Acids and Bases Notes for general chemistry PDF

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Acids and bases pdf for general chemistry...

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Table of 1

Contents Acid & Base Definition

2

Strong & Weak Acids

4

Strong & Weak Bases

5

Neutralization Reaction

6

Conjugate Acids & Bases

9

Amphoteric Species

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10 Autoionization 11 Acidic, Basic & Neutral Solutions + 12 Finding pH, pOH, [H3 O ] & [OH- ] 21 Summary of All Formulas

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1

Acid & Base Definitions There are two different definitions; the Arrhenius and Bronsted-Lowry The Bronsted-Lowry is the best definition used today

Arrhenius Acid: produces H+ions in solution example:

+ H + Br-

HBr

ArrheniusBase: producesOH- ions in solution example: NaOH Na++ OH KNOW THIS ONE

Bronsted-LowryAcid: H+(proton) donor + Loses H example:

HCl + NH3 Acid Loses H

+

NH+ 4+ Cl

+ Bronsted-Lowry Base: H (proton)acceptor Gains H+ example:

HCl + NH3

NH+ + Cl 4

Base Gains H +

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2

Strong & Weak Acids Important!

Strong Acids:

Dissociate

completely dissociate in solution

Ionize Dissociation

example:

HBr

H++ Br-

Ionization All refer to the acid splitting apart into ions. Ions are

Single arrow means complete dissociation

elements with a charge.

(Completely splits apart )

Weak Acids partially dissociate in solution

example: HC2 H3 O2

H+ +C2H3 O 2

Double arrow means partial dissociation

C2 H 3O2

H

+

-

+ H C2 H 3O 2

(Partially splits apart )

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3

Strong & Weak Acids STRONG ACIDS LIST Chemical Formula

Name

Dissociation

HCl

Hydrochloric acid

HCl

H+ + Cl-

HBr

Hydrobromic acid

HBr

H+ + Br-

HI

Hydroiodic acid

HI

HClO4

Perchloric acid

HClO4

HClO 3

Chloric acid

HClO 3

H+ + ClO3

H2SO4

Sulfuric acid

H2SO4

2+ 2 H + SO4

HNO 3

Nitric acid

HNO 3

H++ IH++ ClO4

+ H + NO3

WEAK ACIDS LIST Chemical Formula

Name

Dissociation

Nitrous acid

HNO2

+ H + NO 2

H SO3

Sulfurous acid

H2 SO3

+ H +HSO3-

HClO 2

Chlorous acid

HClO 2

+ H + ClO2

HClO

H++ClO-

HNO2 2

HClO

HC2 H3 O2

Hypochlorous acid Acetic acid

HC2 H3O2

+ H + C2 H3 O 2

H2CO 3

Carbonic acid

H2 CO 3

+ H + HCO 3

H3PO 4

Phosphoric acid

H3 PO4

+ H + H2 PO4

HF

Hydrofluoric acid

HF

H2 S

Hydrosulfuric acid

H2 S

HCN

Hydrocyanic acid

HCN

HCOOH

Formic acid

HCOOH

+ H + F H+ +HS + H +CN + H + HCOO-

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4

Strong & Weak Bases StrongBases:completely dissociate in solution STRONGBASES LIST Chemical Formula

LiOH

Name

Dissociation

lithium hydroxide

NaOH

sodium hydroxide

KOH

potassium hydroxide

RbOH

LiOH

Li++ OH -

NaOH

Na++ OH -

KOH

K++ OH -

rubidium hydroxide

RbOH

Rb+ + OH-

CsOH

cesium hydroxide

CsOH

Cs++ OH -

Ca(OH) 2

calcium hydroxide

Ca(OH) 2

Ca2++ 2 OH-

Sr(OH) 2

strontium hydroxide

Sr(OH) 2

Sr2+  + 2 OH-

Ba(OH) 2

barium hydroxide

Ba(OH)2

Ba2+ + 2 OH -

-

Weak Bases:partially dissociate in solution WEAK BASES LIST Chemical Formula

Name

NH 3

Ammonia

C6 H5 NH2

Aniline

CH3NH2

Methylamine

C5 H5 N

Pyridine

Dissociation

NH3 + H2 O C6 H5NH   2 + H2 O

+ NH4+ OH + C6H5 NH 3 + OH -

CH3NH   2 + H2O

CH3NH3 ++ OH -

C5H  5N + H2 O

C5 H5NH + OH -

+

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5

Acid-Base Neutralization Acid Base N eutralization Reactions Consist of: Acid + Base

Salt + Water

HCl + NaOH 

+

Na Cl

_

H2O

Balanced Equation:

Acid + Base

Salt + Water

HCl + NaOH 

NaCl + H2O Balanc e C harges + 2_ NaSO 4

H 2SO 4 + NaOH 

2_ + NaSO 4 2_

+ NaSO 4

Na 2SO  4

H 2O

Balanced Equation:

H 2SO4 + 2NaOH 

HPO 3 4 + KOH 

Na2SO+ 4 2H2 O

Balanc e C harges + 3 _ K PO4 3_ + K PO4

3_ K PO 4 +

H 2O

K3PO 4

Balanced Equation: HPO 3 4 +3KOH 

K3PO+ 4 3 H2 O

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6

Conjugate Acids & Bases Conjugate Acid-Base Pairs: A pair of substances that differs only by 1 H+ HA + B Acid + Base

Example 1:

+ HB + A

Conjugate Acid + Conjugate Base

Predict the conjugate acid-base pairs for HCl+ NH3

Step 1: Acid donates H+to Base HCl+ NH3 Step 2: Base accepts H+ and becomes new conjugate acid HCl+ NH3

+ NH4

+

Step 3: Acid loses H and becomes new conjugate base HCl+ NH3

+ NH+Cl 4 + Base gainsH

Acids have conjugate base pairs and Bases have conjugate acid pairs

Base HCl+ NH3 Acid

Conjugate Acid + NH+Cl 4 Conjugate Base

Acid loses H+

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Conjugate Acids & Bases Example 2:

Predict the conjugate acid-base pairs for HBr+ H 2 O

Step 1: Acid donates H+to Base HBr+ H2 O Step 2: Base accepts H+ and becomes new conjugate acid + H HBr+ H 3 O 2 O +

Step 3: Acid loses H and becomes new conjugate base + H3 O+Br -

HBr+ H2 O Acid

Example 3:

Base

Conjugate Acid Conjugate Base

Predict the conjugate acid-base pairs for H SO + H2 O 2 4

Step 1: Acid donates H+to Base H2 SO4 + H2 O Step 2: Base accepts H+ and becomes new conjugate acid + H3 O H SO + H O 2

4

2

Step 3: Acid loses H+ and becomes new conjugate base H2 SO4 + H2 O Acid

Base

+ H3 O+ HSO4 Conjugate Acid Conjugate Base

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8

Conjugate Acids & Bases Finding the Conjugate Acid from a Base: + Base gains 1 H to become the conjugate acid B Base

+ HB Conjugate Acid

Charges: Gaining an H+ means you add +1 to the charge

CN-

Base gainsH+

Example 1:

CNBase

HCN Conjugate Acid

+ -1 Charge + 1 H = 0 (no charge)

2-

CO3

Base gainsH+

2-

Example 2: CO3 Base

HCN

HCO 3

+ -2 Charge + 1 H =-1Charge

HCO 3 Conjugate Acid

Finding the Conjugate Base from an Acid: + Acid loses 1 H to become the conjugate base HA A Acid

Conjugate Base

Charges: + means you Losing an H subtract1 to the charge

+ Acid loses H

Example 1:

HNO2 Acid

Conjugate Base

+ Acid loses H

Example 2:

HClO4 Acid

NO-3

ClO 4

HNO2

NO3-

0 (No Charge)- 1 H+ =-1Charge

HClO4

ClO4-

+ 0 (No Charge)- 1 H =-1Charge

Conjugate Base

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9

Amphoteric Species Amphoteric Definition: A substance that can act as either an acid or a base Either loses or gains a H+

Common Amphoteric Species: H2O and HCO3

Example 1: HCO-3 acts as an acid while H2 O acts as a base HCO3 + H2 O Acid

Base

2+ H3 O+CO3

Conjugate Acid

Conjugate Base

Example 2: HCO-3 acts as a base while H2 O acts as an acid H2 O + HCO3Acid

Base

HCO+OH 2 3 Conjugate Acid Conjugate Base

Example 3: H2 O acts an acid H2O+ NH3 Acid

Base

+ NH+OH 4 Conjugate Acid Conjugate Base

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10

Autoionization Autoionization: Occurs when a substance reacts with itself Water can react with itself since it can act as an acid and a base

Water reacting with itself: + H3O+OH

H2O+ H2O Base

Acid

Conjugate Acid

Conjugate Base

Ion-Product Constant of Water (Kw) The equilibrium constant for water when it's reacting with itself + OR Kw= [ H3O+ ] x [OH- ]  Kw= [ H ] x [OH- ]  Numerical value

Note:

-14

Kw=1.0 x 10

Has no units

We can set everything equal to each other -14   = [ H3O+] x [OH-]  Kw=1.0 x 10

[ H3 O+ ] and [ H+ ] both refer to the concentration of acid

This can also mean: +

-7

  [H3O ] = 1.0 x 10

o

In a neutral solution at 25 C

-7   [OH ] = 1.0 x 10

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11

Acidic, Basic & Neutral Solutions pH: A number that represents how acidic, basic or neutral a solution is Typical pH scale is between 0-14 Measures concentration of H3 O+

pH

Ion Concentration   [H3 O+ ]>[OH- ]

Acidic

pH < 7

+

pH= 0 - 6

pH

Neutral

More acid than base

pH=7

  [H3O ] > 1.0 x 10-7 -7   [OH ] 7

Ion Concentration +   [H3O ]< [OH-]

Morebasethanacid +

pH=8-14

-7   [H3O ] < 1.0 x 10 -7   [OH- ] >1.0 x 10

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12

Finding pH of Strong Acid A Strong acid tells us the [ H3 O+ ] or [ H+]

Formula + pH = - log [HO 3 ]

or

pH = - log [H+ ]

[H3 O+ ] to pH Example: What is the pH of a 0.1 M solution of HCl? Step 1:

Since we are given the concentration of HCl which is a strong acid, then that really means we are given [ H3 O+ ] +

Given: [ HCl] = 0.1 M = [ H3 O ] = 0.1 M

Step 2: Plug [ H3 O+ ] into pH formula +

pH = - log [HO 3 ] pH = - log [0.1 ] pH =1

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Finding [H O+] of Strong Acid 3

A Strong acid tells us the [ H3 O+ ] or [ H+]

Formula -pH

+

[ H O ] = 10 3

+

-pH

[ H] = 10

or

+

pH to [H3 O ] Example: What is the [ H3 O+ ] of a solution with a pH of 5.6? Step 1:

Using the formula, plug in pH and solve for [ H3 O+ ]

Given: pH =5.6 +

-pH

+

-5.6

[ H3 O ] = 10 [ H3 O ] = 10

Round to 2 sig figs since the given 5.6

+

-6

[ H O ] =2.5 x 10 3

has 2 sig figs

Step 2: Make sure units are in molarity (M) since this is the concentration of acid

+

-6

[ H3 O ] =2.5 x 10 M

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14

Finding pOH of Strong Base pOH measures the concentration of base [OH- ] A strong base tells us the [OH-]

Formula pOH = - log [OH- ] [ OH- ] to pOH Example: Calculate the pOH of a 0.025 M solution of NaOH Step 1:

Since we are given the concentration of NaOH which is a strong base, then that really means we are given [ OH- ] -

Given: [NaOH] = 0.025 M = [OH ] = 0.025 M

Step 2: Plug [OH- ] into pOH formula pOH = - log [OH- ] pOH = - log [0.025] pOH = 1.6 Round to 2 sig figs since the given 0.025 has 2 sig figs

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-

Finding [OH ] of Strong Base Formula -pOH

[ OH- ] = 10

pOH to [ OH- ] Example: Calculate the [ OH-] of a solution with a pOH of 2.1 Step 1:

Find [OH-]by plugging in pOHinto the formula

Given: pOH = 2.1 -pOH [ OH- ] = 10 -2.1 [ OH- ] = 10 -3 [ OH- ] = 7.9 x 10  Round to 2 sig figs since the given 2.1 has 2 sig figs

Step 2:

Make sure units are in molarity (M) since this is the concentration ofbase

[ OH-] = 7.9 x 10-3M

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-

Finding pH from [OH ] Formulas pOH = - log [OH- ]

pH + pOH = 14

[ OH- ] to pH Example: Calculate the pH of a 0.0020 M KOH solution Step 1:

Since we are given the concentration of KOH, which is a strong base, then that really means we are given [ OH- ]

Given: [KOH] = 0.0020 M = [OH- ] = 0.0020 M

Step 2:

Find pOH by plugging [OH- ] into pOH formula

pOH = - log [OH- ] pOH = - log [0.0020] Round to 2 sig figs since the given

pOH = 2.70

0.0020 has 2 sig figs

Step 3:

Find pH by plugging in pOH into the formula

pH + pOH = 14 pH +2.70= 14 - 2.70 - 2.70

subtract 2.70 from both sides

pH = 11.3

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17

-

Finding [OH ] from pH Formulas -pOH [ OH- ] = 10

pH + pOH = 14 pH to [OH- ]

Example: Find [ OH- ] of a solution with a pH of 8.7 Step 1: Given:

Find the pOH by plugging in pH into the formula

pH = 8.7 pH + pOH = 14 8.7 + pOH = 14 -8.7 -8.7

subtract8.7from both sides

pOH =5.3

Step 2:

Find [OH- ] by plugging in pOH into the formula -pOH [ OH- ] = 10 -5.3 [ OH- ] = 10 -6 [ OH- ] = 5.0 x 10 M

Round to 2 sig figs since the given 8.7 has 2 sig figs

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Finding pH & pOH Formula pH + pOH = 14 pOH to pH Example: Find the pH if the pOH is 6.5 Step 1:

Find the pH by plugging in pOH into the formula

pOH =6.5

Given:

pH + pOH = 14 pH + 6.5 = 14 -6.5 -6.5 pH =7.5

subtract6.5from both sides Round to 2 sig figs since the given 6.5 has 2 sig figs

pH to pOH

Example: Find the pOH if the pH is 9.7 Step 1: Given:

Find the pOH by plugging in pH into the formula

pH =9.7 pH + pOH = 14 9.7 + pOH = 14 -9.7 -9.7 pOH =4.3

subtract9.7from both sides Round to 2 sig figs since the given 9.7 has 2 sig figs

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+

-

Finding [OH ] from [ H3 O ] Formula -14 [ H3 O+ ] x [ OH- ] = 1.0 x 10

[ H3 O+ ] to [ OH- ] Example: Find [ OH- ] of a solution with [ H3O+] = 2.7 x 10-4M Step 1:

Find [OH- ] by plugging in [ H3 O+ ] into the formula

Given: [ H3 O+ ] = 2.7 x 10-4M -14 [ H3 O+] x [ OH- ] = 1.0 x 10 -14 -4 2.7 x 10 x [ OH-] = 1.0 x 10 -14 -4 2.7 x 10 x [ OH- ] = 1.0 x 10

-4

2.7 x 10 

-4

-14

divide 2.7 x 10  from both sides

2.7 x 10 

-11 [ OH- ] =3.7x 10 M

Round to 2 sig figs since the given 2.7 has 2 sig figs

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Finding

[H 3 O

+

-

] from [ OH ]

Formula -14 [ H3 O+ ] x [ OH- ] = 1.0 x 10

[ OH- ] to [ H3 O+ ] Example: Find [ H3O+] of a solution with [ OH-] = 3.4 x 10-8M Step 1:

Find [ H3O+ ] by plugging in [ OH- ] into the formula

Given: [ OH- ] = 3.4 x 10-8M -14 [ H3 O+ ] x [ OH- ] = 1.0 x 10

-8 -14 [ H3 O+ ] x 3.4 x 10 = 1.0 x 10 -8 [ H3 O+ ] x 3.4 x 10 = 1.0 x 10-14

-8

3.4 x 10 

-8

-8

divide3.4x 10 from both sides

3.4 x 10  -7

[ H3O+] =2.9 x 10 M

Round to 2 sig figs since the given 3.4 has 2 sig figs

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