ACIDS

1
ACIDS BASES Acids - acids are sour
tasting - Arrhenius acid Any substance that
when dissolved in water, increases the
concentration of hydronium ion (H3O) -
Bronsted-Lowry acid A proton donor - Lewis
Acid An Electron acceptor Bases - bases are
bitter tasting and slippery - Arrhenius base
Any substance that when dissolved in water,
increases the concentration of hydroxide ion
(OH-) - Bronsted-Lowery base A proton
acceptor - Lewis base An electron donor
2
HBr KOH ? H3O NH3 ? HBr NH3
? NH3 H2O ? ?
3
STRONG VS WEAK - completely
ionized - partially ionized - strong
electrolyte - weak electrolyte - ionic
bonds - some covalent bonds STRONG
ACIDS STRONG BASES HClO4 LiOH H2SO4 N
aOH Hl KOH HBr Ca(OH)2 HCl Sr(OH)2 HNO
3 Ba(OH)2
4
table 18.4 THE CONJUGATE PAIRS IN SOME
ACID-BASE REACTIONS
Conjugate Pair Acid Base ?? Base
Acid Conjugate Pair Reaction 1 HF
H2O ?? F- H3O Reaction 2 HCOOH
CN- ?? HCOO- HCN Reaction 3 NH4
CO32- ?? NH3 HCO3- Reaction
4 H2PO4- OH- ?? HPO42-
H2O Reaction 5 H2SO4 N2H5 ?? HSO4-
N2H62 Reaction 6 HPO42- SO32- ?? PO43-
NSO3-
5
Fig. 16.7 CONJUGATE ACID-BASE PAIRS
ACID BASE HCl Cl-
H2SO4 HSO4- HNO3 NO3- H(aq) H2O
HSO4- SO42- H3PO4 H2PO4 HF F- HC2H3O
2 C2H3O2 H2CO3 HCO3- H2S HS- H2PO4- H
PO42- NH4 NH3 HCO3- CO32- HPO42- PO43
- H2O OH- HS- S2- OH- O2- H2 H-
100 percent ionized in H2O
strong
negligible
Base strength increases ? ? ? ?
? ? ? ? Acid strength increases
weak
weak
100 percent protonated in H2O
negligible
strong
6
The strength of an acid depends on how easily the
proton, H, is lost or removed from an H - x
bond. Greater Acid Strength - more polar
bonds - larger x atom - oxo acids higher
electronegativity - oxo acids more oxygen
atoms - oxo acids more hydrogen atoms List
the following in order of increasing
strength l. HI, HF, HCl 2. H2O, CH4, HF 3.
HIO3, HClO3, HBrO3 4. HBrO, HBrO3, HBrO2 5. HI,
H2SO4, HClO4, HNO3
7
A. H - O - Cl gt H - O - Br gt H -
O - l ? ?- ?
?- ? ?-
O

B. H - O - Cl ltlt H - O - Cl O
? ?- ? ?-
O
8
Acid Base HCl Cl- H2SO4 HSO4- HNO3 NO3- H3
O H2O HSO4- SO42- H2SO3 HSO3- H3PO4 H2PO4
- HF F- CH3COOH CH3COO- H2CO3 HCO3- H2S HS-
HSO3- SO32- H2PO4- HPO42- NH4 NH3 HCO3-
CO32- HPO42- PO43- H2O OH- HS- S2- OH- O2-
A C I D S T R E N G T H
B A S E S T R E N G T H
STRONG
WEAK NEGLIGIBLE
NEGLIGIBLE WEAK STRONG
9
ACID STRENGTH STRONG ACID More polar
? weaker bond larger atoms weaker bond (HF
weak HI strong) strength increases down a
group OXO Acids HIO lt HClO
stronger (more electrones) (more O) (more
H) 1. HF lt HCl lt HI strong 2. CH4 lt H2O
lt HF 3. HIO3 lt HBrO3 lt HClO3 4. HBrO lt
HBr2O lt HBrO3 5. HI lt HNO3 lt H2SO4 lt
HClO4
10
AUTO - IONIZATION A reaction in which
two like molecules react to give Ions. 2 H2O
?? H3O OH- K H3O OH-
but H2O is essentially
H2O2
constant ? KH2O2 H3O OH-
Kw H3O OH- Kw Ion-product constant
for water. Kw 1 x 10-14 at 25C
11
Fig. 16.2 pH of Some Common Solutions
pH H OH- pOH --14 1 x
10-14 1 x 10-0 0
--13 1 x 10-13 1 x 10-1 1
--12 1 x 10-12 1 x 10-2
2 --11 1 x 10-11 1 x
10-3 3 --10 1
x 10-10 1 x 10-4 4
-- 9 1 x 10-9 1 x 10-5 5
-- 8 1 x 10-8 1 x 10-6
6 -- 7 1 x 10-7 1
x 10-7 7 -- 6 1
x 10-6 1 x 10-8 8
-- 5 1 x 10-5 1 x 10-9 9
-- 4 1 x 10-4 1 x 10-10
10 -- 3 1 x 10-3 1 x
10-11 11 -- 2
1 x 10-2 1 x 10-12 12 --
1 1 x 10-1 1 x 10-13 13 --
0 1 x 100 1 x 10-14 14
M O R E B A S I C
NaOH, 0.1 M.. Household bleach.. Househol
d ammonia. Lime Water Milk of
Magnesia.. Borax. Baking
Soda. Egg White, Sea Water.. Human blood,
Tears..
M O R E A C I D I C
Milk. Saliva Rain.. Bl
ack Coffee. Banana. Tomatoes.
Wine. Cola, Vinegar.. Lemon
Juice
Gastric Juice..
12
pH I. Kw H OH- take log Log Kw
Log H OH- Log H log
OH- p Kw pH pOH 14
pH pOH II. pOH -Log OH- 1. A
0.0015M NaOH solution has what pH? pOH?
OH- 2. A soda has pH of 2.94, what is the
H? 3. A solution has pOH of 12.7, what is
the H?
13
WEAK ACIDS HA(aq) ?? H (aq) A-
(aq) Ka H A- Ka acid
dissociation constant HA- The magnitude of
Ka refers to the strength of the acid. Small Ka
value weak acid Q. A student prepared a 0.10M
solution of formic acid HCHO2 and measured its
pH, at 25C, pH 2.38 a) calculate Ka b)
what percent of acid Ionizes?
14
GENERAL STEPS FOR CALCULATING THE pH
(pOH) OF A WEAK
ACID (BASE) Step 1 Write a balanced chemical
equation describing the action. Step 2 Make a
list of given and implied information. Step
3 Write the equilibrium constant equation
associated with the balanced chemical equation
in Step 1. Step 4 An equilibrium table should
be set up since we are dealing with a weak acid
(partially dissociated species). The table
should describe the changes which occurred in
order to establish equilibrium. Step
5 Substitute the equilibrium values from Step 4
into the equilibrium constant equation in Step
3. Solve for x. If the expression can not be
solved with basic algebra, try either the
quadratic equation or the successive-approximatio
n method. Step 6 Calculate the pH (pOH) using
the expression pH -Log H or pOH
-Log OH-
15
CALCULATING pH FOR A WEAK ACID Step
1 Calculate the pH of a 0.20 M
HCN solution. Step 2 Calculate the percent
of HF molecules ionized in a 0.10 M HF
solution. Step 3 Compare the above value to
the percent obtained for a 0.010 M
HF solution.
16
POLYPROTIC ACIDS H2SO3 ?? H HSO3-
Ka1 1.7 x 10-2 HSO3-
?? H SO32- Ka2 6.4 x 10-8 Ka1 gt
Ka2 The solubility of CO2 in pure H2O at 25ºC
and 0.1 atm is 0.0037 M. A) What is the pH of
a 0.0037 M solution of H2CO3? B) What is
the CO32- produced?
17
WEAK BASES B- H2O ?? HB
OH- K HB OH- B- H2O Kb
KH2O HB OH-
B- base dissociation
constant Q. Calculate OH- and pH of a 0.15M
NH3 solution.
18
1. Calculate the pH of a 0.0850 M HC2H3O2
solution. 2. What is the molarity of an aqueous
HCN solution if the pH is 5.7? 3. Calculate
the pOH of a 0.351 M aqueous solution of
NH3. 4. Calculate the pH of a 0.025M solution
of citric acid. Ka (acetic acid) 1.8 x
10-5 Kb (ammonia) 1.8 x 10-5 Ka
(hydrocyanic) 4.9 x 10-10 Ka2 (citric acid)
1.7 x 10-5 Ka1 (citric acid) 7.4 x 10-4 Ka3
(citric acid) 4.0 x 10-7
19
RELATIONSHIP BETWEEN Ka AND
Kb A. NH4 ?? NH3 H B. NH3 H2O
?? NH4 OH- Ka NH3
H NH4 Kb NH4 OH-
NH3 Add equation A to equation B to get
the net reaction H2O ?? H
OH- C. Equation A Equation B Equation
C K1 x K2
K3 KaKb NH3 H NH4 OH-
OH- H Kw NH4
NH3
20
CALCULATE Kb for F- if Ka 6.8 x 10-4 Kb
Kw 1 x 10-14 Ka 6.8 x
10-4 1.5 x 10-11
21
ACID/BASE PROPERTIES OF SALT SOLUTIONS HYDROLYSIS
Ions react with water to generate either
H or OH- A- H2O ?? HA OH- Q.
Predict whether Na2 HPO4 will form an acidic
or basic solution. Q. Predict whether
K2HC7H5O7 will form an acidic or basic
solution.
22
1. SA/SB neither cation nor anion
hydrolyzes 2. SB/WA anion strong CB ?
hydrolyzes to produce OH- Basic 3.
WB/SA cation strong CA ? hydrolyzes to produce
H Acidic 4. NH4CN CN- Kb 2 x
10-5 NH4 Ka 5.6 x 10-10 CN-
hydrolyzes more than NH4 pH gt 7 basic more
basic than acidic FeCO3 Fe3 6 x
10-3 Ka CO32- Ka1 4.5 x 10-7 Ka2
4.7 x 10-11 Kb1 2.2 x
10-8 Kb2 2.13 x 10-4 table 18.7 pg. 786
metal ions
23
SALT SOLUTIONS 1. Salts derived from
strong bases and strong acids have pH
7 NaCl Ca(NO2)2 2. Salts derived from strong
bases and weak acids have pH gt
7 NaClO Ba(C2H3O2)2 3. Salts derived from
weak bases and strong acids have pH lt
7 NH4Cl Al(NO3)3 4. Salts derived from weak
base and weak acids, pH is dependent on
extent NH4CN Fe3(CO3)2 NH4C2H3O2
24
Table 18.8 THE BEHAVIOR OF SALTS IN
WATER Salt Solution pH Nature of Ions Ion
that reacts (Examples) with
water Neutral 7.0 Cation of strong base None
NaCl, KBr, Anion of strong acid
Ba(NO3)2 Acidic lt7.0 Cation of
weak base Cation NH4Cl, NH4NO3, Anion of
strong acid CH3NH3Br Acidic
lt7.0 Small, highly charged Cation
Al(NO3)3, cation CrCl3,
FeBr3 Anion of strong acid Basic
gt7.0 Cation of strong base Anion
CH3COONa, Anion of weak acid KF, Na2CO3
25
CALCULATING Ph OF SALT SOLUTIONS Q.
Household bleach is 5 solution of
sodium hypochlorite NaClO. Calculate the
OH- and pH of a 0.70 M NaClO solution. Kb
2.86 x 10-7
26
COMMON ION EFFECT A shift of an
equilibrium induced by an Ion common to the
equilibrium. HC7H5O2 H2O ?? C7H5O2-
H3O Benzoic Acid Calculate the degree of
ionization of benzoic acid in a 0.15 M solution
where sufficient HCl is added to make 0.010 M
HCl in solution. Compare the degree of
ionization to that of a 0.15 M benzoic Acid
solution Ka 6.3 x 10-5
27
COMMON ION EFFECT HC2H3O2 ?? H
C2H3O2- NaC2H3O2 strong electrolyte HC2H3O2 w
eak electrolyte Addition of NaC2H3O2 causes
equilibrium to shift to the left ?, decreasing
H eq Dissociation of weak acid decreases by
adding strong electrolyte w/common Ion.
Predicted from the Le Chateliers
Principle. Q. What is the pH of a solution
made by adding 0.30 mol of acetic acid and 0.30
mol of sodium acetate to enough water to make
1.0 L of solution?
28
1. Identify the major species in the solution
and consider their acidity or basicity HC2H3O2
Na C2H3O2- H2O
WA neut CB amphoteric
spectator 2. Identify important
equilibrium Rx H2O ltltlt HC2H3O2 acidity ? pH
controlled by HC2H3O2 ? HC2H3O2 H2O ??
H3O C2H3O2- or HC2H3O2 ?? H
C2H3O2- but remember H ? C2H3O2- coz
additional NaCl2
29
Calculate F- and pH of a solution containing
0.10 mol of HCl and 0.20 mol of HF in a 1.0 L
solution. 1. HCl SA HF WA H
Cl- HF H2O eq H F- 2. HF ?? H F-
30
BUFFERS A buffer is a solution
characterized by the ability to resist changes in
pH when limited amounts of acids or bases
are added to it. Buffers contain both an acidic
species to neutralize OH- and a basic species to
neutralize H3O. An important characteristic of
a buffer is its capacity to resist change in pH.
This is a special case of the common Ion effect.
31
Buffer after addition Buffer with
equal Buffer after of H3O
concentrations of addition of OH-
conjugate acid base
CH3COOH
CH3COO-
CH3COO-
CH3COOH
CH3COO-
CH3COOH
H3O ? ?
OH- ??
H2O CH3COOH ? H3O CH3COO-
CH3COOH OH- ? CH3COO- H2O
32
BUFFER 1. What is the pH of a buffer
that is 0.12 M in lactic acid (HC3H5O3) and 0.10
M sodium lactate? Lactic acid Ka 1.4 x
10-4 2. How many moles of NH4Cl must be added
to 2.0 L of 0.10 M NH3 to form a buffer whose
pH is 9.00?
33
Buffer after Buffer with equal Buffer
after addition of concentrations of
addition of OH- weak acid and its H
conjugate base
X-
HX
HX
X-
HX
X-
H
OH-
OH- HX ? H2O X-
H X- ? HX
34
PROCEDURE FOR CALCULATION OF pH (buffer)
Neutralization
Add strong acid
X- H3O ? HX H2O
Use Ka, HX and X- to calculate H
Buffer containing HA and X-
Recalculate HX andX-
pH
Neutralization
HX OH- ? X- H2O
Add strong base
Stoichiometric calculation
Equilibrium calculation
35
ADDITION OF A STRONG ACID OR STRONG BASE TO A
BUFFER
A buffer is made by adding 0.3 mol of acetic acid
and 0.3 mol of sodium acetate to 1.0 L of
solution. If the pH of the buffer is 4.74 A.
Calculate the pH of a solution after 0.02 mol of
NaOH is added B. after
0.02 mol HCl is added.
36
Henderson - Hasselbach Equation
- log H - log Ka HX X-
- log Ka - log HX X- Since
- log H pH and - log Ka pKa pH
pKa - log HX X- pKa log
X- XH pH pKa
log base
acid when base acid pH pKa
37
PROCEDURE FOR CALCULATION OF pH (TITRATION)
Neutralization
Solution containing weak acid and strong base
Calculate HX and X- after reaction
HX OH- ? X- H2O
Use Ka, HX, and X- to calculate H
pH
Stoichiometric calculation Equilibrium
calculation
38
ACID - BASE TITRATION For a strong
acid reacting with a strong base, the point
of neutralization is when a salt and water is
formed ? pH ?. This is also called the
equivalence point. Three types of titration
curves - SA SB - WA SB - SA
WB 1. Calculate the pH if the following
quantities of 0.100 M NaOH is added to 50.0
mL of 0.10 M HCl. A. 49.0 mL B. 50.0 mL C.
51.0 mL
39
STRONG BASE WITH WEAK ACID WA OH-
?? A- H2O for each mole of OH- consumer 1
mol WA to produce 1 mol of A- when WA is in
excess, need to consider proton transfer between
WA and H2O to create A- and H3O WA
H2O ?? A- H3O 1. Stoichiometric
calculation allow SB to react with WA, solution
product WA CB 2. Equilibrium
calculation use Ka and equil. to calculate WA
and CB and H
40
Question If 30.0 ml of 0.200 M acetic acid,
HC2H3O2, is titrated with 15.0 ml of 0.100 M
sodium hydroxide, NaOH, what is the pH of the
resulting solution? Ka for acetic acid is 1.8 x
10-5. Step 1 Write a balanced chemical
equation describing the action HC2H3O2
OH- ? C2H3O2- H2O Step 2 List all
important information under the chemical
equation HC2H3O2 OH- ? C2H3O2-
H20 0.200 M 0.100M 30.0 ml
15.0 ml Step 3 How many moles are
initially present? What are we starting with
before the titration? n(HC2H3O2)i MV
(0.030 L)(0.200 M) 0.006 moles n(OH-)i
MV (0.015 L)(0.100 M) 0.0015
moles
41
Step 4 Since we are dealing with a weak acid,
ie., partially dissociated, an equilibrium can
be established. So we need to set up a table
describing the changes which exist during
equilibrium. HC2H3O2 OH- ? C2H3O2-
H2O i 0.006 0.0015 0
--- ?
-0.0015 -0.0015 0.0015
--- eq 0.0045
0 0.0015 --- HC2H3O2
n/V 0.0045/0.045 L 0.100 M
C2H3O2- n/V 0.0015/0.045 L
0.033 M Step 5 To calculate the pH, we must
first calculate the H HC2H3O2 ? C2H3O2-
H Ka C2H3O2- H/HC2H3O2 1.8 x
10-5 solve for H KaHC2H3O2/C2H3O2-
(1.8 x 10-5)(0.100)/0.033 5.45 x 10-5 M Step
6 Calculate the pH from pH -Log
H pH 4.26
42
Question If 30.0 ml of 0.200 M acetic acid,
HC2H3O2, is titrated with 0.100 M sodium
hydroxide, NaOh, what is the pH at the
equivalence point? Ka for acetic acid is 1.8 x
10-5. Step 1 Calculate the number of moles of
base used to reach the equivalence
point. n(HC2H3O2)i (0.030 L)(0.200 M)
0.006 moles there is 11 mole ratio
between the acid and the base therefore 0.006
moles of base are needed to reach the
equivalence point. This corresponds to 60 ml of
0.10M NaOH. The molarity of the base solution
titrated is equal to the moles of C2H3O2-
produced/total volume of solution 0.006
moles/0.090 L 0.067 M
43
Step 2 At the equivalence point, the solution
contains NaC2H3O2, so we may treat this
problem similar to the calculation for the pH
of a salt solution. NaC2H3O2 H2O ?
HC2H3O2 OH- i 0.067 ---
0 0 ? -x
--- x
x eq 0.067-x --- x
x Kb HOAc OH- / OAc-
5.556 x 10-10 x x/0.067 therefore
by solving for x OH- 6.1 x 10-6 M pOH
-Log OH- 5.21 pKw - pOH
pH 14 - 5.21 8.79
44
If 30.0 ml of 0.200 M acetic acid, HC2H3O2, is
titrated with 15.0 ml of 0.100 M sodium
hydroxide, NaOH, A) What is the pH of the
resulting solution? Ka for acetic acid is 1.8
x 10-5. B) What is the pH at the equivalence
point?
45
Calculate pH is titration of HOAc by NaOH after
30 ml of 0.10 M NaOH has been added to 50 ml of
0.100 M HOAc. Step 1 Stachiometry MAVA
0.10 mol HOAc (.05 L) nA. 5 x 10-3 HOAc
mol 1 L MBVB 0.10 mol
(.030 L) nB. 3 x 10-3 mol NaOH
1 L OH- HOAc ?? OAC-
H2O i 3 x 10-3mol 5 x 10-3
0 --- ? -3 x
10-3mol -3 x 10-3 3 x 10-3
---- eq 0 2 x 10-3
3 x 10-3 ---- total volume
80 mL HOAceq 2 x 10-3/.08 L 0.0250
M OAc-eq 3 x 10-3/.08 L 0.0375 M
46
Step 2 equilibrium Ka H OAc- 1.8
x 10-5 HOAc H Ka HOAc 1.8
x 10-5 .025 1.2 x 10-5
OAc
.0375 if 25 mL is used VT 75 mL NA
1.25 x 10-3 mol HBz NB (0.05 mol/L)(.025L)
1.25 x 10-3 mol OH- Bz- 1.25 x
10-3/.075 0.01667 Bz H2O ??
HBz OH- i .01667 ---
0 0 ? -x
--- x
x eq .01667-x x
x Kb HBzOH- x2
1.54 x 10-10
Bz- .01667-x OH- 1.60 x 10-6
pOH 5.80 pH 8.20
47
Weve seen what happens when a strong acid is
titrated with a strong base but what happens
when a weak acid is titrated? What is the
fundamental difference between a strong acid and
a weak acid? To compare with what we learned
about the titration of a strong acid with a
strong base, lets calculate two points along the
titration curve of a weak acid, HOAc, with a
strong base, NaOH. Q If 30.0 mL of 0.200 M
acetic acid, HC2H3O2, is titrated with 15.0 ml of
0.100 M sodium hydroxide, NaOH, what is the pH of
the resulting solution? Ka for acetic acid is
1.8 x 10-5. Step 1 Write a balanced chemical
equation describing the action HC2H3O2 OH-
? C2H3O2 H2O why did I exclude
Na? Step 2 List all important information
under the chemical equation HC2H3O2
OH- ? C2H3O2 H2O
0.20 M 0.10M
30mL
15mL
48
Step 3 How many moles are initially present?
What are we starting with before the
titration? n(HOAc)i (0.03 L)(0.200M)
0.006 moles n(OH-)i (0.015L)(0.100M)
0.0015 moles Q What does this calculation
represent? A During titration OH- reacts with
HOAc to form 0.0015 moles of
Oac- leaving 0.0045 moles of HOAc left in
solution. Step 4 Since we are dealing with a
weak acid, ie., partially dissociated, an
equilibrium can be established. So we need to
set up a table describing the changes which exist
during equilibrium. HC2H3O2 OH-
? C2H3O2- H2O i
0.006 0.0015 0
--- ? -.0015
-.0015 0.0015 eq
0.0045 0
0.0015 HOAc NN 0.0045/0.045 L
0.100 M Oac- NN 0.0015/0.045 L
0.033 M
49
Step 5 To calculate the pH, we must first
calculate the H Q What is the
relationship between H and pH? A
acid-dissociation expression, products over
reactants. Q Which reaction are we
establishing an equilibrium
acid-dissociation expression for? HC2H3O2 ?
C2H3O2- H Ka Oac- H/HOAc
1.8 x 10-5 solve for H KaHOAc/OAc-
(1.8 x 10-5)(0.100)/0.033 5.45 x 10-5
M Step 6 Calculate the pH from pH -Log
H pH 4.26
50
So at this point, we have a pH of 4.26, Is this
the equivalence point? Is the equivalence point
at pH 7 as with a strong acid titration? Q
By definition, how is the equivalence point
calculated? A moles of base moles of
acid Lets calculate the pH at the equivalence
point. Step 1 Calculate the number of moles of
base used to reach the equivalence
point. n(HOAc)i (0.03 L)(0.200 M) 0.006
moles there is a 11 mole ration between the
acid and the base therefore 0.006 moles of
base are needed. This corresponds to 60 ml of
0.10 M NaOH. The molarity of the base solution
titrated is moles of OAc- produced/total
volume 0.006 moles/0.090 L 0.067 M
51
Step 2 At the equivalence point, the solution
contains NaOAC, so we may treat this problem
similar to the calculation of the pH of a
salt solution. NaC2H3O2 H2O ? HC2H3O2
OH- i 0.067 ---
0 0 ? -x
x
x eq 0.067-x
x x Kb HOAcOH-/OAc-
5.556 x 10-10 x x /0.067 x
OH- 6.1 x 10-6 pOH -LogOH-
5.21 pKw - pOH pH 14 - 5.21 8.79
at the equivalence
point
52
KEY POINTS 1. Weak acid has a higher pH since it
is partially dissociated and less H is
present 2. pH rises rapidly in the beginning
and slowly towards the equivalence point. 3.
The pH at the equivalence point is not 7 (only
applies to strong acid titration).