Transcript General Chemistry

Chapter 17
Additional Aqueous
Equilibria
Lawrence J. Henderson
1878-1942.
Discovered how acid-base
equilibria are maintained
in nature by carbonic acid/
bicarbonate system in the
blood. Developed buffer
equation.
Karl A. Hasselbalch 1874-1962
Developed logarithmic form of
buffer equation.
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Use this “decision tree” to calculate pH
values of solutions of specific solutions.
1. Is it pure water? If yes, pH = 7.00.
2. Is it a strong acid? If yes, pH = -log[HZ]
3. Is it a strong base? If yes, pOH = -log[MOH]
or pOH = -log (2 x [M(OH)2])
4. Is it a weak acid? If yes, use the relationship
Ka = x2/(HZ – x), where x = [H+]
5. Is it a weak base? If yes, use the relationship
Kb = x2/(base – x), where x = [OH-]
6. Is it a salt (MZ)? If yes, then decide if it is neutral, acid,
or base; calculate its K value by the relationship
KaKb = Kw, where Ka and Kb are for a conjugate system;
then treat it as a weak acid or base.
7. Is it a mixture of a weak acid and its weak conjugate base?
It is a buffer; use the Buffer Equation.
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Common Ion Effect
So far, we’ve looked at solutions of weak acids
and solutions of weak bases.
HX  H+ + X- (described by Ka)
Weak acid equilibrium:
Weak base equilibrium:
X- + H2O HX+ OH- (described by Kb)
What if you had both HX and X- in the same solution?
This could be obtained by adding some Na+X- salt to a
solution of HX.
The result is called a buffered solution.
A buffered solution resists changes to pH when an acid or base
is added.
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Buffered Solutions
A buffer consists of a mixture of a weak acid (HX) and its
conjugate base (X-).
An example of preparing a buffered solution: Add 0.10 mole
of lactic acid (H-Z) and 0.14 mole of sodium lactate (Na-Z)
to a liter of solution.
A buffer resists a change in pH when a small amount of OHor H+ is added.
The reason the buffered solution resists pH change becomes
clear when we remember LeChâtlier’s principle.
HX(aq)
H+(aq) + X-(aq)
LeChâtlier says the above equation will
(1) shift to the left if we add HCl or
(2) shift to the right if we add NaOH -thus resisting a change in [H+]
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Buffered Solutions
We can predict the pH of a buffered solution:
[H+][X-]
Ka =
[HX]
[HX]
Rearranging, [ H ] = K a
[X-]
+
Taking –logs of both sides and rearranging:
[HX]
 log[H ] = logKa  log 
[X ]
+
[HX]
pH = pKa  log 
[X ]
[base]
pH = pKa + log
[acid]
or
[X ]
pH = pKa + log
[HX]
Henderson-Hasselbalch equation
or Buffer equation
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Buffered Solutions
What is the pH of a buffered solution of 0.2 M HF and 0.1 M NaF?
(buffer is weak acid HF and its conj. base (salt) F-)
Use:
For HF,
[F ]
pH = pKa + log
[HF]
pKa = -log Ka = -log(6.8 x 10-4) = 3.16
0.1
= 3.16  0.30 = 2.86
So, pH = 3.16 + log
0.2
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One often wants a buffer to have a specific [base]/[acid] ratio.
The H2CO3/HCO3- buffer in blood should be at a pH=7.40.
What must the [base]/[acid] ratio be to obtain this?
For H2CO3  H+ + HCO3- Ka = 4.3x10-7
Henderson-Hasselbalch:
pKa = 6.37
[base]
pH = pKa + log
[acid]
Rearrange and solve for [base]/[acid]:
[base]
log
= pH  pKa = 7.40  6.37 = 1.03
[acid]

[base] [HCO3 ] 11
=

[acid] [H2CO3 ] 1
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Adding strong acids and strong bases to buffers
Assume 1 L of a lactic acid/lactate buffer,
[HLac] = 0.061 M; and [Lac-1] = 0.079 M.
The pKa for lactic acid is 3.85.
The pH is pKa + log[(0.079)/(0.061)] = 3.96
Suppose you add 0.020 mol H+ (strong acid) to this buffer with no
change in volume. Then, there will be a neutralization reaction
between H+ and LacH+ +
Lac- 
HLac
init. 0.020M
0.061M
0.079M
change -0.020M
-0.020M
+0.020 M
0
0.059 M
0.081 M
after rxn
Henderson-Hasselbalch:
0.059
[Lac- ]
pH = pKa + log
= 3.85 + log
= 3.85 + (0.14) = 3.71
[HLac]
0.081
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Adding strong acids and strong bases to buffers
Assume 1 L of the lactic acid/lactate buffer from earlier slide.
[HLac] = 0.061 M; and [Lac-1] = 0.079 M
The pKa for lactic acid is 3.85.
The pH is pKa + log[(0.079)/(0.061)] = 3.96
Suppose you add 0.020 mol OH- (strong base) to this buffer with
no change in volume. Then, there will be a neutralization reaction
between OH- and HLac
OH- +
HLac 
Lacinit. 0.020M
0.079M
0.061M
change -0.020M
-0.020M
+0.020 M
0
0.041 M
0.099 M
after rxn
Henderson-Hasselbalch:
[Lac- ]
0.099
pH = pKa + log
= 3.85 + log
= 3.85 + ( +0.38 ) = 4.23
[HLac]
0.041
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Acid-Base Titrations
Strong Acid-Base Titrations
HCl + NaOH  NaCl
Titration curve
buret
(with 0.10 M
OH- (aq)
beaker
(with 0.10
M H+ (aq)
& indicator)
V=50.0 mL
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Acid-Base Titrations
Strong Base – Weak Acid Titration
HAc + NaOH  NaAc
At the end of the titration,
the pH is determined by
the concentration of
[NaAc] = 0.050 M.
The pH = 8.72
Half-way through the titration,
the pH = pKa = 4.74
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Acid-Base Titrations
Titrations of Polyprotic Acids
• In polyprotic acids, each ionizable proton dissociates
in steps.
• In the titration of H2CO3 with NaOH there are two
equivalence points:
–one for the formation of HCO3H2CO3 + OH-→ HCO3- + H2O
–one for the formation of CO32HCO3- + OH- → CO32- + H2O
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Solubility Equilibria
Solubility-Product Constant, Ksp
• Consider
Ba2+(aq) + SO42-(aq)
BaSO4(s)
•for which
K = [Ba ][SO ]
2+
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= Ksp
•Ksp is the solubility product. (Remember, BaSO4 is
ignored because it is a pure solid so its concentration
is constant.) The larger the solubility product, the
more soluble the salt.
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Solubility Equilibria
Solubility-Product Constant, Ksp
• In general: the solubility product is the molar
concentration of ions raised to their stoichiometric
powers.
Examples: for Ag2CO3 Ksp = [Ag+]2[[CO3-2]
for Al2(SO4)3 is Ksp = [Al+3]2[SO4-2]3
• Solubility is the amount (grams) of substance that
dissolves to form a saturated solution.
• Molar solubility (s) is the number of moles of solute
dissolving to form a liter of saturated solution.
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Solubility and Ksp
Problem solving – always “let s = molar solubility”
Two types of problems:
(1) Calculate Ksp from solubility
(2) Calculate solubility from Ksp
Problem examples:
(a) Calculate Ksp for PbS, if the solubility = 1.73 x 10-14 M
Let s = molar solubility = [PbS(aq)] = 1.73 x 10-14
[Pb+2] = s = 1.73 x 10-14
[S-2] = s = 1.73 x 10-14
Ksp = [Pb+2][S-2] = s2 = (1.73 x 10-14)2 = 2.95 x 10-28
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(b) Calculate Ksp for Co(OH)2 ; solubility = 6.88 x 10-6 M
Let s = molar solubility = [Co(OH)2(aq)] = 6.88 x 10-6
[Co+2] = s = 6.88 x 10-6
[OH-] = 2s = 2(6.88 x 10-6) = 1.38 x 10-5
Ksp = [Co+2][OH-]2 = (6.88 x 10-6)(1.38 x 10-5)2
= 1.31 x 10-15
[Alternatively, Ksp = 4s3 = 4(6.88 x 10-6)3 = 1.30 x 10-15]
(c) Calculate solubility of CdS; Ksp = 8 x 10-28
Let s = molar solubility (unknown) = [CdS(aq)]
Ksp = [Cd+2][S-2] = (s)(s) = s2 = 8 x 10-28
s = 3 x 10-14
(d) Calculate solubility of CaF2; Ksp = 3.9 x 10-11
Let s = molar solubility (unknown) = [CaF2(aq)]
Ksp = [Ca+2][F-]2 = (s)(2s)2 = 4s3 = 3.9 x 10-11
s = 2.1 x 10-4
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Factors That Affect Solubility
Common-Ion Effect
• Solubility is decreased when a common ion is added.
• This is an application of Le Châtelier’s principle:
CaF2(s)
Ca2+(aq) + 2F-(aq)
• As F- (from NaF, say) is added, the equilibrium shifts
to the left.
•Therefore, CaF2(s) is formed and precipitation occurs.
•As NaF is added to the system, the solubility of CaF2
decreases.
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Common-Ion Effect
Example problem. What is solubility of AgCl with and without
NaCl?
Given: Ksp = 1.8 x 10-10 for AgCl
Let s = solubility of AgCl (unknown) = [AgCl(aq)]
Ksp = 1.8 x 10-10 = s2
Solubility = s = (1.8 x 10-10 )1/2 = 1.35 x 10-5 M (without NaCl)
What is solubility of AgCl in a 0.1 M solution of NaCl (which
contains Cl-, a common ion)?
AgCl (s)  Ag+(aq) + Cl- (aq)
x
x+0.1
Ksp = [Ag+][Cl-] = (s)(s+0.1)  (s)(0.1) = 1.8 x 10-10
s = 1.8 x 10-9 M
(with NaCl)
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Solubility and pH
How would we increase the solubility of a sparingly soluble salt
such as CaF2?
In water, the equilibrium is: CaF2(s)  Ca2+(aq) + 2 F- (aq)
LeChatelier says that to increase solubility, we must remove F- ions.
If we add a strong acid, H+, this will cause the reaction:
H+(aq) + F- (aq)
HF (aq)
(Remember, HF is a weak acid)
This reaction removes F- ions thus driving equilibrium in the
top equation to the right which means more CaF2 dissolves.
As acid is added, pH decreases
and solubility of CaF2 increases.
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