Acid and Base Equilibria
Electrolytes StrongConduct electricity WeakPoor conductors of
electricity NonelectrolytesDo not conduct
electricity
Strong Electrolytes Completely ionize or dissociate in
dilute aqueous solutions. Strong acids Strong soluble bases Most soluble salts
HCl + H2O H3O+ + Cl-
KOH K+ + OH-
Concentration of Ions Calculated
directly from molarity of strong electrolyte
What is the concentration of hydrogen ion in a 0.25 M HCl solution?
HCl H+ + Cl-
initial 0.25 Mchange -0.25 M 0.25M 0.25Mfinal 0 M 0.25M 0.25M
Concentration of Ions Calculated
directly from molarity of strong electrolyte
What is the concentration of hydroxide ion in a 0.25 M Ba(OH)2 solution?
Ba(OH)2 Ba2+ + 2OH-
initial 0.25 Mchange -0.25 M 0.25M 2(0.25M)final 0 M 0.25M 0.50M
Weak Electrolyte Slightly ionized in dilute aqueous
solutions. Weak acids Weak bases A few covalent salts
Auto-Ionization of Water
H2O(l) + H2O(l) H3O+(aq) + OH-
(aq)
Kw = [H3O+][OH-]OH-
H3O+
OH-
H3O+
Auto-Ionization of Water
Kw = [H3O+] [OH-] = 1.00 x 10-14 at 25oC
In a neutral solution [H3O+] = [OH-]
and so [H3O+] = [OH-] = 1.00 x 10-7 M
OH-
H3O+
OH-
H3O+
H2O + H2O H3O+ + OH-
Kw = [H3O+] [OH-]
Kw
MxOH
x
xOH
OHxx
9
5
14
514
1000.1
1000.1
1000.1
1000.11000.1
H2O + H2O H3O+ + OH-
The Auto-Ionization of Water Calculate the concentrations of H3O+
and OH- in 0.050 M HCl.
M
MMM
0.050OH thus
0.050 0.050 050.0
Cl OH OH + HCl
+3
+32
The Auto-Ionization of Water Calculate the concentrations of H3O+
and OH- in 0.050 M HCl.
M
M
MMM
13
2
14
+3
14
14+3
+3
+32
100.2OH
100.5100.1
OH100.1
OH
100.1OHOH
0.050OH thus
0.050 0.050 050.0
Cl OH OH+HCl
[H3O+], [OH-] and pH A common way to express acidity
and basicity is with pH
pH = - log [H3O+] In a neutral solution, [H3O+] = [OH-] = 1.00 x 10-7 at 25oCpH = -log (1.00 x 10-7) = - (-7) pH = 7.00
[H3O+], [OH-] and pH
What is the pH of the 0.0010 M NaOH solution?
[H3O+] = 1.0 x 10-11 M
pH = - log (1.0 x 10-11) = 11.00General conclusion — Basic solution pH > 7 Neutral pH = 7 Acidic solution pH < 7
MxH
xH
Hx
11
14
14
1000.1
001.0
1000.1
001.01000.1
If the pH of Coke is 3.12, What is [H3O+]?
Because pH = - log [H3O+] then
log [H3O+] = - pHTake antilog and get
[H3O+] = 10-pH
[H3O+] = 10-3.12
[H3O+] = 7.6 x 10-4 M
[H3O+], [OH-] and pH
Other pX Scales
In general pX = -log Xand so pOH = - log
[OH-] Kw = [H3O+] [OH-] = 1.00 x 10-14
Take the -log of both sides-log (10-14) = - log [H3O+] + (-log
[OH-])14 = pH + pOH
pKa Values The Ka of acetic acid is 1.8 x 10-5,
what is the pKa?
pKa = -log Ka = -log (1.8 x 10-5) = 4.74
pH
pH
[H+][H+][OH-][OH-]
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
The pH and pOH scales Develop familiarity with pH scale by
looking at a series of solutions in which [H3O+] varies between 1.0 M and 1.0 x 10-14 M.[H3O
+] [OH-] pH pOH 1.0 M 1.0 x 10-14 M 0.00 14.00
1.0 x 10-3 M 1.0 x 10-11 M 3.00 11.00 1.0 x 10-7 M 1.0 x 10-7 M 7.00 7.00 2.0 x 10-12 M 5.0 x 10-3 M 11.70 2.30 1.0 x 10-14 M 1.0 M 14.00 0.00
The pH and pOH scales Calculate [H3O+], pH, [OH-], and pOH for
0.020 M HNO3 solution.
70.1100.2-logpH 100.2OH
0.020 0.020 020.0
NOOHOHHNO
223
-33
100%23
MM
MMM
The pH and pOH scales Calculate [H3O+], pH, [OH-], and pOH for
0.020 M HNO3 solution.
30.12100.5logpOH
100.5100.2100.1
OH100.1
OH
100.1OHOH
70.1100.2-logpH 100.2OH
0.020 0.020 020.0
NOOHOHHNO
13
132
14
3
14
143
223
-33
100%23
M
MM
MMM
pH Examples
Ionization Constants for Weak Monoprotic Acids Let’s look at the dissolution of acetic acid, a
weak acid, in water as an example. The equation for the ionization of acetic
acid is:
COOCHOHOH COOHCH -3323
COOHCH
COOCHOHK
3
33c
Ionization Constants for Weak Monoprotic Acids and Bases Write the equation for the ionization of
the weak acid HCN and the expression for its ionization constant.
10-+
a 100.4HCN
CNHK
HCN H+ + CN-
Equilibria Involving Weak Acids and Bases
Consider acetic acid
HOAc + H2O H3O+ + OAc-
Acid Conj. base
Equilibria Involving Weak Acids and Bases
Consider acetic acid
HOAc + H2O H3O+ + OAc-
Acid Conj. base
(K is designated Ka for ACID)Because [H3O+] and [OAc-] are SMALL, Ka << 1.
Ka [H3O+][OAc- ]
[HOAc] 1.8 x 10-5
Ka From Equilibrium Concentrations A solution of a weak acid has the
following concentrations at equilibrium. What is the Ka? [HA] = 0.049 M; [H+] = [A-] = 0.00084 MHA + H2O H3O+ + A-
Ka = [H3O+][A-]/[HA]Ka = (0.00084)2/0.049Ka = 1.4 x 10-5
Ka From Percent Ionization In a 0.0100 M solution, acetic acid
is 4.2% ionized. What is the Ka? HA + H2O H3O+ + A-
initial 0.0100change -0.00042 0.00042 0.00042equil 0.00958 0.00042 0.00042Ka = [H3O+][A-]/[HA]Ka = (0.00042)2/0.00958Ka = 1.8 x 10-5
Ka From pH The pH of a 0.115 M weak acid is
1.92. What is the Ka? HA + H2O H3O+ + A-
[H3O+] = 10-1.92 = 0.012 Minitial 0.115change -0.012 0.012 0.012equil 0.103 0.012 0.012Ka = [H3O+][A-]/[HA]Ka = (0.012)2/0.103Ka = 1.4 x 10-3
Weak AcidYou have 1.00 M HOAc. Calc. the equilibrium
concs. of HOAc, H3O+, OAc-, and the pH.
Step 1. Define equilibrium concs.
[HOAc] [H3O+] [OAc-]
Initial
Change
Equilib
Weak Acid You have 1.00 M HOAc. Calc. the
equilibrium concs. of HOAc, H3O+, OAc-, and the pH.
Step 1. Define equilibrium concs. [HOAc] [H3O+] [OAc-]Initial 1.00 0 0Change -x +x +xEquilib 1.00-x x x
Weak Acid Step 2. Write Ka expression
You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H3O+, OAc-, and the pH.
Weak Acid Step 2. Write Ka expression
Ka 1.8 x 10-5 = [H3O+][OAc- ]
[HOAc]
x2
1.00 - x
You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H3O+, OAc-, and the pH.
Weak Acid Step 2. Write Ka expression
This is a quadratic. Solve using quadratic formula or method of approximations
Ka 1.8 x 10-5 = [H3O+][OAc- ]
[HOAc]
x2
1.00 - x
You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H3O+, OAc-, and the pH.
Weak Acid Step 3. Solve Ka expression
Ka 1.8 x 10-5 = [H3O+][OAc- ]
[HOAc]
x2
1.00 - x
You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H3O+, OAc-, and the pH.
Weak Acid Step 3. Solve Ka expression
First assume x is very small because Ka is so small.
Ka 1.8 x 10-5 = [H3O+][OAc- ]
[HOAc]
x2
1.00 - x
You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H3O+, OAc-, and the pH.
Weak Acid Step 3. Solve Ka expression
First assume x is very small because Ka is so small.
Ka 1.8 x 10-5 = [H3O+][OAc- ]
[HOAc]
x2
1.00 - x
You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H3O+, OAc-, and the pH.
Ka 1.8 x 10-5 = x2
1.00
Weak Acid Step 3. Solve Ka expression
First assume x is very small because Ka is so small.
And so x = [H3O+] = [OAc-] = [Ka • 1.00]1/2
Ka 1.8 x 10-5 = [H3O+][OAc- ]
[HOAc]
x2
1.00 - x
Ka 1.8 x 10-5 = x2
1.00
You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H3O+, OAc-, and the pH.
Weak Acid Step 3. Solve Ka approximate
expression
x = [H3O+] = [OAc-] = [Ka • 1.00]1/2
Ka 1.8 x 10-5 = x2
1.00
You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H3O+, OAc-, and the pH.
Weak Acid Step 3. Solve Ka approximate
expression
x = [H3O+] = [OAc-] = [Ka • 1.00]1/2 x = [H3O+] = [OAc-] = 4.2 x 10-3 M
Ka 1.8 x 10-5 = x2
1.00
You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H3O+, OAc-, and the pH.
Weak Acid Step 3. Solve Ka approximate
expression
x = [H3O+] = [OAc-] = [Ka • 1.00]1/2 x = [H3O+] = [OAc-] = 4.2 x 10-3 M pH = - log [H3O+] = -log (4.2 x 10-3) pH = 2.37
Ka 1.8 x 10-5 = x2
1.00
You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H3O+, OAc-, and the pH.
Percent Ionization Calculate the percent ionization of
a 0.10 M solution of acetic acid.HA + H2O H3O+ + A-
initial 0.10change -x x xequil 0.10 - x x xKa = [H3O+][A-]/[HA] = x2/0.10 - x1.8 x 10-5 = x2/0.10 - x = x2/0.10x = 1.3 x 10-3
Calculations Based on Ionization Constants Calculate the concentrations of the
species in 0.15 M hydrocyanic acid, HCN, solution.
Ka = 4.0 x 10-10 for HCNYou do it!
Calculations Based on Ionization Constants Note that the properly applied
simplifying assumption gives the same result as solving the quadratic equation does.
2a4acbb
c b a
0100.6100.4
100.415.0
2
11102
10
x
xx
Xxx
Calculations Based on Ionization Constants
5-6
1121010
107.7- and 107.7
12100.614100.4100.4
x
x
7.7 x 10-6 is the VALID chemical solution!
Calculations Based on Ionization Constants Let’s look at the percent ionization of two weak acids as a function of their ionization constants.
Note that the [H+] in 0.15 M acetic acid is 210 times greater than for 0.15 M HCN.
Solution Ka [H+] pH % ionization0.15 M
CH3COOH1.8 x 10-5 1.6 x 10-3 2.80 1.1
0.15 MHCN
4.0 x 10-10 7.7 x 10-6 5.11 0.0051
Weak BaseYou have 0.010 M NH3. Calc. the pH.
NH3 + H2O NH4+ + OH-
Kb = 1.8 x 10-5
Step 1. Define equilibrium concs.
[NH3] [NH4+] [OH-]
initial 0.010 0 0
change -x +x +x
equilib 0.010 - x x x
Weak BaseYou have 0.010 M NH3. Calc. the pH.
NH3 + H2O NH4+ + OH-
Kb = 1.8 x 10-5
Step 1. Define equilibrium concs.
[NH3] [NH4+] [OH-]
initial 0.010 0 0
change -x +x +x
equilib 0.010 - x x x
Weak Base
You have 0.010 M NH3. Calc. the pH.
NH3 + H2O NH4+ + OH-
Kb = 1.8 x 10-5
Step 2. Solve the equilibrium expressionKb 1.8 x 10-5 =
[NH4+][OH- ]
[NH3 ] =
x2
0.010 - x
Weak BaseYou have 0.010 M NH3. Calc. the pH.
NH3 + H2O NH4+ + OH-
Kb = 1.8 x 10-5
Step 2. Solve the equilibrium expression
Assume x is small (100•Kb < Co), so x = [OH-] = [NH4
+] = 4.2 x 10-4 M
Kb 1.8 x 10-5 = [NH4
+][OH- ][NH3 ]
= x2
0.010 - x
Weak BaseYou have 0.010 M NH3. Calc. the pH.
NH3 + H2O NH4+ + OH-
Kb = 1.8 x 10-5
Step 2. Solve the equilibrium expression
Assume x is small (100•Kb < Co), so x = [OH-] = [NH4
+] = 4.2 x 10-4 M
and [NH3] = 0.010 - 4.2 x 10-4 = 0.010 M
Kb 1.8 x 10-5 = [NH4
+][OH- ][NH3 ]
= x2
0.010 - x
Weak Base
You have 0.010 M NH3. Calc. the pH.
NH3 + H2O NH4+ + OH-
Kb = 1.8 x 10-5
Step 3. Calculate pH[OH-] = 4.2 x 10-4 Mso pOH = - log [OH-] = 3.37Because pH + pOH = 14,pH = 10.63
Example 18-15 The pH of household ammonia is
11.50. What is its molarity?NH3 + H2O NH4+ + OH-
pOH = 14 - 11.50 = 2.50[OH-] = 10-2.50 = 0.0032 Minitial xchange - 0.0032 0.0032 0.0032 equil x - 0.0032 0.0032 0.0032 Kb = [NH4
+][OH-]/[NH3 ] = (0.0032 )2/x - 0.0032 1.8 x 10-5 = (0.0032 )2/xx = 0.57 M
Polyprotic Acids Many weak acids contain two or more acidic hydrogens.
polyprotic acids ionize stepwise ionization constant for each step
Consider arsenic acid, H3AsO4, which has three ionization constants1 K1=2.5 x 10-4
2 K2=5.6 x 10-8
3 K3=3.0 x 10-13
Polyprotic Acids The first ionization step is
H AsO H H AsO
KH H AsO
H AsO
3 4 2 4
12 4
3 4
2 5 10 4.
Polyprotic Acids The second ionization step is
H AsO H HAsO
KH HAsO
H AsO
2 4-
4
24
2 42-
2
285 6 10.
Polyprotic Acids The third ionization step is
HAsO H AsO
KH AsO
HAsO
42-
4
34
42-
3
3133 0 10.
Polyprotic Acids Notice that the ionization constants vary in the following
fashion:
This is a general relationship.K K K1 2 3
Polyprotic Acids Calculate the concentration of all species
in 0.100 M arsenic acid, H3AsO4, solution.
1 Write the first ionization ionization step and represent the concentrations.
xMxMMx 100.0
AsOHHAsOH 4243
Polyprotic Acids
2 Substitute into the expression for K1.
applynot does assumption gsimplifyin
0105.2105.2
105.210.0
K
105.2AsOH
AsOHHK
542
41
4
43
421
xx
x
xx
Polyprotic Acids
3 Use the quadratic equation to solve for x, and obtain two values
MMx
MxM
MxMx
x
095.0100.0AsOH
109.4AsOHH
109.4 and 101.5
12
105.214105.2105.2
43
342
33
5244
Polyprotic Acids
4 Now we write the equation for the second step ionization and represent the concentrations.
H AsO H + HAsO
from 1st step (4.9 10
algebraically (4.9 10
2 4- +
42-
-3
-3
M
y M yM yM
)
)
Polyprotic Acids
5 Substitute into the second step ionization expression.
K =H O HAsO
H AsO
K =4.9 10
4.9 10
apply assumption
23 4
2
2 4
2
-3
-3
5 6 10 8.
y y
y
Polyprotic Acids
K =H O HAsO
H AsO
K =4.9 10
4.9 10
apply assumption
K =4.9 10
4.9 10
H HAsO
note H H
23 4
2
2 4
2
-3
-3
2
-3
-3
2nd 42
1st 2nd
5 6 10
5 6 10
5 6 10
8
8
8
.
.
.
y y
y
y
y M
Polyprotic Acids
6 Now we repeat the procedure for the third ionization step.
HAsO H AsO
1st and 2nd ionizations 5.6 10 4.9 10 5.6 10
algebraically 5.6 10
42
43
-8 -3 -8
-8
M M
z M z M z M
Polyprotic Acids
7 Substitute into the third ionization expression.
K =H O AsO
HAsO
K =
apply assumption
33 4
3
42
3
3 0 10
4 9 10 5 6 10
5 6 10
13
3 8
8
.
. .
.
z z
z
Polyprotic Acids
K =H O AsO
HAsO
K =
apply assumption
H AsO
33 4
3
42
3
3rd 43
3 0 10
4 9 10 5 6 10
5 6 10
4 9 10
5 6 103 0 10
3 4 10
13
3 8
8
3
813
18
.
. .
.
.
..
.
z z
z
z
z M
Polyprotic Acids Use Kw to calculate the [OH-] in the
0.100 M H3AsO4 solution.
H OH
OHH
OH
10 10
10 10 10 10
4 9 10
2 0 10
14
14 14
3
12
.
. .
.
. M
Polyprotic Acids A comparison of the various species in
0.100 M H3AsO4 solution follows.Species ConcentrationH3AsO4 0.095 M
H+ 0.0049 MH2AsO4
- 0.0049 MHAsO4
2- 5.6 x 10-8 MAsO4
3- 3.4 x 10-18 MOH- 2.0 x 10-12 M
Solvolysis Solvolysis is the
reaction of a substance with the solvent in which it is dissolved.
Hydrolysis refers to the reaction of a substance with water or its ions.
Hydrolysis …the reaction
of a substance with water or its ions
A- + H2O HA + OH-
Ah ha, BASIC!
BH+ + H2O B + H3O+
Hydrolysis Hydrolysis refers to the reaction of
a substance with water or its ions. Combination of the anion of a
weak acid with H3O+ ions from water to form nonionized weak acid molecules.A- + H2O HA + OH-
H2O H+OH-
NaCNNa+ + H2O NR
CN- + H2O HCN + OH-
BasicHCN + NaOH NaCN + H2O
Hydrolysis
Hydrolysis
NH4ClCl- + H2O NR
NH4+ + H2O NH3 + H3O+
Acidic
HCl + NH3 NH4Cl
NH4CNNH4
+ + H2O NH3 + H3O+
CN- + H2O HCN + OH-
Acidic or BasicHCN + NH4OH NH4CN + H2O
Hydrolysis
Salts of Weak Bases and Weak Acids Acidic Solution
Ka > KbNH4F NH4
+ + F-
NH4+ + H2O NH3 + H3O+
F- + H2O HF + OH-
Ka = 7.2 x 10-4 and Kb = 1.8 x 10-5
Salts of Weak Bases and Weak Acids Neutral Solution
Ka = KbNH4CH3COO NH4
+ + CH3COO-
NH4+ + H2O NH3 + H3O+
CH3COO- + H2O CH3COOH + OH-
Ka = 1.8 x 10-5 and Kb = 1.8 x 10-5
Salts of Weak Bases and Weak Acids Basic Solution
Ka < KbNH4CN NH4
+ + CN-
NH4+ + H2O NH3 + H3O+
CN- + H2O HCN + OH-
Ka = 4.0 x 10-10 and Kb = 1.8 x 10-5
Salts of Weak Bases and Weak Acids Acidic Solution
Ka > Kb Basic solution
Kb > Ka Neutral solution
Ka = Kb
Hydrolysis The conjugate base of a strong
acid is a very weak base. The conjugate base of a weak acid
is a stronger base. Hydrochloric acid, a typical strong
acid, is essentially completely ionized in dilute aqueous solutions.
HCl H O H O Cl2 3 ~100%
Hydrolysis The conjugate base of HCl, the Cl-
ion, is a very weak base.
True of all strong acids and their anions.
Cl H O No rxn. in dilute aqueous solutions3
Hydrolysis HF, a weak acid, is only slightly ionized in
dilute aqueous solutions. Its conjugate base, the F- ion, is a much
stronger base than the Cl- ion. F- ions combine with H3O+ ions to form
nonionized HF.HF + H O H O F
only slightly
F + H O HF + H O
nearly completely
2 3+ -
-3
+2
Salts of Strong Soluble Bases and Strong Acids Salts made from strong acids and
strong soluble bases form neutral aqueous solutions.
An example is potassium nitrate, KNO3, made from nitric acid and potassium hydroxide.
K NO K NO
H O H O OH + H O
no rxn. to upset H O OH neutral
+3
in H O +3
2 2-
3+
3+ -
2
( )~
s100%
Salts of Strong Soluble Bases and Weak Acids Salts made from strong soluble bases and
weak acids hydrolyze to form basic solutions. Anions of weak acids (strong conjugate bases) react
with water to form hydroxide ions An example is sodium hypochlorite, NaClO,
made from sodium hydroxide and hypochlorous acid.
Na ClO Na ClO
H O + H O OH + H O
+ - in H O -
2 2-
3+
2( )
~s
100%
Salts of Strong Soluble Bases and Weak Acids
Na ClO Na ClO
H O + H O OH + H O
ClO H O HClO H O
+ - in H O -
2 2-
3+
-3 2
2( )
~s
100%
Combine these equations into one single equation that represents the reaction:
ClO H O HClO OH-2
Acid-Base Properties of Salts
NH4Cl(aq) NH4+(aq) + Cl-(aq)
Reaction of NH4+ with H2O
NH4+ + H2O NH3 + H3O+
acid base base acidNH4
+ ion is a moderate acid because its conjugate base is weak.
Therefore, NH4+ is acidic solution
Salts of Weak Bases and Strong Acids Salts made from weak bases and strong
acids form acidic aqueous solutions. An example is ammonium bromide,
NH4Br, made from ammonia and hydrobromic acid.
acidic issolution OH excess generates
OHNHOHNH
OH OH OHOH
Br NH BrNH
3
23-
4
3-
22
-4
100%~OHs
-4
2
Hydrolysis Constant Relationship of
conjugate acid-base pairs
Kw = KaKb
Ka Kb
Example The Ka for acetic acid, CH3COOH, is
1.8 x 10-5, calculate the Kb for the acetate ion, CH3COO-.
105
14
106.5108.1
1000.1
xx
x
Ka
KwKb
Acid-Base Properties of SaltsCalculate the pH of a 0.10 M solution of NH4Cl.
Cl- + H2O is neutral
NH4++ H2O NH3 + H3O+
acid base base acid Ka = 5.6 x 10-10
Step 1. Set up concentration table [NH4
+] [NH3] [H3O+] initial change equilib
0.10 0 0-x +x +x0.10-x x x
Acid-Base Properties of Salts
x
x
NH
OHNHxKa
10.0106.5
2
4
3310
Calculate the pH of a 0.10 M solution of NH4Cl.
Cl- + H2O is neutral
NH4++ H2O NH3 + H3O+
acid base base acid Ka = 5.6 x 10-10
Calculate the pH of a 0.10 M solution of NH4Cl.
Cl- + H2O is neutral
NH4++ H2O NH3 + H3O+
acid base base acid Ka = 5.6 x 10-10
Assume 0.10-x = 0.10
Acid-Base Properties of Salts
x
x
NH
OHNHxKa
10.0106.5
2
4
3310
Acid-Base Properties of SaltsCalculate the pH of a 0.10 M solution of NH4Cl.
Cl- + H2O is neutral
NH4++ H2O NH3 + H3O+
acid base base acid Ka = 5.6 x 10-10
Assume 0.10-x = 0.10x = [NH3] = [H3O+] = 7.5 x 10-6 M
x
x
NH
OHNHxKa
10.0106.5
2
4
3310
Acid-Base Properties of SaltsCalculate the pH of a 0.10 M solution of NH4Cl.
Cl- + H2O Ž neutral
NH4+ + H2O NH3 + H3O+
acid base base acid Ka = 5.6 x 10-10
Step 3. Calculate the pH
[H3O+] = 7.5 x 10-6 MpH = -log [H3O+] = 5.13
Acid-Base Properties of SaltsCalculate the pH of a 0.10 M solution of NaCH3COO.
Na+ + H2O Ž neutral
CH3COO- + H2O CH3COOH + OH-
base acid acid base Kb = 5.6 x 10-10
Step 1. Set up concentration table [CH3COO-
] [CH3COOH] [OH-] initial change equilib
Acid-Base Properties of SaltsCalculate the pH of a 0.10 M solution of NaCH3COO.
Na+ + H2O Ž neutral
CH3COO- + H2O CH3COOH + OH-
base acid acid base Kb = 5.6 x 10-10
Step 1. Set up concentration table [CH3COO-
] [CH3COOH] [OH-] initial 0.10 0 0 change -x +x +x equilib 0.10-x x x
Acid-Base Properties of SaltsCalculate the pH of a 0.10 M solution of
NaCH3COO.
Na+ + H2O Ž neutral
CH3COO- + H2O CH3COOH + OH-
base acid acid base
Kb = 5.6 x 10-10
Step 2. Solve the equilibrium expression
Acid-Base Properties of SaltsCalculate the pH of a 0.10 M solution of
NaCH3COO.
Na+ + H2O Ž neutral
CH3COO- + H2O CH3COOH + OH-
base acid acid base
Kb = 5.6 x 10-10
Step 2. Solve the equilibrium expression x
x
COOCH
OHCOOHCHxKb
10.0106.5
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3
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Acid-Base Properties of SaltsCalculate the pH of a 0.10 M solution of NaCH3COO.
Na+ + H2O Ž neutral
CH3COO- + H2O CH3COOH + OH-
base acid acid base Kb = 5.6 x 10-10
Step 2. Solve the equilibrium expression
Assume 0.10-x = 0.10
x
x
COOCH
OHCOOHCHxKb
10.0106.5
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3
310
Acid-Base Properties of SaltsCalculate the pH of a 0.10 M solution of NaCH3COO.
Na+ + H2O Ž neutral
CH3COO- + H2O CH3COOH + OH-
base acid acid base Kb = 5.6 x 10-10
Step 2. Solve the equilibrium expression
Assume 0.10-x = 0.10x = [CH3COO] = [OH-] = 7.5 x 10-6 M
x
x
COOCH
OHCOOCHxKb
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3
310
Acid-Base Properties of SaltsCalculate the pH of a 0.10 M solution of NaCH3COO.
Na+ + H2O Ž neutral
CH3COO- + H2O CH3COOH + OH-
base acid acid base Kb = 5.6 x 10-10
Step 3. Calculate the pH[OH-] = = 7.5 x 10-6 MpOH = -log [OH-] = 5.13pH + pOH = 14.00pH = 8.87