Acids and Bases Chapter 7 Zumdahl 6th Ed.
The nature of Acids
(Section 7.5)
pH of an Acid. (Do the strong and weak cases together).
The fraction of dissociation
A weak acid and a weak acid in water
Solve: for pH and fraction dissociation.
Looking ahead: Several weak acids in water.
Strong vs Weak Acids
K large
A
+
−
HA ( aq ) + H 2O ( aq ) H
aq
+
A
(
)
( aq )
K A small
Acid
Base
Acid
Base
Strong Acid
Most Dissociates
 H +   A− 
X2
K A = QA =
=
HA
CA − X
[ ]
Each acid has a conjugate base (partner)
Each base has a conjugate acid (partner).
A strong acid: Mostly Dissociated
Weak acid: Little Dissociated
(
pH = − log10  H + 
)
and
Weak Acid
Little Dissociates
pK A = − log10 ( K A )
Auto-ionization of Water
Equivalent
forms
{
2H 2O ( aq ) H 3O + ( aq ) + OH − ( aq )
H 2O ( aq ) H + ( aq ) + OH − ( aq )
Water at equilibrium on its own will have:
QW =
aH + aOH −
aH 2 O
=  H + ( aq )  OH − ( aq )  = KW = 1.0 ⋅10−14
Pure Water :  H + ( aq )  = OH − ( aq )  ∴
pKW = 14
 H + ( aq )  = KW
But this means that water must always satisfy the equiliribum
when you throw in any other acids or bases into water:
Always :  H +  OH −  = KW
Auto-ionization of Water
The reaction is endothermic (because on net a bond is broken),
therefore the equilibrium moves to the right as the temperature is
increased, more acid is generated. The KW itself increases with
temperature. (LeC’s Principle).
T − 298 )
(
1
pKW (T ) = 14 − 9.5*
pure water: pH = pKW
T
2
At T=50C pH=6.6. It is still neutral water and pH=pOH, but KW
increased with the temperature. Therefore it is not acidic.
http://www.chemguide.co.uk/physical/acidbaseeqia/kw.html#top
What fraction of water molecules ionize? Water is 55 molar and
pH=7.
 +
H 

f =
= 18 ⋅10−10 = 2 ppb
[ H 2O ]
pH and pOH Scales
KW = 1.0 ⋅10−14 =  H +  OH − 
{
}
log10 ( KW ) = log10 (1.0 ⋅10−14 ) = log10  H +  OH − 
pKW = 14 = pH + pOH ⇐ Always
pOH = 14 − pH
⇐ Always
For pure water the proton and hydroxide concentrations
must be equal; charge balance, charge neutral.
 H +  > 10−7
Base  H +  < 10−7
Acid
pH < 7
pOH > 7
pH > 7
pOH < 7
Is an acid hydroxide free?
What is the pOH of 10M Hcl?
What is the pOH of 1e-8M HCl?
Follow the pH via a chemical Reaction
Cr2O7 2− ( aq ) + H 2O ( aq ) 2 H + ( aq ) + 2CrO4 2− ( aq )
orange
yellow
2
 H  CrO4 
KA =
Cr2O7 2− 
+
2−
2
The color is sensitive to the pH and can be used to give us an
idea of the KA of the reaction.
The Universal pH indicator is a set of dyes that change color
over a very wide range of pH
The pH of an Acid (Approximate Treatment)
What is the pH of an acid with concentration CA and
equilibrium constant KA? NICE Table:
HA H + + A−
Species
HA
Initial C A = [ HA]0
H+
0
A−
0
Change
New
X
X
X
X
−X
CA − X
What is the range on X? 0 < X < C A
At Equilibrium:
Acid Conservation:
 H +   A− 
X2
K A = QA =
=
CA − X
[ HA]
 A−  + [ HA] = X + ( C A − X ) = C A = [ HA]o
The pH of an Acid (Approximate Treatment)
+
−



H
=
A
The answer is X because: 
   = X
The range on X:
0 < X < CA
Defines two limits: A) X~0, B) X~CA
Solve the Quadratic
X2
KA =
CA − X
⇒
X=
K A2 + 4 K AC A − K A
2
A) Weak Acid
K A < CA
K A = CA
X =
X = 0.62C A
K AC A
Dissociation: Little
~half
K
Large
B ) Strong Acid
K A > CA
X = CA
Almost All
+
−
HA H
+
A
K Small
pH of Weak Acid Solutions
Example: Calculate the pH of a 0.10 M solution of nitrous acid HNO2
(Ka = 4.0 x 10-4).
+
−
HNO 2 (aq) H (aq) + NO 2 (aq)
This is a weak acid; therefore we will treat this just like a standard
equilibrium problem.
+
−
  H 3O    NO2  
  


 1 M  1 M   H +   NO2− 

 =  

K A = 4 ⋅10−4 = QC = 
 [ HNO2 ] 
[ HNO2 ]


 1M 
[ HNO2 ] = 0.1 − x
2
x
 H  =  NO  = x Final Expression 4 ⋅10 =
.1 − x
+
−
2
−4
pH of Weak Acid Solutions
Example: Calculate the pH of a CA=0.10 M solution of nitrous acid
HNO2 (Ka = 4.0 x 10-4). HNO 2 (aq ) H + (aq ) + NO 2− (aq )
Exact solution of the quadratic
2
x
4 ⋅10−4 =
, x = 0.00613,
.1 − x
pH = px = 2.21
Compare with the approximation: (x<<0.1=CA).
4 ⋅10−4 = x 2 , x = 0.00632,
Is the water contribution negligible?
Excellent Approximations
pH = px = 2.20
 H + 
= 1.0 ⋅10−7
Water
 A−  0.0061
Later: What is the dissociation fraction?
=
= 6.1%
CA
0.1
The pH of an Acid: Specific Cases
A) Weak Acid
K A < CA
X =
K AC A
B ) Strong Acid
K A = CA
K A > CA
X = 0.62C A
X = CA
Determine [H+], and pH for the following examples.
1) HCl, at 0.02M, (stomach acid) KA=10+6
2) Acetic acid, Vinegar, 0.5M, KA=1.6 10-5?
(Why can you put Vinegar on salad but HCL will send
you to the hospital?)
3) HCl at 1e-8 M.
4) HF 1mM = 0.001M, KA=7.4 10-4. ans: (5.6 vs 8.6)e-4
An important observation, weak dissociation:
 H +  = X = K AC A
+

K A ≤  H  ≤ C A
pH =
1
2
( pK A + pC A )
The pH of an Acid: Specific Cases
A) Weak Acid
B ) Strong Acid
K A < CA
X =
K AC A
K A = CA
K A > CA
X = 0.62C A
X = CA
Determine [H+], and pH for the following examples.
1) HCl, at 0.02M, (stomach acid) KA=10+6
 H + 
= 0.0200;
Quad
pH = − log ( 0.02 ) = 1.70
2) Acetic acid, Vinegar, 0.5M, KA=1.6 10-5?
(Why can you put Vinegar on salad but HCL will send
you to the hospital?)
 H +  = K AC A = 1.6 ⋅10−5 ⋅ 0.05 = 9.0 ⋅10−4 ;
Notice : K A <  H +  < C A
and
pH = 3.05
1
pH = 3.05 = ( pK A + pC A )
2
Two cases of pH.
•
•
•
•
•
•
•
•
3) HCl at 1e-8 M.
Using the quadratic solution: pH=8.
Why is this wrong?
Correct Answer:
4) HF 1mM = 0.001M, KA=7.4 10-4.
Quadratic answer ans: (5.6 vs 8.6)e-4
Is this a weak acid? CA~KA.
Outside the ‘5%’ rule
[ H + ]Quad = 5.6 ⋅10−4 ; pH = 3.25
−4
H
+
=
K
C
=
8.6
⋅
10
[ ]
A A
• percent dissociation is 56%.
Effect of Dilution on Percent Dissociation and [H+] of a Weak Acid
Increasing the concentration of an acid
increases all forms, and so the [H+]
goes up.
But how does the equilibrium
partition? One way to see how
concentration effects work is to look at
the fraction of the acid that is
dissociated vs the fraction that is not.
−
The two fractions must sum to one.

 [ HA]
A
−

C A =  A  + [ HA] or 1 =
+
From our observation about the
CA
CA
conservation of acid
−
f A−
 A 
HA]
[
=
and f HA =
CA
CA
Astoundingly, The fraction (or percent) dissociation is :
• INDEPENDENT of the concentration of the acid!!!
• depends ONLY on the hydrogen ion concentration.
• Decreases ALWAYS with increasing hydrogen ion concentration. (LeC)
Effect of Dilution on Percent Dissociation and [H+] of a Weak Acid
K large
A
+
−
HA ( aq ) H
aq
+
A
(
)
( aq )
K small
A
X2
KA =
CA − X
−

 [ HA]
A
−

C A =  A  + [ HA] or 1 =
+
CA
CA
f A−
 A− 
[ HA]
=
and f HA =
CA
CA
LeChatelier tells us that if we remove some of the acid (say the HA
component) then products will shift to have fewer of both products.
But, lowering the concentration is like increasing the volume (which
for a gas means decreasing the pressure). This provides more space
for the molecules, so this will shift the equilibrium to the right to
generate more individual molecules, this means the fraction
dissociated increases even though the concentrations of all three
components decrease.
The fraction of Dissociation of an Acid
HA ( aq ) H + ( aq ) + A− ( aq )
 H +   A− 
 A− 
KA
KA =
or
≡r=
 H + 
[ HA]
[ HA]
How can the fraction of dissociation be independent of the
concentration?
Dissociation fraction depends
 H +   A− 
 A− 
KA
≡r=
KA =
or
only on the proton
+
HA
HA
 H 
[ ]
[ ]
concentration (or pH).
 A− 
 A− 
f A−
CA
r
=
=
⇒ f A− =
r≡
1 − f A−
1+ r
[ HA] [ HA]
CA
Consistent with the equilibrium expression,
and LeChatelier’s Principle: Increase [H+]
KA
r
f A− =
=
1 + r K A +  H +  and the equilibrium moves to left, and
fraction A- goes down.
Dissociation distinguishes weak from strong Acid
pH = pK A
Dissociation: Little
f A− << 0.3
half
Almost All
f A− = 0.5
f A− ~ 1
What is the concentration of Acetic acid (HAc) if the
acid is 30% dissociated?
Is 1mM HAc a weak acid? KA=1.8e-5
Is 1mM HClAc (monochloracetate) weak? KA=1.35e-3
pC A = 3.00;
pK Ac = 4.74;
pK ClAc = 2.87
pH problems, Answer
• Is 1mM HAc a weak acid? KA=1.8e-5
pC A = 3.00;
pH =
1
2
pK Ac = 4.74;
( pC A + pK Ac ) = 12 ( 3.00 + 4.74 ) = 3.87
f A−
KA
1.85 ⋅10−5
=
=
= 12%
−5
−3.87
K A +  H +  1.85 ⋅10 + 10
• Is 1mM HACl (monochloracetate) weak? KA=1.35e-3
• Can’t use the weak acid formula, need to do the exact,
quadratic method, not a weak acid (despite the modest KA)
pC A = 3.00;
pK ClAc = 2.87
 H + 
= 0.67 ⋅10−3 < C A = 1⋅10−3
Quad
f A−
KA
1.38 ⋅10−3
=
=
= 66%
−3
−3
+
K A +  H  1.38 ⋅10 + 0.6710
pH of Weak Acid Solutions
Example: Calculate the pH of a 0.100 M solution of nitrous acid HNO2
(Ka = 4.0 x 10-4).
HNO 2 (aq) H + (aq) + NO 2− (aq)
Exact solution of the quadratic
 A−  =  H +  = 0.00613,
pH = 2.21
What is the dissociation fraction?
f A−
KA
4.0 ⋅10−4
=
=
= 0.0613 = 6.1%
−4
−3
+
K A +  H  4.0 ⋅10 + 6.13 ⋅10
Using the fraction dissociated, estimate the concentration of the
Anion, A-, and compare to exact result.
 A−  = f A− C A = 0.061 ⋅ 0.1 = 0.0061
From the fraction dissociated , we computed the same anion
concentration
A different way to solve for [H+]
• Here is an alternative approach to the quadratic which will
prove to be much more powerful for the problems ahead.
• We combine the charge balance and the dissociation fraction:
• Charge Balance
 H +  =  A− 

• Dissociation Fraction

f A−


KA
=
K A +  H + 
;  A−  = f A− C A
• Substitute the other formulas into the Charge balance:
KA
 H  =  A  = f A− C A =
CA
+
K A +  H 
+
−
• This expression must be a rearrangement of the quadratic
equation we already had. But it is self-limiting and accurate in
all regions. In this form one can have an iterative solution.
An iterative solution to the quadratic
• Solve this expression in the current form. We do not
rearrange it, but (we will see this come into play later)
multiply by [H+] and take the square root of both sides
KA
 H  =
CA =
+
K A +  H 
+
 H +  K A
CA
+
K A +  H 
• To see how this works: plug in a guess on the right, 1e-7 is a
good place to start because it is an acid and can’t be any
lower than this. (We will fix this deficiency soon).
• The value of H+ that comes out is then a better guess to the
answer and can be plugged into the rhs to generate a better
guess.
• Alternatively you can push solver on your TI.
An iterative solution to the quadratic
• Use the iterative form to find the pH of a CA= 1e-4M HAc
solution where KA=1.8e-5.
 H +  =
TI
 H +  ⋅ K A
CA
+
K A +  H 
form :
X ⋅K
C→X
K+X
• Iterative guesses for X (converges to exact quad solution)
1⋅10−7 → 1.64 ⋅10−5 → 2.93 ⋅10−5 → 3.33 ⋅10−5 → 3.42 ⋅10−5 →
3.43 ⋅10−5 → 3.437 ⋅10−5 → same
• This expression (on rhs) is the proton pushing power of the
acid.
• This iterative form works for a strong acid, try 0.01 M HCl.
Acid dissociation
• A base (NaOH) is added to 0.01 M acetic acid
solution (KA=1.8e-5), so that the pH=6.
Neglecting dilution what is the anion (Ac-)
concentration at this pH.
−5
K
1.8
⋅
10
A
 Ac −  = f A− C A =
CA =
0.01
−5
−6
+
1.8 ⋅10 + 1.0 ⋅10
K A +  H 
f A− = 0.947;  Ac −  = 9.47 ⋅10−3