aspects of aqueous equilibria. aspects of aqueous equilibria: the common ion effect recall that...
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Aspects of Aqueous EquilibriaAspects of Aqueous Equilibria
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Aspects of Aqueous Equilibria: Aspects of Aqueous Equilibria: The Common Ion EffectThe Common Ion Effect
Recall that salts like sodium acetate are strong electrolytes
NaC2H3O2(aq) Na+(aq) + C2H3O2-(aq)
…and that the C2H3O2- ion is a conjugate base of a weak acid
HC2H3O2(aq) + H2O H3O+(aq) + C2H3O2-(aq)
Ka = [H3O+] [C2H3O2
-]
[HC2H3O2]
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Aspects of Aqueous Equilibria: Aspects of Aqueous Equilibria: The Common Ion EffectThe Common Ion Effect
Ka = [H3O+] [C2H3O2
-]
[HC2H3O2]
Now, lets think about the problem from the perspective of
LeChatelier’s PrincipleLeChatelier’s PrincipleWhat would happen if the concentration of the acetate ion were increased?
Q > K and the reaction favors reactant
Addition of CAddition of C22HH33OO22-- shifts equilibrium, shifts equilibrium,
reducing Hreducing H++
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HC2H3O2(aq) + H2O H3O+(aq) + C2H3O2-(aq)
Since the equilibrium has shifted to favor the reactant, it would appear as if the dissociation of the weak acid(weak electrolyte) had decreased.
Aspects of Aqueous Equilibria: Aspects of Aqueous Equilibria: The Common Ion EffectThe Common Ion Effect
Ka = [H3O+] [C2H3O2
-]
[HC2H3O2]
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HC2H3O2(aq) + H2O H3O+(aq) + C2H3O2-(aq)
So where might the additional C2H3O2-(aq) come from?
Remember we are not adding H+. So it’s not like we can add more acetic acid.
Aspects of Aqueous Equilibria: Aspects of Aqueous Equilibria: The Common Ion EffectThe Common Ion Effect
How about from the sodium acetate?
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NaC2H3O2(aq) Na+(aq) + C2H3O2-(aq)
HC2H3O2(aq) + H2O H3O+(aq) + C2H3O2-(aq)
Aspects of Aqueous Equilibria: Aspects of Aqueous Equilibria: The Common Ion EffectThe Common Ion Effect
In general, the In general, the dissociation of a weak electrolyte (acetic acid)dissociation of a weak electrolyte (acetic acid)is decreased by adding to the solution a is decreased by adding to the solution a strong electrolyte thatstrong electrolyte thathas an ion in commonhas an ion in common with the weak electrolyte with the weak electrolyte
The shift in equilibrium which occurs is called theCOMMON ION EFFECTCOMMON ION EFFECT
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Aspects of Aqueous Equilibria: Aspects of Aqueous Equilibria: The Common Ion EffectThe Common Ion Effect
Let’s explore the COMMON ION EFFECTCOMMON ION EFFECT in a little more detailSuppose that we add 8.20 g or 0.100 mol sodium acetate, NaC2H3O2, to 1 L of a 0.100 M solution of acetic acetic acid, HC2H3O2. What is the pH of the resultant solution?
HC2H3O2(aq) + H2O H3O+(aq) + C2H3O2-(aq)
NaC2H3O2(aq) Na+(aq) + C2H3O2-(aq)
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Now you try it!
Aspects of Aqueous Equilibria: Aspects of Aqueous Equilibria: The Common Ion EffectThe Common Ion Effect
Calculate the pH of a solution containing 0.06 M formic acid (HCH2O, Ka = 1.8 x 10-4)and 0.03 M potassium formate, KCH2O.
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Calculate the fluoride ion concentration and pH of a solution containing 0.10 mol of
HClHCl and 0.20 mol HF in 1.0 L
Note who the strong electrolyte is this time!
Aspects of Aqueous Equilibria: Aspects of Aqueous Equilibria: The Common Ion EffectThe Common Ion Effect
HF(aq) + H2O H3O+(aq) + F-(aq)
HCl+ H2O(aq) H3O+(aq) + Cl-(aq)
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Aspects of Aqueous Equilibria: Aspects of Aqueous Equilibria: The Common Ion Effect The Common Ion Effect
Kb = [NH4
+] [OH-]
[NH3]
Now, lets think about the problem from the perspective of
LeChatelier’s PrincipleLeChatelier’s PrincipleBut this time lets deal with a weak base and a salt containing its conjugate acid.
Q > K and the reaction favors reactant
Addition of NHAddition of NH44+ shifts equilibrium,+ shifts equilibrium,
reducing OHreducing OH--
NH3(aq) + H2O NH4+(aq) + OH-
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Calculate the pH of a solution produced by mixing 0.10 mol NH4Cl with 0.40 L of 0.10 M NH3(aq), pKb = 4.74?
NH3(aq) + H2O NH4+(aq) + OH-
Aspects of Aqueous Equilibria: Aspects of Aqueous Equilibria: The Common Ion Effect The Common Ion Effect
NH4Cl(aq) NH4+(aq) + Cl- (aq)
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Aspects of Aqueous Equilibria: Common IonsAspects of Aqueous Equilibria: Common IonsGenerated by Acid-Base ReactionsGenerated by Acid-Base Reactions
The common ion that affects a weak-acid or weak-base equilibrium may be present because it is added as a salt, or the common ion can be generated by reacting an acid and base directly (no salt would be necessary….which is kind of convenient if you think about it)
Let’s take a look at a weak acid-strong base combination first:
HC2H3O2(aq) + OH- H2O + C2H3O2
Suppose we react 0.20 mol of acetic acid with 0.10 mol of sodium hydroxide
0.20 mol 0.10 mol
-0.10 mol -0.10 mol-0.10 mol
0
00.10 mol 0.10 mol
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HC2H3O2(aq) + OH- H2O + C2H3O2
Suppose we react 0.20 mol of acetic acid with 0.10 mol of sodium hydroxide
0.20 mol 0.10 mol
-0.10 mol 0.10 mol-0.10 mol
0
00.10 mol 0.10 mol
HC2H3O2(aq) + H2O H3O+(aq) + C2H3O2-(aq)
0.10 M0.10 M
Let’s suppose that all this is occurring in 1.0 L of solution
Aspects of Aqueous Equilibria: Common IonsAspects of Aqueous Equilibria: Common IonsGenerated by Acid-Base ReactionsGenerated by Acid-Base Reactions
0
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Aspects of Aqueous Equilibria: Common IonsAspects of Aqueous Equilibria: Common IonsGenerated by Acid-Base ReactionsGenerated by Acid-Base Reactions
Sample problem: Calculate the pH of a solution produced by mixing 0.60 L of 0.10 MNH4Cl with 0.40 L of 0.10 M NaOH
NH4Cl(aq) NH4+(aq) + Cl- (aq)
NH4+ + OH- NH3 + H2O
0.06 mol 0.04 mol
-0.04 mol 0.04 mol
0
0.04 mol0
-0.04 mol
0.02 mol
Don’t forgetto convert to
MOLARITIESMOLARITIES
0.02 M
NH4+ + H2O H3O+ + NH3
0.04 M0
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Now you try it!
Aspects of Aqueous Equilibria: Common IonsAspects of Aqueous Equilibria: Common IonsGenerated by Acid-Base ReactionsGenerated by Acid-Base Reactions
Calculate the pH of a solution formed by mixing 0.50 L of 0.015 M NaOH with0.50 L of 0.30 M benzoic acid (HC7H5O2, Ka = 6.5 x 10-5)
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[H +] = [HX]
[X-]Ka
Aspects of Aqueous Equilibria: Aspects of Aqueous Equilibria: BUFFERED SOLUTIONSBUFFERED SOLUTIONS
A buffered solution is a solution that resists change in pH upon addition of smallamounts of acid or base.
HERE’S HOW IT WORKS !HERE’S HOW IT WORKS !
Suppose we have a salt: MX M +(aq) + X- (aq)
And we’ve added the salt to a weak acidcontaining the same conjugate base as the salt, HX:
HX +H2O H3O+ + X-
And the equilibrium expression for thisreaction is
Ka = [H+ ] [ X-]
[HX]
Note that the concentration of the H+ isdependent upon the Ka and the ratio between the HX and X- (the conjugateacid-base pair)
CONTINUED ON NEXT SLIDECONTINUED ON NEXT SLIDE
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Aspects of Aqueous Equilibria: Aspects of Aqueous Equilibria: BUFFERED SOLUTIONSBUFFERED SOLUTIONS
Two important characteristics of a buffer are buffering capacity and pH. Bufferingcapacity is the amount of acid or base the buffer can neutralize before the pH beginsto change to an appreciable degree.
•This capacity depends on the amount of acid and base from which thebuffer is made
•The greater the amounts of the conjugate acid-base pair, the more resistant the ratio of their concentrations, and hence the pH, to change
•The pH of the buffer depends upon the Ka
[H +] = [HX]
[X-]Ka
-log[H +] = [HX]
[X-] -log Ka
pH = pKa - log [HX]
[X-]
Henderson-Hasselbalch Equation
pH = pKa + log [X-]
[HX]
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HC2H3O2(aq) + H2O H3O+(aq) + C2H3O2-(aq)
NaC2H3O2(aq) Na+(aq) + C2H3O2-(aq)
0.1 M 0 0.1
0.1 M 0.1 M 0.1 M
-x x x
x0.1 - x 0.1 + x
1.8 x 10-5 = x(0.1 + x )
0.1 - x
x = 1.8 x 10-5
pH = 4.74
Henderson-Hasselbalch Equation
pH = 4.74 + log [.1]
[.1]
Aspects of Aqueous Equilibria: Aspects of Aqueous Equilibria: BUFFERED SOLUTIONSBUFFERED SOLUTIONS
Note that these are initialconcentrations
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Aspects of Aqueous Equilibria: Aspects of Aqueous Equilibria: BUFFERED SOLUTIONSBUFFERED SOLUTIONS
A liter of solution containing 0.100 mol of HC2H3O2 and 0.100 mol NaC2H3O2
forms a buffered solution of pH 4.74. Calculate the pH of this solution (a) afterafter 0.020 mol NaOH is added0.020 mol NaOH is added, (b) after adding 0.020 mol HCladding 0.020 mol HCl is added.
HC2H3O2(aq) + OH- H2O + C2H3O2-(aq)
0.1 M 0.02 M0.02 M
-0.02 M-0.02 M-0.02 M-0.02 M
0.1 M
NaC2H3O2(aq) Na+(aq) + C2H3O2-(aq)
0.1 M 0.1 M 0.1 M
0.02 M0.02 M
0.12 M0.00 M0.00 M0.08 M
Henderson-Hasselbalch Equation
pH = 4.74 + log [.12]
[.08]
Note that these are initialconcentrations
pH = 4.92
Step 1
Step 2
Note that the OH- reacts with the HX
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Aspects of Aqueous Equilibria: Aspects of Aqueous Equilibria: BUFFERED SOLUTIONSBUFFERED SOLUTIONS
A liter of solution containing 0.100 mol of HC2H3O2 and 0.100 mol NaC2H3O2
forms a buffered solution of pH 4.74. Calculate the pH of this solution (a) after 0.020 mol NaOH0.020 mol NaOH is added, (b) after adding 0.020 mol HCl is addedafter adding 0.020 mol HCl is added.
C2H3O2-(aq) + H+ HC2H3O2
Step 1
0.10 M 0.10 M0.02 M0.02 M
-0.02 M-0.02 M
0.00 M0.00 M
-0.02 M-0.02 M 0.02 M0.02 M
0.12 M0.08M
Henderson-Hasselbalch Equation
pH = 4.74 + log [.08]
[.12]
Note that these are initialconcentrations
pH = 4.56
Step 2
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Now consider, for a moment, what would have happened if I had added 0.020 molof NaOH or 0.02 mol HCl to .1 M HC2H3O2.
HC2H3O2(aq) + H2O H3O+(aq) + C2H3O2-(aq)
0.1 M 0 0
-x x x
x0.1 - x x
1.8 x 10-5 = x2
0.1 - x
x = 0.0013
pH = 2.9
Aspects of Aqueous Equilibria: Aspects of Aqueous Equilibria: BUFFERED SOLUTIONSBUFFERED SOLUTIONS
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HC2H3O2(aq) + OH- H2O + C2H3O2-(aq)
0.1 M 0.02 M0.02 M
-0.02 M-0.02 M-0.02 M-0.02 M
0.00
0.02 M0.02 M
0.02 M0.00 M0.00 M0.08 M
Henderson-Hasselbalch Equation
pH = 4.74 + log [.02]
[.08]
pH = 4.13
Aspects of Aqueous Equilibria: Aspects of Aqueous Equilibria: BUFFERED SOLUTIONSBUFFERED SOLUTIONS
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C2H3O2-(aq) + H+ HC2H3O2
0.00 0.10 M0.02 M0.02 M
pH = 1.7
Aspects of Aqueous Equilibria: Aspects of Aqueous Equilibria: BUFFERED SOLUTIONSBUFFERED SOLUTIONS
Complete dissociation
pH = -log [0.02]
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Aspects of Aqueous Equilibria: Aspects of Aqueous Equilibria: BUFFERED SOLUTIONSBUFFERED SOLUTIONS
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Aspects of Aqueous Equilibria: Aspects of Aqueous Equilibria: BUFFERED SOLUTIONSBUFFERED SOLUTIONS
Sample exercise: Consider a litter of buffered solution that is 0.110 M in formic acid(HC2H3O )and 0.100 M in sodium formate (NaC2H3O ). Calculate the pH of the buffer(a) before any acid or base are added, (b) after the addition of 0.015 mol HNO3, (3) afterthe addition of 0.015 mol KOH
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Aspects of Aqueous Equilibria: Aspects of Aqueous Equilibria: Titration CurvesTitration Curves
HCl(aq) + NaOH(aq) H2O + NaCl
Stoichiometrically equivalent quantities of acid and base have reacted
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Titration of a weak acid and a strong base results in pH curves that look similarto those of a strong acid-strong base curve except that the curve (a) begins at a higherpH, (b) the pH rises more rapidly in the early part of the titration, but more slowly near the equivalence point, and (3) the equivalence point pH is not 7.0
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Aspects of Aqueous Equilibria: Aspects of Aqueous Equilibria: Titration Curves of Both Strong and Weak AcidsTitration Curves of Both Strong and Weak Acids
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Aspects of Aqueous Equilibria: Aspects of Aqueous Equilibria: Calculating pH’s from TitrationsCalculating pH’s from Titrations
Calculate the pH in the titration of acetic acid by NaOH after 30.0 ml of 0.100 M NaOH has been added to 50 mL of 0.100 acetic acid
HC2H3O2 (aq) + OH- H2O(l) + C2H3O2
0.005 mol 0.003 mol 0
0.003 mol
0.003 mol
-0.003 mol-0.003 mol
-0.002 mol 0
pH = 4.74 + log [.0370]
[.0250]
pH = 4.91
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Aspects of Aqueous Equilibria: Aspects of Aqueous Equilibria: Determining the KDetermining the Kaa From the Titration Curve From the Titration Curve
pKa = pH = 4.74
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Na2CO3 2Na+(aq) + CO32-
H+(aq) + CO32- HCO3
- (aq)
H+(aq) + HCO- H2CO3 (aq)
Aspects of Aqueous Equilibria: Aspects of Aqueous Equilibria: Titrations of Polyprotic AcidsTitrations of Polyprotic Acids
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