Ch. 16: Aqueous Ionic Ch. 16: Aqueous Ionic EquilibriumEquilibrium

Dr. Namphol Sinkaset

Chem 201: General Chemistry II

I. Chapter OutlineI. Chapter Outline

I. Introduction

II. Buffers

III. Titrations and pH Curves

IV. Solubility Equilibria and Ksp

V. Complex Ion Equilibria

I. Last Aspects of EquilibriaI. Last Aspects of Equilibria

• In this chapter, we cover some final topics concerning equilibria.

• Buffers are designed to take advantage of Le Châtelier’s Principle.

• Solubility can be reexamined from an equilibrium point of view.

• Complex ions are introduced and their formation explained using equilibrium ideas.

II. pH Resistive SolutionsII. pH Resistive Solutions

• A solution that resists changes in pH by neutralizing added acid or base is called a buffer.

• Many biological processes can only occur within a narrow pH range.

• In humans, the pH of blood is tightly regulated between 7.36 and 7.42.

II. Creating a BufferII. Creating a Buffer

• To resist changes in pH, any added acid or base needs to be neutralized.

• This can be achieved by using a conjugate acid/base pair.

II. How the Buffer WorksII. How the Buffer Works

• For a buffer comprised of the conjugate acid/base pair of acetic acid/acetate: OH-

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

H+(aq) + CH3COO-

(aq) CH3COOH(aq)

• As long as we don’t add too much OH- or H+, the buffer solution will not change pH drastically.

II. Calculating the pH of BuffersII. Calculating the pH of Buffers

• The pH of a buffer can be calculated by approaching the problem as an equilibrium in which there are two initial concentrations.

• Since an acid and its conjugate base are both in solution, problem can be solved from a Ka or Kb point of view.

II. Sample ProblemII. Sample Problem

• A solution was prepared in which [CH3COONa] = 0.11 M and [CH3COOH] = 0.090 M. What is the pH of the solution? Note that the Ka for acetic acid is 1.8 x 10-5.

II. Sample ProblemII. Sample Problem

• A student dissolves 0.12 mole NH3 and 0.095 mole NH4Cl in 250 mL of water. What’s the pH of this buffer solution? Note that the Kb for ammonia is 1.8 x 10-5.

II. A Special Buffer EquationII. A Special Buffer Equation

• Buffers are used so widely that an equation has been developed for it.

• This equation can be used to perform pH calculations of buffers.

• More importantly, it can be used to calculate how to make solutions buffered around a specific pH.

II. Henderson-Hasselbalch Eqn.II. Henderson-Hasselbalch Eqn.

II. The H-H EquationII. The H-H Equation

• The Henderson-Hasselbalch equation works for a buffer comprised of conjugate acid/base pairs.

• It works provided the “x is small” approximation is valid.

II. Sample ProblemII. Sample Problem

• A student wants to make a solution buffered at a pH of 3.90 using formic acid and sodium formate. If the Ka for formic acid is 1.8 x 10-4, what ratio of HCOOH to HCOONa is needed for the buffer?

II. Sample ProblemII. Sample Problem

• A researcher is preparing an acetate buffer. She begins by making 100.0 mL of a 0.010 M CH3COOH solution. How many grams of CH3COONa does she need to add to make the pH of the buffer 5.10 if the Ka for acetic acid is 1.8 x 10-5?

II. Upsetting the BufferII. Upsetting the Buffer

• A buffer resists changes to pH, but it is not immune to change.

• Adding strong acid or strong base will result in small changes of pH.

• We calculate changes in pH of a buffer by first finding the stoichiometric change and then performing an equilibrium calculation.

II. Illustrative ProblemII. Illustrative Problem

• A 1.00 L buffer containing 1.00 M CH3COOH and 1.00 M CH3COONa has a pH of 4.74. It is known that a reaction carried out in this buffer will generate 0.15 mole H+. If the pH must not change by more than 0.2 pH units, will this buffer be adequate?

II. Stoichiometric CalculationII. Stoichiometric Calculation

• When the H+ is formed, it will react stoichiometrically with the base.

• We set up a different type of table to find new equilibrium concentrations.

• We use ≈0.00 mol because [H+] is negligible.

H+ + CH3COO- CH3COOH

Initial 0.15 mol 1.00 mol 1.00 mol

Change -0.15 mol -0.15 mol +0.15 mol

Final ≈0.00 mol 0.85 mol 1.15 mol

II. Equilibrium CalculationII. Equilibrium Calculation

• We use the new buffer concentrations in an equilibrium calculation to find the new pH.

• Again, use ≈0.00 M because [H3O+] is negligible.

CH3COOH + H2O H3O+ + CH3COO-

Initial 1.15 M ---- ≈0.00 M 0.85 M

Change -x ---- +x +x

Equil. 1.15 – x ---- x 0.85 + x

Solve to get [H3O+] = 2.44 x 10-5 M, and pH = 4.613.

II. How the Buffer WorksII. How the Buffer Works

II. A Simplification for BuffersII. A Simplification for Buffers

• In buffer problems, # of moles can be used in place of concentration.

• This can be done because all components are in the same solution, and hence have the same volume.

II. Sample ProblemII. Sample Problem

• Calculate the pH when 10.0 mL of 1.00 M NaOH is added to a 1.0-L buffer containing 0.100 mole CH3COOH and 0.100 mole CH3COONa. Note that the Ka for acetic acid is 1.8 x 10-5.

II. Buffers Using a Weak BaseII. Buffers Using a Weak Base

• Up until now, we’ve been creating buffers with a weak acid and it’s conjugate base.

• Can also make a buffer from a weak base and its conjugate acid.

• Henderson-Hasselbalch still applies, but we need to get Ka of the conjugate acid.

II. pKII. pKaa/pK/pKbb Relationship Relationship

II. Sample ProblemII. Sample Problem

• Calculate the pH of a 1.0-L buffer that is 0.50 M in NH3 and 0.20 M in NH4Cl after 30.0 mL of 1.0 M HCl is added. Note that the Kb for NH3 is 1.8 x 10-5.

II. Making Effective BuffersII. Making Effective Buffers

• There are a few parameters to keep in mind when making a buffer: The relative [ ]’s of acid and conjugate base

should not differ by more than factor of 10. The higher the actual [ ]’s of acid and

conjugate base, the more effective the buffer.

The effective range for a buffer system it +/- 1 pH unit on either side of pKa.

II. Buffer CapacityII. Buffer Capacity

• Buffer capacity is defined as the amount of acid or base that can be added to a buffer without destroying its effectiveness.

• A buffer is destroyed when either the acid or conjugate base is used up.

• Buffer capacity increases w/ higher concentrations of buffer components.