Acids and Bases ORIG

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Acids and Bases

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​Class Announcements:

​School Announcements:
Half Day Wednesday

English II STAAR 11th

• ​Naming

• Neutralization Reactions

• pH

• Strength of acids and bases

Acids

Acids: Substances that contain H+ ions when dissolved in water.

​Strong acids - ions completely dissociate (separate) in water

Weak acids - ions partially dissociate in water

Arrhenius Classification of Acids

• Acids: create H+ ions in a solution

  • Solutions with more H⁺ ions

Multiple Choice

Acids produce _________ ions when they break down or dissocciate. 

• Hydroxide (OH-)
• Water
• Chlorine (Cl-)
• Hydrogen (H+)

Naming Binary Acids (2 atoms)

Prefix - at the beginning of a word
Suffix- at the end of a word

For acids without Oxygen the prefix is hydro
The suffix is ic


Examples
Hydrochloric acid (a hydrogen and a chlorine atom)
Hydrofluoric acid (Hydrogen and a Fluorine)
Hydrobromic acid (Hydrogen and Bromine)

Multiple Choice

Binary acids start with the prefix "____________"

• acid

• nitric

• hydraulic

• hydro

Multiple Choice

HBr

• hydrogen bromine acid
• hydrobromide acid
• hydrobromic acid

Naming acid with Oxygen

Suffix endings
-ous - compounds that end with "ite"

-ic- compounds that end in "ate"

*This is over simplified
H + Polyatomic-ite
H + Polyatomic-ate

​Examples:
2H + SO₃ (2 Hydrogen + Sulfite) is Sulfurous acid
H + SO₄ (Hydrogen + Sulfate) is Sulfuric acid

Multiple Choice

What is the formula for nitric acid?

• HNO₂

• HNO₃

• HNO₄

• H₂NO₃

Multiple Choice

carbonic acid

• H₂CO₃

• H₂CrO₄

• H₂C₂O₄

• HCO₃

Multiple Choice

H₃PO₄

• hydrophsophorus acid

• phosphoric acid

• hydrogen phosphorous

• phosphori hydroxide

Examples of Acids

Bases

Bases: Ionic compounds that contain OH^- (hydroxide) ions when dissolved in water

Strong bases - completely dissociate in water
Weak bases - partially dissociate in water

Arrhenius Classification of Bases

• Bases: create OH- ions in a solution

  • Solutions with more OH^-

Multiple Choice

Bases release ____________ ions when they break down or dissociate. 

• Hydrogen (H+)
• Hydroxide (OH-)
• Chlorine (Cl-)
• no

Bases

Multiple Choice

Potassium hydroxide

• KOH
• K2OH
• k(OH)2

Multiple Choice

Mg(OH)₂

• magnesium hydroxide acid
• hydromagnesium acid
• magnesium oxygen hydride
• magnesium hydroxide

Multiple Choice

Is the following compound an acid or a base? 
Ca(OH)₂

• Acid
• Base

Examples of
Bases

Acid Base Reactions

C.10.G DEFINE ACIDS AND BASES AND DISTINGUISH BETWEEN
ARRHENIUS AND BRONSTED‐LOWRY DEFINITIONS AND PREDICT
PRODUCTS IN ACID‐BASE REACTIONS THAT FORM WATER

C.10.H UNDERSTAND AND DIFFERENTIATE AMONG ACID‐BASE
REACTIONS, PRECIPITATION REACTIONS, AND OXIDATION
REDUCTION REACTIONS

Acids and Bases

Generate Ions

Acids:
pH is less than 7

●

Acid: a substance which when added to water produces
hydrogen ions [H+].

●

Acids: react with zinc, magnesium, or aluminum and
form hydrogen gas, H2 (g).

●

react with compounds containing CO3

2- and form carbon

dioxide and water

●

turn litmus red

●

taste sour (lemons contain citric acid, for example)

●

electrolytes

●

Acids Corrode metals

●

DO NOT TASTE ACIDS IN THE LABORATORY!!

Bases
pH greater than 7

● Base: a substance which when added to water

produces hydroxide ions [OH-].

● Bases: feel soapy or slippery

● turn litmus blue

● they react with most cations to precipitate

hydroxides

● Will react with metals

● Most hand soap and drain cleaner are bases

● Dissolve fats and oils

● taste bitter (ever get soap in your mouth?)
DO NOT TASTE BASES IN THE LABORATORY!!

Reactions with Indicators

Indicator

Acid
color

Neutral
color

Base
color

Phenolphthalein Colorless Faint pink Dark pink

Bromothymol

blue

Yellow

Green

Blue

Litmus

Red

-----

Blue

Common Acids

• HCl- hydrochloric- stomach acid
• H2SO4- sulfuric acid - car batteries
• HNO3 – nitric acid - explosives
• HC2H3O2- acetic acid - vinegar
• H2CO3-carbonic acid – sodas
• H3PO4- phosphoric acid -flavorings

Common Bases

• NaOH- (LYE) soaps, drain cleaner

• Mg(OH)2 - antacids

• Al(OH)3- antacids, deodorants

• NH4OH- “ammonia”

What is ionization?

In water HCl, a strong acid will completely break apart into its
ions

HCl (aq) → H+ (aq) + Cl- (aq)

NaOH, a strong base, will completely ionize in water

NaOH (aq) → Na+ (aq) + OH- (aq)

Acetic acid, a weak acid, will only partially ionize in water

CH3COOH (aq) → CH3COO- (aq) + H+ (aq)

K = 1.8 x 10-5

*Small K value tells you reactants are highly favored, very little
products

Weak vs. Strong Acids

● Weak Acids do not ionize completely

● Low [H+] concentrations

○ Acetic, Boric, Nitrous, Phosphoric, Sulfurous acid

● Strong Acids ionize completely

○ High [H+] concentrations

○ Hydrochloric acid (HCl), Nitric acid (HNO3),

Sulfuric Acid (H2SO4), Hydrobromic Acid (HBr),
Perchloric Acid (HClO4). Hydroiodic Acid (HI)

Weak vs. Strong Bases

● Weak Bases: Do not ionize completely in

water
○ Low [OH-] concentration
○ ammonia; potassium carbonate, sodium

carbonate

● Strong Bases: Completely ionize in water

○ High [OH-] concentration
○ NaOH, Ba(OH)2 , KOH

pH of Common Substances

Timberlake, Chemistry 7th Edition, page 335

Arrhenius acid is a substance that produces H+ (H3O+) in water

Arrhenius base is a substance that produces OH- in water

Arrhenius Model

● Basis for Arrhenius model: Action in water

● Arrhenius acid definition: produces H+ or H3O+ in

water solution.

○ H+ and H3O+ can be used interchangeably.

H3O+ is Hydronium

● Arrhenius base definition: produces OH- in water

solution

Bronsted-Lowry Model

●Basis for the model: What the acid or base does, a proton
transfer

●Bronsted-Lowry acid: Substance that donates a proton/hydrogen ion
(H+)

●Bronsted-Lowry base: Substance that accepts a proton/hydrogen ion
(H+)

●Conjugate base definition: the acid becomes the conjugate base
after it donates the proton because it can now accept it back.

●conjugate acid definition: the base becomes the conjugate acid after
it accepts the proton because it can now donate it back.

Conjugate Pairs

A Brønsted-Lowry acid is a proton donor

A Brønsted-Lowry base is a proton acceptor

acid
conjugate

base
base
conjugate

acid

Acid-Base Reactions: Neutralization
Reaction

● Strong acid + strong base:

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

● Complete Ionic Equation:

H+

(aq) + Cl-

(aq) + Na+

(aq) + OH-

(aq) → Na+

(aq) + Cl-

(aq) +

H2O(l)

● Net ionic reaction: (Cancel spectator ions)

○ H+

(aq) + OH-

(aq) → H2O(l)

All acid base reactions will produce H2O and a salt

Calculation Of pH

C.10.I DEFINE PH AND USE THE HYDROGEN OR HYDROXIDE ION
CONCENTRATIONS TO CALCULATE THE PH OF A SOLUTION

Water Dissociation:

●

Water dissociation: H2O(l) → H+

(aq) + OH-

(aq)

●

Self ionization

○ Pure water partially breaks down into charged particles.

●

Equilibrium constant, KW = [H+][OH-]

○ Note: water is not involved in the equilibrium expression because it

is a pure liquid, also, the amount of water not dissociated is so large
compared to that dissociated that we consider it a constant

●

Value for Kw = [H+][OH-] = 1.0 x 10-14

●

Note: 1.0 x 10-14 is a very small number. Which means almost no water
molecules actually dissociate.

Neutralization:

A reaction of Hydronium ions and Hydroxide ions to form
water molecules.

Because Kw = 1.0x10-14 and Kw = [H+][OH-]:

1 x10-7 x 1x10-7 = 1x10-14

[H+] for pure water = 1 x 10-7

[OH-] for pure water = 1 x 10-7

Another way to define Acids

● Definitions of acidic, basic, and neutral solutions

based on [H+]

● acidic: if [H+] is greater than 1 x 10-7 M

basic: if [H+] is less than 1 x 10-7 M
neutral: if [H+] if equal to 1 x 10-7 M

Acid-Base Calculations

Equations you need to know:

pH = -log[H+]

pOH = -log[OH-]

14= pH + pOH

Kw = [H+][OH-]

These equations allow you to solve for pH, pOH, [H+], or [OH-]

Note: Since strong acids and strong bases fully ionize, [HA]=[H+] and
[MOH]=[OH-]. eg. 1.5M HCl = 1.5M H+

Example 1: Given [OH-], find [H+]

➢

What is the [H+] of a sample of lake water with [OH-] of 4.0x10-9
M? Is the lake acidic, basic, or neutral?

●

Solution: Use Kw = [H+][OH-]

1.

Rearrange to solve for [H+]→ Kw

2.

Plug in known values and solve: 1.0x10-14

●

Therefore the lake is slightly acidic (2.5x10-6>1x10-7)

●

Remember: the smaller the negative exponent, the larger the
number is.

●

Therefore:

○ acid solutions should have exponents of [H+] from 0 to -6.

basic solutions will have exponents of [H+] from -8 on.

[OH-] = [H+]

4.0x10-9 = 2.4x10-6 M

Example 2: Given [OH-], find [H+]

● What is the [H+] of human saliva if its [OH-] is 4 x 10-8

M? Is human saliva acidic, basic, or neutral?

Given: [OH-] = 4 x 10-8 M Find: [H+]

● Solution: Use Kw = [H+][OH-]→Kw

○ [H+] = = 2.5 x 10-7 M

● The saliva is nearly neutral.

[OH-] = [H+]

1.0 x 10-14

4 x 10-8

pH Calculations

● pH is defined by the [H+]

● pH = -log10[H+] (The p means "-log")

● Definition of acidic, basic, and neutral solutions based

on pH

○ acidic: if pH is less than 7

○ basic: if pH is greater than 7

○ neutral: if pH is equal to 7

Example 3: Calculating pH from [H+]

➢

The [H+] concentration of a solution is equal to 3.7x10-8 M, what is the
pH of the solution? Is this solution and acid or base?

Given: [H+]=3.7x10-8 M Find: pH

●

Solution: Use pH=-log[H+]

pH =-log(3.7x10-8)

pH=7.43 This solution is slightly basic since pH > 7.

●

pH Significant figure rule:

●

Number of sig figs from concentration is how many decimal places pH
will have.
○

In this problem we had 2 sig figs, therefore pH has 2 decimal
places

Example 4: Finding [H+] from pH

● The [H+] can be calculated from the pH by using algebra to

isolate [H+] in the pH equation.

➢ Calculate the [H+] of a solution of baking soda with a pH of

8.5.

● Solution:

○ 1. Given: pH = 8.5 Find: [H+] Equation: pH=-log[H+]

○ 2. Plug in numbers, use algebra to isolate unknown:

8.5 = -log[H+] → -8.5=log[H+] (divide by -1)

● To remove the "log", raise 10 to both sides.

10^-8.5=10^log[H+] →10-8.5=[H+] = 3.16x10-9 M

● If pH is 8.5, then the [H+] is 3.16 x 10-9 M

Example 5: pH from [OH-] (Using Kw)

●

Calculate the pH of a solution of household ammonia whose [OH-] is
7.93 x 10-3 M.

●

Solution: 1st calculate the [H+] from the [OH-] using Kw = [H+][OH-]

→

●

2nd find the pH using [H+]

●

pH=-log[H+]

○ -log[1.26 x 10-12] = 11.900

[OH-] = [H+]

Kw

7.93x10-3
= 1.26x10-12 M
1.0x10-14

Example 6: pH from [OH-] (Using pOH)

➢

Calculate the pH of a solution of vinegar whose [OH-] is 4.92 x 10-10 M

Given: [OH-]=4.92 x 10-10 MFind: pH

●

Solution: 1st calculate the p[OH-] using pOH =-log[OH-]

○ pOH=-log[OH-]

■ -log(4.92x10-10) = 9.308

●

2nd find the pH using pOH, 14=pH + pOH

●

Rearrange to solve for pH: 14=pH + pOH → pH=14-pOH

○ 14 - 9.308 = 4.692 = pH

Now you try a few by yourself. Attempt all
problems before checking answers.

●

Practice #1. What is the pH of a solution of NaOH that has a
[OH-] of 3.5 x 10-3 M?

●

Practice #2. The [H+] of vinegar that has a pH of 3.2 is what?

●

Practice #3. What is the pH of a 0.001 M HCl solution?

Now you try a few by yourself.

Practice #1. What is the pH of a solution of NaOH that has a [OH-] of
3.5 x 10-3 M?

Kw = [H+][OH-]

pH=-log[H+] -log(2.9x10-12)= 11.54=pH

Practice #2. The [H+] of vinegar that has a pH of 3.2 is what?

pH = -log[H+] 3.2=-log[H+] → -3.2=log[H+] → 10-3.2=[H+]

[H+]=6x10-4M

Practice #3. What is the pH of a 0.001 M HCl solution?

HCl = Strong acid, therefore [HCl]=[H+]
pH=-log[H+] -log(0.001)= 3 =pH

[OH-] = [H+]

Kw

3.5x10-3
= 2.9x10-12M =
[H+]

1.0x10-14

step 2

step 1

Strength of Acids and

Bases

C.10.J DISTINGUISH BETWEEN DEGREES OF DISSOCIATION FOR
STRONG AND WEAK ACIDS AND BASES

Strong Acids:
● completely dissociate in water, forming H+

and an anion. example: HN03 dissociates
completely in water to form H+ and N03

1-.

● The reaction is

● HNO3(aq)→ H+

(aq) + N03

1-

(aq)

● A 0.01 M solution of nitric acid contains

0.01 M of H+ and 0.01 M N03

- ions and

almost no HN03 molecules. The pH of the
solution would be 2.0.

● There are only 6 strong acids: You must learn them.

The remainder of the acids therefore are considered
weak acids.

● HClO4

● HBr

● HI

● HCl

● H2SO4

● HNO3

●

Note: when a strong acid dissociates only one H+
ion is removed. H2S04 dissociates giving H+ and
HS04

- ions.

●

H2SO4→ H+ + HSO4

1-

●

A 0.01 M solution of sulfuric acid would contain
0.01 M H+ and 0.01 M HSO4

1- (bisulfate or

hydrogen sulfate ion)

Weak acids:

●

a weak acid only partially dissociates in water to give H+ and the
anion

●

for example, HF dissociates in water to give H+ and F-. It is a weak
acid. with a dissociation equation that is

●

HF(aq)↔ H+

(aq) + F-

(aq)

●

Note: the use of the double arrow with the weak acid. That is
because an equilibrium exists between the dissociated ions and
the undissociated molecule. In the case of a strong acid
dissociating, only one arrow ( → ) is required since the reaction
goes virtually to completion.

●

An equilibrium expression can be written for this system:

●

Ka = [ H+][F-] / [HF]

● Which are the weak acids? Anything that dissociates in

water to produce H+ and is not one of the 6 strong acids.

○ Molecules containing an ionizable proton. (If the formula starts with

H then it is a prime candidate for being an acid.) Also: organic acids
have at least one carboxyl group, -COOH, with the H being
ionizable.

○ Anions that contain an ionizable proton. ( HSO4

1-→ H+ + SO4

2- )

○ Cations: (transition metal cations and heavy metal cations with high

charge)

○ also NH4

+ dissociates into NH3 + H+

Strong Bases:

●

They dissociate 100% into the cation and OH- (hydroxide ion).

●

example: NaOH(aq)→ Na+

(aq) + OH-

(aq)

●

a. 0.010 M NaOH solution will contain 0.010 M OH- ions (as
well as 0.010 M Na+ ions) and have a pH of 12.

●

Which are the strong bases?

●

The hydroxides of Groups I and II.

●

Note: the hydroxides of Group II metals produce 2 mol of
OH- ions for every mole of base that dissociates. These
hydroxides are not very soluble, but what amount that does
dissolve completely dissociates into ions.

exampIe:

Ba(OH)2(aq)→ Ba2+

(aq) + 2OH-

(aq)

●

a. 0.000100 M Ba(OH)2 solution will be 0.000200 M in OH-
ions (as well as 0.00100 M in Ba2+ ions) and will have a pH
of 10.3.

Weak Bases:

●

What compounds are considered to be weak bases?

●

Most weak bases are anions of weak acids.

●

Weak bases do not produce OH- ions by dissociation. They
react with water to produce the OH- ions.

●

Note that like weak acids, this reaction is shown to be at
equilibrium, unlike the dissociation of a strong base which is
shown to go to completion.

●

When a weak base reacts with water the OH- comes from the
water and the remaining H+ attaches itself to the weak base,
giving a weak acid as one of the products. You may think of it as
a two-step reaction similar to the hydrolysis of water by cations
to give acid solutions. examples:

●

NH3(aq) + H2O(aq)→ NH4

+

(aq) + OH-(aq)

●

methylamine: CH3NH2(aq) + H20(l)→ CH3NH3

+

(aq) + OH-

(aq)

●

acetate ion: C2H3O2

-
(aq) + H2O(aq) → HC2H302(aq) + OH-

(aq)

●

General reaction: weak base(aq) + H2O(aq)→ weak acid(aq) +
OH-

(aq)

●

Since the reaction does not go to completion relatively few OH-
ions are formed.

Neutralization Reactions

Neutralization reactions always produce a salt (ionic
compound) and water.

Ex:
HCl + NaOH 🡪 NaCl + H2O

H2SO4 + 2NH4OH 🡪

(NH4)2SO4 + 2H2O

2HBr + Ba(OH)2 🡪

BaBr2 + 2H2O

What is a SALT?

A salt is a neutral substance produced from
the reaction of an acid and a base.

Composed of the negative ion of an acid and
the positive ion of a base.

Examples: KCl, MgSO4, Na3PO4

[OH-]

[H+]

pOH

pH

10-pOH

10-pH

-Log[H+]

-Log[OH-]

14 - pOH

14 - pH

1.0 x 10-14

[OH-]

1.0 x 10-14

[H+]

HNO3, HCl, H2SO4 and HClO4 are among the
only known strong acids.

Strong and Weak Acids/Bases

The strength of an acid (or base) is
determined by the amount of
IONIZATION.

HONORS ONLY!

Strong and Weak Acids/Bases

●Generally divide acids and bases into STRONG or WEAK ones.

STRONG ACID:HNO3 (aq) + H2O (l) --->

H3O+ (aq)

+ NO3

- (aq)

HNO3 is about 100% dissociated in water.

HONORS ONLY!

● Weak acids are much less than 100% ionized in

water.

One of the best known is acetic acid = CH3CO2H

Strong and Weak Acids/Bases

HONORS ONLY!

● Strong Base:100% dissociated in water.

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

Strong and Weak Acids/Bases

Other common strong
bases include KOH and
Ca(OH)2.

CaO (lime) + H2O -->

Ca(OH)2 (slaked lime)

CaO

HONORS ONLY!

● Weak base:less than 100% ionized in

water

One of the best known weak bases is ammonia

NH3 (aq) + H2O (l) 🡪 NH4

+ (aq) + OH- (aq)

Strong and Weak Acids/Bases

HONORS ONLY!

Weak Bases

HONORS ONLY!

Equilibria Involving
Weak Acids and Bases

Consider acetic acid, HC2H3O2 (HOAc)

HC2H3O2 + H2O 🡪 H3O+ + C2H3O2-

Acid

Conj. base

(K is designated Ka for ACID)

K gives the ratio of ions (split up) to molecules

(don’t split up)

HONORS ONLY!

Ionization Constants for Acids/Bases

Acids

Conjugate

Bases

Increase
strength

Increase
strength

HONORS ONLY!

Equilibrium Constants
for Weak Acids

Weak acid has Ka < 1
Leads to small [H3O+] and a pH of 2 - 7

HONORS ONLY!

Equilibrium Constants
for Weak Bases

Weak base has Kb < 1
Leads to small [OH-] and a pH of 12 - 7

HONORS ONLY!

Relation of

Ka, Kb,

[H3O+] and

pH

HONORS ONLY!

Equilibria Involving A Weak Acid

You have 1.00 M HOAc. Calc. the equilibrium

concs. of HOAc, H3O+, OAc-, and the pH.

Step 1.Define equilibrium concs. in ICE table.

[HOAc] [H3O+]

[OAc-]

initial

change

equilib

1.00

0

0

-x

+x

+x

1.00-x

x

x

HONORS ONLY!

Equilibria Involving A Weak Acid

Step 2.Write Ka expression

You have 1.00 M HOAc. Calc. the equilibrium concs.
of HOAc, H3O+, OAc-, and the pH.

This is a quadratic. Solve using quadratic

formula.

or you can make an approximation if x is very
small! (Rule of thumb: 10-5 or smaller is ok)

HONORS ONLY!

Equilibria Involving A Weak Acid

Step 3.Solve Ka expression

You have 1.00 M HOAc. Calc. the equilibrium concs.
of HOAc, H3O+, OAc-, and the pH.

First assume x is very small because
Ka is so small.

Now we can more easily solve this
approximate expression.

HONORS ONLY!

Equilibria Involving A Weak Acid

Step 3.Solve Kaapproximateexpression

You have 1.00 M HOAc. Calc. the equilibrium concs.
of HOAc, H3O+, OAc-, and the pH.

x =[H3O+] = [OAc-] = 4.2 x 10-3 M

pH = - log [H3O+] = -log (4.2 x 10-3) =2.37

HONORS ONLY!

Equilibria Involving A Weak Acid

Calculate the pH of a 0.0010 M solution of

formic acid, HCO2H.

HCO2H + H2O 🡪 HCO2

- + H3O+

Ka = 1.8 x 10-4
Approximate solution
[H3O+] = 4.2 x 10-4 M,pH = 3.37
Exact Solution

[H3O+] = [HCO2

-] = 3.4 x 10-4 M

[HCO2H] = 0.0010 - 3.4 x 10-4 = 0.0007 M
pH = 3.47

HONORS ONLY!

Equilibria Involving A Weak Base

You have 0.010 M NH3. Calc. the pH.

NH3 + H2O 🡪 NH4

+ + OH-

Kb = 1.8 x 10-5

Step 1.Define equilibrium concs. in ICE table

[NH3]

[NH4

+]

[OH-]

initial

change

equilib

0.010

0

0

-x

+x

+x

0.010 - x

x

x

HONORS ONLY!

Equilibria Involving A Weak Acid

Step 2.Write Ka expression

You have 1.00 M HOAc. Calc. the equilibrium concs.
of HOAc, H3O+, OAc-, and the pH.

This is a quadratic. Solve using quadratic

formula.

or you can make an approximation if x is very
small! (Rule of thumb: 10-5 or smaller is ok)

HONORS ONLY!

Equilibria Involving A Weak Acid

Step 3.Solve Ka expression

You have 1.00 M HOAc. Calc. the equilibrium concs.
of HOAc, H3O+, OAc-, and the pH.

First assume x is very small because
Ka is so small.

Now we can more easily solve this
approximate expression.

HONORS ONLY!

Equilibria Involving A Weak Acid

Step 3.Solve Kaapproximateexpression

You have 1.00 M HOAc. Calc. the equilibrium concs.
of HOAc, H3O+, OAc-, and the pH.

x =[H3O+] = [OAc-] = 4.2 x 10-3 M

pH = - log [H3O+] = -log (4.2 x 10-3) =2.37

HONORS ONLY!

Equilibria Involving A Weak Acid

Calculate the pH of a 0.0010 M solution of

formic acid, HCO2H.

HCO2H + H2O 🡪 HCO2

- + H3O+

Ka = 1.8 x 10-4
Approximate solution
[H3O+] = 4.2 x 10-4 M,pH = 3.37
Exact Solution

[H3O+] = [HCO2

-] = 3.4 x 10-4 M

[HCO2H] = 0.0010 - 3.4 x 10-4 = 0.0007 M
pH = 3.47

HONORS ONLY!

Equilibria Involving A Weak Base

You have 0.010 M NH3. Calc. the pH.

NH3 + H2O 🡪 NH4

+ + OH-

Kb = 1.8 x 10-5

Step 1.Define equilibrium concs. in ICE table

[NH3]

[NH4

+]

[OH-]

initial

change

equilib

0.010

0

0

-x

+x

+x

0.010 - x

x

x

HONORS ONLY!

Equilibria Involving A Weak Base

You have 0.010 M NH3. Calc. the pH.

NH3 + H2O 🡪 NH4

+ + OH-

Kb = 1.8 x 10-5

Step 1.Define equilibrium concs. in ICE table

[NH3]

[NH4

+]

[OH-]

initial

change

equilib

0.010

0

0

-x

+x

+x

0.010 - x

x

x

HONORS ONLY!

Equilibria Involving A Weak Base

You have 0.010 M NH3. Calc. the pH.

NH3 + H2O 🡪 NH4

+ + OH-

Kb = 1.8 x 10-5

Step 2.Solve the equilibrium expression

Assume x is small, so

x = [OH-] = [NH4

+] = 4.2 x 10-4 M

and [NH3] = 0.010 - 4.2 x 10-4 ≈ 0.010 M
The approximation is valid !

HONORS ONLY!

Equilibria Involving A Weak Base

You have 0.010 M NH3. Calc. the pH.

NH3 + H2O 🡪 NH4

+ + OH-

Kb = 1.8 x 10-5

Step 3.Calculate pH

[OH-] = 4.2 x 10-4 M

so pOH = - log [OH-] = 3.37

Because pH + pOH = 14,
pH = 10.63

HONORS ONLY!

Types of Acid/Base Reactions:
Summary

HONORS ONLY!

pH testing

● There are several ways to test pH

○ Blue litmus paper (red = acid)

○ Red litmus paper (blue = basic)

○ pH paper (multi-colored)

○ pH meter (7 is neutral, <7 acid, >7 base)

○ Universal indicator (multi-colored)

○ Indicators like phenolphthalein

○ Natural indicators like red cabbage,

radishes

Paper testing

●Paper tests like litmus paper and pH paper

○ Put a stirring rod into the solution

and stir.

○ Take the stirring rod out, and place a

drop of the solution from the end of
the stirring rod onto a piece of the
paper

○ Read and record the color change.

Note what the color indicates.

○ You should only use a small portion of

the paper. You can use one piece of
paper for several tests.

pH paper

pH meter

● Tests the voltage of the

electrolyte

● Converts the voltage to

pH

● Very cheap, accurate

● Must be calibrated with a

buffer solution

pH indicators

●Indicators are dyes that can be added that will
change color in the presence of an acid or base.

●Some indicators only work in a specific range of
pH

●Once the drops are added, the sample is ruined

●Some dyes are natural, like radish skin or red
cabbage

ACID-BASE REACTIONS
Titrations

H2C2O4(aq) + 2 NaOH(aq) --->

acidbase

Na2C2O4(aq) + 2 H2O(liq)

Carry out this reaction using aTITRATION.

Oxalic acid,
H2C2O4

Setup for titrating an acid with a base

Titration

1. Add solution from the buret.

2. Reagent (base) reacts with compound (acid)

in solution in the flask.

3.

Indicator shows when exact stoichiometric
reaction has occurred. (Acid = Base)

This is called NEUTRALIZATION.

LAB PROBLEM #1: Standardize a
solution of NaOH — i.e., accurately
determine its concentration.

35.62 mL of NaOH is neutralized with 25.2

mL of 0.0998 M HCl by titration to an
equivalence point. What is the
concentration of the NaOH?

PROBLEM: You have 50.0 mL of 3.0 M
NaOH and you want 0.50 M NaOH.
What do you do?

Add water to the 3.0 M solution to lower its

concentration to 0.50 M

Dilute the solution!

PROBLEM: You have 50.0 mL of 3.0 M
NaOH and you want 0.50 M NaOH. What do
you do?

But how much water
do we add?

PROBLEM: You have 50.0 mL of 3.0 M
NaOH and you want 0.50 M NaOH. What do
you do?

How much water is added?

The important point is that --->

moles of NaOH in ORIGINAL solution =
moles of NaOH in FINAL solution

PROBLEM: You have 50.0 mL of 3.0 M NaOH and
you want 0.50 M NaOH. What do you do?

Amount of NaOH in original solution =
M • V=

(3.0 mol/L)(0.050 L) = 0.15 mol NaOH

Amount of NaOH in final solution must also = 0.15 mol NaOH

Volume of final solution =

(0.15 mol NaOH) / (0.50 M) = 0.30 L

or 300 mL

PROBLEM: You have 50.0 mL of 3.0 M
NaOH and you want 0.50 M NaOH. What do
you do?

Conclusion:

add 250 mL

of waterto
50.0 mL of
3.0 M NaOH
to make 300
mL of 0.50 M
NaOH.

Preparing Solutions by
Dilution

A shortcut

M1 • V1 = M2 • V2

You try this dilution problem

●You have a stock bottle of hydrochloric acid, which is 12.1 M. You
need 400 mL of 0.10 M HCl. How much of the acid and how much
water will you need?