Additional Aspects of Aqueous Equilibria

BLB 12th Chapter 17

Buffered Solutions (sections 1-2)Acid/Base Reactions & Titration Curves

(3)Solubility Equilibria (sections 4-5)

Two important points:1. Reactions with strong acids or

strong bases go to completion.2. Reactions with only weak acids and

bases reach an equilibrium.

17.1 The Common Ion EffectWeak acid:HA + H2O ⇌ H3O+ + A-

+Salt of conj. Base:NaA → Na+(aq) + A-(aq)

= two sources of A- Common Ion! What affect does the addition of its conjugate

base have on the weak acid equilibrium? On the pH?

Used in making buffered solutions

Calculate the pH of a solution containing 1.0 M HF and 0.60 M KF. (Recall that the pH of 1.00 HF is 1.58.)

Addition of F- shifts the equilibrium, reducing the [H+].

17.2 Buffered Solutions Resist a change in pH upon the

addition of small amounts of strong acid or strong base

Consist of a weak conjugate acid-base pair

Control pH at a desired level (pKa) Examples: blood (p. 713),

physiological fluids, seawater (p. 728), foods

How do buffers work?

Buffer Calculations

Calculating the pH of a Buffer

][][log

][][loglog]log[

][][][

3

3

acidbasepKpH

AHAKOH

AHAKOH

a

a

a

Henderson-Hasselbalch equation

Calculate the pH of a solution containing 1.0 M HF and 0.60 M KF. (again, but the easy way)

Addition of Strong Acids or Bases to Buffers

Adding acid: H3O+ + HA or A- →

Adding base: OH- + HA or A- →

Calculating pH:1. Stoichiometry of added acid or base2. Equilibrium problem (H-H equation)

Calculate the pH after adding 0.20 mol of HCl to 1.0 L of the 1.0 M HF and 0.60 M KF buffer.

Calculate the pH after adding 0.10 mol of NaOH to 1.0 L of the 1.0 M HF and 0.60 M KF buffer.

Calculate the pH for a 1.0-L solution of 0.25 M NH3 and 0.15 M NH4Br. Kb=1.8x10-5 for NH3

Calculate the pH for a 1.0-L solution of 0.25 M NH3 and 0.15 M NH4Br after the addition of 0.05 mol of RbOH.

Calculate the pH for a 1.0-L solution of 0.25 M NH3 and 0.15 M NH4Br after the addition of 0.35 mol of HCl.

Buffers (wrap up) H-H equation No 5% check When a strong acid or base is

added, start the reaction with that acid or base.

Making buffers of a specific pH? H-H equation

Buffer capacity exceeded – when added acid or base totally consumes a buffer component (p. 726)

How would you prepare a phenol buffer to control pH at 9.50? Ka = 1.3x10-10 for phenol

17.3 Acid-Base Titrations Titration – a reaction used to determine

concentration (acid-base, redox, precipitation) Titrant – solution in buret; usually a strong base

or acid Analyte – solution being titrated; often the

unknown @ equivalence point (or stoichiometric point):

mol acid = mol base Found by titration with an indicator (pp. 721-722) Solution not necessarily neutral pH dependent upon salt formed

pH titration curve – plot of pH vs. titrant volume

Acid-base Titration Reactions and Curves

Type Acid Base

1 strong strong

2 weak strong

3 strong weak

Recognize curve types

Calculate pH at various points on curve.

Type 1: Strong acid + strong base Goes to completion: H3O+ + OH- → 2

H2O Forms a neutral salt Equivalence point - neutral solution,

[H3O+] = 1.0 x 10-7 M, pH = 7.00 pH calculations involve only

stoichiometry and excess H3O+ or OH-

Strong acid – Strong base

Type 1: Strong acid + strong base20.0 mL 0.200 M HClO4 titrated with 0.200 M KOH

mL base

mmol base

added

mmol acid

remain

total mL

[H3O+] pH

0.0010.0020.0030.0040.00

Initial mmol acid =

Another SB/SA titration10.0 mL 0.20 M KOH titrated with 0.10 M HCl

mL acid

mmol acid added

mol base

remain

total mL

[OH-] pH

0.0015.0020.0035.0050.00

Initial mmol base =

Type 2: Weak acid + strong base Titration reaction goes to completion:

HA + OH- → A- + H2O Forms a basic salt (from conj. base of

the weak acid) Equivalence point - basic solution, pH >

7.00 pH calculations involve stoichiometry

and equilibrium

Weak acid – Strong base

Effect of Ka onTitration curve

Polyprotic acid – Strong base

Type 2: Weak acid + strong base25.0 mL 0.100M HC3H5O2 titrated with 0.100 M KOHKa = 1.3x10-5

Calculate the pH at the following points:

A. Initial (0.00 mL KOH)B. 10.00 mL KOHC. Midpoint (12.50 mL KOH)D. Equivalence pt. (25.00 mL KOH)E. 10.00 mL after eq. pt. (35.00 mL

KOH)

Polyprotic Weak acid – Strong base

Type 3: Weak base + strong acid Titration reaction goes to completion:

H3O+ + B → BH+ + H2O Forms an acidic salt (from conj. acid of

the weak base) Equivalence point - acidic solution, pH <

7.00 pH calculations involve stoichiometry

and equilibrium

Strong base

Weak base

Strong base – Strong acidWeak base – Strong acid

Type 3: Weak base + strong acid 25.0 mL 0.150 M NH3 titrated with 0.100 M HCl Kb = 1.8x10-5

Calculate the pH at the following points:

A. Initial (0.00 mL HCl)B. Midpoint (______ mL HCl)C. 25.00 mL HClD. Equivalence pt. (______ mL HCl)E. 10.00 mL after eq. pt. (______ mL

HCl)

Types 2 & 3 pH Calculation Summary

Initial pH – same as weak acid or base problem (chapter 16)

Before equivalence point – Buffer @ midpoint – half of the weak analyte has

been neutralized [weak acid] = [conj. base] or [weak base] = [conj.

acid] [H3O+] = Ka and pH = pKa

@ equivalence point: mol acid = mol base; equilibrium problem with conjugate

Beyond equivalence point – pH based on excess titrant; stoichiometry

Test #2 Summary for Acid/Base problems

1. Weak acid or weak base only (ch. 16)

2. Buffer

3. SA + SB Titration

4. WA + SB or WB + SA Titration

17.4 Solubility Equilibria Solubility – maximum amount of

material that can dissolve in a given amount of solvent at a given temperature; units of g/100 g or M (ch. 13)

Insoluble compound – compound with a solubility less than 0.01 M; also sparingly soluble Solubility rules are given on p. 121 (ch.

4) Dissolution reaches equilibrium in water

between undissolved solid and hydrated ions

Solubility Product Constant, Ksp Equilibrium constant for insoluble

compounds Solid salt nor water included in

expression Appendix D, p. 1063 for values

BaSO4(s) ⇌ Ba2+(aq) + SO42-(aq)

PbCl2(s) ⇌ Pb2+(aq) + 2 Cl-(aq)

Solubility Product Calculations In concentration tables, x =

solubility Problem types

1. solubility → Ksp

2. Ksp → solubility (estimate, see p. 726)

Comparing Salt SolubilitiesGenerally: solubility ↑ Ksp ↑

Can only compare Ksp values if the salts produce the same number of ions

If different numbers of ions are produced, solubility must be compared.

17.5 Factors that Affect Solubility1. Common-Ion Effect

LeChatelier’s Principle revisitedAddition of a product ion causes the solubility of the solid to decrease, but the Ksp remains constant.

2. pH LeChatelier’s Principle again!

Basic salts are more soluble in acidic solution.Acidic salts are more soluble in basic solution.

Environmental example: CaCO3 – limestoneStalactites and stalagmites form due to changing pH in the water and, thus, the solubility of the limestone. (p. 948)

Common-ion Effect

Effect of pH on Solubility