Acid-Base Equilibria and Solubility Equilibria Chapter 16 16.1-16.9

Acid-Base Equilibria and Acid-Base Equilibria and Solubility EquilibriaSolubility Equilibria

Chapter 16Chapter 16

16.1-16.916.1-16.9

Acid-Base Equilibria and Acid-Base Equilibria and Solubility EquilibriaSolubility Equilibria

We will continue to study acid-base We will continue to study acid-base reactionsreactionsBuffersBuffersTitrationsTitrations

We will also look at properties of We will also look at properties of slightly soluble compunds and their slightly soluble compunds and their ions in solutionions in solution

Homogeneous vs. Homogeneous vs. HeterogeneousHeterogeneous

Homogeneous Solution Homogeneous Solution Equilibria- Equilibria- a solution that has the a solution that has the same composition throughout after same composition throughout after equilibrium has been reached.equilibrium has been reached.

Heterogeneous Solution Heterogeneous Solution Equilibria- Equilibria- a solution that after a solution that after equilibrium has been reached, results equilibrium has been reached, results in components in more than one in components in more than one phase.phase.

The Common Ion EffectThe Common Ion Effect

Acid-Base SolutionsAcid-Base SolutionsCommon IonCommon IonThe Common Ion Effect- The Common Ion Effect- the shift the shift

in equilibrium caused by the addition in equilibrium caused by the addition of a compound having an ion in of a compound having an ion in common with the dissolved common with the dissolved substance. substance.

pHpH

The Common Ion EffectThe Common Ion Effect

The presence of a common ion suppresses the ionization of a weak acid or a weak base.

Consider mixture of CH3COONa (strong electrolyte) and CH3COOH (weak acid).

CH3COONa (s) Na+ (aq) + CH3COO- (aq)

CH3COOH (aq) H+ (aq) + CH3COO- (aq)

common ion

Henderson-Hasselbach Henderson-Hasselbach EquationEquation

Relationship between pKRelationship between pKaa and K and Kaa..

pKa = -log Ka

Henderson-Hasselbalch equation

pH = pKa + log[conjugate base]

[acid]

pH CalculationspH Calculations

What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK?

Mixture of weak acid and conjugate base!

HCOOH (aq) H+ (aq) + HCOO- (aq)

Initial (M)

Change (M)

Equilibrium (M)

0.30

-x

0.30 - x

0.00 0.52

+x +x

x 0.52 + x

HCOOH pKa = 3.77

pH CalculationspH Calculations

pH = pKa + log [HCOO-][HCOOH]

pH = 3.77 + log[0.52][0.30]

Common ion effect

0.30 – x 0.30

0.52 + x 0.52

= 4.01

Buffer SolutionsBuffer SolutionsA buffer solution is a solution of:

1. A weak acid or a weak base and

2. The salt of the weak acid or weak base

Both must be present!

A buffer solution has the ability to resist changes in pH upon the addition of small amounts of either acid or base.

Buffer SolutionsBuffer Solutions

IV solutions (~7.4)IV solutions (~7.4)Blood (~7.4)Blood (~7.4)Gastric Juice (~1.5)Gastric Juice (~1.5)


Consider an equal molar mixture of CH3COOH and CH3COONa

Add strong acid

H+ (aq) + CH3COO- (aq) CH3COOH (aq)

Add strong base

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


Effectiveness of Buffer Effectiveness of Buffer SolutionsSolutions

Preparing a Buffer Solution with Preparing a Buffer Solution with a Specific pHa Specific pH

Work BackwardsWork BackwardsChoose a weak acid whose pKa is Choose a weak acid whose pKa is

close to the desired pHclose to the desired pHSubstitute pKa and pH values into Substitute pKa and pH values into

the Henderson-Hasselbach Equationthe Henderson-Hasselbach EquationThis will give a ratio of [conjugate This will give a ratio of [conjugate

Base]/[Acid]Base]/[Acid]Convert ratio to molar quantitiesConvert ratio to molar quantities

Acid-Base TitrationsAcid-Base Titrations

In a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete.

Types of Titrations

Those involving a strong acid and a strong base

Those involving a weak acid and a strong base

Those involving a strong acid and a weak base


Indicator – substance that changes color at (or near) the equivalence point

Slowly add baseto unknown acid

UNTIL

The indicatorchanges color

(pink)

Equivalence point – the point at which the reaction is complete


Strong Acid-Strong Base Strong Acid-Strong Base TitrationsTitrations

NaOH (aq) + HCl (aq) H2O (l) + NaCl (aq)

Strong Acid-Strong Base Titrations

OH- (aq) + H+ (aq) H2O (l)

Strong Acid-Base TitrationsStrong Acid-Base Titrations


Calculate the pH after the addition of 10.0mL of Calculate the pH after the addition of 10.0mL of

0.100M NaOH to 25.0mL of 0.100M HCl.0.100M NaOH to 25.0mL of 0.100M HCl. # moles NaOH# moles NaOH10.0mL 10.0mL xx (0.100mol NaOH/ 1L NaOH) (0.100mol NaOH/ 1L NaOH) xx (1L/ 1000mL) (1L/ 1000mL)

= 1.00 x 10= 1.00 x 10-3-3 mol mol

#moles HCl#moles HCl25.0mL 25.0mL x x (0.100mol HCl/ 1 L HCl) (0.100mol HCl/ 1 L HCl) xx (1L/ 1000mL) (1L/ 1000mL)

= 2.50 x 10= 2.50 x 10-3-3 mol mol

Amount of HCl left after partial neutralizationAmount of HCl left after partial neutralization

2.50 x 102.50 x 10-3-3 mol - 1.00 x 10 mol - 1.00 x 10-3-3 mol = 1.5 x x 10 mol = 1.5 x x 10-3-3 mol mol


Thus, [HThus, [H++] = 1.5 x x 10] = 1.5 x x 10-3-3 mol/ 0.035L mol/ 0.035L

[H[H++] = 0.0429 M] = 0.0429 M

pH = -log [HpH = -log [H++] ]

pH = -log 0.0429pH = -log 0.0429

pH = 1.37pH = 1.37


Calculate pH after the addition of 35.0mL of Calculate pH after the addition of 35.0mL of 0.100M NaOH to 25.0mL of 0.100M HCl.0.100M NaOH to 25.0mL of 0.100M HCl.

# moles NaOH# moles NaOH0.100 mol NaOH / 0.035 L NaOH0.100 mol NaOH / 0.035 L NaOH

= 3.50 x 10= 3.50 x 10-3-3 mol mol#moles HCl#moles HCl0.100 mol HCl / 0.025 L HCl0.100 mol HCl / 0.025 L HCl

= 2.50 x 10= 2.50 x 10-3-3 mol molAmount of NaOH left after full HCl neutralizationAmount of NaOH left after full HCl neutralization3.50 x 103.50 x 10-3-3 mol – 2.50 x 10 mol – 2.50 x 10-3-3 mol = 1.0 x x 10 mol = 1.0 x x 10-3-3 mol mol


Thus, [NaOH] = 1.0 x x 10Thus, [NaOH] = 1.0 x x 10-3-3 mol/ 0.06L mol/ 0.06L[NaOH] = 0.0167 M[NaOH] = 0.0167 M[OH[OH--] = 0.0167 M] = 0.0167 M

pOH = -log [HpOH = -log [H++] ] pOH = -log 0.0167pOH = -log 0.0167pOH = 1.78pOH = 1.78

pH = 14.0-pOHpH = 14.0-pOHpH = 14.0-1.78pH = 14.0-1.78

pH = 12.22pH = 12.22

Weak Acid-Strong Base Weak Acid-Strong Base TitrationsTitrations

Weak Acid-Strong Base TitrationsCH3COOH (aq) + NaOH (aq) CH3COONa (aq) + H2O (l)

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

At equivalence point (pH > 7):

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

Weak Acid-Strong Base Weak Acid-Strong Base TitrationsTitrations

Strong Acid-Weak Base Strong Acid-Weak Base TitrationsTitrations

Strong Acid-Weak Base Titrations

HCl (aq) + NH3 (aq) NH4Cl (aq)

H+ (aq) + NH3 (aq) NH4Cl (aq)

At equivalence point (pH < 7):

NH4+ (aq) + H2O (l) NH3 (aq) + H+ (aq)

Strong Acid-Weak Base Strong Acid-Weak Base TitrationsTitrations

Acid-Base IndicatorsAcid-Base Indicators

Equivalence point occurs when OH- = Equivalence point occurs when OH- = H+ originally present.H+ originally present.

IndicatorsIndicatorsEnd Point- End Point- Occurs when indicator Occurs when indicator

changes colorchanges colorEnd point ~ Equivalence pointEnd point ~ Equivalence point




pH

Solubility EquilibriaSolubility Equilibria

Reactions that produce precipitatesReactions that produce precipitates ImportanceImportance

Tooth Enamel + Acid = tooth decayTooth Enamel + Acid = tooth decayBarium Sulfate = used in x-raysBarium Sulfate = used in x-raysFudge!!!Fudge!!!

Solubility EquilibriaSolubility Equilibria

Solubility Product Constant (KSolubility Product Constant (Kspsp)-)- the the product of the molar concentrations of the product of the molar concentrations of the constituent ions, each raised to the power constituent ions, each raised to the power of its stoichiometric coefficient in the of its stoichiometric coefficient in the equilibrium equation. equilibrium equation.

AgCl (s) Ag+ (aq) + Cl- (aq)

Ksp = [Ag+][Cl-]

Solubility ProductsSolubility Products

What are the correct solubility What are the correct solubility products of the following equations?products of the following equations?

MgF2 (s) Mg2+ (aq) + 2F- (aq) Ksp = [Mg2+][F-]2

Ag2CO3 (s) 2Ag+ (aq) + CO32- (aq) Ksp = [Ag+]2[CO3

2-]

Ca3(PO4)2 (s) 3Ca2+ (aq) + 2PO43- (aq) Ksp = [Ca2+]3[PO3

3-]2

Solubility Product ConstantsSolubility Product Constants

Solubility ConstantsSolubility Constants

What does a large KWhat does a large Kspsp mean? mean?What does a small value mean?What does a small value mean?

Molar Solubility and SolubilityMolar Solubility and Solubility

Molar solubility (mol/L)- is the number of moles of solute dissolved in 1 L of a saturated solution.

Solubility (g/L)- is the number of grams of solute dissolved in 1 L of a saturated solution.



The Ksp of silver bromide is 7.7 x 10The Ksp of silver bromide is 7.7 x 10-13-13. Calculate . Calculate the molar solubility.the molar solubility.

AgBrAgBr(s)(s) ↔ Ag↔ Ag++(aq)(aq) + Br + Br--(aq)(aq)

Initial (M)Initial (M) 0 0 0 0 Change (M)Change (M) -s -s +s +s +s +s Equilibrium (M)Equilibrium (M) s s s s

KKspsp= [Ag= [Ag++][Br][Br--]]

7.7 x 107.7 x 10-13 = -13 = [Ag[Ag++][Br][Br--]]7.7 x 107.7 x 10-13 = -13 = ss22

S= 8.8 x 10S= 8.8 x 10-7-7 M M

KKspsp vs. Q vs. Q

Dissolution of an ionic solid in aqueous solution:

Q < Ksp

Q = Ksp

Q > Ksp

Unsaturated solution

Saturated solution

Supersaturated solution

No precipitate

Precipitate will form

Predicting Precipitation Predicting Precipitation ReactionsReactions

Calculate Q for the reactionCalculate Q for the reaction Is Q larger, smaller or equal to KIs Q larger, smaller or equal to Kspsp??

Separation of Ions by Fractional Separation of Ions by Fractional PrecipitationPrecipitation

Removal of ions from solutionRemoval of ions from solutionUseful in preparation of prescription Useful in preparation of prescription

medicationsmedications Ions can be removed by filtrationIons can be removed by filtration

Fractional PrecipitationFractional Precipitation

Ions + proper reagentIons + proper reagentSmallest Smallest → Largest K→ Largest Kspsp

If AgNo3 is added to a solution containing Cl-, If AgNo3 is added to a solution containing Cl-, Br- and I- ions, which compound will precipitate Br- and I- ions, which compound will precipitate out first?out first?

CompoundCompound KspKsp

AgClAgCl 1.6 x 10 1.6 x 10 -10-10

AgBrAgBr 1.7 x 10 1.7 x 10 -13-13

AgIAgI 8.3 x 10 8.3 x 10 -17-17

The Common Ion Effect and The Common Ion Effect and SolubilitySolubility

What is the molar solubility of AgBr in (a) pure water and (b) 0.0010 M NaBr?

AgBr (s) Ag+ (aq) + Br- (aq)

Ksp = 7.7 x 10-13

s2 = Ksp

s = 8.8 x 10-7

The Common Ion Effect and The Common Ion Effect and SolubilitySolubility

AgBr (s) Ag+ (aq) + Br- (aq)

NaBr (s) Na+ (aq) + Br- (aq)

[Br-] = 0.0010 M

[Ag+] = s

[Br-] = 0.0010 + s 0.0010

Ksp = 0.0010 x s

s = 7.7 x 10-10

pH and SolubilitypH and Solubility

The solubility of many substances The solubility of many substances depends on the pH of the solution. depends on the pH of the solution.

↑ ↑ pHpH↓ ↓ pHpH Insoluble bases dissolve in acidic solutionsInsoluble bases dissolve in acidic solutions Insoluble acids dissolve in basic solutionsInsoluble acids dissolve in basic solutions

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Acid-Base Equilibria and Solubility Equilibria Chapter 16 16.1-16.9