-Ba(OH)2 Ba2+ + 2 OH-

Example Calculate the molarity and normality

of a solution that contains 34.2 g of Ba(OH)2 in 8.00 L of solution.

M = ?

N = ?

answers

Answer (molarity)

M = 34.2 g of Ba(OH)2 x 1 mol = 0.0249 M

8.00 L 171.35g

answer for normality

Answer (normality) N = M x +2 (for Ba2+) = 0.0249 M x 2 =

0.0499 N

Using Normality 1 equivalent of an acid reacts with

1 equivalent of a base, so in titration problems, you can use this equation

NaVa = NbVb

Example 30.0 mL of 0.0750 N HNO3 required

22.5 mL of Ca(OH) 2 for neutralization. Calculate the normality and molarity of the Ca(OH)2 solution.

2 HNO3 + Ca(OH)2 Ca(NO3)2 + 2 H2O

answers

ANSWER (normality) N of Ca(OH)2 = ?

NaVa = NbVb

(0.0750 N)(30.0 mL) = Nb(22.5 mL)

= 0.100 N

Answer (molarity) N = M x (+ oxidation #)

M = N / (+ oxidation #)

M = 0.100 N / (2) because Ca2+

= 0.0500 M