acids bases and salts honors

	acids bases and salts
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Notes

THE ARRHENIUS MODEL

Acids produce H⁺ as the only cation in aqueous solution.

Example

HCl(aq) ---> H⁺(aq) + Cl^-(aq)

In fact H⁺ does not exist in the solution since it reacts with H₂O to produce H₃O⁺

H⁺(aq) + H₂O(l) ---> H₃O⁺(aq)

Bases produce OH^- as the only anion in aqueous solution.

Example

NaOH(aq) ---> OH^-(aq) + Na⁺(aq)

Arrhenius acids:

Taste sour

React with some metals to yield H₂(g)

Generally formed when nonmetal oxides react with water

Example

SO₃(g) + H₂O(l) ---> H₂SO₄(aq)

CO₂(g) + H₂O(l) ---> H₂CO₃(aq)

Note that SO₃and CO₂ are called acid oxides

Arrhenius bases:

Taste bitter

React with acids to yield salts and water (neutralization)

Generally formed when metal oxides react with water

Example

CaO(g) + H₂O(l) ---> Ca(OH)₂(aq)

K₂O(g) + H₂O(l) ---> 2 KOH(aq)

Note that CaOand K₂O are called basic oxides

THE BRONSTED-LOWRY MODEL

Acids donate H⁺ and bases accept H⁺

Example

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

HCl is the acid since it donates a proton to H₂O and becomes Cl^-

H₂O is the base since it accepts a proton from HCl and becomes H₃O⁺

Cl^- become the conjugate base of HCl

H₃O⁺ becomes the conjugate acid of H₂O

In the reverse reaction

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

Cl^- is the base since it accepts a proton from H₃O⁺ and becomes HCl

H₃O⁺ is the acid since it donates a proton to Cl^- and becomes H₂O

HCl becomes the conjugate acid of Cl^-

H₂O becomes the conjugate base of H₃O⁺

Therefore in HCl(aq) + H₂O(l) <---> Cl^-(aq) + H₃O⁺(aq)

There are two conjugate acid base pairs and the difference in the pair is a proton

HCl, Cl^-

H₃O⁺,H₂O

Problems

1. What is the conjugate base of OH^- ?

2. What is the conjugate acid of HPO₄^2- ?

THE LEWIS MODEL

Acids accept an electron pair and base donate an electron pair.

Example

H₂O + H⁺ ---> H₃O⁺ H⁺ is the Lewis acid and H₂O is the Lewis base

Example

H⁺ + NH₃ ---> NH₄⁺ H⁺ is the Lewis acid and NH₃ is the Lewis base

Draw the Lewis structure

Example

H⁺ + OH^- ---> H₂O H⁺ is the Lewis acid and OH^- is the Lewis base

Draw the Lewis structure

Example

BF₃ + NH₃ ---> BF₃NH₃ BH₃ is the Lewis acid and NH₃ is the Lewis base

Draw the Lewis structure

Example

SO₃ + H₂O ---> H₂SO₄ SO₃ is the Lewis acid and H₂O is the Lewis base

Draw the Lewis structure

Example

AlCl₃ + Cl^- ---> AlCl₄^- AlCl₃ is the Lewis acid and Cl^- is the Lewis base

Draw the Lewis structure

The Lewis model is the more encompassing model of acids and bases.

Summary

Model	Acid	Base
Arrhenius	H⁺ producer	OH^- producer
Bronsted-Lowry	H⁺ donor	H⁺ acceptor
Lewis	e- pair acceptor	e- pair donor

IONIZATION CONSTANT

K_eq

For the reaction aB + bB <----> cC + dD

[C]^c [D]^d

K_eq = ------------

[A]^a [B]^b

Note: Solids and liquids have constant concentration therefore they do not appear in K_eq

Example

BF₃(g) + 3H₂O(l) <----> 3HF(aq) + H₃BO₃(aq)

[HF]³ [H₃BO₃]

K_eq = --------------------

[BF₃]

K_a

Ionization of acids

H₂A (aq) <----> 2H⁺ (aq) + A^2- (aq)

[H⁺]² [A^2-]

K_a = --------------

[H₂A]

Note: The larger the K_a the stronger the acid (high ionization)

The strength of an acid can also be expressed as pK_a (pK_a = - log K_a) A small pK_a indicates a strong acid

K_b

Same as for K_a except for bases

K_sp

The solubility product or K_sp is derived when a salt dissolved in a solvent at a given temperature.

Example

For the given reaction at 25^oC

H₂O

CaF₂(s) <-------> Ca²⁺(aq) + 2 F^-(aq)

K_sp = [Ca²⁺] [F^-]² = 4.0 x 10^-11

Strength of acids and bases

A strong acid has a large K_a and its conjugate base is weak (K_b is small)

The major strong acids are

hydrochloric acid, HCl

bromic acid, HBr

iodic acid, HI

nitric acid, HNO₃

chloric acid, HClO₄

sulfuric acid, H₂SO₄

The major strong bases are: NaOH, KOH, Ca(OH)₂, Sr(OH)₂, and Ba(OH)₂

Amphoteric substances

A substance that can act as an acid or a base is called amphoteric.

Example

H₂O + H₂O ---> HO^- + H₃O⁺

NH₃ + NH₃ ---> NH₄⁺ + NH₂^-

THE DISSOCIATION CONSTANT OF WATER

For the reaction H₂O(l) <---> H⁺(aq) + OH^-(aq) at 25^oC

The K_eq or K_w= [H⁺] [OH^-] = 1.0 x 10^-14

And there are equal amount of protons and hydroxide ions

[H⁺] = [OH^-] = 1.0 x 10^-7

Note that the value of K_w is very small therefore water is a very weak electrolyte since very few ions are present in the solution.

When [H⁺] = [OH^-] the solution is neutral

When [H⁺] > [OH^-] the solution is acidic

When [H⁺] < [OH^-] the solution is basic or alkaline

Example

If the hydrogen ion concentration is 1 x 10^-3 what is the hydroxide ion concentration?

Since [H⁺] [OH^-] = 1.0 x 10^-14

Then 1 x 10^-3 [OH^-] = 1.0 x 10^-14

And [OH^-] = 1.0 x 10^-14 / 1.0 x 10^-3 = 1.0 x 10^-11

Therefore, since [H⁺] > [OH^-] the solution is acidic

THE PH SCALE

The pH is a log scale based on 10

pH = -log[H⁺]

Example

When [H⁺] = 1.0 x 10^-7

pH = -log 1.0 x 10^-7 = 7.00

To find [H⁺] from pH take the antilog of negative pH (for antilog press INV then LOG or press “10^x”)

Example

If the pH is 7.41 what is [H⁺]?

7.41 = -log[H⁺] or log[H⁺] = -7.41

And antilog (log[H⁺]) = antilog (-7.41)

Therefore [H⁺] = 3.9 x 10^-8

Magnitude of a pH change

For every change of 1 in pH there is a 10x change in [H⁺]

Example

A pH of 4 has 10x less H⁺ in the solution than a pH of 3

A pH of 2 has 100x more H⁺ in the solution than a pH of 4

Relationship of pH and pOH

pH + pOH = 14

Example

What is the pOH of an aqueous solution if the pH is 5?

Since pH + pOH = 14

Then 5 + pOH = 14

And pOH = 14 - 5 = 9

CALCULATIONS OF THE PH OF STRONG AND WEAK ACIDS

For strong acids (large K_a) an almost complete dissociation occurs therefore [HA] = [H⁺] and the pH is calculated using the concentration of the acid.

Example

HCl(aq) ---> H⁺(aq) + Cl^-(aq) large K_a

For weak acids the pH is calculated as below

Example

What are the pH and pOH of a 1.00 M HF(aq)? K_a = 7.2 x 10^-4 (relatively small K_a)

First, right the dissociation reaction of HF in water

HF(aq) <---> H⁺(aq) + F^-(aq)

Second, write the K_a

[H⁺] [F^-]

K_a = -------------

[HF]

Third, make a table representing initial and final concentrations

	Initial	∆	Final
[HF]	1.00	-x	1.00 – x ≈ 1.00*
[H⁺]	0	+x	x
[F^-]	0	+x	x

* x is so small (small K_a) that it is negligible when subtracted form 1.00

Fourth, substitute in K_a and solve for [H⁺] to calculate pH and pOH

[H⁺] [F^-] x²

K_a = ----------- = ------------ = 7.2 x 10^-4

[HF] 1.00

x² = 7.2 x 10^-4

x = 2.7 x 10^-2

Since [H⁺] = [F^-] = x = 2.7 x 10^-2

pH = -log (2.7 x 10^-2) = 1.57

Since pH + pOH = 14.00

pOH = 14.00 – 1.57 = 12.43

POLYPROTIC ACIDS

A polyprotic acid has more than one proton to donate.

Example:

H₂C₂O₄ has 2 protons to donate.

The first reaction is H₂C₂O₄(aq) <---> H⁺(aq) + HC₂O₄^-(aq) K_a1 = 5.90 x 10 ^-2

The second reaction is HC₂O₄^-(aq) <---> H⁺(aq) + C₂O₄^-(aq) K_a2 = 6.40 x 10 ^-5

Note that the first reaction has the larger K_a. The smaller K_a of the second reaction is due to the negative charge of the ion which attracts H⁺ with a greater force than H₂C₂O₄

ACID-BASE INDICATORS

An indicator is the conjugate pair of a weak acid or base, where each has a different color.

H-Indicator <----> H⁺ + Indicator ^-

(Acid) (Base)

Indicators	Color in acidic solution	Color in basic (alkaline) solution
litmus	red	blue
methyl orange	red	yellow
phenolphthalein	colorless	pink

Example

H-Indicator <----> H⁺ + Indicator^-

(red) (yellow)

When [H⁺] increases equilibrium shifts to the left towards the red

When [H⁺] decreases the equilibrium shifts to the right towards the yellow.

Note: Indicators change color at a pH that is about equal to their pK_a.

ACID-BASE PROPERTIES OF SALTS

When a salt is made up of a cation of a strong base and an anion of a strong acid the aqueous solution is neutral

Example

NaCl (s) + H₂O(l) <---> Na⁺(aq) + Cl^-(aq)

NaCl is composed of Na⁺, and Cl^-

Na⁺has no affinity for H⁺. It is the weak conjugate acid of the strong base NaOH

Cl^- has no affinity for H⁺. It is the weak conjugate base of the strong acid HCl

When a salt is made up of a cation of a strong base and an anion of a weak acid the aqueous solution is basic

Example

NaC₂H₃O₂(s) + H₂O(l) <---> Na⁺(aq) + C₂H₃O₂^-(aq)

NaC₂H₃O₂ is composed of Na⁺, and C₂H₃O₂^-

Na⁺has no affinity for H⁺. It is the weak conjugate acid of the strong base NaOH

C₂H₃O₂^- has affinity for H⁺. It is the strong conjugate base of the weak acid HC₂H₃O₂

Therefore, the strong base C₂H₃O₂^- produces OH^- by removing H⁺ form H₂O

C₂H₃O₂^-(aq) + HOH(l) <---> HC₂H₃O₂(aq) + OH^-(aq)

When a salt is made up of a cation of a weak base and an anion of a strong acid the aqueous solution is acidic

Example

NH₄Cl(s) <---> NH₄⁺(aq) + Cl^-(aq)

NH₄Cl is composed of NH₄⁺, and Cl^-

Cl^- has no affinity for H⁺. It is the weak conjugate base of the strong acid HCl

NH₄⁺has no affinity for H⁺. It is the strong conjugate acid of the weak base NH₄OH

Therefore, being a strong acid it releases of H⁺ in the solution

NH₄⁺(aq) <---> NH₃(aq) + H⁺(aq)

When a salt is made up of a cation of a weak base and an anion of a weak acid the aqueous solution is

Acidic if K_a > K_b

Basic if K_b > K_a

Neutral if K_a = K_b

BUFFERED SOLUTIONS

A buffer solution is one which resists changes in pH when small quantities of H⁺ or OH^- are added to it.

Buffer solutions can be prepared from weak acids or bases and their salts to produce a conjugate acid-base pair.

Strong acids or bases are not used since their reaction with water is complete.

Example

In an acidic buffer solution made of CH₃COOH (aq) and NaCH₃COO(s). The conjugate acid-base pair is CH₃COOH and CH₃COO^-

Example

In a basic buffer solution made of NH₃ and NH₄Cl. The conjugate acid-base pair is NH₃ and NH₄⁺

NORMALITY AND MOLARITY

It is the number of equivalents per liter of solution.

For acid-base reactions, the equivalent is the mass of the acid or base that can donate or accept 1 mole of H⁺

Acid or Base	Molar mass	Equivalent mass	Relationship between Molarity and Normality
HCl	36.5	36.5/1 = 36.5	1 M = 1 N
H₂SO₄	98	98/2 = 49	1 M = 2 N
H₃PO₄	98	98/3 = 32.6	1 M = 3 N
Ca(OH)₂	74	74/2 = 37	1 M = 2 N

Example

What is the normality of a 2 M H₂CO₃(aq)?

For every 1 mole of H₂CO₃ 2 moles of H⁺ can be donated therefore 1 M = 2 N

2 N

2 M ------- = 4 N

1 M

Problem

What is the molarity of a 3N H₃PO₄(aq)?

TITRATION

The method consists of slowly adding an acid or base of known concentration called the titrant to an unknown acid or base to determine their concentration.

At the equivalence point (the end point or the inflection point) just enough titrant has been added to completely neutralize the unknown acid or base.

The equivalence points can be arrived at using two different methods.

1. Using a pH meter

2. Using an indicator introduced previously in the solution.

A pH curve or titration curve illustrates the titration. The end point in a titration curve is the point at which the curve is the steepest. At equivalence point the solution need not be neutral.

The titration curve of a strong base (the titrant) added to a strong acid of unknown concentration

Example

The titration curve of a strong acid (the titrant) added to a strong base of unknown concentration

Example

The titration curve of a strong base (the titrant) added to a weak acid of unknown concentration

Example

The titration curve of a strong acid (the titrant) added to a weak base of unknown concentration

Example

The titration curve of a strong base (the titrant) added to a strong diprotic acid of unknown concentration

Example

Determining Concentration Using Titration Curve

At the end point the moles of titrant and unknown acid or base are equal, therefore

M_aV_a = M_bV_b

When N_A is different than N_B use the formula N_A V_A = N_B V_B

Problem

What is the volume of 2 M H₃PO₄(aq) needed to neutralize 100.0 mL of 1 M Ca(OH)₂(aq)?

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