Unit 12
I.
Chapters 15&16 – Acids and Bases
Page 1
Definitions - Review! (see Unit 3)
A.
Taste
Feel
Litmus (indicator)
Phenolphthalein
(indicator)
Acids
Sour
Sticky
Red
Clear
Bases
Bitter
Slippery
Blue
Pink
Strong
Define –
dissociate
completely
HCl, HBr, HI
HClO4, HNO3, H2SO4
These six acids dissociate
completely.
HCl(g) → H+(aq) + Cl‒(aq)
LiOH, NaOH, KOH, RbOH, CsOH,
Ca(OH)2, Sr(OH)2, Ba(OH)2
These eight bases dissociate
completely.
Weak
Define – can not
dissociate
completely;
<100%
dissociation =
weak
- molecules with ionizable H’s
Ex: HF, HClO
HF(g) ↔ H+(aq) + F‒(aq)
- anions with ionizable H’s
Ex: HSO4–, H2PO4–
- organic acids contain the –COOH
functional group attached to some
carbon chain
Ex: CH3COOH
- molecules with ionizable OH’s
Ex: Fe(OH)3, Mg(OH)2
- anions that are conjugates of weak
acids
Ex: F–, CH3COO–
Concentration
Define – number
of particles in
solution
Arrhenius Model
Brönsted-Lowry
Model
H3COO‒(aq) + H2O(l)↔ OH‒(aq) + H3COOH(aq)
H3COOH (aq) ↔ H+(aq) + H3COO‒(aq)
- organic bases are amines: NH2
attached to a carbon chain, with
unshared e– pair
Ex: CH3NH2
NH4+(aq) + H2O(l) ↔ NH3(aq) + H3O+(aq)
- species that can produce OH ions
by accepting H+ from water
Ex: NH3
- cations, including metal cations,
conjugates of weak bases
Ex: NH4+, Zn2+
2013-2014
v.
CH3NH2(aq) + H2O(l) ↔ OH‒(aq) + CH3NH3+(aq)
–
NH3(g) + H2O(l) ↔ NH4+(aq) + OH‒(aq)
- species with electron pairs that
can be donated
Ex: NH3, Cl‒, OH‒
concentrated means lots of molecules initially
6.0 M HCl or 6.0 M NaOH
dilute means few molecules present initially
0.10 M HBr or 0.0001 M CsOH
dissolves in water to produce
dissolves in water to produce OH—
hydronium ions (H3O+ = H+)
ions
proton donor
notice that an H3O+ is the same as a
“piggybacked” proton on water
one proton is moved at a time
proton acceptor
OH‒ is a common proton acceptor,
forming water; similar to
Arrhenius
Unit 12
Conjugate pairs ‒
Lewis Model
Chapters 15&16 – Acids and Bases
+
NH4
HCO3‒
C4H4O62‒
species formed after base has
accepted a proton is called its
conjugate acid
electron pair acceptor
Ex: metal cations
→ NH3
→ H2CO3
→ HC4H4O6‒
species formed after acid has
donated a proton is called its
conjugate base
electron pair donor; Ligands
H
|
H ‒ O ‒ H H ‒ N ‒ H
,
II. pH -Tracking acidity by way of H+/H3O+ ions (§15.6 - 15.8)
A. Kw and the auto-ionization of water
Kw is the Keq for the dissociation of water:
Kw = [H3O+] [OH‒]
2 H2O(l)  H3O+(aq) + OH–(aq)
[H3O+] = [OH–] = 1.0 x 10–7 M in pure water about 2 molecules per billion dissociate
B. The pH Scale
A logarithmic representation of the H+ (H3O+) ion in a solution.
C. The pOH Scale
D. Calculating pH or pOH
The BOX
2013-2014
Page 2
Unit 12
Chapters 15&16 – Acids and Bases
[H+]
+
‒
Kw = [H ] [OH ]
[H+]=10-pH
pH = ‒ log[H+]
pH
Page 3
[OH‒]
[OH‒] = 10‒pOH
lpOH =+ ‒log[OH‒]
14 = pH + pOH
pOH
1. examples:
a. When concentration, pH, or pOH are known directly…
Calculate the pH of typical adult blood, which has a hydrogen-ion concentration of 4.0x10-8 M.
pH = ‒log[H+] = ‒log[4.0x10-8] = 7.40
b. When concentration of a strong species is known …
Calculate the pOH and pH of a 0.056 M HCl solution.
HCl(g) → H+(aq) + Cl‒(aq)
0.056 M 0.056 M 0.056 M
pH = ‒log[H+] = ‒log[0.056] = 1.25
Calculate the pOH and pH of a 0.0009604 M Ba(OH)2 solution.
2 OH‒(aq)
Ba(OH)2(aq) → Ba2+(aq) +
0.0009604 M 0.0009604 M 0.0019208 M
pOH = ‒log[OH‒] = ‒log[0.0019208] = 2.7165
c. When concentration of a weak species is known …
Calculate the pOH and pH of a 0.056 M HClO2 solution.
HClO2(aq) → H+(aq) + ClO2‒(aq)
Because this is a weak acid, look up its Ka…. Ka = 1.1 x 10-2
 H   ClO2  
Ka 
 HClO2 
0.011 
 x  x
0.056
x = [H+] = 0.0248 M
pH = ‒log[H+] = ‒log[0.0248] = 1.61
pOH = 14 ‒ pH = 14 ‒ 1.61 = 12.39
III. Acid-Base Reactions
A. All of these reactions are from one of the six main types of reactions
acid + metal  H2 + salt
A piece of magnesium ribbon is dropped into a solution of
hydrochloric acid …
Mg(s) + HCl(aq) → MgCl2(aq) + H2(g)
Mg(s) + H+(aq) + Cl‒(aq) → Mg2+(aq) + Cl‒(aq) + H2(g)
Mg(s) + H+(aq) → Mg2+(aq) + H2(g)
metal + H2O  metal-OH + H2
A piece of sodium metal is dropped into water …
Na(s) + H2O(l) → NaOH(aq) + H2(g)
Na(s) + H2O(l) → Na+(aq) + OH‒(aq) + H2(g)
2013-2014
Unit 12
Chapters 15&16 – Acids and Bases
acid + metal oxide  H2O + salt
acid + carbonate  CO2 + H2O + salt
acid + sulfite  SO2 + H2O + salt
Page 4
Solid barium oxide is added to a solution of hydrocyanic
acid …
BaO(s) + HCN(aq) → Ba(CN)2(aq) + H2O(l)
Solutions of sulfuric acid and calcium carbonate were
mixed …
H2SO3(aq) + CaCO3(aq) → CaSO3(s) + H2O(l) + CO2(g)
Solutions of nitrous acid and lithium sulfite are mixed …
HNO2(aq) + Li2SO3(aq) → LiNO3(aq) + H2O(l) + SO2(g)
HNO2(aq) + Li+(aq) + SO32‒(aq) → Li+(aq) + NO2‒(aq) + H2O(l) + SO2(g)
HNO2(aq) + SO32‒(aq) → NO2‒(aq) + H2O(l) + SO2(g)
acid + base  H2O + salt
Solutions of hydrobromic acid and lithium hydroxide are
mixed …
HBr(aq) + LiOH(aq) → LiBr(aq) + H2O(l)
H+(aq) + Br‒(aq) + Li+(aq) + OH‒(aq) → Li+(aq) + Br‒(aq) + H2O(l)
H+(aq) + OH‒(aq) → H2O(l)
Anhydrides - literally means “without
water”
nonmetal oxide + H2O → acid
metal oxide + H2O  metal–OH
Solutions of hydrofluoric acid and potassium hydroxide
are mixed …
HF(aq) + KOH(aq) → KF(aq) + H2O(l)
HF(aq) + K+(aq) + OH‒(aq) → K+(aq) + F‒(aq) + H2O(l)
HF(aq) + OH‒(aq) → F‒(aq) + H2O(l)
Sulfur trioxide is bubbled in a beaker of water …
SO3(g) + H2O(l) → H2SO4(aq)
SO3(g) + H2O(l) → H+(aq) + SO42‒(aq)
Water is mixed with solid barium oxide …
BaO(s) + H2O(l) → Ba(OH)2(aq)
BaO(s) + H2O(l) → Ba2+(aq) + OH‒(aq)
IV. Acid-Base Titration – Strong/Strong
A. What is a titration?
A titration is a reaction that is run paying particularly close attention to the volume measurements of
the species involved. In most titrations, the experiment is run with a buret and the end point is
determined by the indication that one reactant is completely used up at the same time that another
reactant is used up.
B. What is a dominating species?
The dominating species is the acidic or basic species that is in a large enough quantity to dictate the
pH or pOH due to its acidic or basic properties.
At the endpoint of a weak acid with strong base titration, this is the conjugate base.
At the endpoint of a strong acid with strong base titration, this is water.
C. Endpoint
1. Definition:
The endpoint is when the moles of OH‒ are equal to the moles of H+
2. Net reaction for strong acid + strong base: H+ + OH–  H2O
3. pH at endpoint for a strong acid and strong base titration is almost always 7.0 – What species will
dominate the solution at the end point? H2O
D. Simple stoichiometry
E. example:
1. Titrate 25.0 mL of 0.25 M solutions of HCl with a 0.10 M solution of KOH.
a. What is the original pH of the HCl solution?
Since HCl is a strong acid, it undergoes 100% dissociation.
2013-2014
Unit 12
Chapters 15&16 – Acids and Bases
+
Page 5
‒
HCl(aq)
→ H (aq) +
Cl (aq)
I 0.25 M
0
0
C ‒ 0.25 M
+ 0.25 M
+ 0.25 M
E 0
0.25 M
0.25 M
pH = ‒log[H+] = ‒log[0.25] = 0.60
b. What will the pH be after 25.0 mL of KOH is added?
HCl and KOH will react to form water, reducing the amount of these strong acids and bases.
HCl(aq) +
KOH(aq)
→ H2O(l) +
KCl(aq)
I (0.0250L)(0.25M)=
(0.0250L)(0.10M)=
0
0
0.00625 mol
0.0025 mol
C ‒x
‒x
+x
+x
‒ 0.0025
‒ 0.0025
E 0.00375 mol
0
0.0025 mol 0.0025 mol
→ H+(aq) +
Cl‒(aq)
HCl(aq)
I 0.00375 mol/(0.025+0.025)L
0
0
C ‒ 0.075 M
+ 0.075 M + 0.075 M
E 0
0.075 M
0.075 M
pH = ‒log[H+] = ‒log[0.075] = 1.12
c. How much KOH needs to be added to reach the equilibrium point?
In order to reach the equilibrium point, the moles of HCl and KOH must be equal.
HCl(aq) +
KOH(aq)
→ H2O(l) +
KCl(aq)
I (0.0250L)(0.25M)=
(x)(0.10M) =
0
0
0.00625 mol
0.00625 mol
C ‒x
‒x
+x
+x
‒ 0.00625
‒ 0.00625
E 0
0
0.00625 mol 0.00625 mol
0.00625 mol = (x)(0.10M)
x = 0.0625 L = 62.5 mL
d. What will be the pH after 75.0 mL of KOH is added?
HCl and KOH will react to form water, reducing the amount of these strong acids and bases.
HCl(aq) +
KOH(aq)
→ H2O(l) +
KCl(aq)
I (0.0250L)(0.25M)=
(0.0750L)(0.10M)=
0
0
0.00625 mol
0.0075 mol
C ‒x
‒x
+x
+x
‒ 0.00625
‒ 0.00625
E 0
0.00125mol
0.00625 mol 0.00625 mol
KOH(aq)
→
I 0.00125 mol/(0.0750+0.0250)L
C ‒ 0.0125 M
E 0
pOH = ‒log[OH‒] = ‒log[0.0125] = 1.90
2013-2014
K+(aq) +
0
+ 0.0125 M
0.0125 M
OH‒(aq)
0
+ 0.0125 M
0.0125 M
Unit 12
Chapters 15&16 – Acids and Bases
V. Acid-Base Titration Curves - Strong/Strong
Page 6
A. Since the endpoint is where moles H3O+ =
moles OH–, the pH must be 7.
B. Other characteristics:
VI. Hydrolysis - Review! (see Unit 3)
Examples
Comments
pH of solution
Cation is from strong base;
Neither acts as an acid or a base
Neutral
NaCl
Anion is from strong acid
because no measurable
KBr
hydrolysis occurs
LiNO3
These salts do not result in the formation of an acidic or basic solution because strong acids and strong bases remain
dissociated.
KBr(aq) + H2O(l) → KOH(aq) + HBr(aq) → K+(aq) + OH‒(aq) + H+(aq) + Br‒(aq)
K+(aq) + Br‒(aq) + H2O(l) → K+(aq) + Br‒(aq) + H2O(l)
LiNO3(aq) + H2O(l) → LiOH(aq) + HNO3(aq) → Li+(aq) + OH‒(aq) + H+(aq) + NO3‒(aq)
Li+(aq) + NO3‒(aq) + H2O(l) → Li+(aq) + NO3‒(aq) + H2O(l)
Cation is from strong base;
Anion hydrolyzes with water,
Basic
KClO
Anion is conjugate base of WEAK acid
acting as a base;
KC2H3O2
Cation has no effect
Na2CO3
These salts result in the formation of OH‒, creating a basic solution.
KClO(aq) + H2O(l) → KOH(aq) + HClO(aq) → K+(aq) + OH‒(aq) + HClO(aq)
K+(aq) + ClO‒(aq) + H2O(l) → K+(aq) + OH‒(aq) + HClO(aq)
Na2CO3(aq) + H2O(l) → NaOH(aq) + HCO3‒(aq) → Na+(aq) + OH‒(aq) + HCO3‒(aq)
Na+(aq) + CO32‒(aq) + H2O(l) → Na+(aq) + OH‒(aq) + HCO3‒ (aq)
Cation is conjugate acid of WEAK base;
Cation hydrolyzes with water,
Acidic
NH4Cl
Anion is from strong acid
acting as a base;
NaHSO3
Anion has no effect
+
+
These salts result in the formation of H (H3O ), creating an acidic solution.
NH4Cl(aq) + H2O(l) → NH3(aq) + H3O+(aq) + Cl‒(aq)
NH4+(aq) + Cl‒(aq) + H2O(l) → NH3(aq) + H3O+(aq) + Cl‒(aq)
NaHSO3(aq) + H2O(l) → NaOH(aq) + SO32‒(aq) + H3O+(aq)
Na+(aq) + HSO32‒(aq) + H2O(l) → Na+(aq) + OH‒(aq) + SO32‒(aq) + H3O+(aq)
Cation is highly charged metal ion;
Hydrated cation acts as an acid;
Acidic
Al(NO3)3
Anion is from strong acid
Anion has no effect on pH
FeCl3
These salts result in the formation of H+ by accepting the Lewis base (ligand) of OH‒, creating an acidic solution.
Al(NO3)3(aq) + H2O(l) → AlOH2+(aq) + H+(aq) + 3 NO3‒(aq)
Al3+(aq) + 3 NO3‒(aq) + H2O(l) → AlOH2+(aq) + H+(aq) + 3 NO3‒(aq)
FeCl3(aq) + H2O(l) → FeOH2+(aq) + H+(aq) + 3 Cl‒(aq)
Fe3+(aq) + 3 Cl‒(aq) + H2O(l) → FeOH2+(aq) + H+(aq) + 3 Cl‒(aq)
A. Type of Salt
2013-2014
Unit 12
Chapters 15&16 – Acids and Bases
Page 7
Salts that contain an anion of a WEAK acid
Solution’s pH is dependent on
Acidic, Basic, or NH4ClO
and a cation of a WEAK base; both ions
Ka and Kb values of the cation
Al2(CO3)3
Neutral
hydrolyze
and anion
NH4ClO(aq) + 2 H2O(l) → NH3(aq) + HClO(aq) + H3O+(aq) + OH‒(aq)
The Ka of NH4+ = 5.6x10‒10, while the Kb of ClO‒ = 2.9x10‒7. Because the Kb value of ClO‒ is larger than the Ka value
of NH4+, the basic nature of ClO‒ will take over the solution, making the solution basic.
‒
‒
NH4+(aq) + ClO (aq) + 2 H2O(l) → NH3(aq) + HClO(aq) + H3O+(aq) + OH (aq)
Al2(CO3)3(aq) + H2O(l) → AlOH2+(aq) + HCO3‒(aq) + H3O+(aq) + OH‒(aq)
The Ka of Al3+ = 1.0x10‒5, while the Kb of CO32‒ = 2.1x10‒4. Because the Kb value of CO32‒ is larger than the Ka value
of Al3+, the basic nature of CO32‒ will take over the solution, making the solution basic.
2‒
‒
‒
Al3+(aq) + CO3 (aq) + H2O(l) → AlOH2+(aq) + HCO3 (aq) + H3O+(aq) + OH (aq)
D. An understanding of these acid-base properties of salts in water (hydrolysis) can also explain the
strength of any ion.
1. Basic properties of conjugate bases – The weaker the acid, the stronger the basic properties of
the conjugate base.
2. Acidic properties of conjugate acids – The weaker the base, the stronger the acidic properties
of the conjugate acid.
VII. Acid-Base Equilibria (§16.1 - §16.3)
Hydrolysis reactions do not react all the way – they are PARTIAL dissociations.
Therefore, all hydrolysis reactions have an equilibrium constant, Keq.
A. Write the hydrolysis
reaction for …
Write the
equilibrium constant
for this reaction …
Is this acid-base
species acidic or
basis?
What is the
conjugate to this
acid-base species?
Write the hydrolysis
reaction for the
conjugate …
Write the
equilibrium constant
for the conjugate …
NaF
NH4NO3
NaF(aq) + H2O(l) → Na+(aq) + OH‒(aq) + HF(aq)
Na+(aq) + F‒(aq) + H2O(l) → Na+(aq) + OH‒(aq)
+ HF(aq)
F‒(aq) + H2O(l) → OH‒(aq) + HF(aq)
NH4NO3(aq) + H2O(l) → NH3(aq) + NO3‒(aq) +
H3O+(aq)
NH4+(aq) + NO3‒(aq) + H2O(l) → NH3(aq) +
NO3‒(aq) + H3O+(aq)
OH    HF 
Kb  
 F  
Basic
 H    NH 3 
Ka    
 NH 4 
Acidic
The conjugate of F‒ is HF.
The conjugate of NH4+ is NH3.
HF(aq) + H2O(l) → H3O+(aq) + F‒(aq)
NH3(aq) + H2O(l) → NH4+(aq) + H3O+(aq)
 H    F  
Ka 
 HF 
OH    NH 4  
Kb 
 NH 3 
NH4+(aq) + H2O(l) → NH3(aq) + H3O+(aq)
How do the reactions and their equilibrium constants compare?
OH    HF   H    F  
Kb  K a 

  H   OH    K w

 F 
 HF 
 H    NH 3  OH    NH 4  
K a  Kb 

  H   OH    K w

 NH 4 
 NH 3 
2013-2014
Unit 12
Chapters 15&16 – Acids and Bases
Page 8
1. examples:
a. Write the reaction that causes a pH change in a solution of nitrous acid. Does this reaction
represent a Ka, Kb, or neither?
HNO2(aq) + H2O(l) → H3O+(aq) + NO2‒(aq)
This represents Ka.
b. Write the reaction that causes a pH change in a solution of sodium hypochlorite. Does this
reaction represent a Ka, Kb, or neither?
Na+(aq) + ClO‒(aq) + H2O(l) → Na+(aq) + OH‒(aq) + HClO(aq)
ClO‒(aq) + H2O(l) → OH‒(aq) + HClO(aq)
This represents Kb.
c. For each of the following reactants, predict the products, then determine whether the reaction
represents an acid dissociation (Ka), base dissociation (Kb), or neutralization.
i. CH3NH3+ + H2O → H3O+(aq) + CH3NH2(aq)
Ka
ii. HNO2 + H2O → H3O+(aq) + NO2‒(aq)
Ka
iii. NH4+ + F‒ → HF + NH3
Neither
iv. H2PO4‒ + H2O → H3O+(aq) + HPO42–(aq)
Ka
OR
v. H2PO4‒ + H2O → OH‒(aq) + H3PO4(aq)
Kb
B. Calculating with Ka and Kb
1. Calculations will often involve either the
a. quadratic equation OR
b. neglecting x values that are added or subtracted (OK if < 5% dissociation) OR
2. examples:
c. Household bleach is a 5% solution of sodium hypochlorite, NaClO. This corresponds to a
molar concentration of about 0.70 M NaClO (Kb = 2.86x10–7). What is the concentration of
the OH–, and the pH of such a solution?
0.70 M NaClO = 0.70 M Na+ = 0.70 M ClO‒
Na+(aq) + ClO‒(aq) + H2O(l) → Na+(aq) + OH‒(aq) + HClO(aq)
ClO‒(aq) +
H2O(l) → OH‒(aq) +
HClO(aq)
I 0.70 M
-0
0
C ‒x
-+x
+x
E (0.7‒x)
-(x)
(x)

OH   HClO 
 x  x 
 2.86 x107 
Kb  

ClO 
 0.7  x 
2.002 x10 7  2.86 x10 7 x  x 2  0
Using polyRoots: x = ‒4.48x10‒4, 4.48x10‒4
‒
[OH ] = 4.48x10‒4 M
 1014 
 10.65
pH   log 
7 
 2.86 x10 
d. Sore throat medications sometimes contain the weak acid phenol, HC6H5O. A 0.10 M solution
of phenol has a pH of 5.43 at 25°C. What is the acid-ionization constant, Ka, for this acid at
25°C?
2013-2014
Unit 12
Chapters 15&16 – Acids and Bases
+
HC6H5O + H2O → H3O (aq) + C6H5O (aq)
HC6H5O (aq) +
H2O(l) → H3O+(aq) +
I 0.10 M
-0
C ‒x
-+x
E (0.1‒x) ≈
-10‒5.43 =
0.1 M
3.7x10‒6 M
 H    F    3.7 x106  3.7 x106 

 1.4 x1010
Ka 
HF
0.1
 
 
C.
Page 9
‒
C6H5O‒(aq)
0
+x
3.7x10‒6 M
Kw = Ka * Kb = 1.0x10–14
1. examples:
a. HCN(aq) + H2O(l)  H3O+(aq) + CN–(aq) with a Ka of 6.2x10–10.
Write the dissociation reaction of the conjugate base and calculate its Kb. Based on the Ka and
Kb, which conjugate is stronger?
K
1014
 1.6 x105
CN‒(aq) + H2O(l)  OH‒(aq) + HCN(aq) Kb  w 
10
K a 6.2 x10
Since the value of the Kb is larger, the conjugate base (CN‒) is the stronger species.
b. Obtain the Kb for the F– ion, the ion added to public water supplies to protect teeth.
For HF, Ka = 6.8x10–4.
Kw
1014

 1.5 x1011
Kb 
4
K a 6.8 x10
c. Predict whether an aqueous solution of each of the following salts will be acidic, basic, or
neutral. (Ka of NH4+ = 5.6x10–10; Kb of C2H3O2– = 5.6x10–10; Kb for CN– = 1.6x10–5)
i. KCl
KCl(aq) + H2O(l) → KOH(aq) + HCl(aq)
K+(aq) + Cl‒(aq) + H2O(l) → K+(aq) + Cl‒(aq) + H2O(l)
Neutral
ii. NH4C2H3O2
NH4C2H3O2(aq) + 2 H2O(l) → NH3(aq) + HC2H3O2(aq) + H3O+(aq) + OH‒(aq)
The Ka of NH4+ = 5.6x10‒10, while the Kb of C2H3O2‒ = 5.6x10‒10. Because Ka is equal to
Kb, neither acidic nor basic species is stronger. Therefore, these species will cancel and
result in a neutral solution.
NH4+(aq) + C2H3O2‒(aq) + H2O(l) → NH4+(aq) + C2H3O2‒(aq) + H2O(l)
iii. NH4CN
NH4CN(aq) + 2 H2O(l) → NH3(aq) + HCN(aq) + H3O+(aq) + OH‒(aq)
The Ka of NH4+ = 5.6x10‒10, while the Kb of CN‒ = 1.6x10‒5. Because Kb of CN‒ is larger
than the Ka of NH4+, the basic properties of CN‒ will dominate the solution, making it
basic.
NH4+(aq) + CN‒(aq) + H2O(l) → NH4+(aq) + HCN(aq) + OH‒(l)
d. Calculate the pH of a 0.10 M NH4Cl solution. The Kb for NH3 is 1.8x10–5.
NH4+(aq) + Cl‒(aq) + H2O → H3O+(aq) + Cl‒(aq) + NH3(aq)
NH4+(aq) +
H2O(l) → H3O+(aq) +
NH3(aq)
I 0.10 M
-0
0
C ‒x
-+x
+x
E (0.1‒x)
-(x)
(x)
2013-2014
Unit 12
Chapters 15&16 – Acids and Bases
Ka 
Page 10
Kw
1014

 5.6 x1010
Kb 1.8 x105
 H    NH 3 
 x  x 
 5.6 x1010 
Ka 

 NH 4 
 0.1  x 
5.6 x10 11  5.6 x10 10 x  x 2  0
Using polyRoots: x = ‒7.48x10‒6, 7.48x10‒6
+
[H ] = 7.48x10‒6 M
pH   log  7.48 x10 6   5.12
D. Amphiprotic species
1. species that can gain or lose H+, depending on other reactant
Ex: HCO3–, HOH
2. Amphoteric species can also act as acids or bases, but don’t contain H+’s
ZnO + 2H+ → Zn2+ + H2O
ZnO + H2O + 2OH- → [Zn(OH)4]2Al(OH)3 + 3HCl → AlCl3 + 3H2O
Al(OH)3 + NaOH → Na[Al(OH)4]
E. Predicting strength of weak acidic species
As with many physical properties, the strength of an acid (or base) is determined by its structure.
So, draw a Lewis dot structure.
1. As effective nuclear charge on central atom increases, acid strength increases
Higher effective nuclear charge of an atom increases e-‒nucleus attraction thus lowering the
energy of its e-'s.
Fluorine's extremely high effective nuclear charge makes it the most electronegative element.
3. As electronegativity of central atom increases, acid strength increases
2013-2014
Unit 12
Chapters 15&16 – Acids and Bases
Page 11
2. As number of lone oxygen atoms (oxygen atoms not bonded to hydrogen) increases, acid strength
increases
4. Loss of H+ by a neutral acid molecule reduces acid strength; Ka of H2SO3 > Ka of HSO3–
2013-2014
Unit 12
Chapters 15&16 – Acids and Bases
Page 12
VIII. Reactions of Weak Species (§16.5)
In this course, WEAK species will only be reacted against a STRONG species.
A. Reactions to consider
Write a net ionic reaction for the following aqueous combinations:
HCN(aq) + NaOH(aq) →
Molecular: HCN(aq) + NaOH(aq) → H2O(l) + NaCN(aq)
Identify strong and weak electrolytes
Ionic: HCN(aq) + Na+(aq) + OH‒(aq) → H2O(l) + Na+(aq) + CN‒(aq)
Net Ionic: HCN(aq) + OH‒(aq) → H2O(l) + CN‒(aq)
HNO3(aq) + NH3(aq) →
Molecular: HNO3(aq) + NH3(aq) → NO3‒(aq) + NH4+(aq)
Ionic: H+(aq) + NO3‒(aq) + NH3(aq) → NO3‒(aq) + NH4+(aq)
Net Ionic: H+(aq) + NH3(aq) → NH4+(aq)
NaOCl(aq) + HClO4(aq) →
Molecular: NaOCl(aq) + HClO4(aq) → HOCl(aq) + NaClO4(aq)
Ionic: Na+(aq) + OCl‒(aq) + H+(aq) + ClO4‒(aq) → HOCl(aq) + Na+(aq) + ClO4‒(aq)
Net Ionic: OCl‒(aq) + H+(aq) → HOCl(aq)
B. WEAK acid with STRONG base
1. 0.3984 g of solid NaOH are added to a 35.68 mL sample of 0.7251 M NH3OH+, a weak acid with a
Ka of 9.09x10‒7
What is the acid dissociation reaction for NH3OH+?
Who is the conjugate to this
In an acid dissociation reaction, begin with an acid [NH3OH+]
acidic species?
+
+
and create H (or H3O ) and a base.
Conjugate base = NH3O (aq)
NH3OH+ (aq) + H2O(l) → NH3O (aq) + H3O+(aq)
Initially, what is occurring?
What is the initial pH for the NH3OH+ solution?
NH3OH+(aq) +
H2O(l)
→ NH3O(aq) +
H3O+(aq)
The acid is dissociating.
I
0.7251 M
0
0
What must be the overriding
C ‒x
+x
+x
equation?
E (0.7251 ‒ x)
(x)
(x)
The acid dissociation reaction



 NH 3O   H 3O 
 x  x 
Ka 
 9.09 x107 

 NH 3OH 
0.7251  x 
x = [NH3O] = [H3O+] = 8.11x10‒4 M
pH = ‒log(8.11x10‒4 M) = 3.09
After the NaOH has dissolved and reacted, what is the resulting
pH?
 1mol  9.96 x103 mol
 0.2791M NaOH
0.3984 g NaOH 

0.03568L
 40.00 g 
+
I
C
E
NH3OH (aq) +
0.7251 M
‒x
‒ 0.2791
0.4460 M
NaOH(aq)
Na+ + OH‒
0.279 M
‒x
0.279
0 = LR
→ NH3O(aq) +
0
+x
+ 0.2791
0.2791 M
Notice that Na+ is a spectator, so it was removed.
NH3OH+(aq) +
I 0.4460 M
C ‒x
2013-2014
H2O(l)
→ NH3O(aq) +
0.2791 M
+x
H2O(l)
H3O+(aq)
0
+x
As strong base is added to the
weak acid, what will
immediately happen?
The strong base will
dissociate and react
completely with the weak
acid.
Then what will happen?
An equilibrium will be
reached between the conjugate
acid-base pair.
Unit 12
Chapters 15&16 – Acids and Bases
E
(0.4460 ‒ x)
 NH 3O   H 3O
(0.2791 + x)
Page 13
(x)

 0.2791  x  x 
 9.09 x107 

 NH 3OH 
0.4460  x 
x = [H3O+] = 1.453x10‒6 M
pH = ‒log(1.453x10‒6 M) = 5.838
2. 50.00 mL of 1.000 M acetic acid, HC2H3O2, is titrated with 1.000 M NaOH. Ka = 1.8x10–5. Find
the pH of the solution after the following volumes of 1.000 M NaOH have been added:
0.00 mL
Initially, what is occurring?
+
‒
H3O (aq)
The acid is dissociating.
HC2H3O2(aq) + H2O(l) → C2H3O2 (aq) +
What must be the overriding
I 1.000 M
0
0
C ‒x
+x
+x
equation?
E (1 ‒ x)
(x)
(x)
The acid dissociation reaction


C2 H 3O2   H 3O 
 x x
Ka  
 1.8 x105 
 HC2 H 3O2 
1  x
‒3
x = 4.234x10 M
pH = ‒log(4.234x10‒3 M) = 2.373
As strong base is added to the
10.00 mL
weak acid, what will
 0.01000L 1.000M NaOH   0.010000mol
immediately happen?
 0.05000L 1.000M HC2 H 3O2   0.05000mol HC2 H 3O2
The strong base will react
NaOH(aq)
completely with the weak
HC2H3O2(aq) + Na+ + OH‒
→ C2H3O2‒(aq) + H2O(l)
acid.
0.01000 mol
0
I 0.05000 mol
‒x
+x
C ‒x
Then what will happen?
‒ 0.01
0.01
+ 0.01
An equilibrium will be
0 = LR
0.01 mol
E 0.04 mol
reached between the conjugate
These aren’t MOLARITY values! Vtotal = 0.01+0.05 L = 0.06 L
acid-base pair.
H2O(l) → C2H3O2‒(aq) +
H3O+(aq)
HC2H3O2(aq) +
0
I 0.04mol
0.01mol
0.06 L
0.06 L
+x
+x
C ‒x
E  0.04mol

 0.01mol
 (x)
 x
 x


 0.06 L

 0.06 L



C2 H 3O2   H 3O 
Ka  

 HC2 H 3O2 
Ka 

 0.01mol

 0.01mol 
 x   x 
 x

0.06 L
0.06 L 



5

1.8 x10 
 0.04mol

 0.04mol 

x
 0.06 L

 0.06 L 
+
‒5
x = [H3O ] = 7.196x10 M
pH = ‒log(7.196x10‒5 M) = 4.143
25.00 mL
 0.02500L 1.000M NaOH   0.025000mol
The initial amount of HC2H3O2 isn’t going to change!
2013-2014
Always start your calculations
OVER, as if no base were yet
added.
As strong base is added to the
Unit 12
Chapters 15&16 – Acids and Bases
HC2H3O2(aq) +
I 0.05000 mol
C ‒x
‒ 0.025
NaOH(aq)
Na+ + OH‒
0.02500 mol
‒x
0.025
0 = LR
→
C2H3O2‒(aq)
+ H2O(l)
0
+x
+ 0.025
0.025 mol
E 0.025 mol
Hey! Notice anything?
Right, the moles of the conjugate acid (HC2H3O2) and the moles
of conjugate base (C2H3O2‒) are the same! This is called the
half-titration point. Looking over this next step, see if anything
stands out…
Vtotal = 0.025+0.05 L = 0.075 L
H2O(l) → C2H3O2‒(aq) +
H3O+(aq)
HC2H3O2(aq) +
0
I 0.025mol
0.025mol
0.075 L
0.075 L
+x
+x
C ‒x
(x)
E  0.025mol

 0.025mol

 x
 x


 0.075 L

 0.075 L

 0.025mol

 0.025mol 

x
x


 0.075L

 0.075L   x 


5
1.8 x10 

 0.025mol

 0.025mol 
x

 0.075L

 0.075L 
+
‒5
x = [H3O ] = 1.8x10 M
pH = ‒log(1.8x10‒5 M) = 4.745
50.00 mL
 0.05000L 1.000M NaOH   0.050000mol
The initial amount of HC2H3O2 STILL isn’t going to change!
HC2H3O2(aq) +
I 0.05000 mol
C ‒x
‒ 0.05
E 0 = LR
NaOH(aq)
Na+ + OH‒
0.05000 mol
‒x
0.05
0 = LR
→ C2H3O2‒(aq) + H2O(l)
0
+x
+ 0.05
0.05 mol
BOTH are limiting reactants. This is going to change things!!
Vtotal = 0.05+0.05 L = 0.1 L
H2O(l) → HC2H3O2(aq)
H3O+(aq)
C2H3O2‒(aq) +
+
0
0
I 0.05mol
0.1L
+x
+x
C ‒x
(x)
(x)
E  0.05mol

 x

 0.1L

Don’t you dare try to use Ka!!
Kw
1014

 5.6 x1010
Kb 
5
K a 1.8 x10
2013-2014
Page 14
weak acid, what will
immediately happen?
The strong base will react
completely with the weak
acid.
What species dominates?
Both the conjugate acid and
the conjugate base.
What does this mean will
happen?
Half-titration point…
 Did you see it?
If [HA] = [A‒],
then Ka = [H+]
Always start your calculations
OVER, as if no base were yet
added.
As strong base is added to the
weak acid, what will
immediately happen?
The strong base will react
completely with the weak
acid.
What species NOW
dominates?
The conjugate base (C2H3O2‒)
Unit 12
Chapters 15&16 – Acids and Bases
Kb 
 HC2 H 3O2  OH  
C2 H 3O 

2
 5.6 x1010 
Page 15
 x  x 
 0.05mol

 0.1L  x 
x = [OH‒] = 1.67x10‒5 M
pOH = ‒log(1.67x10‒5 M) = 4.776
pH = 14 ‒ 4.776 = 9.224
C. WEAK base with STRONG acid
3. 50.00 mL of 1.000 M methylamine, CH3NH2, is titrated with 0.500 M HCl. Kb for methylamine is
4.4x10–4.
What is the pH before any HCl is added?
Initially, what is occurring?
CH3NH2(aq) +
H2O(l) → CH3NH3+(aq) + OH‒(aq)
Dissociation of the weak base
I
1.000 M
0
0
What must be the overriding
C ‒x
+x
+x
equation?
E (1.000 ‒ x)
(x)
(x)
The base-dissociation
reaction, Kb.
CH 3 NH 3  OH  
 x x
 4.4 x104 
Kb 
CH 3 NH 2 
1.000  x 
‒2
+
x = [CH3NH3 ] = [OH‒] = 2.076x10 M
pH = ‒log(2.076x10‒2 M) = 1.683
What is the pH after 10.00 mL of HCl has been added?
As strong acid is added to the
weak base, what will
 0.01000 L  0.500M HCl   0.00500mol
immediately happen?
 0.05000 L 1.000M CH 3 NH 2   0.05000mol CH 3 NH 2
The strong acid will react
HCl(aq)
completely with the weak
CH3NH2(aq) + H+ + Cl‒
→ CH3NH3+(aq)
base.
0.00500 mol
0
I 0.05000 mol
‒x
+x
C ‒x
‒ 0.005
0.045 mol
0.005
0 = LR
+ 0.005
0.005 mol
E
These aren’t MOLARITY values! Vtotal = 0.01+0.05 L = 0.06 L
H2O(l) → CH3NH3+(aq) +
OH‒(aq)
CH3NH2(aq) +
0
I 0.045mol
0.005mol
0.06 L
0.06 L
+x
+x
C ‒x
E  0.045mol

 0.005mol
 (x)
 x
 x


 0.06 L

 0.06 L

 0.005mol

 x   x

CH 3 NH 3  OH  
0.06 L

Kb 
 4.4 x104  
mol
0.045
C
H
N
H


 3 2
 0.06 L  x 
x = [OH‒] = 3.770x10‒3 M
pOH = ‒log(3.770x10‒3 M) = 2.424
pH = 14 ‒ 2.424 = 11.576
What is the pH at the equivalence point, when 100.00 mL of
HCl has been added?
Reaching the “equivalence point” means that the moles of H+
must equal moles of OH‒. P.S. being generous on identifying the
2013-2014
Then what will happen?
An equilibrium will be
reached between the conjugate
acid-base pair.
Always start your calculations
OVER, as if no acid were yet
added.
As strong acid is added to the
Unit 12
Chapters 15&16 – Acids and Bases
amount of volume. Can you figure it out?
The initial amount of CH3NH2 STILL isn’t going to change!
CH3NH2(aq) +
I 0.05000 mol
C ‒x
‒ 0.05
E 0 = LR
HCl(aq)
H+ + Cl‒
0.05000 mol
‒x
0.05
0 = LR
Vtotal = 0.1+0.05 L = 0.15 L
H2O(l) →
CH3NH3+(aq) +
I 0.05mol
0.15 L
C ‒x
E  0.05mol

 x

 0.15L

Don’t you dare try to use Kb!!
K w 1x1014

 2.3x1011
Ka 
4
Kb 4.4 x10
Ka 
CH 3 NH 2   H 3O  
CH 3 NH 3 
→ CH3NH3+(aq)
0
+x
+ 0.05
0.05 mol
CH3NH2(aq) +
0
H3O+(aq)
0
+x
(x)
+x
(x)
 2.3 x1011 
HCl(aq)
H+ + Cl‒
0.05750 mol
‒x
0.05
0.00750
Vtotal = 0.115+0.05 L = 0.165 L
HCl(aq)
I
0.05mol
0.165 L
→ H+(aq) +
0.05mol
0.165 L
+x
C ‒x
(x)
E  0.05mol

 x

 0.15L

Don’t you dare try to use Kb!!
K
1x1014
 2.3x1011
Ka  w 
Kb 4.4 x104
2013-2014
weak base, what will
immediately happen?
The strong acid will react
completely with the weak
base.
What species dominates?
The weak acid, since the weak
base has completely reacted.
What does this mean will
happen?
Use the acid-dissociation
reaction
 x  x 
 0.05mol

 0.15 L  x 
x = [H3O+] = 2.75x10‒6 M
pH = ‒log(2.75x10‒6 M) = 5.560
What is the pH when 115.00 mL of HCl has been added?
The initial amount of CH3NH2 STILL isn’t going to change!
 0.11500 L  0.500M HCl   0.05750mol
CH3NH2(aq) +
I 0.05000 mol
C ‒x
‒ 0.05
E 0 = LR
Page 16
→ CH3NH3+(aq)
0
+x
+ 0.05
0.05 mol
H3O+(aq)
0
+x
(x)
Always start your calculations
OVER, as if no base were yet
added.
As strong base is added to the
weak acid, what will
immediately happen?
The strong acid will react
completely with the weak
base.
What species NOW
dominates?
Left over strong acid
Unit 12
Chapters 15&16 – Acids and Bases
Ka 
CH 3 NH 2   H 3O  
CH 3 NH 

3
 2.3 x1011 
x = [H3O+] = 2.75x10‒6 M
pH = ‒log(2.75x10‒6 M) = 5.560
Page 17
 x  x 
 0.05mol

 0.15 L  x 
IX. Titration Curves of Weak-Strong Reactions (§16.7)
A. pH at the equivalence point
1. On weak acid/strong base titrations, the pH at the equivalence point is dependent on the acid and
base species involved (mainly the acid).
2. On weak base/strong acid titrations, the pH at the equivalence point is dependent on the acid and
base species involved (mainly the base).
B. pH at the half-equivalence point
1. On weak acid/strong base titrations, the pH at the equivalence point is equal to the pKa of the weak
acid. At the half-way point, the moles of HA = moles of A–. Therefore, when looking at the acid’s
 A    H 3O 
+
–
  H 3O 
dissociation, HA  H + A , the Ka expression becomes K a 
HA


2. On weak base/strong acid titrations, the pOH at the equivalence point is equal to the pKb of the
weak acid. At the half-way point, the moles of BOH = moles of B+. Therefore, when looking at
the bases dissociation, BOH  OH– + B+, the Kb expression becomes
 HB   OH  
Kb 
  OH  
B
 
To find pH then take 14 - pOH = 14 - pKb
2013-2014
Unit 12
Chapters 15&16 – Acids and Bases
Page 18
C. Using the titration curve
The curves below show a monoprotic acid being titrated with 0.10 M NaOH.
How is the equivalence point determined on each graph?
If 0.394 g of the acid were titrated, what would be the molar mass of this acid?
 0.10M  0.037 L   0.0037mol NaOH  0.0037mol HA
0.394 g
 106 g mol
0.0037 mol HA
Equivalence point →
37 mL
How would the titration curve differ if it had been a STRONG acid being titrated with NaOH?
The pH would begin lower and more horizontal. The jump at the eq pt would be more vertical. The pH of the eq
pt would be at 7.00.
2013-2014
Unit 12
Chapters 15&16 – Acids and Bases
Page 19
X. Indicators (§15.8)
A. An indicator is a weak acid or a base, which changes color at a particular pH when the dominant
species changes. The most common indicators are weak acids with complex structures.
B. Predicting information about a titration using an indicator…
1. Obtain the best information about a titration depends on selecting the most appropriate indicator
2. From the list of indicators below, which would be the best choice to determine the most
information from the titration curve on the previous page?
Methyl violet
yellow 0.5 ‒ 1.8 violet
Methyl orange
red 1.0 ‒ 2.8 yellow
Bromthymol blue
yellow 6.0 ‒ 7.6 blue
Alizarin yellow R
yellow 9.9 ‒ 12.0 red
C. Predicting information about the indicator…
The point at which the indicator will change color is related to its Ka.
1. Take a hypothetical indicator, HIn, which has a Ka of 1.0x10–8
2. Its dissociation equation is
H+ +
In–
HIn 
 H    In  
Ka 
red
blue
 HIn 
 In  
Ka
3. Rearranging the Ka expression gives

 H    HIn 
4. If a few drops of this indicator is added to an acidic solution of pH 1.0, then [H+] = 0.1 M and
 In –  1.0x10 8
1
Ka


1.0x10 7 

0.1
10, 000, 000
 H  HIn 
This indicates that the HIn is the dominant species and therefore the color of the indicator will be
red.
5. example;
a. Two drops of indicator HIn (Ka = 1.0x10–9), where HIn is yellow and In– is blue, are placed in
100.0 mL of 0.10 M HCl.
i. What color is the solution initially?
The HCl will dissociate, leaving 0.10 M H+ in solution with the indicator
 In  
 In   108
Ka
109
blue






1
1
yellow
 H   HIn 
10   HIn 
The solution will be blue.
ii. The solution is titrated with 0.10 M NaOH. At what pH will the color change (yellow to
greenish yellow) occur?
In order for the pH to be at the intermediate color, the amount of yellow [HIn] to blue [In‒]
must be equal:  In     HIn 
 In  
Ka
109 1


   H    109


 H   HIn 
 H  1
 log  H     log 109   9
2013-2014
Unit 12
Chapters 15&16 – Acids and Bases
Page 20
XI. Buffers (§16.6)
A. Made of a weak species and its conjugate partner
1. example:
a. What is the pH of a buffer made by mixing 1.00 L of 0.20 M benzoic acid, HC7H5O2, with 3.00
L of 0.060 M sodium benzoate, NaC7H5O2? The Ka for benzoic acid is 6.3x10–5.
Vtotal = 1.00+3.00 L = 4.00 L
Don’t you dare try reach the benzoic acid with the benzoate ion!!
H2O(l) → C7H5O2‒(aq) +
H3O+(aq)
HC7H5O2(aq) +
I 1.00 L  0.20M 
 3.00 L  0.060M  0
4.00 L
4.00 L
‒
x
+
x
+x
C
(x)
E  0.05  x 
 0.045  x 
C7 H 5O2    H 3O  
 0.045  x  x 
 6.3 x105 
Ka  
 HC7 H 5O2 
 0.05  x 
+
‒5
x = [H3O ] = 6.99x10 M
pH = ‒log(6.99x10‒5 M) = 4.156
B. Buffer solution can tolerate addition of acid or base without major change in pH ‒ Buffering action
1. acid component neutralizes added base, producing more of the conj. base component
a. HB + added OH–  HOH + B–
b. acid component [HB] decreases by x, while the base component [B–] increases by x
K HB   x 
c.  H 3O  new  a 
 B   x
d. example:
i. If 1.000 g of NaOH is added to the buffer above, what will be the resulting pH?
Vtotal = 1.00+3.00 L = 4.00 L
React the strong base with the weak acid.

I
HC7H5O2 (aq) +
0.20 mol
C ‒x
‒ 0.025
0.175

NaOH(aq)
Na+ + OH‒
1.00 g  1mol 


 40.00 g 
 0.025mol
‒x
0.025
0 = LR
→ C7H5O2‒(aq) +
0.18 mol
+x
+ 0.025
0.205 mol
H2O(l)
E
Don’t you dare try reach the benzoic acid with the benzoate ion!!
H2O(l) → C7H5O2‒(aq) +
H3O+(aq)
HC7H5O2(aq) +
0
I 0.175mol
0.205mol
4.00 L
4.00 L
+x
+x
C ‒x
(x)
E  0.175

 0.205

 x
 x


 4.00

 4.00

2013-2014
Unit 12
Chapters 15&16 – Acids and Bases
Page 21
 0.205

 x   x

C7 H 5O2    H 3O  
4

 6.3 x105  
Ka 
 0.175 
 HC7 H 5O2 
 x
 4

+
‒5
x = [H3O ] = 5.37x10 M
pH = ‒log(5.37x10‒5 M) = 4.270
2. base component neutralizes added acid, producing more of the conj. acid component
a. B‒ + added H+  HB
b. example:
i. If 0.0250 moles of HCl were added to the original buffer above, what will be the resulting
pH?
Vtotal = 1.00+3.00 L = 4.00 L
React the strong acid with the weak base.
C7H5O2‒(aq) +
I 0.18 mol
C ‒x
‒ 0.025
E 0.155
HCl(aq)
H+ + Cl‒
0.025 mol
‒x
0.025
0 = LR
→ HC7H5O2 (aq) +
0.20 mol
+x
+ 0.025
0.225 mol
H2O(l)
Don’t you dare try reach the benzoic acid with the benzoate ion!!
H2O(l) → C7H5O2‒(aq) +
H3O+(aq)
HC7H5O2(aq) +
0
I 0.225mol
0.155mol
4.00 L
4.00 L
+x
+x
C ‒x
(x)
E  0.225

 0.155

 x
 x


 4.00

 4.00

 0.155

 x   x

C7 H 5O2    H 3O  
4

 6.3 x105  
Ka 
 0.225 
 HC7 H 5O2 
 x
 4

+
‒5
x = [H3O ] = 9.11x10 M
pH = ‒log(9.11x10‒5 M) = 4.040
3. “Better” Buffers
a. Buffering action depends on two things:
i. Ratio of conj. acid to its conj. base
K HA 
HA ratio is crucial
where the
from Ka:  H 3O   a 
 A 
 A  
ii. Concentration of the buffer ‒ number of moles of conj. acid and its conj. base
4. note that addition of large amount of acid or base at one time would overwhelm and destroy the
buffer; buffer survives alternating additions of acid and base
5. example: bicarbonate buffer system in the bloodstream, which controls pH and rate of breathing
C. While a buffer contains both acidic and basic species, both Ka and Kb still true
D. Choice of buffer system – Which buffer is better for a given situation?
for equimolar sol’n, [HA] = [A–] so ratio = 1
1. [H3O+] = Ka
2. pH = pKa
2013-2014
Unit 12
Chapters 15&16 – Acids and Bases
Page 22
pKa gives info for choice of suitable buffer system for a pH need
HA ratio to fine-tune pH
3. manipulate the
 A  
E. example:
1. When 21.0 mL of 0.125 M hydroxylamine, NH2OH, is mixed with 14.7 mL of 0.155 M
hydroxylammonium chloride, what is the pH? The Kb for hydroxylamine is 1.1x10–8.
Vtotal = 21.0+14.7 L = 35.7 mL
The reaction shown is not of the Kb, although the same type of calculations could be done
with Kb…
H2O(l) → NH2OH(aq) +
H3O+(aq)
NH2O‒(aq) +
I  0.0147 L  0.155M 
 0.0210 L  0.125M  0
0.0357 L
0.0357 L
+x
+x
C ‒x
(x)
E  0.0638  x 
 0.0735  x 
Ka 
Ka 
K w 1x1014

 9.09 x107
Kb 1.1x108
 NH 2OH   H 3O  
 9.09 x107 
0.0735  x  x 
0.0638  x 
 NH 2O 
x = [H3O+] = 7.89x10‒7 M
pH = ‒log(7.89x10‒7 M) = 6.103
Add 5.0 mL of 0.010 M KOH to this buffer. What is the new pH?
Vtotal = 21.0+14.7+5.0 L = 40.7 mL
React the strong base with the weak acid.
I

NH2O‒(aq) +
NaOH(aq)
Na+ + OH‒
 0.0228mol
 0.00005mol
 0.02625mol
‒ 0.00005
0.0022285 mol
‒x
0.00005
0 = LR
+x
+ 0.00005
0.002675 mol
NH2O‒(aq) +
H2O(l) → NH2OH(aq) +
 0.0147 L  0.155M   0.0050 L  0.010M 
C ‒x
E
I
0.0022285mol
0.0407 L
C ‒x
E  0.0022285

 x

 0.0407

→ NH2OH(aq) +
 0.0210 L  0.125M 
0.002675
0.0407L
+x
 0.002675

 x

 0.0407

 0.002675

 x   x


 NH 2OH   H 3O 
0.00407

 9.09 x107  
Ka 

0.0022285
 NH 2O 


 0.00407  x 
x = [H3O+] = 7.57x10‒7 M
pH = ‒log(7.57x10‒7 M) = 6.121
2013-2014
H3O+(aq)
0
+x
(x)
H2O(l)
Unit 12
Chapters 15&16 – Acids and Bases
Page 23
2. How many grams of NaC2H3O2 need to be added to a 13.0 mL sample of 0.134 M acetic acid in
order to make a buffer solution with a pH of 5.0? Assume no volume change.
The Ka for acetic acid is 1.8x10–5.
Since the mass of NaC2H3O2 is requested, solve for the moles of its anion, the conjugate base,
C2H3O2‒.
The pH represents the concentration of the H+ at equilibrium.
HC2H3O2(aq) +
I 0.134M
C ‒x
E (0.134 ‒ 10‒5)
H2O(l)
→ C2H3O2‒(aq) +
Y
+x
(Y + 10‒5)
H3O+(aq)
0
+x
10‒5
C2 H 3O2   H  
Y  105  105 
Ka  
 1.8 x105  
5
 HC2 H 3O2 
 0.134  10 
Y = 0.241 M C2H3O2‒ = 0.241 M NaC2H3O2
 0.0130 L   0.241mol   82.04 g 



  0.257 g
1L


  1mol 
XII. Polyprotic acids
A. Definition:
An acidic species that has more than one protonizable hydrogen. Remember that not all hydrogen
atoms are protonizable – they typically must be directly bonded to an oxygen atom.
B. Ka of polyprotic acids
1. example:
a. Tartaric acid, H2C4H4O6, is a diprotic acid used in food products. Write the dissociation
reactions that correspond with the following dissociation constants:
Ka1 = 9.2x10–4 and Ka2 = 4.3x10–5.
H2C4H4O6 + H2O → HC4H4O6‒ + H3O+
Ka1 = 9.2x10–4
HC4H4O6‒ + H2O → C4H4O62‒ + H3O+
Ka2 = 4.3x10–5
‒
Notice how HC4H4O6 is acting amphiprotically.
C. Polyprotic acids in titration
2013-2014
Unit 12
Chapters 15&16 – Acids and Bases
nd
Page 24
rd
D. first ionization energy proceeds to much greater extent than the 2 and 3 , so strength of acid depends
only on the first ionization
1. example:
a. Tartaric acid, H2C4H4O6, is a diprotic acid used in food products. What is the pH of a 0.10 M
solution and the concentration of the C4H4O62– ion?
Ka1 = 9.2x10–4 and Ka2 = 4.3x10–5.
H2C4H4O6 + H2O → HC4H4O6‒ + H3O+
Ka1 = 9.2x10–4
‒
H2C4H4O6 (aq) +
H2O(l)
→ HC4H4O6 (aq) +
H3O+(aq)
0
0
I 0.10M
+x
+x
C ‒x
(x)
(x)
E (0.10 ‒ x)
 HC4 H 4O6   H  
 x  x 
 9.2 x104 
K a1 
 H 2C4 H 4O6 
 0.10  x 
‒
‒3
x = [HC4H4O6 ] = [H3O+] = 9.1x10 M
Notice that the H+ concentration is going to be the same between Ka1 and Ka2.
HC4H4O6‒ + H2O → C4H4O62‒ + H3O+
Ka2 = 4.3x10–5
2‒
HC4H4O6‒ (aq) +
H3O+(aq)
H2O(l)
→ C4H4O6 (aq) +
0
I 9.1x10‒3 M
9.1x10‒3 M
+x
+x
C ‒x
‒3
(x)
E (9.1x10 ‒ x)
(9.1x10‒3 + x)
C4 H 4O62   H  
0.0091  x  x 
K a1 
 4.3 x105 

 HC4 H 4O6 
0.0091  x 
x = [C4H4O62‒] = 4.3x10‒5 M
pH = ‒log[H3O+] = ‒log(9.1x10‒3 M) = 2.04
2013-2014