Acids, Bases, and Salts CHM 1010 PGCC Barbara A. Gage

Acids, Bases, and Salts

CHM 1010 PGCC Barbara A. Gage

Acid Base

Litmus color

Phenolphthalein color

pH range

Reaction with active metal (like

Mg)

Taste

Formula component

Characteristics of Acids and Bases


AcidsStrong

hydrochloric acid, HClhydrobromic acid, HBrhydroiodic acid, HInitric acid, HNO3

sulfuric acid, H2SO4

perchloric acid, HClO4

Weak

hydrofluoric acid, HFphosphoric acid, H3PO4acetic acid, CH3COOH (or HC2H3O2)

ionizes completely in water ionizes partially in water

carbonic acid, H2CO3


Strong acid: HA(g or l) + H2O(l) H3O+(aq) + A-

(aq)

The extent of dissociation for strong acids.

H+ and H2O H3O+ (hydronium ion)


The extent of dissociation for weak acids.

Weak acid: HA(aq) + H2O(l) H3O+(aq) + A-

(aq)


Bases (or alkalis)Strong

Weak

sodium hydroxide, NaOH

calcium hydroxide, Ca(OH)2

potassium hydroxide, KOH

strontium hydroxide, Sr(OH)2

barium hydroxide, Ba(OH)2

ammonia, NH3 (NH4OH)

Moderate

Dissociates completely Dissociates completely but is not very soluble

aluminum hydroxide, Al(OH)3

magnesium hydroxide, Mg(OH)2

Dissociates partially

carbonates, CO32-

bicarbonates, HCO31-


An aqueous strong acid-strong base reaction on the atomic scale.

MX is a “salt” – an electrolyte that is not an acid or baseCHM 1010 PGCC

Barbara A. Gage

Acid and Base Definitions

• Arrhenius• Acid = compound that forms

hydrogen (H+) ions in water• Base = compound that forms

hydroxide (OH-) ions in water


Acid and Base Definitions

• Bronsted-Lowry• Acid = proton donor (H+ is a proton)• Base = proton acceptor

An acid-base reaction can now be viewed from the standpoint of the reactants AND the products.

An acid reactant will produce a base product and the two will constitute an acid-base conjugate pair.


Proton transfer as the essential feature of a Brønsted-Lowry acid-base reaction.

(acid, H+ donor) (base, H+ acceptor)

HCl H2O

+

Cl- H3O+

+

Lone pair binds H+

(base, H+ acceptor) (acid, H+ donor)

NH3 H2O

+

NH4+ OH-

+

Lone pair binds H+


The Conjugate Pairs in Some Acid-Base Reactions

Base Acid+Acid Base+

Conjugate Pair

Conjugate Pair

Reaction 4

H2PO4- OH-+

Reaction 5

H2SO4 N2H5++

Reaction 6

HPO42- SO3

2-+

Reaction 1

HF H2O+ F- H3O++

Reaction 3

NH4+ CO3

2-+

Reaction 2

HCOOH CN-+ HCOO- HCN+

NH3 HCO3-+

HPO42- H2O+

HSO4- N2H6

2++

PO43- HSO3

-+


SAMPLE PROBLEM Identifying Conjugate Acid-Base Pairs

PROBLEM: The following reactions are important environmental processes. Identify the conjugate acid-base pairs.

(a) H2PO4-(aq) + CO3

2-(aq) HPO42-(aq) + HCO3

-(aq)

(b) H2O(l) + SO32-(aq) OH-(aq) + HSO3

-(aq)

SOLUTION:

Identify proton donors (acids) and proton acceptors (bases).

(a) H2PO4-(aq) + CO3

2-(aq) HPO42-(aq) + HCO3

-(aq)

proton

donor

proton accepto

r

proton accepto

r

proton

donor

conjugate pair1conjugate pair2

(b) H2O(l) + SO32-(aq) OH-(aq) + HSO3

-(aq)

conjugate pair2conjugate pair1

proton

donor

proton accepto

r

proton accepto

r

proton

donorCHM 1010 PGCC Barbara A. Gage

Molecules as Lewis Acids

F

B

F F

H

N

H H

+

F

B

F F

H

N

H H

acid base adduct

An acid is an electron-pair acceptor.A base is an electron-pair donor.

M2+

H2O(l)

M(H2O)42+(aq)

adductCHM 1010 PGCC

Barbara A. Gage

SAMPLE PROBLEM Identifying Lewis Acids and Bases

PROBLEM: Identify the Lewis acids and Lewis bases in the following reactions:

(a) H+ + OH- H2O

(b) Cl- + BCl3 BCl4-

(c) K+ + 6H2O K(H2O)6+

SOLUTION:

PLAN: Look for electron pair acceptors (acids) and donors (bases).

(a) H+ + OH- H2Oacceptor

donor

(b) Cl- + BCl3 BCl4-

donor

acceptor

(c) K+ + 6H2O K(H2O)6+

acceptor

donorCHM 1010 PGCC

Barbara A. Gage

Acid Anhydrides

• Non-metal oxides react with water to form acidic solutions

CO2 (g) + H2O (l) H2CO3 (aq)

N2O5 (s) + H2O (l) 2 HNO3 (aq)

SO3 (g) + H2O (l) H2SO4 (aq)

Dissolved non-metal oxides cause acid rain.


Basic Anhydrides

• Metal oxides react with water to form alkaline solutions

Na2O (s) + H2O (l) 2 NaOH (aq)CaO (s) + H2O (l) Ca(OH)2 (aq)Al2O3 (s) + 3 H2O (l) 2 Al(OH)3 (aq)

Lime (CaO) is used on lawns and is converted to Ca(OH)2 when it rains. CaO is less hazardous to handle.


An acid-base titration.

Start of titrationExcess of acid

Point of neutralization

Slight excess of base


Sample Problem Finding the Concentration of Acid from an Acid-Base Titration

PROBLEM: You perform an acid-base titration to standardize an HCl solution by placing 50.00 mL of HCl in a flask with a few drops of indicator solution. You put 0.1524 M NaOH into the buret, and the initial reading is 0.55 mL. At the end point, the buret reading is 33.87 mL. What is the concentration of the HCl solution?

SOLUTION:

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

(33.87-0.55) mL x

1L

103

mL

= 0.03332 L

0.03332 LX 0.1524 M = 5.078x10-3 molNaOH

Molar ratio is 1:15.078x10-3 mol HCl

0.05000 L= 0.1016 M HCl


Kc = [H3O+][OH-]

[H2O]2

Kc[H2O]2 = [H3O+][OH-]

The Ion-Product Constant for Water

Kw =

A change in [H3O+] causes an inverse change in [OH-].

= 1.0 x 10-14 at 250C

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

In an acidic solution, [H3O+] > [OH-]

In a basic solution, [H3O+] < [OH-]

In a neutral solution, [H3O+] = [OH-]


The relationship between [H3O+] and [OH-] and the

relative acidity of solutions.

[H3O+] [OH-]Divide into Kw

ACIDIC SOLUTION

BASIC SOLUTION

[H3O+] > [OH-]

[H3O+] = [OH-]

[H3O+] < [OH-]

NEUTRAL SOLUTION


SAMPLE PROBLEM Calculating [H3O+] and [OH-] in an Aqueous Solution

PROBLEM: A research chemist adds a measured amount of HCl gas to pure water at 250C and obtains a solution with [H3O+] = 3.0x10-4M. Calculate [OH-]. Is the solution neutral, acidic, or basic?

SOLUTION:

Use the Kw at 250C and the [H3O+] to find the corresponding [OH-].

Kw = 1.0x10-14 = [H3O+] [OH-] so

[OH-] = Kw/ [H3O+] = 1.0x10-14/3.0x10-4 =

[H3O+] is > [OH-] and the solution is acidic.

3.3x10-

11M


The pH values of some familiar

aqueous solutions.

pH = -log [H3O+]


pOH = -log [OH-]pH + pOH = 14

The relations among [H3O+], pH, [OH-], and pOH.


SAMPLE PROBLEM Calculating [H3O+], pH, [OH-], and pOH

PROBLEM: In an art restoration project, a conservator prepares copper-plate etching solutions by diluting concentrated HNO3 to 2.0M, 0.30M, and 0.0063M HNO3. Calculate [H3O+], pH, [OH-], and pOH of the three solutions at 250C.

SOLUTION:

PLAN: HNO3 is a strong acid so [H3O+] = [HNO3]. Use Kw to find the [OH-] and then convert to pH and pOH.

For 2.0M HNO3, [H3O+] = 2.0M and -log [H3O+] = -0.30 = pH[OH-] = Kw/ [H3O+] = 1.0x10-14/2.0 = 5.0x10-15M; pOH = 14.30

[OH-] = Kw/ [H3O+] = 1.0x10-14/0.30 = 3.3x10-14M; pOH = 13.48

For 0.3M HNO3, [H3O+] = 0.30M and -log [H3O+] = 0.52 = pH

[OH-] = Kw/ [H3O+] = 1.0x10-14/6.3x10-3 = 1.6x10-12M; pOH = 11.80

For 0.0063M HNO3, [H3O+] = 0.0063M and -log [H3O+] = 2.20 = pH


Buffers

• Solutions that resist change in pH• Can maintain any pH value between 0

and 14 (not just neutral pH 7)• Composed of a weak acid and a salt

made from the weak acid or weak base and salt made from the weak base

• Examples: HC2H3O2 and NaC2H3O2

NH4OH and NH4Cl


Buffers

Reaction with acid:HC2H3O2 + C2H3O2

- + H+ HC2H3O2 + HC2H3O2

Reaction with base:HC2H3O2 + C2H3O2

- + OH- C2H3O2

- + C2H3O2- + HOH

A buffer regenerates it’s own components. The pH it maintains depends on the ratio of salt to acid (or base) and the nature of the acid (or base).


Documents

Acids, Bases, and Salts CHM 1010 PGCC Barbara A. Gage