Assess the relative strengths of acids and bases according to their ionization constants
Rationalize trends in acid–base strength in relation to molecular structure
Carry out equilibrium calculations for weak acid–base systems
We can rank the strengths of acids by the extent to which they ionize in aqueous solution. The reaction of an acid with water is given by the general expression:
HA(aq)+H2O(l)⇌H3O+(aq)+A−(aq)
Water is the base that reacts with the acid HA, A− is the conjugate base of the acid HA, and the hydronium ion is the conjugate acid of water. A strong acid yields 100% (or very nearly so) of
H3O+
and A− when the acid ionizes in water; Table 1 lists several strong acids. A weak acid gives small amounts of
H3O+
and A−.
Table 1. Some Common Strong acids and Strong Bases
Strong Acids
Strong Bases
HClO4 perchloric acid
LiOH lithium hydroxide
HCl hydrochloric acid
NaOH sodium hydroxide
HBr hydrobromic acid
KOH potassium hydroxide
HI hydroiodic acid
Ca(OH)2 calcium hydroxide
HNO3 nitric acid
Sr(OH)2 strontium hydroxide
H2SO4 sulfuric acid
Ba(OH)2 barium hydroxide
The relative strengths of acids may be determined by measuring their equlibrium constants in aqueous solutions. In solutions of the same concentration, stronger acids ionize to a greater extent, and so yield higher concentrations of hydronium ions than do weaker acids. The equilibrium constant for an acid is called the acid-ionization constant, Ka. For the reaction of an acid HA:
HA(aq)+H2O(l)⇌H3O+(aq)+A−(aq)
,
we write the equation for the ionization constant as:
Ka=[HA][H3O+][A−]
where the concentrations are those at equilibrium. Although water is a reactant in the reaction, it is the solvent as well, so we do not include [H2O] in the equation. The larger the Ka of an acid, the larger the concentration of
H3O+
and A− relative to the concentration of the nonionized acid, HA. Thus a stronger acid has a larger ionization constant than does a weaker acid. The ionization constants increase as the strengths of the acids increase. (A table of ionization constants of weak acids appears in Ionization Constants of Weak Acids, with a partial listing in Table 1.)
The following data on acid-ionization constants indicate the order of acid strength CH3CO2H < HNO2 <
Another measure of the strength of an acid is its percent ionization. The percent ionization of a weak acid is the ratio of the concentration of the ionized acid to the initial acid concentration, times 100:
% ionization=[HA]0[H3O+]eq×100
Because the ratio includes the initial concentration, the percent ionization for a solution of a given weak acid varies depending on the original concentration of the acid, and actually decreases with increasing acid concentration.
Example 1: Calculation of Percent Ionization from pH
Calculate the percent ionization of a 0.125-M solution of nitrous acid (a weak acid), with a pH of 2.09.
Show Answer
The percent ionization for an acid is:
[HNO2]0[H3O+]eq×100
The chemical equation for the dissociation of the nitrous acid is:
HNO2(aq)+H2O(l)⇌NO2−(aq)+H3O+(aq)
. Since 10−pH =
[H3O+]
, we find that 10−2.09 = 8.1 × 10−3M, so that percent ionization is:
0.1258.1×10−3×100=6.5%
Remember, the logarithm 2.09 indicates a hydronium ion concentration with only two significant figures.
Check Your Learning
Calculate the percent ionization of a 0.10-M solution of acetic acid with a pH of 2.89.
Show Answer
1.3% ionized
We can rank the strengths of bases by their tendency to form hydroxide ions in aqueous solution. The reaction of a Brønsted-Lowry base with water is given by:
B(aq)+H2O(l)⇌HB+(aq)+OH−(aq)
Water is the acid that reacts with the base, HB+ is the conjugate acid of the base B, and the hydroxide ion is the conjugate base of water. A strong base yields 100% (or very nearly so) of OH− and HB+ when it reacts with water; Table 1 lists several strong bases. A weak base yields a small proportion of hydroxide ions. Soluble ionic hydroxides such as NaOH are considered strong bases because they dissociate completely when dissolved in water.
As we did with acids, we can measure the relative strengths of bases by measuring their base-ionization constant, (Kb) in aqueous solutions. In solutions of the same concentration, stronger bases ionize to a greater extent, and so yield higher hydroxide ion concentrations than do weaker bases. A stronger base has a larger ionization constant than does a weaker base. For the reaction of a base, B:
B(aq)+H2O(l)⇌HB+(aq)+OH−(aq)
,
we write the equation for the ionization constant as:
Kb=[B][HB+][OH−]
where the concentrations are those at equilibrium. Again, we do not include [H2O] in the equation because water is the solvent. The chemical reactions and ionization constants of the three bases shown are:
A table of ionization constants of weak bases appears in Ionization Constants of Weak Bases (with a partial list in Table 2). As with acids, percent ionization can be measured for basic solutions, but will vary depending on the base ionization constant and the initial concentration of the solution.
Consider the ionization reactions for a conjugate acid-base pair, HA − A−:
As shown in the previous chapter on equilibrium, the K expression for a chemical equation derived from adding two or more other equations is the mathematical product of the other equations’ K expressions. Multiplying the mass-action expressions together and cancelling common terms, we see that:
For example, the acid ionization constant of acetic acid (CH3COOH) is 1.8 × 10−5, and the base ionization constant of its conjugate base, acetate ion (CH3COO−), is 5.6 × 10−10. The product of these two constants is indeed equal to Kw:
Ka×Kb=(1.8×10−5)×(5.6×10−10)=1.0×10−14=Kw
The extent to which an acid, HA, donates protons to water molecules depends on the strength of the conjugate base, A−, of the acid. If A− is a strong base, any protons that are donated to water molecules are recaptured by A−. Thus there is relatively little A− and
H3O+
in solution, and the acid, HA, is weak. If A− is a weak base, water binds the protons more strongly, and the solution contains primarily A− and
H3O+
—the acid is strong. Strong acids form very weak conjugate bases, and weak acids form stronger conjugate bases (Figure 1).
Figure 1. This diagram shows the relative strengths of conjugate acid-base pairs, as indicated by their ionization constants in aqueous solution.
Figure 2 lists a series of acids and bases in order of the decreasing strengths of the acids and the corresponding increasing strengths of the bases. The acid and base in a given row are conjugate to each other.
Figure 2. The chart shows the relative strengths of conjugate acid-base pairs.
The first six acids in Figure 2 are the most common strong acids. These acids are completely dissociated in aqueous solution. The conjugate bases of these acids are weaker bases than water. When one of these acids dissolves in water, their protons are completely transferred to water, the stronger base.
Those acids that lie between the hydronium ion and water in Figure 2 form conjugate bases that can compete with water for possession of a proton. Both hydronium ions and nonionized acid molecules are present in equilibrium in a solution of one of these acids. Compounds that are weaker acids than water (those found below water in the column of acids) in Figure 3 exhibit no observable acidic behavior when dissolved in water. Their conjugate bases are stronger than the hydroxide ion, and if any conjugate base were formed, it would react with water to re-form the acid.
The extent to which a base forms hydroxide ion in aqueous solution depends on the strength of the base relative to that of the hydroxide ion, as shown in the last column in Figure 2. A strong base, such as one of those lying below hydroxide ion, accepts protons from water to yield 100% of the conjugate acid and hydroxide ion. Those bases lying between water and hydroxide ion accept protons from water, but a mixture of the hydroxide ion and the base results. Bases that are weaker than water (those that lie above water in the column of bases) show no observable basic behavior in aqueous solution.
Example 2: The Product Ka ×Kb = Kw
Use the Kb for the nitrite ion,
NO2−
, to calculate the Ka for its conjugate acid.
Show Answer
Kb for
NO2−
is given in this section as 2.22 × 10−11. The conjugate acid of
NO2−
is HNO2; Ka for HNO2 can be calculated using the relationship:
and HCN by comparing their ionization constants. The ionization constant of HCN is given in Ionization Constants of Weak Acids as 4 × 10−10. The ionization constant of
NH4+
is not listed, but the ionization constant of its conjugate base, NH3, is listed as 1.8 × 10−5. Determine the ionization constant of
NH4+
, and decide which is the stronger acid, HCN or
NH4+
.
Show Answer
NH4+
is the slightly stronger acid (Ka for
NH4+
= 5.6 × 10−10).
Exercises
Explain why the neutralization reaction of a strong acid and a weak base gives a weakly acidic solution.
Explain why the neutralization reaction of a weak acid and a strong base gives a weakly basic solution.
Use this list of important industrial compounds (and Figure 2) to answer the following questions regarding: CaO, Ca(OH)2, CH3CO2H, CO2, HCl, H2CO3, HF, HNO2, HNO3, H3PO4, H2SO4, NH3, NaOH, Na2CO3.
Identify the strong Brønsted-Lowry acids and strong Brønsted-Lowry bases.
List those compounds in (a) that can behave as Brønsted-Lowry acids with strengths lying between those of
H3O+
and H2O.
List those compounds in (a) that can behave as Brønsted-Lowry bases with strengths lying between those of H2O and OH−.
Explain why the ionization constant, Ka, for H2SO4 is larger than the ionization constant for H2SO3.
Explain why the ionization constant, Ka, for HI is larger than the ionization constant for HF.
What is the ionization constant at 25 °C for the weak acid
CH3NH3+
, the conjugate acid of the weak base CH3NH2, Kb = 4.4 × 10−4.
What is the ionization constant at 25 °C for the weak acid
(CH3)2NH2+
, the conjugate acid of the weak base (CH3)2NH, Kb = 7.4 × 10−4?
Show Selected Answers
2. The salt ionizes in solution, but the anion slightly reacts with water to form the weak acid. This reaction also forms OH−, which causes the solution to be basic. An example is NaCN. The CN− reacts with water as follows:
CN−(aq)+H2O(l)⇌HCN(aq)+OH−(aq)
4. The oxidation state of the sulfur in H2SO4 is greater than the oxidation state of the sulfur in H2SO3.
6. Kw = Ka × Kb; thus,
KaKa==KbKw4.4×10−41.0×10−14=2.3×10−11
The Ionization of Weak Acids and Weak Bases
Many acids and bases are weak; that is, they do not ionize fully in aqueous solution. A solution of a weak acid in water is a mixture of the nonionized acid, hydronium ion, and the conjugate base of the acid, with the nonionized acid present in the greatest concentration. Thus, a weak acid increases the hydronium ion concentration in an aqueous solution (but not as much as the same amount of a strong acid).
Acetic acid, CH3CO2H, is a weak acid. When we add acetic acid to water, it ionizes to a small extent according to the equation:
CH3CO2H(aq)+H2O(l)⇌H3O+(aq)+CH3CO2−(aq)
,
giving an equilibrium mixture with most of the acid present in the nonionized (molecular) form. This equilibrium, like other equilibria, is dynamic; acetic acid molecules donate hydrogen ions to water molecules and form hydronium ions and acetate ions at the same rate that hydronium ions donate hydrogen ions to acetate ions to reform acetic acid molecules and water molecules. We can tell by measuring the pH of an aqueous solution of known concentration that only a fraction of the weak acid is ionized at any moment (Figure 3). The remaining weak acid is present in the nonionized form.
For acetic acid, at equilibrium:
Ka=[CH3CO2H][H3O+][CH3CO2−]=1.8×10−5
Figure 3. pH paper indicates that a 0.l-M solution of HCl (beaker on left) has a pH of 1. The acid is fully ionized and [H3O+] = 0.1 M. A 0.1-M solution of CH3CO2H (beaker on right) is has a pH of 3 ([H3O+] = 0.001 M) because the weak acid CH3CO2H is only partially ionized. In this solution, [H3O+] < [CH3CO2H]. (credit: modification of work by Sahar Atwa)
Table 2. Ionization Constants of Some Weak Acids
Ionization Reaction
Ka at 25 °C
HSO4−+H2O⇌H3O++SO42−
1.2 × 10−2
HF+H2O⇌H3O++F−
7.2 × 10−4
HNO2+H2O⇌H3O++NO2−
4.5 × 10−4
HNCO+H2O⇌H3O++NCO−
3.46 × 10−4
HCO2H+H2O⇌H3O++HCO2−
1.8 × 10−4
CH3CO2H+H2O⇌H3O++CH3CO2−
1.8 × 10−5
HCIO+H2O⇌H3O++CIO−
3.5 × 10−8
HBrO+H2O⇌H3O++BrO−
2 × 10−9
HCN+H2O⇌H3O++CN−
4 × 10−10
Table 2 gives the ionization constants for several weak acids; additional ionization constants can be found in Ionization Constants of Weak Acids.
At equilibrium, a solution of a weak base in water is a mixture of the nonionized base, the conjugate acid of the weak base, and hydroxide ion with the nonionized base present in the greatest concentration. Thus, a weak base increases the hydroxide ion concentration in an aqueous solution (but not as much as the same amount of a strong base).
For example, a solution of the weak base trimethylamine, (CH3)3N, in water reacts according to the equation
(CH3)3N(aq)+H2O(l)⇌(CH3)3NH+(aq)+OH−(aq)
,
giving an equilibrium mixture with most of the base present as the nonionized amine. This equilibrium is analogous to that described for weak acids.
We can confirm by measuring the pH of an aqueous solution of a weak base of known concentration that only a fraction of the base reacts with water (Figure 4). The remaining weak base is present as the unreacted form. The equilibrium constant for the ionization of a weak base, Kb, is called the ionization constant of the weak base, and is equal to the reaction quotient when the reaction is at equilibrium. For trimethylamine, at equilibrium:
Kb=[(CH3)3N][(CH3)3NH+][OH−]
Figure 4. pH paper indicates that a 0.1-M solution of NH3 (left) is weakly basic. The solution has a pOH of 3 ([OH−] = 0.001 M) because the weak base NH3 only partially reacts with water. A 0.1-M solution of NaOH (right) has a pOH of 1 because NaOH is a strong base. (credit: modification of work by Sahar Atwa)
Both HF and HCN ionize in water to a limited extent. Which of the conjugate bases, F− or CN−, is the stronger base? See Table 3.
[practice-area rows="4"][/practice-area]
Exercises
Calculate the equilibrium concentration of the nonionized acids and all ions in a solution that is 0.25 M in HCO2H and 0.10 M in HClO.
Calculate the equilibrium concentration of the nonionized acids and all ions in a solution that is 0.134 M in HNO2 and 0.120 M in HBrO.
Calculate the equilibrium concentration of the nonionized bases and all ions in a solution that is 0.25 M in CH3NH2 and 0.10 M in C5H5N (Kb = 1.7 × 10−9).
Calculate the equilibrium concentration of the nonionized bases and all ions in a solution that is 0.115 M in NH3 and 0.100 M in C6H5NH2.
Solving for x gives 7.77 × 10−3M. Because this value is 5.8% of 0.134, our assumption is incorrect. Therefore, we must use the quadratic formula. Using the data gives the following equation: x2 + 4.5 × 10−7x − 6.03 × 10−5 = 0
Using the quadratic formula gives (a = 1, b = 4.5 × 10−4, and c = −6.03 × 10−5)
Solving for x gives 1.44 × 10−3M. Because this value is less than 5% of 0.115 M, our assumption is correct. The equilibrium concentrations are therefore:
Example 3: Determination of Ka from Equilibrium Concentrations
Acetic acid is the principal ingredient in vinegar (Figure 5); that's why it tastes sour. At equilibrium, a solution contains [CH3CO2H] = 0.0787 M and
[H3O+]=[CH3CO2−]=0.00118M
. What is the value of Ka for acetic acid?
Figure 5. Vinegar is a solution of acetic acid, a weak acid. (credit: modification of work by “HomeSpot HQ”/Flickr)
Show Answer
We are asked to calculate an equilibrium constant from equilibrium concentrations. At equilibrium, the value of the equilibrium constant is equal to the reaction quotient for the reaction:
What is the equilibrium constant for the ionization of the
HPO42−
ion, a weak base:
HPO42−(aq)+H2O(l)⇌H2PO4−(aq)+OH−(aq)
In a solution containing a mixture of NaH2PO4 and Na2HPO4 at equilibrium, [OH−] = 1.3 × 10−6M;
[H2PO4−]=0.042M
; and
[HPO42−]=0.341M
.
Show Answer
Kb for
HPO42−=1.6×10−7
Example 5: Determination of Ka or Kb from pH
The pH of a 0.0516-M solution of nitrous acid, HNO2, is 2.34. What is its Ka?
HNO2(aq)+H2O(l)⇌H3O+(aq)+NO2−(aq)
Show Answer
We determine an equilibrium constant starting with the initial concentrations of HNO2,
H3O+
, and
NO2−
as well as one of the final concentrations, the concentration of hydronium ion at equilibrium. (Remember that pH is simply another way to express the concentration of hydronium ion.)
We can solve this problem with the following steps in which x is a change in concentration of a species in the reaction:
We can summarize the various concentrations and changes as shown here (the concentration of water does not appear in the expression for the equilibrium constant, so we do not need to consider its concentration):
To get the various values in the ICE (Initial, Change, Equilibrium) table, we first calculate
[H3O+]
, the equilibrium concentration of
H3O+
, from the pH:
[H3O+]=10−2.34=0.0046M
The change in concentration of
H3O+
,
x[H3O+]
, is the difference between the equilibrium concentration of
H3O+
, which we determined from the pH, and the initial concentration,
[H3O+]i
. The initial concentration of
H3O+
is its concentration in pure water, which is so much less than the final concentration that we approximate it as zero (~0).
The change in concentration of
NO2−
is equal to the change in concentration of
[H3O+]
. For each 1 mol of
H3O+
that forms, 1 mol of
NO2−
forms. The equilibrium concentration of HNO2 is equal to its initial concentration plus the change in its concentration.
Now we can fill in the ICE table with the concentrations at equilibrium, as shown here:
Finally, we calculate the value of the equilibrium constant using the data in the table:
Step 1: Determine x and equilibrium concentrations
The equilibrium expression is
HCO2H(aq)+H2O(l)⇌H3O+(aq)+HCO2−(aq)
The concentration of water does not appear in the expression for the equilibrium constant, so we do not need to consider its change in concentration when setting up the ICE table.The table shows initial concentrations (concentrations before the acid ionizes), changes in concentration, and equilibrium concentrations follows (the data given in the problem appear in color):
Step 2: Solve for x and the equilibrium concentrations
Now solve for x. Because the initial concentration of acid is reasonably large and Ka is very small, we assume that x << 0.534, which permits us to simplify the denominator term as (0.534 − x) = 0.534. This gives:
To check the assumption that x is small compared to 0.534, we calculate:
0.534x=0.5349.8×10−3=1.8×10−2(1.8% of 0.534)
x is less than 5% of the initial concentration; the assumption is valid.
We find the equilibrium concentration of hydronium ion in this formic acid solution from its initial concentration and the change in that concentration as indicated in the last line of the table:
[H3O+]==0+x=0+9.8×10−3M9.8×10−3M
The pH of the solution can be found by taking the negative log of the
[H3O+]
, so:
−log(9.8×10−3)=2.01
Check Your Learning
Only a small fraction of a weak acid ionizes in aqueous solution. What is the percent ionization of acetic acid in a 0.100-M solution of acetic acid, CH3CO2H?
at equilibrium.) Recall that the percent ionization is the fraction of acetic acid that is ionized × 100, or
[CH3CO2H]initial[CH3CO2−]×100
.
Show Answer
percent ionization = 1.3%
Example 7 shows that the concentration of products produced by the ionization of a weak base can be determined by the same series of steps used with a weak acid.
Example 7: Equilibrium Concentrations in a Solution of a Weak Base
Find the concentration of hydroxide ion in a 0.25-M solution of trimethylamine, a weak base:
This problem requires that we calculate an equilibrium concentration by determining concentration changes as the ionization of a base goes to equilibrium. The solution is approached in the same way as that for the ionization of formic acid in Example 6. The reactants and products will be different and the numbers will be different, but the logic will be the same:
Step 1: Determine x and equilibrium concentrations
The table shows the changes and concentrations:
Step 2: Solve for x and the equilibrium concentrations
If we assume that x is small relative to 0.25, then we can replace (0.25 − x) in the preceding equation with 0.25. Solving the simplified equation gives
x=4.3×10−3
.
This change is less than 5% of the initial concentration (0.25), so the assumption is justified.Recall that, for this computation, x is equal to the equilibrium concentration of hydroxide ion in the solution (see earlier tabulation):
[OH−]=0+x=x=4.3×10−3M
=4.3×10−3M
Then calculate pOH as follows:
pOH=−log(4.3×10−3)=2.37
.
Using the relation introduced in the previous section of this chapter, we get
pH+pOH=pKw=14.00
, which permits the computation of pH:
pH=14.00−pOH=14.00−2.37=11.63
.
Step 3: Check the work
A check of our arithmetic shows that Kb = 7.4 × 10−5.
Check Your Learning
Show that the calculation in Step 2 of this example gives an x of 4.3 × 10−3 and the calculation in Step 3 shows Kb = 7.4 × 10−5.
Find the concentration of hydroxide ion in a 0.0325-M solution of ammonia, a weak base with a Kb of 1.76 × 10−5. Calculate the percent ionization of ammonia, the fraction ionized × 100, or
[NH3][NH4+]×100
Show Answer
7.56 × 10−4M
2.33%
Some weak acids and weak bases ionize to such an extent that the simplifying assumption that x is small relative to the initial concentration of the acid or base is inappropriate. As we solve for the equilibrium concentrations in such cases, we will see that we cannot neglect the change in the initial concentration of the acid or base, and we must solve the equilibrium equations by using the quadratic equation.
Example 8: Equilibrium Concentrations in a Solution of a Weak Acid
Sodium bisulfate, NaHSO4, is used in some household cleansers because it contains the
HSO4−
ion, a weak acid. What is the pH of a 0.50-M solution of
We need to determine the equilibrium concentration of the hydronium ion that results from the ionization of
HSO4−
so that we can use
[H3O+]
to determine the pH. As in the previous examples, we can approach the solution by the following steps:
Step 1: Determine x and equilibrium concentrations
This table shows the changes and concentrations:
Step 2: Solve for x and the concentrations
As we begin solving for x, we will find this is more complicated than in previous examples. As we discuss these complications we should not lose track of the fact that it is still the purpose of this step to determine the value of x. At equilibrium:
If we assume that x is small and approximate (0.50 − x) as 0.50, we find
x=7.7×10−2
. When we check the assumption, we calculate:
[HSO4−]ix
0.50x=0.507.7×10−2=0.15(15%)
The value of x is not less than 5% of 0.50, so the assumption is not valid. We need the quadratic formula to find x. The equation
Ka=1.2×10−2=0.50−x(x)(x)
gives
6.0×10−3−1.2×10−2x=x2+
or
x2++1.2×10−2x−6.0×10−3=0
.
This equation can be solved using the quadratic formula. For an equation of the form
ax2++bx+c=0
, x is given by the equation
x=2a−b±b2+−4ac
.
In this problem, a = 1, b = 1.2 × 10−3, and c = −6.0 × 10−3.Solving for x gives a negative root (which cannot be correct since concentration cannot be negative) and a positive root:
x=7.2×10−2
.
Now determine the hydronium ion concentration and the pH:
[H3O+]=0+x=0+7.2×10−2M=7.2×10−2M
. The pH of this solution is
pH=−log[H3O+]=−log7.2×10−2=1.14
.
Check Your Learning
Show that the quadratic formula gives x = 7.2 × 10−2.
Calculate the pH in a 0.010-M solution of caffeine, a weak base:
Calculate the concentration of all solute species in each of the following solutions of acids or bases. Assume that the ionization of water can be neglected, and show that the change in the initial concentrations can be neglected. Ionization constants can be found in Ionization Constants of Weak Acids and Ionization Constants of Weak Bases.
Solution 1
0.0092 M HClO, a weak acid
Show Answer
The reaction is:
HClO(aq)+H2O(l)⇌H3O+(aq)+ClO−(aq)
The equilibrium expression is
Ka=[HClO][H3O+][ClO−]=3.5×10−8
The initial and equilibrium concentrations for this system can be written as follows:
[HClO]
[H3O+]
[ClO−]
Initial concentration (M)
0.0092
0
0
Change (M)
−x
x
x
Equilibrium (M)
0.0092 − x
x
x
Substituting the equilibrium concentrations into the equilibrium expression and making the assumption that (0.0092 − x) ≈ 0.0092 gives:
Solving for x gives 1.79 × 10−5M. This value is less than 5% of 0.0092, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:
Solving for x gives 6.01 × 10−6M. This value is less than 5% of 0.0784, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:
Solving for x gives 5.69 × 10−6M. This value is less than 5% of 0.0810, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:
[H3O+]
= [CN−] = 5.7 × 10−6M
[HCN] = 0.0810 − 5.69 × 10−6 = 0.08099 = 0.0810 M
[OH−]=[H3O+]Kw=5.69×10−61.0×10−14=1.8×10−9M
Solution 4
0.11 M (CH3)3N, a weak base
Show Answer
The reaction is
(CH3)3N(aq)+H2O(l)⇌(CH3)3NH+(aq)+OH−(aq)
The equilibrium expression is
Kb=[(CH3)3N][(CH3)3NH+][OH−]=7.4×10−5
The initial and equilibrium concentrations for this system can be written as follows:
[(CH3)3N]
[(CH3)3NH+]
[OH−]
Initial concentration (M)
0.11
0
0
Change (M)
−x
x
x
Equilibrium (M)
0.11 − x
x
x
Substituting the equilibrium concentrations into the equilibrium expression and making the assumption that (0.11 − x) ≈ 0.11 gives:
Solving for x gives 2.85 × 10−3M. This value is less than 5% of 0.11, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are: