Chapter 16: Aqueous Ionic Equilibria

•Common Ion Effect

•Buffer Solutions

•Titrations

•Solubility

•Precipitation

•Complex Ion Equilibria

Acid Base ReactionsNeutralization Reactions

•Strong base + Strong acidReacts completely (neutral at eq. pt.)

•Strong base + Weak acidAll base reacts (basic at eq. pt.)

•Weak base + Strong acidAll acid reacts (acidic at eq. pt.)

•Equivalence point: equal moles acid and base have reacted

Common Ion Effect HA(aq) + H2O(l) A−

(aq) + H3O+(aq)

• Adding a salt containing the anion, NaA, that is the conjugate base of the acid (the common ion) shifts the position of equilibrium to the left

• Therefore, the pH will be higher than the pH of the acid solutionlowering the H3O+ ion concentration

Calculate the pH of a solution which is 1.0 M HF and 1.0M NaF.

Note: Last chapter we calculated pH of 1.0 M solution of HF only. We calculated the pH to be 1.57.

Buffers

• Buffers: Solutions that resist changes in pH when a small amount of acid or base is added.

• They act by neutralizing the added acid or base

• Many buffers are made by mixing: A weak acid with its conjugate base anion

HF + NaF acidic buffer A weak base with it’s conjugate acid cation

NH3 + NH4+ basic buffer

How Acid Buffers WorkHA(aq) + H2O(l) A−

(aq) + H3O+(aq)

• Acid species (HA) reacts with added base

• Base species (A−) reacts with added acid

• Therefore, small amounts of acid or base are neutralized

H2O

How Buffers Work

HA + H3O+A−A−

AddedH3O+

newHA

HA

H2O

HA

How Buffers Work

HA + H3O+

A−

AddedHO−

newA−

A−

Calculate the pH of a buffer which is 0.50M in acetic acid (CH3CO2H) and 0.50M in sodium acetate (NaCH3CO2).

Calculate the pH after 0.01 mol HCl is added to one liter of the above buffer.

Compare with the pH of one liter of water after 0.01 mol of HCl is added to it.

Henderson-Hasselbalch Equation• Calculating the pH of a buffer solution can be

simplified by using an equation derived from the Ka expression called the Henderson-Hasselbalch Equation

• The equation calculates the pH of a buffer from the Ka and initial concentrations of the weak acid and salt of the conjugate baseNote: the “x is small” approximation must be valid

Deriving the Henderson-Hasselbalch Equation

[ ]

⎟⎟⎠

⎞⎜⎜⎝

⎛=

=

+

+

][A

[HA]]OH[

HA]OH][[A

-3

3-

a

a

K

K

€

pH = log[A−]

[HA]+ pKa

()()()

Do I Use the Full Equilibrium Analysis or the Henderson-Hasselbalch Equation?

• The Henderson-Hasselbalch equation is generally good enough when the “x is small” (conjugate acid/base initial concentrations high, Ka small)

• Rule of thumb: the initial acid and salt concentrations should be over 1000x larger than the value of Ka

Basic BuffersB:(aq) + H2O(l) H:B+

(aq) + OH−(aq)

• buffers can also be made by mixing a weak base, (B:), with a soluble salt of its conjugate acid, H:B+Cl−

H2O(l) + NH3 (aq) NH4+

(aq) + OH−(aq)

The Henderson-Hasselbalch Equation for Basic Buffers

€

Ka =[B][H3O

+]

BH+[ ]

pH = log[B]

[BH+]+ pKa

Calculate the pH of a buffer solution containing 0.25 M NH3 and 0.40 M NH4Cl. Kb(NH3)= 1.8 x10-5

Then, if we add 20mL of 1.0M NaOH to 500mL of the above buffer, what would the final pH be?

Note: the volume of the buffer changes in this problem, so we need to work in moles and calculate new concentrations.

We can predict whether a buffer will be acidic or basic by comparing values of Ka and Kb.

Acetic acid CH3CO2H Ka=1.8 x 10-5

Acetate ion CH3CO2- Kb= 5.6 x 10-10

Ka > Kb so buffer will be acidic

Ammonia NH3 Kb=1.8 x 10-5

Ammonium ion NH4+ Ka= 5.6 x 10-10

Kb > Ka so buffer will be basic