Additional Aspects of Aqueous Equilibria BLB 11 th Chapter 17.

Post on 24-Dec-2015

218 views 1 download

Tags:

transcript

Additional Aspects of Additional Aspects of Aqueous EquilibriaAqueous Equilibria

BLB 11BLB 11thth Chapter 17 Chapter 17

Buffered Solutions (sections 1-2)Buffered Solutions (sections 1-2)

Acid/Base Reactions & Titration Curves Acid/Base Reactions & Titration Curves (3)(3)

Solubility Equilibria (sections 4-5)Solubility Equilibria (sections 4-5)

Two important points:Two important points:

1.1. Reactions with strong acids or Reactions with strong acids or strong bases go to completion.strong bases go to completion.

2.2. Reactions with only weak acids and Reactions with only weak acids and bases reach an equilibrium.bases reach an equilibrium.

17.1 The Common Ion Effect17.1 The Common Ion Effect

Weak acid:Weak acid:

HA + HHA + H22O ⇌ HO ⇌ H33OO++ + + AA--

++Salt of conj. Base:Salt of conj. Base:

NaA NaA → Na→ Na++(aq) + A(aq) + A--

(aq)(aq)

= two sources of A= two sources of A-- Common Ion!Common Ion!

What affect does the addition of its conjugate What affect does the addition of its conjugate base have on the weak acid equilibrium? On the base have on the weak acid equilibrium? On the pH?pH?

Used in making buffered solutionsUsed in making buffered solutions

Calculate the pH of a 0.60 M HF Calculate the pH of a 0.60 M HF solution. The Ksolution. The Kaa of HF is 7.2×10 of HF is 7.2×10-4.-4.

Calculate the pH of a solution Calculate the pH of a solution containing 0.60 M HF and 1.00 M KF.containing 0.60 M HF and 1.00 M KF.

17.2 Buffered Solutions17.2 Buffered Solutions Resist a change in pH upon the Resist a change in pH upon the

addition of small amounts of strong addition of small amounts of strong acid or strong baseacid or strong base

Consist of a weak conjugate acid-base Consist of a weak conjugate acid-base pairpair

Control pH at a desired level (pControl pH at a desired level (pKKaa)) Examples: blood (p. 729), Examples: blood (p. 729),

physiological fluids, seawater, foodsphysiological fluids, seawater, foods

How do buffers work?How do buffers work?

Calculating pH of a BufferCalculating pH of a Buffer

][

][log

][

][loglog]log[

][

][][

3

3

acid

basepKpH

A

HAKOH

A

HAKOH

a

a

a

Henderson-Hasselbalch equation

Calculate the pH of a solution containing Calculate the pH of a solution containing 0.60 M HF and 1.00 M KF. (0.60 M HF and 1.00 M KF. (again, but the again, but the easy way)easy way)

Adding strong acid or base to a Adding strong acid or base to a bufferbuffer

Adding acid: HAdding acid: H33OO++ + HA or A + HA or A- - →→

Adding base: OHAdding base: OH-- + HA or A + HA or A-- →→

Calculating pH:Calculating pH:

1.1. Stoichiometry of added acid or baseStoichiometry of added acid or base

2.2. Equilibrium problem (H-H equation)Equilibrium problem (H-H equation)

Calculate the pH after adding 0.20 mol of Calculate the pH after adding 0.20 mol of HCl to 1.0 L of the 0.60 M HF and 1.00 M HCl to 1.0 L of the 0.60 M HF and 1.00 M KF buffer.KF buffer.

Calculate the pH after adding 0.10 mol of Calculate the pH after adding 0.10 mol of NaOH to 1.0 L of the 0.60 M HF and 1.00 M NaOH to 1.0 L of the 0.60 M HF and 1.00 M KF buffer.KF buffer.

Calculate the pH for a 1.0-L solution that Calculate the pH for a 1.0-L solution that contains 0.25 M NHcontains 0.25 M NH33 and 0.15 M NH and 0.15 M NH44Br. Br. KKbb=1.8x10=1.8x10-5-5 for NH for NH3 3

Calculate the pH for a 1.0-L solution that Calculate the pH for a 1.0-L solution that contains 0.25 M NHcontains 0.25 M NH33 and 0.15 M NH and 0.15 M NH44Br Br after the addition of 0.05 mol of RbOH.after the addition of 0.05 mol of RbOH.

Calculate the pH for a 1.0-L solution that Calculate the pH for a 1.0-L solution that contains 0.25 M NHcontains 0.25 M NH33 and 0.15 M NH and 0.15 M NH44Br Br after the addition of 0.35 mol of HCl.after the addition of 0.35 mol of HCl.

Buffers (wrap up)Buffers (wrap up) H-H equationH-H equation No 5% checkNo 5% check When When strongstrong acid or base is added, acid or base is added,

start start reactionreaction with that acid or base. with that acid or base. Making buffers of a specific pH? H-H Making buffers of a specific pH? H-H

equationequation Buffer capacity exceeded – when Buffer capacity exceeded – when

added acid or base totally consumes added acid or base totally consumes a buffer component (p. 726)a buffer component (p. 726)

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

17.3 Acid-Base Titrations17.3 Acid-Base Titrations Titration – a reaction used to determine Titration – a reaction used to determine

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

base or acidbase or acid Analyte – solution being titrated; often the Analyte – solution being titrated; often the

unknownunknown @ equivalence point (or stoichiometric point): @ equivalence point (or stoichiometric point):

mol acid = mol basemol acid = mol base Found by titration with an indicatorFound by titration with an indicator Solution not necessarily neutralSolution not necessarily neutral pH dependent upon salt formedpH dependent upon salt formed

pH titration curve – plot of pH vs. titrant pH titration curve – plot of pH vs. titrant volumevolume

Acid-base Titration Reactions and Acid-base Titration Reactions and CurvesCurves

TypeType AcidAcid BaseBase

11 strongstrong strongstrong

22 weakweak strongstrong

33 strongstrong weakweak

Recognize curve Recognize curve typestypes

Calculate pH at Calculate pH at various points on various points on curve.curve.

Type 1: Strong acid + strong baseType 1: Strong acid + strong base

Goes to completionGoes to completion Forms a neutral saltForms a neutral salt Equivalence point - neutral solution, Equivalence point - neutral solution,

[H[H33OO++] = 1.0 x 10] = 1.0 x 10-7-7 M, pH = 7.00 M, pH = 7.00 pH calculations involve only pH calculations involve only

stoichiometry and excess Hstoichiometry and excess H33OO++ and and OHOH--

Strong acid – Strong base

Type 1: Strong acid + strong baseType 1: Strong acid + strong base20.0 mL 0.200 M HClO20.0 mL 0.200 M HClO44 titrated with 0.200 M titrated with 0.200 M KOHKOH

mL mL basebase

mmol mmol base base

addedadded

mmol mmol acid acid

remainremain

total total mLmL

[H[H33OO++]] pHpH

0.000.00

10.0010.00

20.0020.00

30.0030.00

40.0040.00

Initial mmol acid =

Another SA/SB titrationAnother SA/SB titration10.0 mL 0.20 M KOH titrated with 0.10 M 10.0 mL 0.20 M KOH titrated with 0.10 M HClHCl

mL mL acidacid

mmol acid mmol acid addedadded

mol mol base base

remainremain

total total mLmL

[OH[OH--]] pHpH

0.000.00

15.015.000

20.020.000

35.035.000

50.050.000

Initial mmol base =

Type 2: Weak acid + strong baseType 2: Weak acid + strong base

Titration reaction goes to completionTitration reaction goes to completion Forms a basic salt (from conj. base of Forms a basic salt (from conj. base of

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

7.007.00 pH calculations involve stoichiometry pH calculations involve stoichiometry

and equilibriumand equilibrium

Weak acid – Strong base

Type 2: Weak acid + strong baseType 2: Weak acid + strong base25.0 mL 0.100M HC25.0 mL 0.100M HC33HH55OO22 titrated with 0.100 M titrated with 0.100 M KOHKOHKKaa = 1.3x10 = 1.3x10-5-5

Calculate the pH at the following points:Calculate the pH at the following points:

A.A. Initial (0.00 mL KOH)Initial (0.00 mL KOH)

B.B. 10.00 mL KOH10.00 mL KOH

C.C. Midpoint (12.50 mL KOH)Midpoint (12.50 mL KOH)

D.D. Equivalence pt. (25.00 mL KOH)Equivalence pt. (25.00 mL KOH)

E.E. 10.00 mL after eq. pt. (35.00 mL 10.00 mL after eq. pt. (35.00 mL KOH)KOH)

Polyprotic Weak acid – Strong base

Type 3: Weak base + strong acid Type 3: Weak base + strong acid

Titration reaction goes to completionTitration reaction goes to completion Forms an acidic salt (from conj. acid of Forms an acidic salt (from conj. acid of

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

7.007.00 pH calculations involve stoichiometry pH calculations involve stoichiometry

and equilibriumand equilibrium

Strong base

Weak base

Strong base – Strong acidWeak base – Strong acid

Type 3: Weak base + strong acid Type 3: Weak base + strong acid 25.0 mL 0.150 M NH25.0 mL 0.150 M NH33 titrated with 0.100 M HCl titrated with 0.100 M HCl K Kbb = 1.8x10 = 1.8x10-5-5

Calculate the pH at the following points:Calculate the pH at the following points:

A.A. Initial (0.00 mL HCl)Initial (0.00 mL HCl)

B.B. Midpoint (______ mL HCl)Midpoint (______ mL HCl)

C.C. 25.00 mL HCl25.00 mL HCl

D.D. Equivalence pt. (______ mL HCl)Equivalence pt. (______ mL HCl)

E.E. 10.00 mL after eq. pt. (______ mL HCl)10.00 mL after eq. pt. (______ mL HCl)

Types 2 & 3 pH CalculationsTypes 2 & 3 pH Calculations Initial pH – same as weak acid or base problem Initial pH – same as weak acid or base problem

(chapter 16)(chapter 16) Before equivalence point – BufferBefore equivalence point – Buffer @ midpoint – half of the weak analyte has @ midpoint – half of the weak analyte has

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

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

@ equivalence point: mol acid = mol base@ equivalence point: mol acid = mol base Beyond equivalence point – pH based on Beyond equivalence point – pH based on

excess titrantexcess titrant

Test #2 Summary for Acid/Base Test #2 Summary for Acid/Base problemsproblems

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

2.2. BufferBuffer

3.3. SA + SB TitrationSA + SB Titration

4.4. WA + SB or WB + SA TitrationWA + SB or WB + SA Titration

17.4 Solubility Equilibria17.4 Solubility Equilibria

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

Insoluble compoundInsoluble compound – compound with a – compound with a solubility less than 0.01 M; also sparingly solubility less than 0.01 M; also sparingly solublesoluble Solubility rules are given on p. 125 (ch. 4)Solubility rules are given on p. 125 (ch. 4) Dissolution reaches equilibrium in water Dissolution reaches equilibrium in water

between undissolved solid and hydrated ionsbetween undissolved solid and hydrated ions

Solubility Product Constant, Solubility Product Constant, KKspsp

Equilibrium constant for insoluble Equilibrium constant for insoluble compoundscompounds

Solid salt nor water included in expressionSolid salt nor water included in expression Appendix D, p. 1116 for valuesAppendix D, p. 1116 for values

BaSO4(s) ⇌ Ba2+(aq) + SO42-(aq)

PbCl2(s) ⇌ Pb2+(aq) + 2 Cl-(aq)

Solubility Product CalculationsSolubility Product Calculations

In concentration tables, x = In concentration tables, x = solubilitysolubility

Problem typesProblem types1.1. solubility solubility → → KKspsp

2.2. KKspsp → solubility → solubility

Comparing Salt SolubilitiesComparing Salt Solubilities

Generally: solubility Generally: solubility ↑ ↑ KKspsp ↑ ↑

Can only compare Can only compare KKspsp values if the salts values if the salts

produce the same number of ionsproduce the same number of ions If different numbers of ions are produced, If different numbers of ions are produced,

solubility must be compared.solubility must be compared.

17.5 Factors that Affect Solubility17.5 Factors that Affect Solubility

1.1. Common-Ion EffectCommon-Ion Effect LeChatelier’s Principle revisitedLeChatelier’s Principle revisited

Addition of a product ion causes the solubility of Addition of a product ion causes the solubility of the solid to decrease, but the the solid to decrease, but the KKspsp remains constant. remains constant.

2.2. pHpH LeChatelier’s Principle again!LeChatelier’s Principle again!

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

Environmental example: CaCOEnvironmental example: CaCO33 – limestone – limestoneStalactites and stalagmites form due to changing Stalactites and stalagmites form due to changing pH in the water and thus solubility of the pH in the water and thus solubility of the limestone. (p. 964)limestone. (p. 964)