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Chapter 15
Chemical Equilibrium
Chemical Equilibrium
15
15.1 The Concept of Equilibrium
15.2 The Equilibrium Constant
Calculating Equilibrium Constants
Magnitude of the Equilibrium Constant
15.3 Equilibrium Expressions
Heterogeneous Equilibria
Manipulating Equilibrium Expressions
Gaseous Equilibria
15.4 Using Equilibrium Expressions to Solve Problems
Predicting the Direction of a Reaction
Calculating Equilibrium Concentrations
15.5 Factors that Affect Chemical Equilibrium
Addition or Removal of a Substance
Changes in Volume and Pressure
Changes in Temperature
Catalysis
The Concept of Equilibrium
The decomposition of N2O4 is a reversible process, meaning the products of the reaction can react to form reactants.
The system is in equilibrium when the rates of the forward reaction and the reverse reaction are the same.
rate forward = kf[N2O4] and rate reverse = kr[NO2]2
15.1
N2O4(g) 2NO2(g)
The Concept of Equilibrium
Starting with N2O4
The Concept of Equilibrium
Starting with NO2
The Concept of Equilibrium
Some important things to remember about equilibrium are:
Equilibrium is a dynamic state—both the forward and reverse reactions continue to occur, although there is no net change in reactant and product concentration over time.
At equilibrium, the rates of the forward and reverse reactions are equal.
Equilibrium can be established starting with only reactants, with only products, or with any mixture of reactants and products.
The Equilibrium Constant
rate forward = rate reverse kf[N2O4] eq = kr[NO2]2eq
The subscript “eq” denotes a concentration at equilibrium.
Rearranging
The ratio of two constants (kf/kr) is also a constant:
15.2
equilibrium expression
N2O4(g) 2NO2(g)
equilibrium constant
The Equilibrium Constant
Note the relationship between the equilibrium constant and the balanced chemical equation:
N2O4(g) 2NO2(g)
The Equilibrium Constant
The reaction quotient (Qc ) is a fraction with product concentrations in the numerator and reactant concentrations in the denominator.
Each concentration is raised to a power equal to the corresponding stoichiometric coefficient in the balanced chemical equation.
aA + bB cC + dD
This law of mass action applies to not only elementary reactions, but also to more complex reactions.
(at equilibrium)
Worked Example 15.1
Strategy Use the law of mass action to write the equilibrium expression and plug in the equilibrium concentrations of all three species to evaluate Kc.
Carbonyl chloride (COCl2), also called phosgene, is a highly poisonous gas that was used on the battlefield of World War I. It is produced by the reaction of carbon monoxide with chlorine gas:
CO(g) + Cl2(g) COCl2(g)
In an experiment conducted at 74°C, the equilibrium concentrations of the species involved in the reaction were as follows: [CO] = 1.2×10-2 M, [Cl2] = 0.054 M, and [COCl2] = 0.14 M. (a) Write the equilibrium expression, and (b) determine the value of the equilibrium constant for this reaction at 74°C.
Solution (a) Kc =
(b) Kc = = 216 or 2.2×102
[COCl2]
[CO][Cl2]
(0.14)
(1.2×10-2)(0.054)
Think About It When putting the equilibrium concentrations into the equilibrium expression, we leave out the units. It is common practice to express equilibrium constants without units.
Worked Example 15.2
Strategy Use the law of mass action to write reaction quotients.
Write reaction quotients for the following reactions:
(a) N2(g) + 3H2(g) 2NH3(g)
(b) H2(g) + I2(g) 2HI(g)
(c) Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)
(d) 2O3(g) 3O2(g)
(e) Cd2+(aq) + 4Br-(aq) CdBr42-(aq)
(f) 2NO(g) + O2(g) 2NO2(g)
Solution (a) Qc =
(d) Qc =
[NH3]2
[N2][H2]3
(b) Qc =
[HI]2
[H2][I2]
(c) Qc =
[Ag(NH3)2+]
[Ag+][NH3]2
[O2]3
[O3]2
(e) Qc =
[CdBr42-]
[Cd2+][Br-]4
(f) Qc =
[NO2]2
[NO]2[O2]
Think About It With practice, writing reaction quotients becomes second nature. Without sufficient practice, it will seem inordinately difficult. It is important that you become proficient at this. It is very often the first step in solving equilibrium problems.
The Equilibrium Constant
At any point during the progress of a reaction:
aA + bB cC + dD
The Equilibrium Constant
The value of the reaction quotient, Q, changes as the reaction progresses
N2O4(g) 2NO2(g)
The Equilibrium Constant
The equilibrium constant gives the extent a reaction will proceed at a particular temperature.
Three outcomes are possible:
1) The reaction will go essentially to completion and the equilibrium mixture will consist predominately of products.
Ag+(aq) + 2NH3(aq) Ag(NH3)2+(g) Kc = 1.5 x 107 (at 25°C)
Large Kc, product favored
The Equilibrium Constant
The equilibrium constant gives the extent a reaction will proceed at a particular temperature.
Three outcomes are possible:
2) The reaction will not occur to any significant degree, and the equilibrium mixture will consist predominantly of reactant.
N2(g) + O2(g) 2NO(g) Kc = 4.3 x 10–25 (at 25°C)
3) The reaction will proceed a significant degree but will not go to completion, and the equilibrium mixture will contain comparable amounts of both reactants and products.
Small Kc, reactant favored
Equilibrium Expressions
When the species in a reversible chemical reaction are not all in the same phase, the equilibrium is heterogeneous.
Only gaseous species and aqueous species appear in equilibrium expressions, pure solids and pure liquids do not.
15.3
CO2(g) + C(s) 2CO(g)
2Fe(s) + 3H2O(l) Fe2O3(s) + 2H2(g)
Worked Example 15.3
Strategy Use the law of mass action to write the equilibrium expressions for each reaction. Only gases and aqueous species appear in the expression.
Write equilibrium expressions for each of the following reactions:
(a) CaCO3(s) CaO(s) + CO2(g)
(b) Hg(l) + Hg2+(aq) Hg22+(aq)
(c) 2Fe(s) + 3H2O(l) Fe2O3(s) + 2H2(g)
(d) O2(g) + 2H2(g) 2H2O(l)
Solution (a) Kc = [CO2]
(b) Kc =
[Hg22+]
[Hg2+]
(d) Kc =
1
[O2][H2]2
(c) Kc = [H2]2
Think About It Like writing equilibrium expressions for homogeneous equilibria, writing equilibrium expressions for heterogeneous equilibria becomes second nature if you practice. The importance of developing this skill now cannot be overstated. Your ability to understand the principles and to solve many of the problems in and Chapters 16 to 19 depends on your ability to write equilibrium expressions correctly and easily.
Equilibrium Expressions
When a reversible chemical equation is manipulated, it is also necessary to make appropriate changes in the equilibrium expression and the equilibrium constant.
Worked Example 15.4
Strategy Begin by writing the equilibrium expressions for the reactions that are given. Then, determine the relationship of each equation’s equilibrium expression to the equilibrium expression of the original equations, and make the corresponding change to the equilibrium constant for each.
Kc = and Kc =
The following reactions have the indicated equilibrium constants at 100°C:
(1) 2NOBr(g) 2NO(g) + Br2(g) Kc = 0.014
(2) Br2(g) + Cl2(g) 2BrCl(g) Kc = 7.2
Determine the value of Kc for the following reactions at 100°C:
(a) 2NO(g) + Br2(g) 2NOBr(g) (d) 2NOBr(g) + Cl2(g) 2NO(g) + 2BrCl(g)
(b) 4NOBr(g) 4NO(g) + 2Br2(g) (e) NO(g) + BrCl(g) NOBr(g) + Cl2(g)
(c) NOBr(g) NO(g) + Br2(g)
1
2
1
2
[NO]2[Br2]
[NOBr]2
[BrCl]2
[Br2][Cl2]
Worked Example 15.4 (cont.)
Solution (a) This equation is the reverse of original equation 1. Its equilibrium expression is the reciprocal of that for the original equation:
Kc = = 1/0.014 = 71
(b) This is original equation 1 multiplied by a factor of 2. Its equilibrium expression is the original expression squared:
Kc = = (0.014)2 = 2.0×10-4
(c) This is the original equation 1 multiplied by . Its equilibrium expression is the square root of the original
Kc = or Kc = = (0.014)1/2 = 0.12
[NOBr]2
[NO]2[Br2]
[NO]2[Br2]
[NOBr]2
2
1
2
[NO]2[Br2]
[NOBr]2
1/2
Worked Example 15.4 (cont.)
Solution (d) This is the sum of the original equations 1 and 2. Its equilibrium expression is the product of the two individual expressions:
Kc = = (0.014)(7.2) = 0.10
(e) Probably the simplest way to analyze this reaction is to recognize that it is the reverse of the reaction in part (d), multiplied by . Its equilibrium expression is the square root of the reciprocal of the expression in part (d):
Kc = = (1/0.10)1/2 = 3.2
[NO]2[BrCl]2
[NOBr]2[Cl2]
[NO]2[BrCl]2
[NOBr]2[Cl2]
1/2
1
2
Worked Example 15.4 (cont.)
Think About It The magnitude of an equilibrium constant reveals whether products or reactants are favored, so the reciprocal relationship between Kc values of forward and reverse reactions should make sense. A very large Kc value means that products are favored. In the reaction of hydrogen ion and hydroxide ion to form water, the value of Kc is very large, indicating that the product, water, is favored.
H+(aq) + OH-(aq) H2O(l) Kc = 1.0×1014 (at 25°C)
Simply writing the equation backward doesn’t change the fact that water is the predominate species. In the reverse reaction, therefore, the favored species is on the reactant side:
H2O(l) H+(aq) + OH-(aq) Kc = 1.0×10-14 (at 25°C)
As a result, the magnitude of Kc should correspond to reactants being favored; that is, it should be very small.
Gaseous Equilibria
When an equilibrium expression contains only gases, we can write an alternate form of the expression in which the concentrations of gases are expressed as partial pressures (atm). Thus, for the equilibrium
we can write as
Kc = or KP =
The relationship between Kc and KP can be expressed as
where Δn = moles of gaseous products – moles of gaseous reactants.
N2O4(g) 2NO2(g)
[NO2]2
[N2O4]
(PNO2)2
PN2O4
KP = Kc[(0.08206
L atm/K mol)×T]Δn
Worked Example 15.5
Strategy Write equilibrium expressions for each equation, expressing the concentrations of the gases in partial pressures.
Write KP expressions for (a) PCl3(g) + Cl2(g) PCl5(g), (b) O2(g) + 2H2(g) 2H2O(l), and (c) F2(g) + H2(g) 2HF(g).
Solution (a) All the species in this equation are gases, so they will all appear in the KP expression.
KP =
(b) Only the reactants are gases. KP =
(c) All species are gases. KP =
(PPCl5)
(PPCl3)(PCl2)
1
(PO2)(PH2)2
(PHF)2
(PF2)(PH2)
Think About It It isn’t necessary for every species in the reaction to be a gas–only those species that appear in the equilibrium expression.
Worked Example 15.6
Strategy Use KP = Kc[(0.08206 L atm/K mol)×T]Δn. Be sure to convert temperature in degrees Celsius to kelvins. Using Δn = moles of gaseous products – moles of gaseous reactants, Δn = 2(NO2) – 1(N2O4) = 1. T = 298K.
The equilibrium constant, Kc, for the reaction
N2O4(g) 2NO2(g)
is 4.63×10-3 at 25°C. What is the value of KP at this temperature.
Solution
KP = Kc ×T
= (4.63×10-3)(0.08206 × 298)
= 0.113
0.08206 L atm
K mol
Think About It Note that we have essentially disregarded the units of R and T so that the resulting equilibrium constant, KP, is unitless. Equilibrium constants commonly are treated as unitless quantities.
Using Equilibrium Expressions to Solve Problems
The equilibrium expression may be used to predict the direction of a reaction and to calculate equilibrium concentrations.
Predictions are made based on comparisons between Qc and Kc.
There are three possibilities:
Q < K The ratio of initial concentrations of products to reactants is too small. To reach equilibrium, reactants must be converted to products. The system proceeds in the forward direction.
Q = K The initial concentrations are equilibrium concentrations. The system is at equilibrium.
Q > K The ratio of initial concentrations of products to reactants is too large. To reach equilibrium products must be converted to reactants. The system proceeds in the reverse direction.
15.4
Worked Example 15.7
Strategy Use the initial concentrations to calculate Qc, and then compare Qc with Kc.
Qc = = = 0.61
At 375°C, the equilibrium constant for the reaction
N2(g) + 3H2(g) 2NH3(g)
is 1.2. At the state of a reaction, the concentrations of N2, H2, and NH3 are 0.071 M, 9.2×10-3 M, and 1.83×10-4 M, respectively. Determine whether the system is at equilibrium, and if not, determine in which direction it must proceed to establish equilibrium.
[NH3]i2
[N2]i[H2]i3
(1.83×10-4)2
(0.071)(9.2×10-3)3
Strategy The calculated value of Qc is less than Kc. Therefore, the reaction is not at equilibrium and must proceed to the right to establish equilibrium.
Think About It In proceeding to the right, a reaction consumes reactants and produces more products. This increases the numerator in the reaction quotient and decreases the denominator. The result is an increase in Qc until it is equal to Kc, at which point equilibrium will be established.
Calculating Equilibrium Concentrations
Equilibrium concentrations can be calculated from initial concentrations if the equilibrium constant is known.
Kc = 24.0 (200°C)
Initial concentration (M) 0.850 0
Change in concentration (M)
Equilibrium concentration (M)
–x
0.850 – x
+x
x
cis-Stilbene
trans-Stilbene
cis-Stilbene
trans-Stilbene

Using Equilibrium Expressions to Solve Problems
Initial concentration (M) 0.850 0
Change in concentration (M)
Equilibrium concentration (M)
–x
0.850 – x
+x
x
cis-Stilbene
trans-Stilbene

Use the equilibrium concentrations, defined in terms of x, in the equilibrium expression:
x = 0.816 M (Represents the change in initial concentrations)
Using Equilibrium Expressions to Solve Problems
Initial concentration (M) 0.850 0
Change in concentration (M)
Equilibrium concentration (M)
–0.816
0.850 – 0.816
+0.816
0.816
cis-Stilbene
trans-Stilbene

Calculate the equilibrium concentrations of cis- and trans-stilbene:
[cis-stilbene] = (0.850 – x) M = (0.850 – 0.816) M = 0.034 M
[trans-stilbene] = x M = 0.816 M
Worked Example 15.8
Strategy Insert the starting concentrations that we know into the equilibrium table:
Kc for the reaction of hydrogen and iodine to produce hydrogen iodide,
H2(g) + I2(g) 2HI(g)
is 54.3 at 430°C. What will the concentrations be at equilibrium if we start with 0.240 M concentrations of both H2 and I2
Initial concentration (M) 0.240 0.240 0
Change in concentration (M)
Equilibrium concentration (M)
H2
I2
2HI
+

Worked Example 15.8 (cont.)
Solution We define the change in concentration of one of the reactants as x. Because there is no product at the start of the reaction, the reactant concentration must decrease; that is, this reaction must proceed in the forward direction to reach equilibrium. According to the stoichiometry of the chemical reaction, the reactant concentrations will both decrease by the same amount (x), and the product concentration will increase by twice that amount (bining the initial concentration and the change in concentration for each species, we get expressions (in terms of x) for the equilibrium concentrations.
Initial concentration (M) 0.240 0.240 0
Change in concentration (M) –x –x +2x
Equilibrium concentration (M) 0.240 – x 0.240 – x 2x
H2
I2
2HI
+

Worked Example 15.8 (cont.)
Solution Next, we insert these expressions for the equilibrium concentrations into the equilibrium expression and solve for x.
Kc =
54.3 = =
=
x = 0.189
Use the calculated value of x, we can determine the equilibrium concentration of each species as follows:
[H2] = (0.240 – x) M = 0.051 M
[I2] = (0.240 – x) M = 0.051 M
[HI] = 2x = 0.378 M
[HI]2
[H2][I2]
(2x)2
(0.240 – x)(0.240 – x)
(2x)2
(0.240 – x)2
2x
0.240 – x
Think About It Always check your answer by inserting the calculated concentrations into the equilibrium expression:
The small difference between the calculated Kc and the one given in the problem statement is due to rounding.
= 54.9 ≈ Kc
[HI]2
[H2][I2]
(0.378)2
(0.051)2
=
Worked Example 15.9
Strategy Using the initial concentrations, calculate the reaction quotient, Qc, and compare it to the value of Kc (given in the problem statement of Worked Example 15.8) to determine which direction the reaction will proceed to establish equilibrium. Then, construct an equilibrium table to determine the equilibrium concentrations.
For the same reaction and temperature as in Worked Example 15.8, calculate the equilibrium concentrations of all three species if the starting concentrations are as follows: [H2] = 0.00623 M, [I2] = 0.00414 M, [HI] = 0.0424 M.
[HI]2
[H2][I2]
(0.0424)2
(0.00623)(0.00414)
=
= 69.7
Worked Example 15.9 (cont.)
Strategy Therefore, Qc > Kc, so the system will proceed to the left (reverse) to reach equilibrium. The equilibrium table is
Initial concentration (M) 0.00623 0.00414 0.0424
Change in concentration (M)
Equilibrium concentration (M)
H2
I2
2HI
+

Worked Example 15.9 (cont.)
Solution Because we know the reaction must proceed from right to left, we know that the concentration of HI will decrease and the concentrations of H2 and I2 will increase. Therefore, the table should be filled in as follows:
Next, we insert these expressions for the equilibrium concentration into the equilibrium expression and solve for x.
Initial concentration (M) 0.00623 0.00414 0.0424
Change in concentration (M) +x +x –2x
Equilibrium concentration (M) 0.00623 + x 0.00414 + x 0.0424 – 2x
H2
I2
2HI
+

[HI]2
[H2][I2]
Kc =
(0.0424 – 2x)2
(0.00623 + x)(0.00414 + x)
54.3 =
Worked Example 15.9 (cont.)
Solution It isn’t possible to solve this equation the way we did in Worked Example 15.8 (by taking the square root of both sides) because the concentrations of H2 and I2 are unequal. Instead, we have to carry out the multiplications.
54.3(2.58×10-5 + 1.04×10-2x + x2) = 1.80×10-3 – 1.70×10-1x + 4x2
Collecting terms we get
50.3x2 + 0.735x – 4.00×10-4 = 0
This is a quadratic of the form ax2 + bx + c = 0. The solution for the quadratic equation [Appendix 1] is
x =
Here we have a = 50.3, b = 0.735, and c = -4.00×10-4, so
x =
Worked Example 15.9 (cont.)
Solution x = 5.25×10-4 or x = –0.0151
Only the first of these values, 5.25×10-4, makes sense because concentration cannot be a negative number. Using the calculated value of x, we can determine the equilibrium concentration of each species as follows:
[H2] = (0.00623 + x) M = 0.00676 M
[I2] = (0.00414 + x) M = 0.00467 M
[HI] = (0.0424 – 2x) M = 0.0414 M
Think About It Checking this result gives
[HI]2
[H2][I2]
(0.0414)2
(0.00676)(0.00467)
=
= 54.3
Kc =
Worked Example 15.10
Strategy Construct an equilibrium table to determine the equilibrium partial pressures.
A mixture of 5.75 atm of H2 and 5.75 atm of I2 is contained in a 1.0-L vessel at 430°C. The equilibrium constant (KP) for the reaction
H2(g) + I2(g) 2HI(g)
at this temperature is 54.3. Determine the equilibrium partial pressures of H2, I2, and HI.
Initial partial pressure (atm) 5.75 5.75 0
Change in partial pressure (atm) –x –x +2x
Equilibrium partial pressure (atm) 5.75 – x 5.75 – x 2x
H2
I2
2HI
+

Worked Example 15.10 (cont.)
Solution Setting the equilibrium expression equal to KP,
54.3 =
Taking the square root of both sides of the equation gives
=
The equilibrium partial pressures are PH2 = PI2 = 5.75 – 4.52 = 1.23 atm, and PHI = 9.04 atm.
(2x)2
(5.75 – x)2
2x
5.75 – x
2x
5.75 – x
7.369(5.75 – x) = 2x
42.37 – 7.369x = 2x
42.37 = 9.369x
x = 4.52
7.369 =
Think About It Plugging the calculated partial pressures into the equilibrium expression gives
The small difference between this result and the equilibrium constant given in the problem is due to rounding.
= 54.0
(PHI)2
(PH2)(PI2)
(9.04)2
(1.23)2
=
Factors That Affect Chemical Equilibrium
Le Ch telier’s principle states that when a stress is applied to a system at equilibrium, the system, will respond by shifting in the direction that minimizes the effect of the stress.
Stress refers to any of the following:
The addition of a reactant or product
The removal of a reactant or product
A change in volume of the system, resulting in a change in concentration or partial pressure of the reactants and products
A change in temperature
15.5
Factors That Affect Chemical Equilibrium
Consider the Haber process at 700 K:
N2(g) + 3H2(g) 2NH3(g)
At equilibrium:
[N2] = 2.05 M [H2] = 1.56 M [NH3] = 1.52 M
Applying stress by the addition of N2 to give the following concentrations:
[N2] = 3.51 M [H2] = 1.56 M [NH3] = 1.52 M
The reaction shifts to the right.
N2(g) + 3H2(g) 2NH3(g)
Factors That Affect Chemical Equilibrium
N2(g) + 3H2(g) 2NH3(g)
Factors That Affect Chemical Equilibrium
Addition of a reactant or removal of a product will cause an equilibrium to shift to the right.
Addition of a product or removal of a reactant will cause an equilibrium to shift to the left.
Worked Example 15.11
Strategy Use Le Ch telier’s principle to predict the direction of shift in each case. Remember that the position of the equilibrium is only changed by the addition or removal of a species that appears in the reaction quotient expression.
Because sulfur is a solid, it does not appear in the expression.
Hydrogen sulfide (H2S) is a contaminant commonly found in natural gas. It is removed by reaction with oxygen to produce elemental sulfur.
2H2S(g) + O2(g) 2S(s) + 2H2O(g)
For each of the following scenarios, determine whether the equilibrium will shift to the right, shift to the left, or neither: (a) addition of O2(g), (b) removal of H2S(g), (c) removal of H2O(g), and (d) addition of S(s).
[H2O]2
[H2S][O2]
Qc =
Worked Example 15.11 (cont.)
Solution
Changes in concentration of any of the other species will cause a change in the equilibrium position. Addition of a reactant or removal of a product that appears in the expression Qc will shift the equilibrium to the right:
2H2S(g) + O2(g) 2S(s) + 2H2O(g)
Removal of a reactant or addition of a product that appears in the expression Qc will shift the equilibrium to the left:
2H2S(g) + O2(g) 2S(s) + 2H2O(g)
(a) Shift to the right
(d) No change
[H2O]2
[H2S][O2]
Qc =
addition
addition
removal
removal
removal
addition
(b) Shift to the left
(c) Shift to the right
Think About It In each case, analyze the effect the change will have on the value of Qc. In part (a), for example, O2 is added, so its concentration increases. Looking at the reaction quotient expression, we can see that a larger concentration of oxygen corresponds to a larger overall denominator – giving the overall fraction a smaller value. Thus, Q will temporarily be smaller that K and the reaction will have to shift to the right, consuming some of the added O2 (along with some of the H2S in the mixture) to reestablish equilibrium.
Factors That Affect Chemical Equilibrium
When volume is decreased, the equilibrium is driven toward the side with the smallest number of moles of gas.
N2O4(g) 2NO2(g)
Equilibrium mixture:
[N2O4] = 0.643 M
[NO2] = 0.0547 M
Volume decreases by half, concentrations are initially doubled:
[N2O4] = 1.286 M
[NO2] = 0.1094 M
The reaction shifts to the left.
N2O4(g) 2NO2(g)
Worked Example 15.12
Strategy Determine which direction minimized the number of moles of gas in the reaction. Count only moles of gas.
For each reaction, predict in what direction the equilibrium will shift when the volume of the reaction vessel is decreased.
(a) PCl5(g) PCl3(g) + Cl2(g)
(b) 2PbS(s) + 3O2(g) 2PbO(s) + 2SO2(g)
(c) H2(g) + I2(g) 2HI(g)
Solution (a) We have 1 mole of gas on the reactant side and 2 moles of gas on the product side, so it will shift to the left.
(b) 3 moles of gas on the reactant side and 2 moles of gas on the product side, so it will shift to the right.
(c) 2 moles of gas on each side, so no shift.
Think About It When there is a difference in the number of moles of gas, changing the volume of the reaction vessel will change the concentrations of reactant(s) and product(s)–but the system will remain at equilibrium. (Q will remain equal to K.)
Factors That Affect Chemical Equilibrium
Changes in volume and concentration do not change the value of the equilibrium constant.
A change in temperature can alter the value of the equilibrium constant.
Heat + N2O4(g) 2NO2(g) ΔH° = 58.0 kJ/mol
Because the processes is endothermic, adding heat shifts the equilibrium toward products
Factors That Affect Chemical Equilibrium
For any endothermic reaction, heat is a reactant:
Adding heat shifts the reaction towards products, Kc increases
Removing heat shifts the reaction towards reactants, Kc decreases.
For any exothermic reaction, heat is a product:
Adding heat shifts the reaction towards reactants, Kc decreases
Removing heat shifts the reaction towards products, Kc increases
heat + reactants products ΔH° > 0 kJ/mol
reactants products + heat ΔH° < 0 kJ/mol
Factors That Affect Chemical Equilibrium
CoCl42- + 6H2O Co(H2O)62+ + 4Cl- + heat
blue
pink
Factors That Affect Chemical Equilibrium
Catalysts act by reducing the activation energy of a reaction (Section 14.8) which occurs to the same extent for both the forward and reverse reactions.
As a result the addition of a catalyst reduces the time required to reach equilibrium but has no effect on equilibrium constants of the position of equilibrium.
Key Points
15
The Concept of Equilibrium
The Equilibrium Constant
Calculating Equilibrium Constants
Magnitude of the Equilibrium Constant
Equilibrium Expressions
Heterogeneous Equilibria
Manipulating Equilibrium Expressions
Using Equilibrium Expressions to Solve Problems
Predicting the Direction of a Reaction
Calculating Equilibrium Concentrations
Factors that Affect Chemical Equilibrium
Addition or Removal of a Substance
Changes in Volume and Pressure
Changes in Temperature
Chemistry: Atoms First
Julia Burdge & Jason Overby
Chapter 16
Acids and Bases
Acids and Bases
16
16.1 Br nsted Acids and Bases
16.2 Molecular Structure and Acid Strength
Hydrohalic Acids
Oxoacids
Carboxylic Acids
16.3 The Acid-Base Properties of Water
16.4 The pH Scale
16.5 Strong Acids and Bases
Strong Acids
Strong Bases
16.6 Weak Acids and Acid Ionization Constants
The Ionization Constant, Ka
Calculating pH from Ka
Percent Ionization
Using pH to Determine Ka
16.7 Weak Bases and Base Ionization Constants
The Ionization Constant, Kb
Calculating pH from Kb
Using pH to Determine Kb
Acids and Bases
16
16.8 Conjugate Acid-Base Pairs
The Strength of a Conjugate Acid or Base
The Relationship Between Ka and Kb of a Conjugate Acid-Base Pair
16.9 Diprotic and Polyprotic Acids
16.10 Acid-Base Properties of Salt Solutions
Basic Salt Solutions
Acidic Salt Solutions
Neutral Salt Solutions
Salts in Which Both the Cation and the Anion Hydrolyze
16.11 Acid-Base Properties of Oxides and Hydroxides
Oxides of Metals and Nonmetals
Basic and Amphoteric Hydroxides
16.12 Lewis Acids and Bases
Br nsted Acids and Bases
When a Br nsted acid donates a proton, what remains of the acid is known as a conjugate base.
The two species HCl and Cl– are known as a conjugate acid-base pair or simply a conjugate pair.
16.1
HCl(aq) + H2O(l) H3O+(aq) + Cl–(aq)
Gains a proton
Loses a proton
acid
base
conjugate acid
conjugate base
Br nsted Acids and Bases
When a Br nsted base accepts a proton, the newly formed protonated species is known as a conjugate acid.
NH3(aq) + H2O(l) NH4+(aq) + OH–(aq)
acid
base
conjugate acid
conjugate base
Gains a proton
Loses a proton
Worked Example 16.1
Strategy To find the conjugate base of a species, remove a proton from the formula. To find the conjugate acid of a species, add a proton to the formula. The word proton, in this context, refers to H+. Thus, the formula and the charge will both be affected by the addition or removal of H+.
What is (a) the conjugate base on HNO3, (b) the conjugate acid of O2-, (c) the conjugate base of HSO4-, and (d) the conjugate acid of HCO3-.
Solution (a) NO3-
(b) OH-
(c) SO42-
(d) H2CO3
Think About It A species does not need to be what we think of as an acid in order for it to have a conjugate base. For example, we would not refer to the hydroxide ion (OH-) as an acid – but it does have a conjugate base, the oxide ion (O2-). Furthermore, a species that can either lose or gain a proton, such as HCO3-, has both a conjugate base (CO32-) and a conjugate acid (H2CO3).
Worked Example 16.2
Strategy In each equation, the reactant that loses a proton is the acid and the reactant that gains the proton is the base. Each product is the conjugate of one of the reactants. Two species that differ only by a proton constitute a conjugate pair.
Label each of the species in the following equations as an acid, base, conjugate base, or conjugate acid:
(a) HF(aq) + NH3(aq) F-(aq) + NH4+(aq)
(b) CH3COO-(aq) + H2O(l) CH3COOH(aq) + OH-(aq)
Solution (a) HF loses a proton a becomes F-; NH3 gains a proton and becomes NH4+.
HF(aq) + NH3(aq) F-(aq) + NH4+(aq)
(b) CH3COO- gains a proton to become CH3COOH; H2O loses a proton to become OH-.
CH3COO-(aq) + H2O(l) CH3COOH(aq) + OH-(aq)
acid
base
conjugate base
conjugate acid
acid
base
conjugate base
conjugate acid
Think About It In a Br nsted acid-base reaction, there is always an acid and a base, and whether a substance behaves as an acid or a base depends on what it is combined with. Water, for example, behaves as a base when combined with HCl but behaves as an acid when combined with NH3.
Molecular Structure and Acid Strength
The strength of an acid is measured by its tendency to ionize.
Two factors influence ionization:
1) The strength of the H—X bond
2) The polarity of the H—X bond
16.2
HX → H+ + X–
H—X
δ+
δ–
Molecular Structure and Acid Strength
Hydrohalic acid strength:
Biggest factor is bond strength.
Only HF is a weak acid.
HF << HCl < HBr < HI
Molecular Structure and Acid Strength
Oxoacids:
An oxoacid contains hydrogen, oxygen, and a central nonmetal atom.
Molecular Structure and Acid Strength
To compare oxoacid strength, it is convenient to divide the oxoacids into two groups:
Oxoacids having different central atoms that are from the same group of the periodic table and that have the same oxidation number.
HClO3 > HBrO3
Cl is more electronegative; the O—H bond is more polar.
Molecular Structure and Acid Strength
To compare oxoacid strength, it is convenient to divide the oxoacids into two groups:
2) Oxoacids having the same central atom, but different numbers of oxygen atoms.
HClO4 > HClO3 > HClO2 > HClO
Worked Example 16.3
Strategy In each group, compare the electronegativies or oxidation numbers of the central atoms to determine which O–H bonds are the most polar. The more polar the O–H bond, the more readily it is broken and the stronger the acid.
Predict the relative strengths of the oxoacids in each of the following groups:
(a) HClO, HBrO, and HIO; (b) HNO3 and HNO2.
Solution (a) In a group with different central atoms, we must compare electronegativities. The electronegativities of the central atoms in this group decrease as follows: Cl > Br > I.
Acid strength decreases as follows: HClO > HBrO > HIO
(b) These two acids have the same central atom but differ in the number of oxygen atoms. In a group such as this, the greater the number of attached oxygen atoms, the higher the oxidation number and the stronger the acid.
HNO3 is stronger than HNO2.
Think About It Four of the strong acids are oxoacids: HNO3, HClO4, HClO3, and H2SO4.
The Acid-Base Properties of Water
16.3
A species that can behave either as a Br nsted acid or a Br nsted base is called amphoteric.
The acid-base properties of water produces H3O+ and OH– ions in equilibrium with water in a reaction known as the autoionization of water.
The equilibrium expression for the autoionization of water is given by:
Kw = [H3O+][OH–] = 1.0 x 10–14 (at 25°C)
Molecular Structure and Acid Strength
An important group of organic acids is the carboxylic acids:
The strength of the acid depends on the nature of the R group.
Acetic acid (Ka = 1.8 x 10–5)
Chloroacetic acid (Ka = 1.4 x 10–3)
The Acid-Base Properties of Water
Since the product of the concentrations of H3O+ and OH– is equal to a constant, the relative amount of each obeys a fixed relationship.
Depending on which ion concentration is in excess, the solution will be considered acidic or basic.
When [H3O+] = [OH–], the solution is neutral
When [H3O+] > [OH–], the solution is acidic
When [H3O+] < [OH–], the solution is basic
Kw = [H3O+][OH–] = 1.0 x 10–14 (at 25°C)
Worked Example 16.4
Strategy Use the value of Kw to determine [OH-] when [H3O+] = 0.10 M.
The concentration of hydronium ions in stomach acid is 0.10 M. Calculate the concentration of hydroxide ions in stomach acid at 25°C.
Solution Kw = [H3O+][OH-] = 1.0×10-14 at 25°C. Rearranging to solve for [OH-],
[OH-] =
1.0×10-14
[H3O+]
[OH-] =
1.0×10-14
0.10
[OH-] = 1.0×10-13 M
Think About It Remember that the equilibrium constants are temperature dependent. The value of Kw = 1.0×10-14 only at 25°C.
The pH Scale
The acidity of an aqueous solution depends on the concentration of hydronium ions, [H3O+].
The pH of a solution is defined as the negative base-10 logarithm of the hydronium ion concentration (in mol/L)
In pure water at 25°C, [H3O+] = log1.0 x 10–7
pH = –log(1.0 x 10–7) = 7.00
pH is a dimensionless quantity.
16.4
pH = –log[H3O+]
[H3O+] = 10–pH
The pH Scale
The pH Scale
Worked Example 16.5
Strategy Given [H3O+], use pH = –log[H3O+] to solve for pH.
Determine the pH of a solution at 25°C in which the hydronium ion concentration is (a) 3.5×10-4 M, (b) 1.7×10-7 M, and (c) 8.8×10-11 M.
Solution
(a) pH = –log(3.5×10-4) = 3.46
(b) pH = –log(1.7×10-7) = 6.77
(c) pH = –log(8.8×10-11) = 10.06
Think About It When a hydronium ion concentration falls between two “benchmark” concentrations in Table 16.4, the pH falls between the two corresponding pH values. In part (c), for example, the hydronium ion concentration (8.8×10-11 M) is greater than 1.0×10-11 M but less than 1.0×10-10 M. Therefore, we expect the pH to be between 11.00 and 10.00.
Worked Example 16.6
Strategy Given pH, use [H3O+] = 10-pH to calculate [H3O+].
Calculate the hydronium ion concentration in a solution at 25°C in which the pH is (a) 4.76, (b) 11.95, and (c) 8.01.
Solution
(a) [H3O+] = 10-4.76 = 1.7×10-5 M
(b) [H3O+] = 10-11.95 = 1.1×10-12 M
(c) [H3O+] = 10-8.01 = 9.8×10-9 M
Think About It Think About It If you use the calculated hydronium ion concentrations to recalculate pH, you will get numbers slightly different from those given in the problem. In part (a), for example, log(1.7×10-5) = 4.77. The small difference between this and 4.76 (the pH given in the problem) is due to a rounding error. Remember that a concentration derived from a pH with two digits to the right of the decimal point can have only two significant figures. Note also that the benchmarks can be used equally well in this circumstance. A pH between 4 and 5 corresponds to a hydronium ion concentration between 1.7×10-4 M and 1.0×10-5 M.
The pH Scale
A pOH scale analogous to the pH scale can be defined as the negative base-10 logarithm of the hydroxide ion concentration.
From the definition of pH and pOH:
pOH = –log[OH–]
[OH–] = 10–pOH
pH + pOH = 14.00
The pH Scale
Worked Example 16.7
Strategy Given [OH-], use pOH = –log[OH-] to calculate pOH.
Determine the pOH of a solution at 25°C in which the hydroxide ion concentration is (a) 3.7×10-5 M, (b) 4.1×10-7 M, and (c) 8.3×10-2 M.
Solution
(a) pOH = –log(3.7×10-5) = 4.43
(b) pOH = –log(4.1×10-7) = 6.39
(c) pOH = –log(8.3×10-2) = 1.08
Think About It Remember that the pOH scale is, in essence, the reverse of the pH scale. On the pOH scale, numbers below 7 indicate a basic solution, whereas number above 7 indicate an acidic solution. The pOH benchmarks (abbreviated in Table 16.6) work the same way the pH benchmarks do. In part (a), for example, a hydroxide ion concentration between 1×10-4 M and 1×10-5 M corresponds to a pOH between 4 and 5.
Worked Example 16.8
Strategy Given pOH, use [OH-] = 10-pOH to calculate [OH-].
Calculate the hydroxide ion concentration in a solution at 25°C in which the pOH is (a) 4.91, (b) 9.03, and (c) 10.55.
Solution
(a) [OH-] = 10-4.91 = 1.2×10-5 M
(b) [OH-] = 10-9.03 = 9.3×10-10 M
(c) [OH-] = 10-10.55 = 2.8×10-11 M
Think About It Use the benchmark pOH values to determine whether these solutions are reasonable. In part (a), for example, the pOH between 4 and 5 corresponds to [OH-] between 1×10-4 M and 1×10-5 M.
Strong Acids and Bases
Strong acid dissociations are not treated as equilibria, rather as processes that go to completion.
16.5
Hydrochloric acid
HCl(aq) + H2O(l)
H3O+(aq) + Cl–(aq)
HBr(aq) + H2O(l)
H3O+(aq) + Br–(aq)
Hydrobromic acid
HI(aq) + H2O(l)
H3O+(aq) + I–(aq)
Hydroiodic acid
HNO3(aq) + H2O(l)
H3O+(aq) + NO3–(aq)
Nitric acid
HClO3(aq) + H2O(l)
H3O+(aq) + ClO3–(aq)
Chloric acid
HClO4(aq) + H2O(l)
H3O+(aq) + ClO4–(aq)
Perchloric acid
H2SO4(aq) + H2O(l)
H3O+(aq) + HSO4–(aq)
Sulfuric acid
Worked Example 16.9
Strategy HI, HNO3, and HClO4 are all strong acids, so the concentration of hydronium ions in each solution is the same as the stated concentration of the acid. Use pH = –log[H3O+] to calculate pH.
Calculate the pH of an aqueous solution at 25°C that is (a) 0.035 M in HI, (b) 1.2×10-4 M in HNO3, and (c) 6.7×10-5 M in HClO4.
Solution
(a) [H3O+] = 0.035 M
pH = –log(0.035) = 1.46
(b) [H3O+] = 1.2×10-4 M
pH = –log(1.2×10-4) = 3.92
(c) [H3O+] = 6.7×10-5 M
pH = –log(6.7×10-5) = 4.17
Think About It Again, note that when a hydronium ion concentration falls between two of the benchmark concentrations in Table 16.4, the pH falls between the two corresponding pH values. In part (b), for example, the hydronium ion concentration of 1.2×10-4 M is greater than 1×10-4 M and less than 1×10-3 M. Therefore, we expect the pH to be between 4.00 and 3.00.
Worked Example 16.10
Strategy Use [H3O+] = 10-pH to convert from pH to [H3O+]. In a strong acid solution, [H3O+] is equal to the acid concentration.
Calculate the concentration on HCl in a solution at 25°C that has pH (a) 4.95, (b) 3.45, and (c) 2.78.
Solution
(a) [HCl] = [H3O+] = 10-4.95 = 1.1×10-5 M
(b) [HCl] = [H3O+] = 10-3.45 = 3.5×10-4 M
(c) [HCl] = [H3O+] = 10-2.78 = 1.7×10-3 M
Think About It As pH decreases, acid concentration increases.
Strong Acids and Bases
The list of strong bases consists of the hydroxides of alkali metals and the heaviest alkaline earth metals.
LiOH(aq)
Li+(aq) + OH–(aq)
NaOH(aq)
Na+(aq) + OH–(aq)
KOH(aq)
K+(aq) + OH–(aq)
RbOH(aq)
Rb+(aq) + OH–(aq)
CsOH(aq)
Cs+(aq) + OH–(aq)
Ca(OH)2(aq)
Ca2+(aq) + 2OH–(aq)
Sr(OH)2(aq)
Sr2+(aq) + 2OH–(aq)
Ba(OH)2(aq)
Ba2+(aq) + 2OH–(aq)
Group 1A hydroxides
Group 2A hydroxides
Worked Example 16.11
Strategy LiOH, Ba(OH)2, and KOH are all strong bases. Use reaction stoichiometry to determine the hydroxide ion concentration and pOH = –log[OH-] to determine pOH.
Calculate the pOH of the following aqueous solutions at 25°C: (a) 0.013 M LiOH, (b) 0.013 M Ba(OH)2, and (c) 9.2×10-5 M KOH.
Solution (a) The hydroxide ion concentration is simply equal to the concentration of the base. Therefore, [OH-] = [LiOH] = 0.013 M.
pOH = –log(0.013) = 1.89
(b) The hydroxide ion concentration is twice that of the base:
Ba(OH)2(aq) → Ba2+(aq) + 2OH-(aq)
Therefore, [OH-] = 2[Ba(OH)2] = 2(0.013 M) = 0.026 M.
pOH = –log(0.026) = 1.59
(c) The hydroxide ion concentration is equal to the concentration of the base. Therefore, [OH-] = [KOH] = 9.2×10-5 M.
pOH = –log(9.2×10-5) = 4.04
Think About It These are basic pOH values, which is what we should expect for the solutions described in the problem. Note that while the solutions in parts (a) and (b) have the same base concentration, they do not have the same hydroxide concentration and therefore do not have the same pOH.
Worked Example 16.12
Strategy Use pH + pOH = 14.00 to convert from pH to pOH and [OH-] = 10-pOH to determine the hydroxide ion concentration. Consider the stoichiometry of dissociation in each case to determine the concentration of the base itself.
An aqueous solution of a strong base has pH 8.15 at 25°C. Calculate the original concentration of base in the solution (a) if the base is NaOH and (b) if the base is Ba(OH)2.
Solution
pOH = 14.00 – 8.15 = 5.85
[OH-] = 10-5.85 = 1.41×10-6 M
(a) The dissociation of 1 mole of NaOH produces 1 mole of OH-. Therefore, the concentration of base is equal to the concentration of hydroxide ion.
[NaOH] = [OH-] = 1.41×10-6 M
(b) The dissociation of 2 mole of Ba(OH)2 produces 2 moles of OH-. Therefore, the concentration of base is only one-half the concentration of hydroxide ion.
[Ba(OH)2] = [OH-] = 7.1×10-7 M
1
2
Think About It Alternatively, we could determine the hydroxide ion concentration using [H3O+] = 10-8.15 = 7.1×10-7 M and
Once [OH-] is known, the solution is the same as shown previously.
[OH-] =
1.0×10-14
7.1×10-9
= 1.4×10-6 M
Weak Acids and Acid Ionization Constants
The ionization of a weak monoprotic acid HA in water is represented by:
Ka is called the acid ionization constant.
The larger the value of Ka, the stronger the acid.
16.6
HA(aq) + H2O(l) H3O+(aq) + A–(aq)
Solution (at 25 °C) Ka pH
0.10 M HF 7.1 x 10–4 2.09
0.10 M CH3COOH 1.8 x 10–5 2.87
Weak Acids and Acid Ionization Constants
Weak Acids and Acid Ionization Constants
Calculate the pH of a 0.50 M HF solution at 25°C.
HF(aq) + H2O(l) H3O+(aq) + F–(aq)
Initial concentration (M)
Change in concentration (M)
Equilibrium concentration (M)
HF(aq) + H2O(l) H3O+(aq) + F–(aq)
–x
0.50 – x
+x
+x
x
x
0.50
0
0
Weak Acids and Acid Ionization Constants
Initial concentration (M) 0.50 0 0
Change in concentration (M) –x +x +x
Equilibrium concentration (M) 0.50 – x x x
HF(aq) + H2O(l) H3O+(aq) + F–(aq)
Use quadratic formula to solve
–or –
Since HF is a weak acid, x could be small compared to 0.50
Weak Acids and Acid Ionization Constants
Initial concentration (M) 0.50 0 0
Change in concentration (M) –x +x +x
Equilibrium concentration (M) 0.50 – x x x
HF(aq) + H2O(l) H3O+(aq) + F–(aq)
x2 = (0.50)(7.1 x 10–4) = 3.55 x 10–4
x = 1.9 x 10–2
simplifies
Weak Acids and Acid Ionization Constants
Initial concentration (M) 0.50 0 0
Change in concentration (M) –1.9 x 10–2 +1.9 x 10–2 +1.9 x 10–2
Equilibrium concentration (M) 0.48 1.9 x 10–2 1.9 x 10–2
HF(aq) + H2O(l) H3O+(aq) + F–(aq)
pH = –log(0.019) = 1.72
The shortcut is acceptable to use if the calculate value of x is less than 5% of the initial acid concentration
Worked Example 16.13
Strategy Construct an equilibrium table, and express the equilibrium and concentration of each species in terms of x. Solve for x using the approximation shortcut, and evaluate whether or not the approximation is valid. Use pH = –log[H3O+] to determine pH.
The Ka of hypochlorous acid (HClO) is 3.5×10-8. Calculate the pH of a solution at 25°C that is 0.0075 M in HClO.
Initial concentration (M) 0.0075 0 0
Change in concentration (M) –x +x +x
Equilibrium concentration (M) 0.0075 – x x x
HClO(aq) + H2O(l) H3O+(aq) + ClO–(aq)
Worked Example 16.13 (cont.)
Solution These equilibrium concentrations are then substituted into the equilibrium expression to give
Assuming that 0.0075 – x ≈ 0.0075,
Solving for x, we get
x = = 1.62×10-5 M
According to the equilibrium table, x = [H3O+]. Therefore,
pH = –log(1.62×10-5) = 4.79
Ka =
(x)(x)
0.0075 – x
= 3.5×10-8
x2
0.0075
= 3.5×10-8
x2
= (3.5×10-8)(0.0075)
Think About It We learned in Section 16.3 that the concentration of hydronium ion in pure water at 25°C is 1.0×10-7 M, yet we use 0 M as the starting concentration to solve for the pH of a weak acid. The reason for this is that the actual concentration of hydronium ion in pure water is insignificant compared to the amount produced by the ionization of the weak acid. We could use the actual concentration of hydronium as the initial concentration, but doing so would not change the result because (x + 1.0×10-7) M ≈ x M. In solving problems of this type, we neglect the small concentration of H+ due to the autoionization of water.
Weak Acids and Acid Ionization Constants
Initial concentration (M) 0.50 0 0
Change in concentration (M) –1.9 x 10–2 +1.9 x 10–2 +1.9 x 10–2
Equilibrium concentration (M) 0.48 1.9 x 10–2 1.9 x 10–2
HF(aq) + H2O(l) H3O+(aq) + F–(aq)
A quantitative measure of the degree of ionization is percent ionization.
Weak Acids and Acid Ionization Constants
Initial concentration (M) 1.00 0 0
Change in concentration (M)
Equilibrium concentration (M)
HF(aq) + H2O(l) H3O+(aq) + F–(aq)
Solution (at 25 °C) pH % ionization
0.5 M HF 1.72 3.8
1.0 M HF 1.57 2.7
Calculate the percent ionization of a 1.0 M HF solution at 25°C.
–2.7 x 10–2
+2.7 x 10–2
+2.7 x 10–2
2.7 x 10–2
2.7 x 10–2
0.97
Weak Acids and Acid Ionization Constants
HF(aq) + H2O(l) H3O+(aq) + F–(aq)
Solution (at 25 °C) pH % ionization
0.5 M HF 1.72 3.8
1.0 M HF 1.57 2.7
Worked Example 16.14
Strategy Using the procedure described in Worked Example 16.13, we construct an equilibrium table and for each concentration of acetic acid, we solve for the equilibrium concentration of H+. We use pH = –log[H3O+] to find pH, and the equation below to find percent ionization. Ka for acetic acid is 1.8×10-5.
Determine the pH and percent ionization for acetic acid solutions at 25°C with concentrations (a) 0.15 M, (b) 0.015 M, and (c) 0.0015 M.
Worked Example 16.14 (cont.)
Solution (a)
Solving for x gives [H3O+] = 0.0016 M and pH = –log(0.0016) = 2.78.
(b) Solving the same way as part (a) gives [H3O+] = 5.2×10-4 M and pH = 3.28.
Initial concentration (M) 0.15 0 0
Change in concentration (M) –x +x +x
Equilibrium concentration (M) 0.15 – x x x
CH3COOH(aq) + H2O(l) H3O+(aq) + CH3COO–(aq)
0.0016 M
0.15 M
percent ionization
=
× 100% = 1.1%
5.2×10-4 M
0.015 M
percent ionization
=
× 100% = 3.5%
Worked Example 16.14 (cont.)
Solution (c) Solving the quadratic equation, or using successive approximation [Appendix 1] gives [H3O+] = 1.6×10-4 M and pH = 3.78.
1.6×10-4 M
0.0015 M
percent ionization
=
× 100% = 11%
Think About It Check your work by using the calculated value of Ka to solve for the pH of a 0.10-M solution of aspirin.
Weak Acids and Acid Ionization Constants
Determine the Ka of a weak acid that has a concentration of 0.25 M and a pH of 3.47 at 25°C.
HA(aq) + H2O(l) H3O+(aq) + A–(aq)
Initial concentration (M) 0.25 0 0
Change in concentration (M)
Equilibrium concentration (M)
HA(aq) + H2O(l) H3O+(aq) + A–(aq)
0.2497
3.39 x 10–4
3.39 x 10–4
+3.39 x 10–4
+3.39 x 10–4
–3.39 x 10–4
H3O+ = 10–3.47 = 3.39 x 10–4 M
Weak Acids and Acid Ionization Constants
Determine the Ka of a weak acid that has a concentration of 0.25 M and a pH of 3.47 at 25°C.
Initial concentration (M) 0.25 0 0
Change in concentration (M)
Equilibrium concentration (M)
HA(aq) + H2O(l) H3O+(aq) + A–(aq)
0.2497
3.39 x 10–4
3.39 x 10–4
+3.39 x 10–4
+3.39 x 10–4
–3.39 x 10–4
Worked Example 16.15
Strategy Determine the hydronium ion concentration from the pH. Use the hydronium ion concentration to determine the equilibrium concentrations of the other species, and plug the equilibrium concentrations into the equilibrium expressions to evaluate Ka.
Aspirin (acetylsalicylic acid, HC9H7O4) is a weak acid. It ionizes in water according to the equation
HC9H7O4(aq) + H2O(l) H3O+(aq) + C9H7O4-(aq)
A 0.10-M aqueous solution of aspirin has a pH of 2.27 at 25°C. Determine the Ka of aspirin.
Worked Example 16.15 (cont.)
Solution [H3O+] = 10–2.27 = 5.37×10-3 M
To calculate Ka, though, we also need the equilibrium concentrations of C9H7O4- and HC9H7O4. The stoichiometry of the reaction tells us that [C9H7O4-] = [H3O+]. Furthermore, the amount of aspirin that has ionized is equal to the amount of hydronium ion in solution. Therefore, the equilibrium concentration of aspirin is (0.10 – 5.37×10-3) M = 0.095 M.
The Ka of aspirin is 3.0×10-4.
Initial concentration (M) 0.10 0 0
Change in concentration (M) –0.005 +5.37×10-3 +5.37×10-3
Equilibrium concentration (M) 0.095 5.37×10-3 5.37×10-3
HC9H7O4(aq) + H2O(l) H3O+(aq) + C9H7O4-(aq)
Ka =
[H3O+][C9H7O4-]
[HC9H7O4]
= 3.0×10-4
=
(5.37×10-3)2
0.095
Think About It Check your work by using the calculated value of Ka to solve for the pH of a 0.10-M solution of aspirin.
Weak Bases and Base Ionization Constants
The ionization of a weak base is incomplete and is treated in the same way as the ionization of a weak acid.
Kb is called the base ionization constant.
The larger the value of Kb, the stronger the base.
16.7
B(aq) + H2O(l) HB+(aq) + OH–(aq)
Worked Example 16.16
Strategy Construct an equilibrium table, and express equilibrium concentrations in terms of the unknown x. Plug these equilibrium concentrations into the equilibrium expression, and solve for x. From the value of x, determine the pH.
What is the pH of a 0.040 M ammonia solution at 25°C.
Initial concentration (M) 0.040 0 0
Change in concentration (M) –x +x +x
Equilibrium concentration (M) 0.040 – x x x
NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)
Worked Example 16.16 (cont.)
Solution The equilibrium concentrations are substituted into the equilibrium expression to give
Assuming that 0.040 – x ≈ 0.040 and solving for x gives
x2 = (1.8×10-5)(0.040) = 7.2×10-7
x = = 8.5×10-4 M
According to the equilibrium table, x = [OH-]. Therefore, pOH = – log(x):
–log(8.5×10-4) = 3.07
and pH = 14.00 – pOH = 14.00 – 3.07 – 10.93. The pH of a 0.040-M solution of NH3 at 25°C is 10.93.
Kb =
[NH4+][OH-]
[NH3]
= 1.8×10-5
=
(x)(x)
0.040 – x
= 1.8×10-5
(x)(x)
0.040 – x
≈
(x)(x)
0.040
Think About It It is a common error in Kb problems to forget that x is hydroxide ion concentration rather than the hydronium ion concentration. Always make sure that the pH you calculate for a solution of base is a basic pH, that is, a pH greater than 7.
Weak Bases and Base Ionization Constants
Worked Example 16.17
Strategy Use pH to determine pOH, and pOH to determine the hydroxide ion concentration. From the hydroxide ion concentration, use reaction stoichiometry to determine the other equilibrium concentrations and plus those concentrations into the equilibrium expression to evaluate Kb.
Caffeine, the stimulant in coffee and tea, is a weak base that ionizes in water according to the equation
C8H10N4O2(aq) + H2O(l) HC8H10N4O2+(aq) + OH-(aq)
A 0.15-M solution of caffeine at 25°C has a pH of 8.45. Determine the Kb of caffeine.
Worked Example 16.17 (cont.)
Solution pOH = 14.00 – 8.45 – 5.55; [OH-] = 10-5.55 = 2.82×10-6 M
Based on the reaction stoichiometry, [HC8H10N4O2+] = [OH-], and the amount of hydroxide ion in solution at equilibrium is equal to the amount of caffeine that has ionized. At equilibrium, therefore,
[C8H10N4O2] = (0.15 – 2.82×10-6) M ≈ 0.15 M
Kb =
[HC8H10N4O2+][OH-]
[C8H10N4O2]
= 5.3×10-11
=
(2.82×10-6)2
0.15
Initial concentration (M) 0.15 0 0
Change in concentration (M) –2.82×10-6 +2.82×10-6 +2.82×10-6
Equilibrium concentration (M) 0.15 2.82×10-6 2.82×10-6
C8H10N4O2(aq) + H2O(l) HC8H10N4O2+(aq) + OH-(aq)
Think About It Check your answer using the calculated Kb to determine the pH ofa 0.15-M solution.
Conjugate Acid-Base Pairs
A strong acid ionizes completely in water:
16.8
HCl(aq) H+(aq) + Cl–(aq)
No affinity for the H+ ion
Cl–(aq) + H2O(l) HCl(aq) + OH–(aq)
X
The chloride ion is a weak conjugate base.
Conjugate Acid-Base Pairs
A weak acid ionizes to a limited degree in water:
HF(aq) H+(aq) + F–(aq)
Strong affinity for the H+ ion
F–(aq) + H2O(l) HF(aq) + OH–(aq)
The fluoride ion is a strong conjugate base.
Conjugate Acid-Base Pairs
A strong acid has a weak conjugate base.
A weak acid has a strong conjugate base.
A strong base has a weak conjugate acid.
A weak base has a strong conjugate base.
Conjugate Acid-Base Pairs
A simple relationship between the ionization constant of a weak acid (Ka) and the ionization constant of a weak base (Kb) can be derived:
CH3COOH(aq) H+(aq) + CH3COO–(aq)
CH3COO– (aq) + H2O(l) CH3COOH(aq) + OH–(aq)
H2O(l) H+(aq) + OH–(aq)
Ka x Kb = Kw
Worked Example 16.18
Strategy Each species listed is either a conjugate base or a conjugate acid. Determine the identity of the acid corresponding to each conjugate base and the identity of the base corresponding to each conjugate acid; then, consult Table 16.7 and 16.8 for their ionization constants. Use the tabulated ionization constants and Kw = Ka×Kb to calculate each indicated K value.
Determine (a) Kb of the acetate ion (CH3COO-), (b) Ka of the methylammonium ion (CH3NH3+), (c) Kb of the fluoride (F-), and (d) Ka of the ammonium ion (NH4+).
Solution (a) A Kb value is requested, indicating that the acetate ion is a conjugate base. To identify the corresponding Br nsted acid, add a proton to the formula to get CH3COOH (acetic acid). The Ka of acetic acid is 1.8×10-5.
Conjugate base CH3COO-: Kb =
= 5.6×10-10
1.0×10-14
1.8×10-5
Kw
Kb
Ka =
Kw
Ka
Kb =
and
Worked Example 16.18 (cont.)
Solution (b) A Ka value is requested, indicating that the methylammonium ion is a conjugate acid. Determine the identity of the corresponding Br nsted base by removing a proton from the formula to get CH3NH2 (methylamine). The Kb of methylamine is 4.4×10-4.
Conjugate acid CH3NH3+: Ka =
(c) F- is the conjugate base of HF; Ka = 7.1×10-4.
Conjugate base F-: Kb =
(d) NH4+ is the conjugate acid of NH3; Kb = 1.8×10-5.
Conjugate acid NH4+: Ka =
= 2.3×10-11
1.0×10-14
4.4×10-4
= 1.4×10-11
1.0×10-14
7.1×10-4
= 5.6×10-10
1.0×10-14
1.8×10-5
Think About It Because the conjugates of weak acids and bases have ionization constants, salts containing these ions have an effect on the pH of a solution. In Section 16.10 we will use the ionization constants of conjugate acids and conjugates bases to calculate pH for solutions containing dissolved salts.
Diprotic and Polyprotic Acids
Diprotic and polyprotic acids undergo successive ionizations, losing one proton at a time, and each has a Ka associate with it.
Ka1 > Ka2
For a given acid, the first ionization constant is much larger than the second, and so on.
16.9
H2CO3(aq) H+(aq) + HCO3–(aq)
HCO3– (aq) H+(aq) + CO32– (aq)
Diprotic and Polyprotic Acids
Worked Example 16.19
Strategy Follow the same procedure for each ionization as for the determination of equilibrium concentrations for a monoprotic acid. The conjugate base resulting from the first ionization is the acid for the second ionizations, and its starting concentration is the equilibrium concentration from the first ionization.
H2C2O4(aq) H+(aq) + HC2O4–(aq) Ka1 = 6.5 x 10–2
HC2O4– (aq) H+(aq) + C2O42– (aq) Ka2 = 6.1 x 10–5
Construct an equilibrium table for each ionization, using x as the unknown in the first ionization and y as the unknown in the second ionization.
Oxalic acid (H2C2O4) is a poisonous substance used mainly as a bleaching agent. Calculate the concentrations of all species present at equilibrium in a 0.10-M solution at 25°C.
Worked Example 16.19 (cont.)
Strategy
The equilibrium concentration of the hydrogen oxalate (HC2O4-) after the first ionization becomes the starting concentration for the second ionization. Additionally, the equilibrium concentration of H+ is the starting concentration for the second ionization.
Initial concentration (M) 0.10 0 0
Change in concentration (M) –x +x +x
Equilibrium concentration (M) 0.10 – x x x
H2C2O4(aq) H+(aq) + HC2O4–(aq)
Initial concentration (M) x x 0
Change in concentration (M) –y +y +y
Equilibrium concentration (M) x – y x + y y
HC2O4– (aq) H+(aq) + C2O42– (aq)
Worked Example 16.19 (cont.)
Solution
Applying the approximation and neglecting x in the denominator of the expression gives
Testing the approximation,
Clearly the approximation is not valid, so we must solve the following quadratic equation:
x2 + 6.5×10-3x – 6.5×10-3 = 0
6.5×10-2 =
[H+][HC2O4-]
[H2C2O4]
Ka1 =
x2
0.10 – x
6.5×10-2 ≈
x2
0.10
x2 = 6.5×10-3
x = 8.1×10-2 M
8.1×10-2 M
0.10 M
×100% = 81%
Worked Example 16.19 (cont.)
Solution The result x = 0.054 M. Thus, after the first ionization, the concentrations of species in solution are
[H+] = 0.054 M
[HC2O4-] = 0.054 M
[H2C2O4] = (0.10 – 0.054) M = 0.046 M
Rewriting the equilibrium table for the second ionization, using the calculated value of x, gives the following:
6.1×10-5 =
[H+][C2O42-]
[HC2O4-]
Ka2 =
(0.054 + y)(y)
0.054 – y
Initial concentration (M) 0.054 0.054 0
Change in concentration (M) –y +y +y
Equilibrium concentration (M) 0.054 – y 0.054 + y y
HC2O4– (aq) H+(aq) + C2O42– (aq)
Worked Example 16.19 (cont.)
Solution Assuming that y is very small and applying the approximations 0.054 + y ≈ 0.054 and 0.054 – y ≈ 0.054 gives
We must test the approximation as follows to see if it is valid:
This time, because the ionization constant is much smaller, the approximation is valid. At equilibrium, the concentrations of all species are
[H2C2O4] = 0.046 M
[HC2O4-] = (0.054 – 6.1×10-5) = 0.054 M
[H+] = (0.054 + 6.1×10-5) = 0.054 M
[C2O42-] = 6.1×10-5 M
= y = 6.1×10-5
(0.054)(y)
0.054
6.1×10-5 M
0.054 M
×100% = 0.11%
Think About It Note that the second ionization did not contribute significantly to the H+ concentration. Therefore, we could determine the pH of this solution by considering only the first ionization. This is true in general for polyprotic acids where Ka1 is at least 1000×Ka2. [Note that it is necessary to consider the second ionization to determine the concentration of oxalate ion (C2O42-).]
Acid-Base Properties of Salt Solutions
Salt hydrolysis occurs when ions produced by the dissociation of a salt react with water to produce either hydroxide ions or hydronium ions.
Basic salts (conjugates of weak acids):
Acidic salts (conjugates of weak bases)
16.10
F–(aq) + H2O(l) HF(aq) + OH–(aq)
NH4+(aq) + H2O(l) NH3(aq) + H3O+(aq)
Worked Example 16.20
Strategy A solution of NaF contains Na+ ions and F- ions. The F- ion is the conjugate base of the weak acid, HF. Use the Ka value for HF (7.1×10-4) and Kw = Kb×Ka to determine Kb for F-:
Then, solve this pH problem like any equilibrium problem, using an equilibrium table.
F-(aq) + H2O(l) HF(aq) + OH-(aq)
Calculate the pH of a 0.10-M solution of sodium fluoride (NaF) at 25°C.
= 1.4×10-11
1.0×10-14
7.1×10-4
Kw
Ka
Kb =
=
[HF][OH-]
[F-]
Kb =
F-(aq) + H2O(l) HF(aq) + OH-(aq)
Initial concentration (M) 0.10 0 0
Change in concentration (M) –x +x +x
Equilibrium concentration (M) 0.10 – x x x
Worked Example 16.20 (cont.)
Solution Substituting the equilibrium concentrations into the equilibrium expression and using the shortcut to solve x, we get
According to our equilibrium table, x = [OH-]. In this case, the autoionization of water makes a significant contribution to the hydroxide ion concentration so the total concentration will be the sum of 1.2×10-6 M (from the ionization of F-) and 1.0×10-7 M (from the autoionization of water). Therefore, we calculate the pOH first as
pOH = –log(1.2×10-6 + 1.0×10-7) = 5.95
and then the pH,
pH = 14.00 – pOH = 14.00 – 5.95 = 8.05
The pH of a 0.10-M solution of NaF at 25°C is 8.05.
1.4×10-11 =
x2
0.10 – x
x2
0.10
≈
x = = 1.2×10-6 M
Think About It It’s easy to mix up pH and pOH in this type of problem. Always make a qualitative prediction regarding the pH of a salt solution first, and then check to make sure that your calculated pH agrees with your prediction. In this case, we would predict a basic pH because the anion in the salt (F-) is the conjugate base of a weak acid (HF). The calculated pH, 8.05, is indeed basic.
Worked Example 16.21
Strategy A solution of NH4Cl contains NH4+ ions and Cl- ions. The NH4+ ion is the conjugate acid of the weak base, NH3. Use the Kb value for NH3 (1.8×10-5) and Kw = Kb×Ka to determine Ka for F-:
Again, we write the balanced chemical equation and the equilibrium expression:
NH4+ (aq) + H2O(l) NH3(aq) + H3O+(aq)
Calculate the pH of a 0.10-M solution of ammonium chloride (NH4Cl) at 25°C.
= 5.6×10-10
1.0×10-14
1.8×10-5
Kw
Kb
Ka =
=
[NH3][H3O+]
[NH4+ ]
Kb =
NH4+ (aq) + H2O(l) NH3(aq) + H3O+(aq)
Initial concentration (M) 0.10 0 0
Change in concentration (M) –x +x +x
Equilibrium concentration (M) 0.10 – x x x
Worked Example 16.21 (cont.)
Solution Substituting the equilibrium concentrations into the equilibrium expression and using the shortcut to solve x, we get
According to our equilibrium table, x = [H3O+]. The pH can be calculated as follows:
pH = –log(7.5×10-6) = 5.12
The pH of a 0.10-M solution of ammonium chloride (at 25°C) is 5.12.
5.6×10-10 =
x2
0.10 – x
x2
0.10
≈
x = = 7.5×10-6 M
Think About It In this case, we would predict an acidic pH because the cation in the salt (NH4+) is the conjugate acid of a weak base (NH3). The calculated pH is acidic.
Acid-Base Properties of Salt Solutions
Small, highly charged metal ions can react with water to produce an acidic solution.
Acid-Base Properties of Salt Solutions
The pH of salt solutions can be qualitatively predicted by determining which ions facilitate hydrolysis.
Examples
A cation that will make a solution acidic is
The conjugate acid of a weak base
A small, highly charged metal ion (other than Group 1A or 2A)
NH4+ , CH3NH3+ , C2H5NH3+
Al3+ , Cr3+ , Fe3+ , Bi3+
An anion that will make a solution basic is
The conjugate base of a weak acid
CN– , NO2– , CH3COO–
A cation that will not affect the pH of a solution is
A Group 1A or heavy Group 2A cation (except Be2+)
Li+ , Na+ , Ba2+
An anion that will not affect the pH of a solution is
The conjugate base of a strong acid
Cl– , NO3– , ClO4–
Worked Example 16.22
Strategy Identify the ions present in each solution, and determine which, if any, will impact the pH of the solution.
Predict whether a 0.10-M solution of each of the following salts will be basic, acidic, or neutral: (a) LiI, (b) NH4NO3, (c) Sr(NO3)2, (d) KNO2, (e) NaCN.
Solution (a) Ions in solution: Li+ and I-. Li+ is a Group 1A cation; I- is the conjugate base of the strong acid HI. Therefore, neither ion hydrolyzes to any significant degree. Solution will be neutral.
(b) Ions in solution: NH4+ and NO3-. NH4+ is the conjugate acid of the weak base NH3; NO3- is the conjugate base of the strong acid HNO3. In this case, the cation will hydrolyze, making the pH acidic:
NH4+ (aq) + H2O(l) NH3(aq) + H3O+(aq)
Worked Example 16.22 (cont.)
Solution (c) Ions in solution: Sr2+ and NO3-. Sr2+ is a heavy Group 2A cation; NO3- is the conjugate base of the strong acid, HNO3. Neither ion hydrolyzes to any significant degree.
(d) Ions in solution: K+ and NO2-. K+ is a Group 1A cation; NO2- is the conjugate base of the weak acid HNO2. In this case, the anion hydrolyzes, thus making the pH basic:
NO2- (aq) + H2O(l) HNO2(aq) + OH-(aq)
(e) Ions in solution: Na+ and CN-. Na+ is a Group 1A cation; CN- is the conjugate base of the weak acid HCN. In this case, too, the anion hydrolyzes, thus making the pH basic:
CN- (aq) + H2O(l) HCN(aq) + OH-(aq)
Think About It It’s very important that you be able to identify the ions in solution correctly. If necessary, review the formulas and charges of the common polyatomic ions.
Acid-Base Properties of Salt Solutions
The pH of a solution that contains a salt in which both the cation and the anion hydrolyze depends on the relative strengths of the weak acid and base.
Qualitative predictions can be made using the Kb (of the salts anion) and the Ka (of the salts cation).
When Kb > Ka, the solution is basic
When Kb < Ka, the solution is acidic
When Kb ≈ Ka, the solution is neutral or nearly neutral
Acid-Base Properties of Oxides and Hydroxides
16.11
Acid-Base Properties of Oxides and Hydroxides
Basic metallic oxides react with water to form metal hydroxides:
Na2O(s) + H2O(l) → 2NaOH(aq)
BaO(s) + H2O(l) → Ba(OH)2(aq)
Acidic oxides reaction with water as follows:
CO2(g) + H2O(l) H2CO3(aq)
SO3(g) + H2O(l) H2SO4(aq)
Reactions between acidic oxides and bases and those between basic oxides and acids resemble normal acid-base reactions that produce a salt and water.
CO2(g) + 2NaOH(aq) → Na2CO3(aq) + H2O(l)
BaO(s) + 2HNO3(aq) → Ba(NO3)2 (aq) + H2O(l)
Acid-Base Properties of Oxides and Hydroxides
Aluminum oxide (Al2O3) is amphoteric.
It can act as an acid:
Al2O3(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2O(l)
Or it can act as a base:
Al2O3(s) + 2NaOH(aq) + 3H2O(l) → 2NaAl(OH)4(aq)
Acid-Base Properties of Oxides and Hydroxides
All the alkali and alkaline earth metal hydroxides, except Be(OH)2, are basic.
Be(OH)2
Al(OH)3
Sn(OH)2
Pb(OH)2
Cr(OH)3
Cu(OH)2
Zn(OH)2
Cd(OH)2
Acid:
Be(OH)2(s) + 6H+(aq) → 2Be2+(aq) + 2H2O(l)
Base:
Be(OH)2(s) + 2OH–(aq) → Be(OH)42– (aq)
amphoteric
Lewis Acids and Bases
A Lewis base is a substance that can donate a pair of electrons.
A Lewis acid is a substance that can accept a pair of electrons.
16.12
Ammonia,
a Lewis base
Boron trifluoride
a Lewis

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