pH = - log [H+]

or

pH = - log [H3O+]

Example: If [H+] = 1 X 10-10

pH = - log 1 X 10-10

pH = - (- 10)pH = 10

What would be the pH of a 0.000018 M HNO3 solution?

Example: If [H+] = 1.8 X 10-5

pH = - log 1.8 X 10-5

pH = - (- 4.74)pH = 4.74

pH Problems

If the pH of Coke is 3.12, [H+] = ???

[H+] = 10-3.12 = 7.6 x 10-4 M

More pH Problems

Identify as SA, SB, WA, WB

Big or Small K value?

H2O can function as both

an ACID and a BASE.In pure water there can

be AUTOIONIZATION

Pure Water – Acid or Base?

HH22O + HO + H22O O H H33OO++ + OH + OH--

WB SAWA SB

Equilibrium constant for water = KwKw = [H3O+] [OH-] = 1.00 x 10-14 at 25 oC

Kw = [H3O+] [OH-]

In pure water, [H3O+] = [OH-]

so Kw = [x][x] = [x]2

and so, [H3O+] = [OH-] = 1.00 x 10-7

THEREFORE, the pH is…

pH of Pure Water

•What would be the pH of a 1.0 M HCl solution? Of a 0.01 M HCl solution?

[H+] and [OH-] have an inverse relationship in aqueous solution

pOH is from the base perspectivepOH = - log [OH-]

Since we are dealing with aqueous solution…

[H+] [OH-] = 1.00 x 10-14

pH + pOH always equals14

pOH

pH pOH [H+] [OH-] Acid or Base

2.45

4.75

3.5 x 10-10

0.00084

Why is the pH of a 0.1 or 10-1 M acetic acid not 1?

A strong acid ionize 100% but a weak acid does not!

Weak acid has Ka < 1

Leads to small [H3O+]

Strong vs Weak Acids

Weak base has Kb < 1

Leads to small [OH-]

Strong vs Weak Bases

Complete ionization!

Virtually no reactants left

No equilibrium

The equivalence point is the point where the number of moles of base equal the number of moles of acid.

http://www.youtube.com/watch?v=ILn79QpYwPc

http://www.youtube.com/watch?v=HnGy8Um6ibM

http://www.youtube.com/watch?v=H63dHo-T1TM

At equivalence point, the pH > 7

A polyprotic acid has two or more hydrogen which can ionize in multiple steps.

H2CO3 (aq) H+ + HCO3- K1 = 4.5 x10-7

HCO3 (aq) H+ + CO3-2 K2 = 4.7 x10-11

H2CO3 (aq) 2 H+ + CO3-2 Ka = ?

(Overall Ka = K1 x K2)

What is the Ka of the equation below?

H3PO4 (aq) 3 H+ + PO4

-3

H3PO4 H+ + H2PO4

- K1 = 7.1 x 10-3

H2PO4- H+

+ HPO4-2

K2 = 6.3 x 10-8

HPO4-2

H+ + PO4

-3 K3 = 4.5 x 10-

13

When dissolved, the salt created from a strong acid and a weak base will be acidic.

NH4Cl (s) NH4+ + Cl-

Why? NH4

+ + H2O NH4OH + H3O+

The ammonium ion acts as an acid. It will have a Ka value.

conjugate of NH3

conjugate of HCl

When dissolved, the salt created from a strong base and a weak acid will be basic.

NaC2H3O2 (s) Na+ + C2H3O2-

So?C2H3O2

- + H2O HC2H3O2 + OH-

The acetate ion acts as a base. It will have a Kb value.

How will the products react with H+ and OH-?

For an acid-base conjugate pair:

The conjugate of a strong is weak, and the conjugate of a weak is strong. Why?

(Ka)(Kb) = Kw = 1.4 x 10-14

Ka is the acid ionization constantKb is the base ionization constantKw is the ionization constant of water

Acetic acid has a Ka value of 1.7 x 10-

5.

• What is the conjugate base?

• What is the Kb value of the conjugate base?

Carboxyl or

organic acid group

(-COOH)

Methylamine has a Kb value of 4.38 x 10-4.

• What is the conjugate acid?

• What is the Ka value of the conjugate base?

Amine group (-NH2)

Methyl group

(-CH3)

Information Given Concentration of salt Ka or Kb value

Additional Information Needed Identify as Acidic, Basic , or Neutral Equation of ions with water Keq expression

Find the pH of a 0.500 M solution of KCN. The Ka value of HCN is 5.8 x 10-

10.

Dissociation of Salt

KCN + H2O K+ + CN-

Identify base and acid

KOH (strong base) + HCN (weak acid)

The salt must be basic. CN- is a strong conjugate base

Reaction between CN- and water

CN- + H2O OH-

+ HCN

Write Kb expression

Kb = [HCN][OH-] [CN-]

Use Ka of HCN to find Kb of CN-

Ka•Kb = Kw(5.8 x 10-10)•(Kb) = (1.0 x 10-14)Kb = 1.7 x 10-5

Use Kb and Kb expression to solve for [OH-] (set up ice table first)

CN- + H2O OH-

+ HCN

I .500M n/a 0 0

C -x +x +x

E .500 - x x x

1.7 x 10-5 = [x][x] [.500 - x]

X = 2.9 x 10-3

Using [OH-], find pOH and pH

pOH = -log[2.9 x 10-3]pOH = 2.54

pOH + pH = 142.54 + pH = 14pH = 11.46

What would be the pH of a 0.200 M ammonium chloride solution? Kb of ammonia is 1.7 x 10-5.• Write dissociation of ammonium chloride• Write reaction of ammonium and water• Write Ka expression for ammonium• Calculate Ka value using Kb and Kw• Solve for [H+]• Find pH

What would be the pH of a 0.200 M ammonium chloride solution? Kb of ammonia is 1.7 x 10-5.

Ka of NH4+ =

[H3O+] [NH3]

[NH4+]

5.88 x 10-10 =__x2__(.200)

x = 1.08 x 10-5 pH = 4.96

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

A buffer contains:• A weak acid or weak base, AND• The salt of the weak acid or base

NH3 and NH4Cl (Weak Base and Acidic Salt)

HC2H3O2 and NaC2H3O2 (Weak Acid and Basic Salt)

The NH3 (base) will neutralize any acid i.e. HCl (extra H+ ions) by combining with the extra H+ ions to form NH4

+ ions.

The NH4+ (conjugate acid) will

neutralize any base i.e. KOH (extra OH- ions) by donating its H+ ion to form HOH.

HCl + CH3COO- CH3COOH + Cl-

BUFFER: some H+,some C2H3O2-, HC2H3O2 and

Na+,C2H3O2- ADD: HCl

ADD: KOH

KOH + CH3COOH

CH3COOK + HOH

Consider mixture of HC2H3O2 and NaC2H3O2

The common ion (C2H3O2-) suppresses the

ionization of the weak acid. This is called the common ion effect.

NaC2H3O2 (aq) Na+ + C2H3O2-

HC2H3O2 (aq) H+ + C2H3O2

-

Which of the following are buffer systems?

(a) KCl/HF (b) NH4NO3/NH3

(c) KCl/HCl(d) NaHCO3/H2CO3

(e) Ca(OH)2/CaSO4

(f) NH3/HNO2

Which buffers have a pH above 7?

-------------- No Common Ion

--------------- Strong Acid (not weak)

----------------------------- Strong Base (not weak)-------------------- Weak Base and Acid (no salt)

What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK?Ka for HCOOH = 1.8 x 10 -4

Mixture of weak acid and conjugate base!

Initial (M)

Change (M)

Equilibrium (M)

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

-x +x

0.30 - x

+x

x 0.52 + x

0.30 0.00 0.52

x = 1.04 X 10 -4 pH = 4.0

Henderson-Hasselbach equation (on reference sheet)

For weak acid and its salt

For weak base and its salt

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

pOH = pKb + log [HB+][B]

What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK?Ka for HCOOH = 1.8 x 10 -4

pH = 3.77 + log[0.52][0.30] = 4.0

Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system.

The Kb of NH3 is 1.8 x 10-5

NH3 + HOH NH4+ + OH-

What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution?

NH3 + HOH NH4+ + OH-

(.30)(.08) n/a (.36)(.08) (.050)(.020)

I .024 moles .0288 .0010 moles C + .001 - .0010 - .0010 E .025 .0278 0

E . 25 M .278 M

At equivalence point, the pH > 7

Halfway to equivalence, [H+] = Kathis means that the pH = pKa

Prior to equivalence, [H+] > Kathis means that the pH < pKa

Between halfway to equivalence, [H+] < Kathis means that the pH >pKa

Which indicator would you use for a titration of HNO3 with NH3 ?

Strong Acid vs. Strong Base• 100 % ionized! pH = 7 No equilibrium!

Weak Acid vs. Strong Base• Acid is neutralized; Need Kb for conjugate base equilibrium

Strong Acid vs. Weak Base• Base is neutralized; Need Ka for conjugate acid equilibrium

Weak Acid vs. Weak Base• Depends on the strength of each; could be conjugate acid,

conjugate base, or pH 7

These are usually formed from a transition metal surrounded by ligands (polar molecules or negative ions).

As a "rule of thumb" you place twice the number of ligands around an ion as the charge on the ion... example: the dark blue Cu(NH3)4

2+ (ammonia is used as a test for Cu2+ ions), and Ag(NH3)2

+. Memorize the common ligands.

Ligands Names used in the ion

H2O aqua

NH3 ammine

OH- hydroxy

Cl- chloro

Br- bromo

CN- cyano

SCN- thiocyanato (bonded through sulphur) isothiocyanato (bonded through nitrogen)

Names: ligand first, then cationExamples:• tetraamminecopper(II) ion: Cu(NH3)4

2+

• diamminesilver(I) ion: Ag(NH3)2+.

• tetrahydroxyzinc(II) ion: Zn(OH)4 2-

The charge is the sum of the parts (2+) + 4(-1)= -2.

Aluminum also forms complex ions as do some post transitions metals. Ex: Al(H2O)6

3+

Transitional metals, such as Iron, Zinc and Chromium, can form complex ions.

The odd complex ion, FeSCN2+, shows up once in a while

Acid-base reactions may change NH3 into NH4+ (or vice

versa) which will alter its ability to act as a ligand. Visually, a precipitate may go back into solution as a

complex ion is formed. For example, Cu2+ + a little NH4OH will form the light blue precipitate, Cu(OH)2. With excess ammonia, the complex, Cu(NH3)4

2+, forms. Keywords such as "excess" and "concentrated" of

any solution may indicate complex ions. AgNO3 + HCl forms the white precipitate, AgCl. With excess, concentrated HCl, the complex ion, AgCl2-, forms and the solution clears.

Total number of bonds from the ligands to the metal atom.

Coordination numbers generally range between 2 and 12, with 4 (tetracoordinate) and 6 (hexacoordinate) being the most common.

molecular formula

Lewis base/ligand

Lewis acid donor atom

coordination number

Ag(NH3)2+ NH3 Ag+ N 2

[Zn(CN)4]2- CN- Zn2+ C 4

[Ni(CN)4]2- CN- Ni2+ C 4

[PtCl6] 2- Cl- Pt4+ Cl 6

[Ni(NH3)6]2+ NH3 Ni2+ N 6