Acid-Base Equilibria and Solubility Equilibria Chapter 16 16.1-16.9.
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Transcript of 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)
Buffer SolutionsBuffer Solutions
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)
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
Acid-Base TitrationsAcid-Base Titrations
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
Strong Acid-Base TitrationsStrong Acid-Base Titrations
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
Strong Acid-Base TitrationsStrong Acid-Base Titrations
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
Strong Acid-Base TitrationsStrong Acid-Base Titrations
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)
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)
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
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 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.
Molar Solubility and SolubilityMolar Solubility and Solubility
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