Unit 6 - Chpt 15 - Acid/Base Equilibria

• Common Ion Effect• Buffers / Buffer Capacity• Titration / pH curves• Acid / Base Indicators• HW set1: Chpt 15 - pg. 736-742 # 17, 19, 21, 23,

25, 34, 38, 40, 44 Use Appendix 5 for Ka Kb values -

• HW set2: Chpt 15 #49, 50, 52, 59, 103

Common Ion Effect• Shift in equilibrium position that occurs because of the

addition of an ion already involved in the equilibrium reaction.

• An application of Le Châtelier’s principle.

HCN(aq) + H2O(l) H3O+(aq) + CN-(aq)

• Addition of NaCN will shift the equilibrium to the left because of the addition of CN-, which is already involved in the equilibrium reaction.

• A solution of HCN and NaCN is less acidic than a solution of HCN alone.

Ka Problem with common ion

Calculate the pH of a 0.50 M aqueous solution of the weak acid HF (Ka = 7.2 x 10–4) and 0.10 M NaF (a strong electrolyte).

HF(aq) + H2O H3O+(aq) + F–(aq)Initial 0.50 M ~ 0 0.10MChange –x +x +xEquilibrium0.50 – x x 0.10 + x

[H3O+]= 3.6 x 10–3

pH = 2.44 What was pH before adding NaF?

Buffered Solutions

• Buffered Solution – resists a change in pH.

• They are weak acids or bases containing a common ion.

• After addition of strong acid or base, deal with stoichiometry first, then the equilibrium.

Buffer Problems flow chart

Buffers - How do they work?

Buffers - How do they work? (2)

Henderson–Hasselbalch Equation

• For a particular buffering system (conjugate acid–base pair), all solutions that have the same ratio [A–] / [HA] will have the same pH.

Henderson–Hasselbalch Derivation

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

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

can use this formula for calculating [H+] in buffer solutions

Take -log of everything

pH = pKa - log ( [HA] / [A-] )

pH = pKa + log ([A-] / [HA] ) base / acid

Buffer Problem

What is the pH of a buffer solution that is 0.45 M acetic acid (HC2H3O2) and 0.85 M sodium acetate (NaC2H3O2)? The Ka for acetic acid is 1.8 × 10–5.

pH = 5.02

Buffers graphic

Buffers - Recap

• Buffers contain relatively large amounts of weak acid and corresponding conjugate base.

• Added H+ reacts to completion with the conjugate base.

• Added OH- reacts to completion with the weak acid.

Buffers - Recap (2)

• The pH in the buffered solution is determined by the ratio of the concentrations of the weak acid and weak base. As long as this ratio remains virtually constant, the pH will remain virtually constant. This will be the case as long as the concentrations of the buffering materials (HA and A– or B and BH+) are large compared with amounts of H+ or OH– added.

Buffer Capacity

• The amount of protons or hydroxide ions the buffer can absorb without a significant change in pH.

• Determined by the magnitudes of [HA] and [A–].

• A buffer with large capacity contains large concentrations of the buffering components.

Buffer Capacity• Optimal buffering occurs when [HA] is equal

to [A–].• It is for this condition that the ratio [A–] / [HA]

is most resistant to change when H+ or OH– is added to the buffered solution.

• pKa of the weak acid to be used in the buffer should be as close as possible to the desired pH.

pKa = -log Ka

Titration Curves• Plotting the pH of the

solution being analyzed as a function of the amount of titrant added.

• Equivalence (Stoichiometric) Point – point in the titration when enough titrant has been added to react exactly with the substance in solution being titrated. The pH Curve for the Titration of 50.0 mL

of 0.200 M HNO3 with 0.100 M NaOH

Weak acid / strong base problem

Step 1: A stoichiometry problem (reaction is assumed to run to completion) then determine remaining species.

Step 2: An equilibrium problem (determine position of weak acid equilibrium and calculate pH).

Concept check - example 1

Consider a solution made by mixing 0.10 mol of HCN (Ka = 6.2 x 10–10) with 0.040 mol NaOH in 1.0 L of aqueous solution.

What are the major species immediately upon mixing (that is, before a reaction)?

HCN, Na+, OH–, H2O

Concept Check 1 (cont)

The possibilities for the dominant reaction are:

1. H2O(l) + H2O(l) H3O+(aq) + OH–(aq)

2. HCN(aq) + H2O(l) H3O+(aq) + CN–(aq)

3. HCN(aq) + OH–(aq) CN–(aq) + H2O(l)

– Na+(aq) + OH–(aq) NaOH

1. Na+(aq) + H2O(l) NaOH + H+(aq)

Concept check 1 (cont)

• How do we decide which reaction controls the pH?

H2O(l) + H2O(l) H3O+(aq) + OH–(aq)

HCN(aq) + H2O(l) H3O+(aq) + CN–(aq)

HCN(aq) + OH–(aq) CN–(aq) + H2O(l)

• Check out Ka’s and Kb’s

Concept check 2

Calculate the pH of a solution made by mixing 0.20 mol HC2H3O2 (Ka = 1.8 x 10–5) with 0.030 mol NaOH in 1.0 L of aqueous solution.

Concept check 2 (cont)

• What are the major species in solution?

Na+, OH–, HC2H3O2, H2O

• Why isn’t NaOH a major species?

• Why aren’t H+ and C2H3O2– major

species?

Concept check 2 (cont)

• What are the possibilities for the dominant reaction?

1.H2O(l) + H2O(l) H3O+(aq) + OH–(aq)

2.HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2–(aq)

3.HC2H3O2(aq) + OH–(aq) C2H3O2–(aq) + H2O(l)

4.Na+(aq) + OH–(aq) NaOH

5.Na+(aq) + H2O(l) NaOH + H+(aq)

1.Which of these reactions really occur?

Concept check 2 (cont)

• Which reaction controls the pH?

H2O(l) + H2O(l) H3O+(aq) + OH–(aq)

HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2–(aq)

HC2H3O2(aq) + OH–(aq) C2H3O2–(aq) + H2O(l)

• How do you know? • Ka’s and Kb’s

Concept check 2 (cont)

HC2H3O2(aq) + H2O --> H3O+ + C2H3O2-(aq)

Initial 0.170 M ~0 0.030 MChange –x +x +xEquilibrium0.170 – x x 0.030 + x

Ka = 1.8 x 10–5

pH = 3.99

Can also use pH = pKa + log [A-]/[HA]

Concept check 3

Calculate the pH of a solution at the equivalence point when 100.0 mL of a 0.100 M solution of acetic acid (HC2H3O2), which has a Ka value of 1.8 x 10–5, is titrated with a 0.10 M NaOH solution.

pH = 8.72

Acetic Acid / Strong Base Titration Curve

• The pH Curve for the Titration of 50.0 mL of 0.100 M HC2H3O2 with 0.100 M NaOH

Weak acid / Strong base general titration curve

• The pH Curves for the Titrations of 50.0-mL Samples of 0.10 M Acids with Various Ka Values with 0.10 M NaOH

Weak Base / Strong Acid Titration Curve

• The pH Curve for the Titration of 100.0mL of 0.050 M NH3 with 0.10 M HCl

Molarity as mmol/mL

Molarity M = moles/Liter (each divide/1000)

because the concentrations tend to be low for titration

problems and we titrate small volumes (mL)

Molarity M = mmol/mL

or mmol = Molarity x mL

Titration Table for weak base with strong acid using mmol

Indicators

• Typically a weak acid or weak base• Marks the end point of a titration by

changing color.• The equivalence point is not necessarily

the same as the end point (but they are ideally as close as possible).

Phenolphthalein Indicator

The Acid and Base Forms of the Indicator Phenolphthalein (clear in acid, pink in basic solution)