Acids & Bases

Properties of an Acid•Tastes sour•Turns litmus paper red•Has a pH of less than 7• Lemon juice and vinegar are good examples.

Properties of a Base•Turns litmus paper blue•Has a pH greater than 7•taste bitter and have a slippery feel•Bases that are soluble (dissolve in water) are calledalkalis

Most hand soaps and drain cleaners are bases

What is an Acid and a Base?

Key Concepts: Acids and Bases

produce producegives Neutralization

100% Small % 100% Small % product Salt & Water

Acid Base Ionization in water undergo

H + ions OH- ionsH+ OH-

to form

Strongacid

Weakacid

Strongbase

Weakbase

[H+] x [OH-]

is

pH

• Acidic solution [H+ ] > [OH- ]

• Neutral solution [H+ ] = [OH- ]

• Basic solution [H+ ] < [OH- ]

On the pH scale,values below 7 areacidic, a value of 7is neutral, and valuesabove 7 are basic.

How Do We Measure pH?

– Litmus paper• “Red” paper turns blue above ~pH = 8• “Blue” paper turns red below ~pH = 5

– An indicator• Compound that changes color in solution.

Johannes Nicolaus Brønsted (February 22, 1879-December 17, 1947)Danish physical chemist

Svante August Arrhenius (February 19, 1859 – October 2, 1927)Swedish chemist; Nobel Prize in Chemistry, 1903* Arrhenius equation (activation energy)

* Greenhouse effect

Thomas Martin Lowry (October 26, 1874–November 2, 1936)English organic chemistGilbert Newton Lewis (October 23, 1875-March 23, 1946)American physical chemist

Acids & BasesAcids & Bases

Acids and Bases

Arrhenius

Acids producehydrogen ions inaqueous solutions

Bases producehydroxide ions

Acids are proton(H+) donors

Bases are protonacceptors

Strengths ofacids & bases

pH = -log [H+]

Lewis

Acids accept apair of electronsto form acovalent bond

Bronsted -Lowry pH

Some Definitions

• Arrhenius acids and bases– Acid: Substance that, when dissolved in

water, increases the concentration ofhydrogen ions (protons, H+).

– Base: Substance that, when dissolved inwater, increases the concentration ofhydroxide ions.

Some Definitions• Brønsted–Lowry: must have both

1. an Acid: Proton donor

and

2. a Base: Proton acceptor

The Brønsted-Lowry acid donates a proton,

while the Brønsted-Lowry base accepts it.

Brønsted-Lowry acids and bases are always paired.

Which is the acid and which is the base in each of these rxns?

A Brønsted–Lowry acid…

…must have a removable (acidic) proton.

HCl, H2O, H2SO4

A Brønsted–Lowry base…

…must have a pair of nonbonding electrons.

NH3, H2O

O

H

H + O

H

H O

H

H H OH-+[ ] +

Acid-Base Properties of Water

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

H2O + H2O H3O+ + OH-

acid conjugatebase

base conjugateacid

autoionization of water

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

The Ion Product of Water

Kc =[H+][OH-]

[H2O][H2O] = constant

Kc[H2O] = Kw = [H+][OH-]

The ion-product constant (Kw) is the product of the molarconcentrations of H+ and OH- ions at a particular temperature.

At 250CKw = [H+][OH-] = 1.0 x 10-14

[H+] = [OH-]

[H+] > [OH-]

[H+] < [OH-]

Solution Is

neutral

acidic

basic

What is the concentration of OH- ions in a HCl solutionwhose hydrogen ion concentration is 1.3 M?

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

[H+] = 1.3 M

[OH-] =Kw

[H+]

1 x 10-14

1.3= = 7.7 x 10-15 M

pH – A Measure of Acidity

pH = -log [H+]

[H+] = [OH-]

[H+] > [OH-]

[H+] < [OH-]

Solution Is

neutral

acidic

basic

[H+] = 1 x 10-7

[H+] > 1 x 10-7

[H+] < 1 x 10-7

pH = 7

pH < 7

pH > 7

At 250C

pH [H+]

pOH = -log [OH-]

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

-log [H+] – log [OH-] = 14.00

pH + pOH = 14.00

The pH of rainwater collected in a certain region of thenortheastern United States on a particular day was 4.82.What is the H+ ion concentration of the rainwater?

pH = -log [H+]

[H+] = 10-pH = 10-4.82 = 1.5 x 10-5 M

The OH- ion concentration of a blood sample is 2.5 x 10-7 M.What is the pH of the blood?

pH + pOH = 14.00

pOH = -log [OH-] = -log (2.5 x 10-7) = 6.60

pH = 14.00 – pOH = 14.00 – 6.60 = 7.40

Strong Electrolyte – 100% dissociation

NaCl (s) Na+ (aq) + Cl- (aq)H2O

Weak Electrolyte – not completely dissociated

CH3COOH CH3COO- (aq) + H+ (aq)

Strong Acids are strong electrolytes

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

HNO3 (aq) + H2O (l) H3O+ (aq) + NO3- (aq)

HClO4 (aq) + H2O (l) H3O+ (aq) + ClO4- (aq)

H2SO4 (aq) + H2O (l) H3O+ (aq) + HSO4- (aq)

HF (aq) + H2O (l) H3O+ (aq) + F- (aq)

Weak Acids are weak electrolytes

HNO2 (aq) + H2O (l) H3O+ (aq) + NO2- (aq)

HSO4- (aq) + H2O (l) H3O+ (aq) + SO4

2- (aq)

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

Strong Bases are strong electrolytes

NaOH (s) Na+ (aq) + OH- (aq)H2O

KOH (s) K+ (aq) + OH- (aq)H2O

Ba(OH)2 (s) Ba2+ (aq) + 2OH- (aq)H2O

F- (aq) + H2O (l) OH- (aq) + HF (aq)

Weak Bases are weak electrolytes

NO2- (aq) + H2O (l) OH- (aq) + HNO2 (aq)

Conjugate acid-base pairs:

• The conjugate base of a strong acid has no measurablestrength.

• H3O+ is the strongest acid that can exist in aqueoussolution.

• The OH- ion is the strongest base that can exist in aqeoussolution.

Strong Acid Weak Acid

How Do We Measure pH?

pH meters

measure the voltage inthe solution

What is the pH of a 2 x 10-3 M HNO3 solution?

HNO3 is a strong acid – 100% dissociation.

HNO3 (aq) + H2O (l) H3O+ (aq) + NO3- (aq)

pH = -log [H+] = -log [H3O+] = -log(0.002) = 2.7

Start

End

0.002 M

0.002 M 0.002 M0.0 M

0.0 M 0.0 M

What is the pH of a 1.8 x 10-2 M Ba(OH)2 solution?

Ba(OH)2 is a strong base – 100% dissociation.

Ba(OH)2 (s) Ba2+ (aq) + 2OH- (aq)

Start

End

0.018 M

0.018 M 0.036 M0.0 M

0.0 M 0.0 M

pH = 14.00 – pOH = 14.00 + log(0.036) = 12.56

HA (aq) + H2O (l) H3O+ (aq) + A- (aq)

Weak Acids (HA) and Acid Ionization Constants

HA (aq) H+ (aq) + A- (aq)

Ka =[H+][A-]

[HA]

Ka is the acid ionization constant

Kaweak acidstrength

What is the pH of a 0.5 M HF solution (at 250C)?

HF (aq) H+ (aq) + F- (aq) Ka =[H+][F-]

[HF]= 7.1 x 10-4

HF (aq) H+ (aq) + F- (aq)

Initial (M)

Change (M)

Equilibrium (M)

0.50 0.00

-x +x

0.50 - x

0.00

+x

x x

Ka =x2

0.50 - x= 7.1 x 10-4

Ka ≈x2

0.50= 7.1 x 10-4

0.50 – x ≈ 0.50Ka << 1

x2 = 3.55 x 10-4 x = 0.019 M

[H+] = [F-] = 0.019 M pH = -log [H+] = 1.72

[HF] = 0.50 – x = 0.48 M

When can I use the approximation?

0.50 – x ≈ 0.50Ka << 1

When x is less than 5% of the value from which it is subtracted.

x = 0.0190.019 M0.50 M

x 100% = 3.8%Less than 5%

Approximation ok.

What is the pH of a 0.05 M HF solution (at 250C)?

Ka ≈x2

0.05= 7.1 x 10-4 x = 0.006 M

0.006 M0.05 M

x 100% = 12%More than 5%

Approximation not ok.

Must solve for x exactly using quadratic equation or method ofsuccessive approximation.

Solving weak acid ionization problems:

1. Identify the major species that can affect the pH.

• In most cases, you can ignore the autoionization ofwater.

• Ignore [OH-] because it is determined by [H+].

2. Use ICE to express the equilibrium concentrations in termsof single unknown x.

3. Write Ka in terms of equilibrium concentrations. Solve for xby the approximation method. If approximation is not valid,solve for x exactly.

4. Calculate concentrations of all species and/or pH of thesolution.

What is the pH of a 0.122 M monoprotic acid whoseKa is 5.7 x 10-4?

HA (aq) H+ (aq) + A- (aq)

Initial (M)

Change (M)

Equilibrium (M)

0.122 0.00

-x +x

0.122 - x

0.00

+x

x x

Ka =x2

0.122 - x= 5.7 x 10-4

Ka ≈x2

0.122= 5.7 x 10-4

0.122 – x ≈ 0.122Ka << 1

x2 = 6.95 x 10-5 x = 0.0083 M

0.0083 M0.122 M

x 100% = 6.8%More than 5%

Approximation not ok.

Ka =x2

0.122 - x= 5.7 x 10-4 x2 + 0.00057x – 6.95 x 10-5 = 0

ax2 + bx + c =0-b ± b2 – 4ac √

2ax =

x = 0.0081 x = - 0.0081

HA (aq) H+ (aq) + A- (aq)

Initial (M)

Change (M)

Equilibrium (M)

0.122 0.00

-x +x

0.122 - x

0.00

+x

x x

[H+] = x = 0.0081 M pH = -log[H+] = 2.09

The common ion effect is the shift in equilibrium caused by theaddition of a compound having an ion in common with thedissolved substance.

The presence of a common ion suppressesthe ionization of a weak acid or a weak base.

Consider mixture of CH3COONa (strong electrolyte) andCH3COOH (weak acid).

CH3COONa (s) Na+ (aq) + CH3COO- (aq)

CH3COOH (aq) H+ (aq) + CH3COO- (aq)

commonion

Consider mixture of salt NaA and weak acid HA.

HA (aq) H+ (aq) + A- (aq)

NaA (s) Na+ (aq) + A- (aq)

Ka =[H+][A-]

[HA]

[H+] =Ka [HA]

[A-]

-log [H+] = -log Ka - log[HA][A-]

-log [H+] = -log Ka + log[A-][HA]

pH = pKa + log[A-][HA]

pKa = -log Ka

Henderson-Hasselbalchequation

pH = pKa + log[conjugate base]

[acid]

What is the pH of a solution containing 0.30 M HCOOHand 0.52 M HCOOK?

HCOOH (aq) H+ (aq) + HCOO- (aq)

Initial (M)

Change (M)

Equilibrium (M)

0.30 0.00

-x +x

0.30 - x

0.52

+x

x 0.52 + x

Common ion effect

0.30 – x ≈ 0.30

0.52 + x ≈ 0.52

pH = pKa + log[HCOO-][HCOOH]

HCOOH pKa = 3.77

pH = 3.77 + log[0.52][0.30]

= 4.01

Mixture of weak acid and conjugate base!

A buffer solution is a solution of:

1. A weak acid or a weak base and

2. The salt of the weak acid or weak base

Both must be present!

A buffer solution has the ability to resist changes in pH uponthe addition of small amounts of either acid or base.

Add strong acid

H+ (aq) + CH3COO- (aq) CH3COOH (aq)

Add strong base

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

Consider an equal molar mixture of CH3COOH and CH3COONa

HCl H+ + Cl-

HCl + CH3COO- CH3COOH + Cl-

Which of the following are buffer systems? (a) KF/HF(b) KBr/HBr, (c) Na2CO3/NaHCO3

(a) KF is a weak acid and F- is its conjugate basebuffer solution

(b) HBr is a strong acidnot a buffer solution

(c) CO32- is a weak base and HCO3

- is it conjugate acidbuffer solution

TitrationsIn a titration a solution of accurately known concentration isadded gradually added to another solution of unknownconcentration until the chemical reaction between the twosolutions is complete.

Equivalence point – the point at which the reaction is complete

Indicator – substance that changes color at (or near) the equivalence point

Slowly add baseto unknown acid

UNTIL

The indicatorchanges color

(pink)

Strong Acid-Strong Base Titrations

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

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

Weak Acid-Strong Base Titrations

CH3COOH (aq) + NaOH (aq) CH3COONa (aq) + H2O (l)

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

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

At equivalence point (pH > 7):

Strong Acid-Weak Base Titrations

HCl (aq) + NH3 (aq) NH4Cl (aq)

NH4+ (aq) + H2O (l) NH3 (aq) + H+ (aq)

At equivalence point (pH < 7):

H+ (aq) + NH3 (aq) NH4Cl (aq)

Exactly 100 mL of 0.10 M HNO2 are titrated with a 0.10 MNaOH solution. What is the pH at the equivalence point ?

HNO2 (aq) + OH- (aq) NO2- (aq) + H2O (l)

start (moles)

end (moles)

0.01 0.01

0.0 0.0 0.01

NO2- (aq) + H2O (l) OH- (aq) + HNO2 (aq)

Initial (M)

Change (M)

Equilibrium (M)

0.05 0.00

-x +x

0.05 - x

0.00

+x

x x

[NO2-] =

0.010.200 = 0.05 MFinal volume = 200 mL

Kb =[OH-][HNO2]

[NO2-]

=x2

0.05-x= 2.2 x 10-11

0.05 – x ≈ 0.05 x ≈ 1.05 x 10-6 = [OH-]

pOH = 5.98

pH = 14 – pOH = 8.02

Acid-Base Indicators

HIn (aq) H+ (aq) + In- (aq)

≥ 10[HIn][In-]

Color of acid (HIn) predominates

≤ 10[HIn][In-]

Color of conjugate base (In-) predominates

The titration curve of a strong acid with a strong base.

Which indicator(s) would you use for a titration of HNO2

with KOH ?

Weak acid titrated with strong base.

At equivalence point, will have conjugate base of weak acid.

At equivalence point, pH > 7

Use cresol red or phenolphthalein

Solubility Equilibria

AgCl (s) Ag+ (aq) + Cl- (aq)

Ksp = [Ag+][Cl-] Ksp is the solubility product constant

MgF2 (s) Mg2+ (aq) + 2F- (aq) Ksp = [Mg2+][F-]2

Ag2CO3 (s) 2Ag+ (aq) + CO32- (aq) Ksp = [Ag+]2[CO3

2-]

Ca3(PO4)2 (s) 3Ca2+ (aq) + 2PO43- (aq) Ksp = [Ca2+]3[PO3

3-]2

Dissolution of an ionic solid in aqueous solution:

Q = Ksp Saturated solution

Q < Ksp Unsaturated solution No precipitate

Q > Ksp Supersaturated solution Precipitate will form

Molar solubility (mol/L) is the number of moles of solutedissolved in 1 L of a saturated solution.

Solubility (g/L) is the number of grams of solute dissolved in1 L of a saturated solution.

What is the solubility of silver chloride in g/L ?

AgCl (s) Ag+ (aq) + Cl- (aq)

Ksp = [Ag+][Cl-]Initial (M)

Change (M)

Equilibrium (M)

0.00

+s

0.00

+s

s s

Ksp = s2

s = Ksp√s = 1.3 x 10-5

[Ag+] = 1.3 x 10-5 M [Cl-] = 1.3 x 10-5 M

Solubility of AgCl = 1.3 x 10-5 mol AgCl

1 L soln143.35 g AgCl

1 mol AgClx = 1.9 x 10-3 g/L

Ksp = 1.6 x 10-10

If 2.00 mL of 0.200 M NaOH are added to 1.00 L of0.100 M CaCl2, will a precipitate form?

The ions present in solution are Na+, OH-, Ca2+, Cl-.

Only possible precipitate is Ca(OH)2 (solubility rules).

Is Q > Ksp for Ca(OH)2?

[Ca2+]0 = 0.100 M [OH-]0 = 4.0 x 10-4 M

Ksp = [Ca2+][OH-]2 = 8.0 x 10-6

Q = [Ca2+]0[OH-]02 = 0.10 x (4.0 x 10-4)2 = 1.6 x 10-8

Q < Ksp No precipitate will form