15-1 CHEM 102 Summer 2015, LA TECH Instructor: Dr. Upali Siriwardane e-mail: [email protected]...

15-1CHEM 102 Summer 2015, LA TECH

Instructor: Dr. Upali Siriwardane

e-mail: [email protected]

Office: CTH 311

Phone 257-4941

Office Hours: M,Tu, W,Th,F 9:00-11:00 am or by appointment..;

Test Dates:

Chemistry 102(060) Summer 2015

July 20, 2015 (Test 1): Chapter 13

July 27, 2015 (Test 2): Chapter 14

August 4, 2015 (Test 3): Chapter 15

August 12, 2015 (Test 4): Chapter 17

August 13, 2015 (Make-up test) comprehensive:

Chapters 13-17


Chapter 15. Acids and Bases

15.1 Heartburn 659

15.2 The Nature of Acids and Bases 660

15.3 Definitions of Acids and Bases 662

15.4 Acid Strength and the Acid Dissociation Constant (Ka) 665

15.5 Autoionization of Water and pH 668

15.6 Finding the and pH of Strong and Weak Acid Solutions 673

15.7 Base Solutions 682

15.8 The Acid–Base Properties of Ions and Salts 685

15.9 Polyprotic Acids 693

15.1 0 Acid Strength and Molecular Structure 698

15.1 1 Lewis Acids and Bases 700

15.1 2 Acid Rain 701


Types of Reactionsa) Precipitation Reactions. Reactions of ionic compounds or

saltsb) Acid/base Reactions. Reactions of acids and basesc) Redox Reactions. reactions of oxidizing & reducing

agents


What are Acids &Bases?

Definition?

a) Arrhenius

b) Bronsted-Lowry

c) Lewis


Arrhenius, Svante August (1859-1927), Swedish chemist, 1903 Nobel Prize in chemistry

•Acid Anything that produces hydrogen ions in a water solution.

HCl (aq) H+

( aq) + Cl-

( aq)

•Base Anything that producs hydroxide ions in a water solution.

NaOH (aq) Na+

( aq) + OH

- ( aq)

•Arrhenius definitions are limited proton acids and hydroxide bases

to aqueous solutions.

Arrhenius Definitions


Expands the Arrhenius definitions to include many bases other than hydroxides and gas phase reactions

Acid Proton donor

Base Proton acceptor

This definition explains how substances like ammonia can act as bases.

Eg. HCl(g) + NH3(g) ------> NH4Cl(s)

HCl (acid), NH3 (base).

NH3(g) + H2O(l) NH4+

+ OH-

Brønsted-Lowry definitions


Lewis DefinitionG.N. Lewis was successful in including acid and bases

without proton or hydroxyl ions.

Lewis Acid: A substance that accepts an electron pair.

Lewis base: A substance that donates an electron pair.

E.g. BF3(g) + :NH3(g) F3B:NH3(s)

the base donates a pair of electrons to the acid forming a

coordinate covalent bond common to coordination

compounds. Lewis acids/bases will be discussed later in

detail


1) Acids and bases can be defined in several ways. Which definitions of the bases that fits the description below?

a) a compound that produces more OH- ions in water:

b) a proton acceptor:

c) an electron pair donor:


Types of Acids and Bases

Binary acids

Oxyacid

Organic acids

Acidic oxides

Basic oxides

Amine

Polyprotic acids


Types of Acids and BasesBinary acids: HCl, HBr, HI, H2S

More than two elements: HCN

Oxyacid: HNO3, H2SO4, H3PO4

Polyprotic acids: H2SO4, H3PO4

Organic acids: R-COOH, R= CH3-, CH3CH2-

Acidic oxides: SO3, NO2, CO2,

Basic oxides: Na2O, CaO

Amine: NH3. R-NH2, R= CH3-, CH3CH2- : primary

R2-NH : secondary, R3-N: tertiary

Lewis acids & bases: BF3 and NH3


Acidic Oxides

These are usually oxides of non-metallic elements such as P, S and N.

E.g. NO2, SO2, SO3, CO2

They produce oxyacids when dissolved in waterSO3 + H2O ---> H2SO4

CO2 + H2O ---> H2CO3

NO2 + H2O ---> HNO3


Basic Oxides

Oxides oxides of metallic elements such as Na, K, Ca. They produce hydroxyl bases when dissolved in water.

e.g. CaO + H2O ---> Ca(OH)2

Na2O + H2O ---> 2 NaOH


Protic Acids

Monoprotic Acids: The form protic refers to acidity due to protons. Monoprotic acids have only one acidic proton. e.g. HCl.

Polyprotic Acids: They have more than one acidic proton.

e.g. H2SO4 - diprotic acid

H3PO4 - triprotic acid.


Polyprotic Acidsacids where more than one hydrogen per molecule

is released


Amines

Class of organic bases derived from ammonia NH3 by replacing hydrogen by organic groups. They are defined as bases similar to NH3 by Bronsted-Lowery or Lewis acid/base definitions.


Amines


2) Identify types of acids/bases as: binary acids, oxy acids, organic acids, acidic oxides, basic oxides, amine and polyprotic acids.

a) HF b) HBr c) H3PO4

d) H2SO4 e) HNO3 f) R-COOH

g) NO2 h) SO3 j)CaO

k) R-NH2


Influence of Molecular Structure on Acid Strength

Binary Hydrides• hydrogen & one other element

Bond Strengths• weaker the bond, the stronger the acid

Stability of Anion• higher the electronegativity, stronger the acid


Binary Acids

Compounds containing acidic protons bonded to a more electronegative atom.

e.g. HF, HCl, HBr, HI, H2S

The acidity of the haloacid (HX; X = Cl, Br, I, F)Series increase in the following order: HF < HCl < HBr < HI


Oxyacids

Compounds containing acidic - OH groups in the molecule.

Acidity of H2SO4 is greater than H2SO3 because of the extra O (oxygens)

The order of acidity of oxyacids from the a halogen (Cl, Br, or I) shows a similar trend.

HClO4 > HClO3 > HClO2 > HClO

perchloric chloric chlorus hyphochlorus


Influence of Molecular Structure on Acid Strength

Oxyacids• hydrogen, oxygen, & one other element

H-O-E• higher the electronegativity on E, stronger the acid as

this weakens the bond between the O and H


< <<

<

Oxo Acid


3) Which of the following is stronger acid or base:

a) H2SO4 or H2SO3:

b) HCl or HI:

c) HClO or HClO3:

d) H2S or HF:

e) CF3COOH or CH3COOH:

f) CH3COOH or CH3CH2COOH


3) Which of the following is stronger acid or base:

a) H2SO4 or H2SO3: Ka H2SO4>> 1 ; H2SO3= 1.4 x 10-2

b) HCl or HI:

c) HClO or HClO3:

d) H2S or HF: H2S: Ka:H2S=6.3 x 10-8 ; HF= 6.3 x 10-4

e) CF3COOH or CH3COOH: 2.5 x 10-3 and 1.8 x 10-5

f) CH3COOH or CH3CH2COOH: 1.8 x 10-5 and 1.2 x 10-5


DissociationStrong Acids:

HCl(aq) + H2O(l) H3+O(aq) + Cl-(aq)

H2SO4(aq) + H2O(l) H3+O(aq) + HSO4

-(aq)

Dissociation Equilibrium Weak Acid/base:

H2O(l) + H2O(l) H3+O(aq) + OH-(aq)

This dissociation is called autoionization of water.

HC2H3O2(aq) + H2O(l) H3+O(aq) + C2H3O2

-(aq)

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

Equilibrium constants: Ka, Kb and Kw


4) Write equations for the dissociation equilibrium reactions for the following acids and bases in water. Which of these are acid or dissociations?

a) HCl:

b) H2SO4 :


4) Write equations for the dissociation equilibrium reactions for the following acids and bases in water. Which of these are acid or dissociations?

c) H2O (autoionization): What is auto ionization?

d) HC2H3O(acetic acid):

e) NH3:


Bronsted acid/conjugate base and base/conjugate acid pairs inacid/base equilibria


HCl(aq): acid

H2O(l): base

H3+O(aq): conjugate acid

Cl-(aq): conjugate base

H2O/ H3+O: base/conjugate acid pair

HCl/Cl-: acid/conjugate base pair


Conjugate acid-base pairs.

Acids and bases that are related by loss or gain of H+ as H3O+

and H2O.

Examples. Acid Base

H3O +

H2O

HC2H3O2 C2H3O2-

NH4 +

NH3

H2SO4 HSO4-

HSO4- SO4

2-

Brønsted-Lowry Definitions


Select acid, base, acid/conjugate base pair,base/conjugate acid pair

H2SO4(aq) + H2O(l) H 3+O(aq) + HSO4

-(aq)acid

base

conjugate acid

conjugate base

base/conjugate acid pair

acid/conjugate base pair


5) For HOCl write:

a) Dissociation equilibrium reaction for the HOCl:

b) Identify the acid/conjugate base pair:

c) Identify the base/conjugate acid pair:

d) The equilibrium constant expression:


Strong Acid vs. Weak AcidsStrong acidcompletely ionizedHydrioidic HI Ka ~ 1011 pKa = -11Hydrobromic HBr Ka ~ 109 pKa = -9Perchloric HClO4 Ka ~ 107 pKa = -7Hyrdrochloric HCl Ka ~ 107 pKa = -7Chloric HClO3 Ka ~ 103 pKa = -3Sulfuric H2SO4 Ka ~ 102 pKa = -2Nitric HNO3 Ka ~ 20 pKa = -1.3Weak acidpartially ionizedHydrofluoric acid HF Ka = 6.6x10-4 pKa = 3.18Formic acid HCOOH Ka = 1.77x10-4 pKa = 3.75Acetic acid CH3COOH Ka = 1.76x10-5 pKa = 4.75Nitrous acid HNO2 Ka = 4.6x10-4 pKa = 3.34Acetyl Salicylic acid C9H8O4 Ka = 3x10-4 pKa = 3.52Hydrocyanic acid HCN Ka = 6.17x10-10 pKa = 9.21


Strong Base vs. Weak BaseStrong Basecompletely ionizedLithium hydroxide LiOHSodium hydroxide NaOH

Potassium hydroxide KOH Kb~ 102-103

Rubidium hydroxide RbOHCesium hydroxide CsOHBoarder-line Bases

Magnesium hydroxide Mg(OH)2

Calcium hydroxide Ca(OH)2

Strotium hydroxide Sr(OH)2 Kb~ 0.01 to0.1

Barium hydroxide Ba(OH)2

Weak Basepartially ionized

Ammonia NH3 Kb=1.79x10-5 pKb = 4.74

Ethyl amine CH3CH2NH2 Kb=5.6x10-4 pKb = 3.25


•Strong acids Ionize completely in water. HCl, HBr, HI,

HClO3,

HNO3, HClO4, H2SO4.

•Weak acids Partially ionize in water.

Most acids are weak.

•Strong bases Ionize completely in water.

Strong bases are metal hydroxides - NaOH,

KOH

•Weak bases Partially ionize in water.

Acid and Base Strength


Common Acids and BasesAcids Formula Molarity*

nitric HNO3 16

hydrochloric HCl 12

sulfuric H2SO4 18

acetic HC2H3O2 18

Bases

ammonia NH3(aq) 15

sodium hydroxide NaOH solid

*undiluted.


6. Identify stronger and weaker acids:


Autoionization When water molecules react with one another to form ions.

Acids and bases alter the dissociation equilibrium of water based on Le Chaterlier’s principle

Kw = [ H3O+

] [ OH- ]

= 1.0 x 10-14

at 25oC

Note: [H2O] is constant and is included in Kw.

ion product

of water

ion product

of water

H2O(l) + H2O(l) H3O+(aq) + OH-(aq)

(10-7

M) (10-7M)

Autoionization of Water


What is pH?

Kw = [H3+O][OH-] = 1 x 10-14

[H3+O][OH-] = 10-7 x 10-7

Extreme cases:

Basic medium

[H3+O][OH-] = 10-14 x 100

Acidic medium

[H3+O][OH-] = 100 x 10-14

pH value is -log[H+]

spans only 0-14 in water.


We need to measure and use acids and bases over a very large

concentration range.

pH and pOH are systems to keep track of these very large ranges.

pH = -log[H3O+

]

pOH = -log[OH-]

pH + pOH = 14

Substance pH

1 M HCl 0.0

Gastric juices 1.0 - 3.0

Lemon juice 2.2 - 2.4

Classic Coke 2.5

Coffee 5.0

Pure Water 7.0

Blood 7.35 - 7.45

Milk of Magnesia 10.5

Household ammonia 12.0

1M NaOH 14.0

Substance pH

1 M HCl 0.0

Gastric juices 1.0 - 3.0

Lemon juice 2.2 - 2.4

Classic Coke 2.5

Coffee 5.0

Pure Water 7.0

Blood 7.35 - 7.45

Milk of Magnesia 10.5

Household ammonia 12.0

1M NaOH 14.0

pH and other “p” scales


A logarithmic scale used to keep track of the large changes in [H+].

0 7 14

10 0

M 10-7

M 10-14

M

Very Neutral Very

acidic Basic

When you add an acid to, the pH gets smaller.

When you add a base to, the pH gets larger.

pH scale


pH of Aqueous Solutions


pH, pKw and pOHThe relation of pH, Kw and pOH Kw = [H+][OH-] log Kw = log [H+] + log [OH-] -log Kw= -log [H+] -log [OH-] ; previous equation multiplied by -1 pKw = pH + pOH; pKw = 14 since Kw =1 x 10-14

14 = pH + pOH

pH = 14 - pOH

pOH = 14 - pH

14 = pH + pOH

pH = 14 - pOH

pOH = 14 - pH


Measuring pH

Arnold Beckman

inventor of the pH meter

father of electronic instrumentation


7. Identify the following as acidic/basic/neutral and estimate/calculate pH.

Solution Acidic/basic/neutral pH of the solution

a) [ H+] > [OH-] and [H+] > 1.0 x 10-7 M : :

b) [H+] < [OH-] and [H+] < 1.0 x 10-7 M : :

c) [H+] = [OH-] = 1.0 x 10-7 M : :

d) [H+] > [OH-] = 1.0 × 10-14 M : :

e) [H+]< [OH-] = 1.0 x 10+7 M : :


pH and pOH calculations of acid and base solutions

a) Strong acids/bases

dissociation is complete for strong acid such as HNO3 or base NaOH

[H+] is calculated from molarity (M) of the solution

b) weak acids/bases

needs Ka , Kb or percent(%)dissociation


pH of 0.5 M H2SO4 Solution


-(aq)

HSO4-(aq) + H2O(l) H3

+O(aq) + SO42-(aq)

[H3+O][HSO4

-]

H2SO4 ; Ka1 = -------------------

[H2SO4]

[H3+O][SO4

2-]

H2SO4 ; Ka2 = ------------------- ; Ka2 ignored

[HSO4-]


H2SO4(aq) + H2O(l) H3+

O(aq) + HSO4-(aq)

the moles of H+

ions in the solution is equal to moles of H2SO4 at the beginning.

[H2SO4] = [H+

] = 0.5 mole/L

pH = -log [H+

]

pH = -log(0.5)

pH = 0.30

pH of 0.5 M H2SO4 Solution


1.5 x 10-2 M NaOH.1.5 x 10-2 M NaOH.

NaOH is also a strong base dissociates completely in water.

[NaOH] = [HO- ] = 1.5 x 10-2 mole/L

pOH = -log[HO-]= -log(1.5 x 10-2)

pOH = 1.82

As defined and derived previously:

pKw= pH + pOH; pKw= 14

pH = pKw + pOH

pH = 14 - pOH

pH = 14 - 1.82 ; pH = 12.18


8) For a 0.10 M solution of Ba(OH)2.Is it a strong base?

a) How many OH- are in the compound?

b) Calculate the [OH-] and [H+]:

c) pH of the solution:


9) Calculate the pH of the strong acid 0.2 M H2SO4.

a) Is it a strong acid?

b) Is it polyprotic acid?

c) Dissociation equilibria:

d) Why second dissociation equilibria is not considered for [H+] concentration?

e) Calculate the [H+]

f) pH of the solution:


pH of Mixtures of Strong and Weak Acidsthe presence of the strong acid retards the

dissociation of the weak acid

The pH of the solution is mainly based on the strong acid

Eg. 1.0 M HCl and 1.0 HC2H3O2


HC2H3O2(aq) + H2O(l) H3+

O(aq) + C2H3O2-(aq)


pH of Mixtures of Acids and BasesThe pH of the solution is mainly based on the

excess acid or base present

Eg. 10 mL of 1.0 M HCl and 20 mL 1.0 NaOH

Moles of excess NaOH ( M x L) = 1.0 x 0.001= 0.001

Mixed together volume = 30 mL = 0.030 L

Molarity of excess NaOH = 0.001/0.030= 0.030

Calculate pOH and then pH


Equilibrium, Constant, Ka & Kb

Ka: Acid dissociation constant for a equilibrium reaction.

Kb: Base dissociation constant for a equilibrium reaction.

Acid: HA + H2O H3+O + A-

Base: BOH + H2O B+ + OH-

[H3+O][ A-] [B+ ][OH-]

Ka = --------------- ; Kb = -----------------

[HA] [BOH]



[H3+O][Cl-]

Ka= ----------------- [HCl]

[H+][Cl-]

Ka= ----------------- [HCl]

Acid Dissociation Constant


Base Dissociation Constant

NH3 + H2O NH4+ + OH-

[NH4+

][OH-]

K =

[NH3]


Comparing Kw and Ka & Kb

Any compound with a Ka value greater than Kw of water will be a an acid in water.

Any compound with a Kb value greater than Kw of water will be a base in water.


Ionization Constants for Acids


WEAKER/STRONGER Acids and Bases & Ka and Kb values

A larger value of Ka or Kb indicates an equilibrium favoring product side.

Acidity and basicity increase with increasing Ka or Kb.

pKa = - log Ka and pKb = - log Kb

Acidity and basicity decrease with increasing pKa or pKb.


Which is weaker?a. HNO2 ; Ka= 4.0 x 10-4.

b. HOCl2 ; Ka= 1.2 x 10-2.

c. HOCl ; Ka= 3.5 x 10-8.

d. HCN ; Ka= 4.9 x 10-10.


What is Ka1 and Ka2?H2SO4(aq) + H2O(l) H3

+O(aq) + HSO4-(aq)


+O(aq) + SO42-(aq)



-(aq)


+O(aq) + SO42-(aq)

[H3+O][HSO4

-]

H2SO4 ; Ka1 = -------------------

[H2SO4]

[H3+O][SO4

2-]

H2SO4 ; Ka2 = -------------------

[HSO4-]

Ka Examples


HC2H3O2(aq) + H2O(l) H3+O(aq) + C2H3O2

-(aq)

[H+][C2H3O2

-]

H C2H3O2; Ka= ------------------

[H C2H3O2]

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

[NH4+][OH-]

NH3; Kb= --------------

[ NH3]

Ka Examples


% Dissociation gives x (amount dissociated) need for pH calculation

Amount dissociated

% Dissoc. = ------------------------- x 100

Initial amount/con.

x

% Dissoc. = --------------------------- x 100

concentration


How do you calculate pH of weak acids/bases?

a) From % dissociation

b) From Ka or Kb

What is % dissociation

Amount dissociated

% Dissoc. = ------------------------- x 100

Initial amount


How do you calculate % dissociation from Ka or Kb

1.00 M solution of HCN; Ka = 4.9 x 10-10

What is the % dissociation for the acid?



First write the dissociation equilibrium equation:

HCN(aq) + H 2O(l) <===> H 3+O(aq) + CN-(aq)

[HCN] [H+ ] [CN- ]Ini. Con. 1.00 M 0.0 M 0.00 M

Cha. Con -x x x

Eq. Con. 1.0 - x x x

[H 3+O ][CN-] x2

Ka = ------------------- = ----------------

[HCN] 1.0 - x



1.0 - x ~ 1.00 since x is small

x2

Ka = -----------; Ka = 4.9 x 10-10 = x2

1.0

x = 4.9 x 10-10 = 2.21 x 10 -5

Amount disso. 2.21 x 10 -5

----------------- x 100 =- ------------- x 100 Ini. amount 1.00

% Diss. =2.21 x 10 -5 x 100 = 0.00221 %



10) Calculate the % dissociation of 2.00 M solutions of HCN (Ka= 4.9 x 10-10)

a) Dissociation equilibria:

b) ICE setup:

c) Amount dissociated:

d) % dissociation::


1 M HF, 2.7% dissociated

Notice the conversion of % dissociation to a fraction (x): 2.7/100=0.027) x=0.027

Calculate the pH of a weak acid from % dissociation


HF(aq) + H 2O(l) <===> H 3+O(aq) + F-(aq)

[H+][F-]

Ka = ----------- [HF] [HF] [H+ ] [F- ]Ini. Con. 1.00 M 0.0 M 0.00 MChg. Con -x x xEq.Con. 1.0-0.027 0.027 0.027 pH = -log [H+] pH = -log(0.027) pH = 1.57

Calculate the pH of a weak acid from % dissociation


11) Calculate the Ka of if 5.0 M HF, 2.7% dissociated:

a) Dissociation equilibrium:

b) ICE setup:

c) Amount dissociated:

d) Ka :


Weak acid Equilibria

Example

Determine the pH of a 0.10 M benzoic acid solution at 25 oC if Ka = 6.28 x 10-5

HBz(aq) + H2O(l) H3O+(aq) + Bz-(aq)

The first step is to write the equilibrium expression

Ka = [H3O+

][Bz-]

[HBz]


Weak acid Equilibria

HBz H3O+ Bz-

Initial conc., M 0.10 0.00 0.00

Change, DM -x x x

Eq. Conc., M0.10 - x x x

[H3O+] = [Bz-] = x

We’ll assume that [Bz-] is negligible compared to [HBz]. The contribution of H3O+ from water is also negligible.


Weak Acid Equilibria

Solve the equilibrium equation in terms of x

Ka = 6.28 x 10-5 =

x = (6.28 x 10-5 )(0.10)

H3O+ = 0.0025 M

pH = 2.60

x2

0.10


pH from Ka or Kb


First write the dissociation equilibrium equation:

HCN(aq) + H 2O(l) H 3+O(aq) + CN-(aq)

[HCN] [H+ ] [CN- ]

Ini. Con. 1.00 M 0.0 M 0.00 M

Chg. Con -x x x

Eq. Con. 1.0 - x x x


[H 3+O ][CN-] x2

Ka = --------------- = ----------------

[HCN] 1.0 - x

1.0 - x ~ 1.00 since x is small x2 Ka = -----------; Ka = 4.9 x 10-10 = x2

1.0

x = 4.9 x 10-10 = 2.21 x 10 -5

pH = -log [H+]

pH = -log(2.21 x 10-5)

pH = 4.65

Weak Acid Equilibria


The Conjugate Partners of Strong Acids and BasesThe conjugate acid/base of a strong base/acid has

no net effect on the pH of a solution

The conjugate base of a weak acid hydrolyze in water and basic or

pH of a solution > 7.00 E.g. Na+C2H3O2- sodium

acetate

The conjugate acid of a weak base hydrolyze in water and acidic or

pH of a solution < 7.00 E.g NH4Cl


12) Calculate the [H+], [OH-] and pH of 0.90 M HC2H3O2; Ka= 1.8 x 10-5.


b) ICE setup:

c) [H+] and [OH-]:

d) pH:


13) Calculate the [H+], [OH-] pOH and pH 5.0 M NH3; Kb = 1.8 x 10-5


b) ICE setup:

c) [H+] and [OH-]:

d) pOH and pH:


14) Calculate the pH of a 0.015 M solution of lactic acid. The Ka for lactic acid is 1.4 x 10-4.


15) Calculate the pH of a 0.14 M solution of an acid with Ka = 6.2 x 10-8 (pH = 4.03)


Acid-Base Properties of Typical Ions


Reaction of a basic anion or acidic cation with water is an ordinary Brønsted-Lowry acid-

base reaction.

CH3COO-(aq) + H2O(l) CH3COOH(aq) + OH-(aq)

NH4+

(aq) + H2O(l) NH3 (aq) + H3O+

(aq)

This type of reaction is given a special name.

Hydrolysis

The reaction of an anion with water to produce the conjugate acid and OH-.

The reaction of a cation with water to produce the conjugate base and H3O+

.

Hydrolysis


What salt solutions would be acidic, basic and neutral?

1) strong acid + strong base = neutral

2) weak acid + strong base = basic

3) strong acid + weak base = acidic

4) weak acid + weak base = neutral,

basic or an acidic solution depending on the relative strengths of the acid and the base.


What pH? Neutral, basic or acidic?

•a)NaCl • neutral•b) NaC2H3O2

• basic•c) NaHSO4 • acidic•d) NH4Cl• acidic


1) If the following substance is dissolved in pure water, will the solution be acidic, neutral, or basic?

a) Solid sodium carbonate-(Na2CO3):

b) Sodium chloride- (NaCl):

c) Sodium acetate- (NaC2H3O2):

d) Ammonium sulfate-((NH4)2SO4):


How do you calculate pH of a salt solution?Find out the pH, acidic or basic?

If acidic it should be a salt of weak base

If basic it should be a salt of weak acid

if acidic calculate Ka from Ka= Kw/Kb

if basic calculate Kb from Kb= Kw/Ka

Do a calculation similar to pH of a weak acid or base


What is the pH of 0.5 M NH4Cl salt solution? (NH 3; Kb = 1.8 x 10-5)

Find out the pH, acidic

if acidic calculate Ka from Ka= Kw/Kb

Ka= Kw/Kb = 1 x 10-14 /1.8 x 10-5)

Ka= 5.56. X 10-10

Do a calculation similar to pH of a weak acid


Continued

NH4+ + H2O H 3

+O + NH3

[NH4+] [H3

+O ] [NH3 ]Ini. Con. 0.5 M 0.0 M 0.00 MChange -x x xEq. Con. 0.5 - x x x [H 3+O ] [NH3 ]

Ka(NH4+) = -------------------- =

[NH 4+] x2

---------------- ; appro.:0.5 - x . 0.5 (0.5 - x)


x2 Ka(NH4

+) = ----------- = 5.56 x 10 -10

0. 5 x2

= 5.56 x 10 -10 x 0.5 = 2.78 x 10 -10

x= 2.78 x 10 -10 = 1.66 x 10-5

[H+ ] = x = 1.66 x 10-5 MpH = -log [H+ ] = - log 1.66 x 10-5

pH = 4.77

pH of 0.5 M NH4Cl solution is 4.77 (acidic)

Continued

15-1 CHEM 102 Summer 2015, LA TECH Instructor: Dr. Upali Siriwardane e-mail: [email protected]...

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Transcript of 15-1 CHEM 102 Summer 2015, LA TECH Instructor: Dr. Upali Siriwardane e-mail: [email protected]...