Example 14.14 The Effect of a Concentration Change on Equilibrium
Consider the following reaction at equilibrium.
What is the effect of adding additional CO2 to the reaction mixture? What is the effect of adding additional
CaCO3?
SOLUTION
Adding additional CO2 increases the concentration of CO2 and causes the reaction to shift to the left. Adding
additional CaCO3, however, does not increase the concentration of CaCO3 because CaCO3 is a solid and
therefore has a constant concentration. Thus, adding additional CaCO3 has no effect on the position of the
equilibrium. (Note that, as we saw in Section 14.5, solids are not included in the equilibrium expression.)
For Practice 14.14
Consider the following reaction in chemical equilibrium:
What is the effect of adding additional Br2 to the reaction mixture? What is the effect of adding additional
BrNO?
Example 14.15 The Effect of a Volume Change on Equilibrium
Consider the following reaction at chemical equilibrium:
What is the effect of decreasing the volume of the reaction mixture? Increasing the volume of the reaction
mixture? Adding an inert gas at constant volume?
SOLUTION
The chemical equation has 3 mol of gas on the right and zero moles of gas on the left. Decreasing the volume of
the reaction mixture increases the pressure and causes the reaction to shift to the left (toward the side with
fewer moles of gas particles). Increasing the volume of the reaction mixture decreases the pressure and causes
the reaction to shift to the right (toward the side with more moles of gas particles). Adding an inert gas has no
effect.
For Practice 14.15
Consider the following reaction at chemical equilibrium:
What is the effect of decreasing the volume of the reaction mixture? Increasing the volume of the reaction
mixture?
Example 14.16 The Effect of a Temperature Change on Equilibrium
The following reaction is endothermic.
What is the effect of increasing the temperature of the reaction mixture? Decreasing the temperature?
SOLUTION
Since the reaction is endothermic, we can think of heat as a reactant:
Raising the temperature is equivalent to adding a reactant, causing the reaction to shift to the right. Lowering
the temperature is equivalent to removing a reactant, causing the reaction to shift to the left.
For Practice 14.16
The following reaction is exothermic.
What is the effect of increasing the temperature of the reaction mixture? Decreasing the temperature?
Arrhenius Definition: The Simplest
•
What is an acid?
– Acids ionize in water to produce H+ ions and anions.
HCl(aq) → H+(aq) + Cl−(aq)
CH3CO2H(aq) → H+(aq) + CH3CO2−(aq)
•
What is a base?
– Bases dissociate in water to produce OH− ions and cations.
NaOH(aq) → Na+(aq) + OH−(aq)
•
Bronsted–Lowry
Definition
The most general theory for common aqueous acids and bases
•
What is an acid?
– ACIDS DONATE H+ IONS.
– Similar definition as presented in Arrhenius theory
•
What is a base?
– BASES ACCEPT H+ IONS.
– This broadens the Arrhenius definition for a base.
– Answers the questions about nonhydroxide bases that the Arrhenius
theory can not explain.
Bronsted–Lowry:
Conjugate Acid/Base Pair
• NH3 is a BASE in water and water is an ACID.
• NH3 / NH4+ is a conjugate pair.
• Related by the gain or loss of H+
• Every acid has a conjugate base.
• Every base has a conjugate acid.
Example 15.1 Identifying Brønsted–Lowry Acids and Bases and Their
Conjugates
Identify the Brønsted–Lowry acid, the Brønsted–Lowry base, the conjugate acid, and the conjugate base in each
reaction.
SOLUTION
(a) Since H2SO4 donates a proton to H2O in this reaction,
it is the acid (proton donor). After H2SO4 donates the
proton, it becomes HSO4−, the conjugate base. Since
H2O accepts a proton, it is the base (proton acceptor).
After H2O accepts the proton it becomes H3O+, the
conjugate acid.
(b) Since H2O donates a proton to HCO3− in this reaction,
it is the acid (proton donor). After H2O donates the
proton, it becomes OH–, the conjugate base. Since
HCO3– accepts a proton, it is the base (proton
acceptor). After HCO3– accepts the proton it becomes
H2CO3, the conjugate acid.
Acid Strength: Strong vs. Weak
Strong
Weak
• COMPLETE (~100%) dissociation
in water
– Complete ionization
– Strong electrolyte
• PARTIAL (< 5%) dissociation in
water
– Weak electrolyte
• Acid
HNO3(aq) + H2O(l) 
H3O+ + NO3-
• Acid
CH3COOH(aq) + H2O(l)  
H3O+ + CH3COO-
• Base
Ca(OH)2(aq) +  2 OH- + Ca2+
• Base
NH3(aq) + H2O(l)  
NH4+ + OH-
Ka Values for Weak Acids
Weak Acid
Equation
Ka
acetic acid
HC2H3O2   H+ + C2H3O2-
1.8 x 10-5
benzoic acid
C6H5CO2H   H+ + C6H5CO2-
6.4 x 10-5
formic acid
hydrocyanic acid
HCHO2  
HCN  
H+ + CHO2-
1.8 x 10-4
H+ + CN-
6.2 x 10-10
hydrofluoric acid
HF   H+ + F-
7.2 x 10-4
hypochlorous acid
HOCl   H+ + OCl-
3.5 x 10-8
lactic acid
CH3CH(OH)CO2H  
H+ + CH3CH(OH)CO2-
1.4 x 10-4
phenol
HOC6H5   H+ + OC6H5-
1.6 x 10-10
Problem: What is the percent ionization of a 2.5 M HNO2
solution?
Equation:
HNO2 + H2O  NO2 + H3O+
HNO2

NO2
H3O+
I
2.5
0
0
C
-x
+x
+x
E
2.5 – x
+x
+x
Assume x <<<< 2.5
Ka = 4.6 x 10-4 = (x2/2.5)
(4.6 x 10-4)1/2 = (x2/2.5)1/2
x = 3.4 x10-2
Percent ionization = (3.4 x 10-2/2.5) x 100 = 1.4%
Ka = 4.6 x 10-4
Strength of Conjugate Acid-Base Pairs
Water, Strong Acid, Strong Base
•
Water is an extremely weak electrolyte (K = 1 x 10-14).
– About 1 out of every 10 million water molecules forms ions through a process called
auto ionization.
2 H2O(l)   H3O+ + OH–
•
Equilibrium constant for water’s auto ionization is given as Kw.
– Kw = [H3O+] [OH-] = 1.00 x 10-14 at 25 oC
•
As [H3O+] increases, the [OH–] MUST decrease so that the product stays constant.
– Inversely proportional relationship between H+ and OH- ion.
Summary of Mathematical Relationships
• pH = - log [H+]
• pOH = - log [OH-]
[H+] = 10-pH
[OH-] = 10-pOH
• pKw = -log (Kw)
• pKa = -log (Ka)
• pKb = -log (Kb)
Kw = 10-pKw
Ka = 10-pKa
Kb = 10-pKb
• pKw = pKa + pKb
Kw = Ka x Kb
• pH + pOH = 14
[H+] [OH-] = 1 x 10-14
Example 15.2 Using Kw in Calculations
Calculate [OH−] at 25 ºC for each solution and determine whether the solution is acidic, basic, or neutral.
SOLUTION
(a) To find [OH–] use the ion product constant. Substitute
the given value for [H3O+] and solve the equation for
[OH−].
Since [H3O+] > [OH−], the solution is acidic.
(b) Substitute the given value for [H3O+] and solve the acid
ionization equation for [OH−].
Since [H3O+] < [OH−], the solution is basic.
Example 15.2 Using Kw in Calculations
Continued
(c) Substitute the given value for [H3O+] and solve the
acid ionixation equation for [OH–].
Since [H3O+] = 1.0 10–7 and [OH–] = 1.0 10–7. the
solution is neutral.
For Practice 15.2
Calculate [H3O+] at 25 °C for each solution and determine whether the solution is acidic, basic, or neutral.
Example 15.3 Calculating pH from [H3O+] or [OH–]
Calculate the pH of each solution at 25 °C and indicate whether the solution is acidic or basic.
(a) [H3O+] = 1.8
10–4 M
(b) [OH–] = 1.3
10–2 M
SOLUTION
(a) To calculate pH, substitute the given [H3O+] into the pH
equation.
Since pH < 7, this solution is acidic.
(b) First use Kw to find [H3O+] from [OH–].
Then substitute [H3O+] into the pH expression to find pH.
Since pH > 7, this solution is basic.
For Practice 15.3
Calculate the pH of each solution at 25 °C and indicate whether the solution is acidic or basic.
(a) [H3O+] = 9.5 10–9 M
(b) [OH–] = 7.1 10–3 M
Example 15.4 Calculating [H3O+] from pH
Calculate the [H3O+] concentration for a solution with a pH of 4.80.
SOLUTION
To find the [H3O+] from pH, start with the equation that defines pH.
Substitute the given value of pH and then solve for [H3O+]. Since the given
pH value is reported to two decimal places, the [H3O+] is written to two
significant figures. (Remember that 10log x = x (see Appendix I). Some
calculators use an inv log key to represent this function.)
For Practice 15.4
Calculate the H3O+ concentration for a solution with a pH of 8.37.
Example 15.5 Finding the [H3O+] of a Weak Acid Solution
Find the [H3O+] of a 0.100 M HCN solution.
1. Write the balanced equation for the ionization of the acid and
use it as a guide to prepare an ICE table showing the given
concentration of the weak acid as its initial concentration.
Leave room in the table for the changes in concentrations and for
the equilibrium concentrations.
(Note that the H3O+ concentration is listed as approximately zero
because the autoionization of water produces a negligibly small
amount of H3O+.)
2. Represent the change in the concentration of H3O+ with the
variable x. Define the changes in the concentrations of the
other reactants and products in terms of x, always keeping
in mind the stoichiometry of the reaction.
Example 15.5 Finding the [H3O+] of a Weak Acid Solution
Continued
3. Sum each column to determine the equilibrium
concentrations in terms of the initial
concentrations and the variable x.
4. Substitute the expressions for the equilibrium
concentrations (from step 3) into the
expression for the acid ionization constant
(Ka). In many cases, you can make the
approximation that x is small (as discussed in
Section 14.8). Substitute the value of the acid
ionization constant (from Table 15.5) into the
Ka expression and solve for x.
Confirm that the x is small approximation is
valid by calculating the ratio of x to the
number it was subtracted from in the
approximation. The ratio should be less than
0.05 (or 5%).
Example 15.5 Finding the [H3O+] of a Weak Acid Solution
Continued
5. Determine the H3O+ concentration from the
calculated value of x and calculate the pH if
necessary.
6. Check your answer by substituting the
calculated equilibrium values into the acid
ionization expression. The calculated value of
Ka should match the given value of Ka. Note that
rounding errors and the x is small approximation
could cause a difference in the least significant
digit when comparing values of Ka.
Since the calculated value of Ka matches the given
value, the answer is valid.
For Practice 15.5
Find the H3O+ concentration of a 0.250 M hydrofluoric acid solution.
Example 15.6 Finding the pH of a Weak Acid Solution
Find the pH of a 0.200 M HNO2 solution.
1. Write the balanced equation for the ionization of the
acid and use it as a guide to prepare an ICE table
showing the given concentration of the weak acid as its
initial concentration. Leave room in the table for the
changes in concentrations and for the equilibrium
concentrations.
(Note that the H3O+ concentration is listed as
approximately zero because the autoionization of water
produces a negligibly small amount of H3O+.)
2. Represent the change in the concentration of H3O+ with
the variable x. Define the changes in the concentrations
of the other reactants and products in terms of x,
always keeping in mind the stoichiometry of the
reaction.
Example 15.6 Finding the pH of a Weak Acid Solution
Continued
3. Sum each column to determine the equilibrium
concentrations in terms of the initial concentrations
and the variable x.
4. Substitute the expressions for the equilibrium
concentrations (from step 3) into the expression for
the acid ionization constant (Ka).
In many cases, you can make the approximation that x is
small (as discussed in Section 14.8). Substitute the value
of the acid ionization constant (from Table 15.5 ) into the
Ka expression and solve for x.
Confirm that the x is small approximation is valid by
calculating the ratio of x to the number it was
subtracted from in the approximation. The ratio
should be less than 0.05 (or 5%).
Example 15.6 Finding the pH of a Weak Acid Solution
Continued
5. Determine the H3O+ concentration from the
calculated value of x and calculate the pH if
necessary.
6. Check your answer by substituting the
calculated equilibrium values into the acid
ionization expression. The calculated value of
Ka should match the given value of Ka. Note that
rounding errors and the x is small approximation
could cause a difference in the least significant
digit when comparing values of Ka.
For Practice 15.6
Find the pH of a 0.0150 M acetic acid solution.
Missing text: Since the calculated value of Ka
matches the given value, the answer is valid.