### Learning Objective

To usage Ka and Kb values to calculation the percent ionization and the pH the a equipment of an mountain or a base.

This ar presents a quantitative technique to analyzing acid–base equilibriums. You will learn just how to recognize the worths of Ka and Kb, exactly how to use Ka or Kb to calculation the percent ionization and the pH of an aqueous equipment of an mountain or a base, and how to calculate the equilibrium consistent for the reaction of an acid with a basic from the Ka and Kb of the reactants.

You are watching: Which of the following will decrease the percentage ionization of 1.0 m acetic acid, ch3co2h(aq)?

## Determining Ka and Kb

The ionization constants Ka and Kb space equilibrium constants that space calculated indigenous experimentally measure concentrations, similar to the equilibrium constants discussed in thing 15 "Chemical Equilibrium". Prior to proceeding further, the is important to understand specifically what is meant once we describe the concentration of one aqueous equipment of a weak mountain or a weak base. Suppose, because that example, we have actually a party labeled 1.0 M acetic mountain or 1.0 M ammonia. Together you learned in thing 4 "Reactions in Aqueous Solution", such a systems is usually all set by dissolving 1.0 mol that acetic mountain or ammonia in water and adding enough water to provide a final volume of precisely 1.0 L. If, however, we were to list the actual concentration of all the types present in either solution, us would discover that none of the worths is exactly 1.0 M since a weak acid such together acetic acid or a weak basic such together ammonia always reacts with water to part extent. The extent of the reaction counts on the Ka or the Kb, the concentration that the acid or the base, and the temperature. Consequently, only the total concentration the both the ionized and also unionized varieties is equal to 1.0 M.

The analytical concentration (C) is defined as the total concentration of all forms of an mountain or a base that are present in solution, nevertheless of their state of protonation. Hence a “1.0 M” solution of acetic acid has actually an analytical concentration that 1.0 M, i beg your pardon is the amount of the actual concentrations of unionized acetic mountain (CH3CO2H) and the ionized type (CH3CO2−):

Equation 16.38

C CH 3 CO 2 H = < CH 3 CO 2 H > + < CH 3 CO 2 − >

As us shall view shortly, if we understand the analysis concentration and also the Ka, we can calculate the actual worths of and .

The equilibrium equations because that the reaction that acetic acid and ammonia through water room as follows:

Equation 16.39

K a = < H + > < CH 3 CO 2 − > < CH 3 CO 2 H >

Equation 16.40

K b = < NH 4 + > < oh − > < NH 3 >

where Ka and Kb space the ionization constants for acetic acid and also ammonia, respectively. In addition to the analysis concentration of the acid (or the base), we must have a means to measure the concentration the at least one the the varieties in the equilibrium continuous expression to identify the Ka (or the Kb). There space two typical ways to obtain the concentrations: (1) measure the electrical conductivity of the solution, which is pertained to the complete concentration of ion present, and (2) measure the pH the the solution, which provides or .

Example 6 and also Example 7 show the procedure because that determining Ka because that a weak acid and Kb because that a weak base. In both cases, we will certainly follow the procedure developed in thing 15 "Chemical Equilibrium": the analysis concentration of the mountain or the basic is the initial concentration, and the stoichiometry of the reaction through water determines the change in concentrations. The final concentration of all varieties are calculated from the early concentrations and also the changes in the concentrations. Inserting the final concentrations into the equilibrium consistent expression allows us to calculate the Ka or the Kb.

### Example 6

Electrical conductivity measurements suggest that 0.42% of the acetic mountain molecules in a 1.00 M solution are ionized in ~ 25°C. Calculate Ka and pKa for acetic mountain at this temperature.

Given: analytical concentration and percent ionization

Asked for: Ka and also pKa

Strategy:

A compose the balanced equilibrium equation for the reaction and derive the equilibrium consistent expression.

B usage the data given and the stoichiometry of the reaction to construct a table mirroring the initial concentrations, the transforms in concentrations, and the final concentrations for all varieties in the equilibrium consistent expression.

C instead of the last concentrations right into the equilibrium continuous expression and also calculate the Ka. Take it the an adverse logarithm that Ka to acquire the pKa.

Solution:

A The well balanced equilibrium equation because that the dissociation of acetic acid is together follows:

CH 3 CO 2 H(aq) ⇌ H + (aq) + CH 3 CO 2 − (aq)

and the equilibrium consistent expression is as follows:

K a = < H + > < CH 3 CO 2 − > < CH 3 CO 2 H >

B To calculate the Ka, we require to understand the equilibrium concentration of CH3CO2H, CH3CO2−, and also H+. The most direct means to carry out this is to construct a table the lists the early concentrations and the alters in concentrations the occur during the reaction to provide the last concentrations, making use of the procedure presented in thing 15 "Chemical Equilibrium". The early concentration of unionized acetic mountain (i) is the analytical concentration, 1.00 M, and also the early acetate concentration (i) is zero. The early stage concentration of H+ is no zero, however; i is 1.00 × 10−7 M due to the autoionization the water. The measure up percent ionization tells united state that 0.42% of the acetic acid molecules space ionized in ~ equilibrium. Consequently, the adjust in the concentration that acetic acid is Δ = −(4.2 × 10−3)(1.00 M) = −0.0042 M. Vice versa, the change in the acetate concentration is Δ = +0.0042 M because every 1 mol of acetic acid that ionizes gives 1 mol of acetate. Due to the fact that one proton is created for every acetate ion formed, Δ = +0.0042 M together well. These results are summary in the following table.

CH 3 CO 2 H (aq) ⇌ H + (aq) + CH 3 CO 2 − (aq)
initial 1.00 1.00 × 10−7 0
change −0.0042 +0.0042 +0.0042
final

The final concentrations that all types are as such as follows:

< CH 3 CO 2 H > f = < CH 3 CO 2 H > ns + Δ < CH 3 CO 2 H > = 1.00  M + ( − 0.0042  M ) = 1.00  M < CH 3 CO 2 − > f = < CH 3 CO 2 − > ns + Δ < CH 3 CO 2 − > = 0  M + ( + 0.0042  M ) = 0.0042  M < H + > f = < H + > i + Δ < H + > = 1.00 × 10 − 7  M + ( + 0.0042  M ) = 0.0042  M

C We can now calculate Ka by inserting the last concentrations into the equilibrium continuous expression:

K a = < H + > < CH 3 CO 2 − > < CH 3 CO 2 H > = ( 0.0042 ) ( 0.0042 ) 1.00 = 1.8 × 10 − 5

The pKa is the an unfavorable logarithm of Ka: pKa = −log Ka = −log(1.8 × 10−5) = 4.74.

Exercise

Picric acid is the common name because that 2,4,6-trinitrophenol, a derivative that phenol (C6H5OH) in which three H atoms are replaced by nitro (–NO2) groups. The visibility of the nitro groups removes electron density from the phenyl ring, making picric mountain a much stronger acid than phenol (pKa = 9.99). The nitro groups additionally make picric acid potentially explosive, together you might expect based on its chemistry similarity to 2,4,6-trinitrotoluene, far better known as TNT. A 0.20 M solution of picric mountain is 73% ionized in ~ 25°C. Calculate Ka and pKa for picric acid.

### Example 7

A 1.0 M aqueous equipment of ammonia has a pH of 11.63 in ~ 25°C. Calculate Kb and pKb for ammonia.

Given: analytical concentration and also pH

Asked for: Kb and also pKb

Strategy:

A compose the well balanced equilibrium equation because that the reaction and also derive the equilibrium continuous expression.

B usage the data given and the stoichiometry that the reaction to construct a table mirroring the initial concentrations, the changes in concentrations, and the last concentrations for all species in the equilibrium constant expression.

C instead of the final concentrations into the equilibrium constant expression and calculate the Kb. Take it the an unfavorable logarithm that Kb to achieve the pKb.

Solution:

A The balanced equilibrium equation for the reaction the ammonia with water is together follows:

NH 3 (aq) + H 2 O(l) ⇌ NH 4 + (aq) + five − (aq)

and the equilibrium continuous expression is together follows:

K b = < NH 4 + > < oh − > < NH 3 >

Remember the water does not appear in the equilibrium constant expression because that Kb.

B To calculation Kb, we require to understand the equilibrium concentration of NH3, NH4+, and also OH−. The early concentration of NH3 is the analytical concentration, 1.0 M, and also the initial concentrations of NH4+ and OH− are 0 M and also 1.00 × 10−7 M, respectively. In this case, we are given the pH that the solution, which allows us to calculation the final concentration of one varieties (OH−) directly, fairly than the change in concentration. Recall the pKw = pH + pOH = 14.00 in ~ 25°C. Hence pOH = 14.00 − pH = 14.00 − 11.63 = 2.37, and also f = 10−2.37 = 4.3 × 10−3 M. Ours data thus much are noted in the following table.

NH 3 (aq) ⇌ NH 4 + (aq) + oh − (aq)
initial 1.0 0 1.00 × 10−7
change
final 4.3 × 10−3

The last is much higher than the early stage , therefore the readjust in is as follows:

Δ = (4.3 × 10−3 M) − (1.00 × 10−7 M) ≈ 4.3 × 10−3 M

The stoichiometry that the reaction tells us that 1 mol of NH3 is converted to NH4+ for each 1 mol that OH− formed, so

Δ = +4.3 × 10−3 M and Δ = −4.3 ×10−3 M

We can now insert these worths for the transforms in concentrations into the table, which permits us to finish the table.

H 2 O(l) + NH 3 (aq) ⇌ NH 4 + (aq) + oh − (aq)
initial 1.0 0 1.00 × 10−7
change −4.3 × 10−3 +4.3 × 10−3 +4.3 × 10−3
final 1.0 4.3 × 10−3 4.3 × 10−3

C Inserting the final concentrations right into the equilibrium continuous expression offers Kb:

K b = < NH 4 + > < five − > < NH 3 > = ( 4.3 × 10 − 3 ) 2 1.0 = 1.8 × 10 − 5

and pKb = −log Kb = 4.74.

The Kb and the pKb for ammonia are almost exactly the same as the Ka and also the pKa because that acetic mountain at 25°C. In other words, ammonia is practically exactly as solid a base together acetic mountain is an acid. Consequently, the extent of the ionization reaction in one aqueous systems of ammonia at a provided concentration is the same as in one aqueous solution of acetic acid at the very same concentration.

Exercise

The pH of a 0.050 M systems of pyridine (C6H5N) is 8.96 at 25°C. Calculation Kb and also pKb for pyridine.

One means to identify the concentration of species in solutions of weak acids and bases is a variation of the tabular method we provided previously to recognize Ka and also Kb values. As a demonstration, we will calculate the concentrations of all varieties and the percent ionization in a 0.150 M equipment of formic mountain at 25°C. The data in Table 16.2 "Values that " display that formic acid (Ka = 1.8 × 10−4 in ~ 25°C) is a slightly more powerful acid 보다 acetic acid. The equilibrium equation for the ionization the formic acid in water is together follows:

We collection the early concentration that HCO2H equal to 0.150 M, and also that that HCO2− is 0 M. The initial concentration of H+ is 1.00 × 10−7 M because of the autoionization the water. Due to the fact that the equilibrium continuous for the ionization reaction is small, the equilibrium will lie come the left, favoring the unionized form of the acid. Thus we can define x as the quantity of formic mountain that dissociates.

If the change in is −x, climate the readjust in and is +x. The last concentration that each species is the amount of its initial concentration and the readjust in concentration, as summarized in the complying with table.

HCO 2 H(aq) ⇌ H + (aq) + HCO 2 − (aq)
initial 0.150 1.00 × 10−7 0
change x +x +x
final (0.150 − x) (1.00 × 10−7 + x) x

We deserve to calculate x by substituting the final concentrations indigenous the table right into the equilibrium consistent expression:

K a = < H + > < HCO 2 − > < HCO 2 H > = ( 1.00 × 10 − 7 + x ) x 0.150 − x

Because the ionization consistent Ka is small, x is likely to be tiny compared with the early concentration of formic acid: (0.150 − x) M ≈ 0.150 M. Moreover, as result of the autoionization the water (1.00 × 10−7 M) is likely to be negligible compared with as result of the dissociation of formic acid: (1.00 × 10−7 + x) M ≈ x M. Inserting these values into the equilibrium continuous expression and also solving for x,

K a = x 2 0.150 = 1.8 × 10 − 4 x = 5.2 × 10 − 3

We can now calculation the concentrations of the varieties present in a 0.150 M formic acid systems by inserting this value of x into the expressions in the last heat of the table:

< HCO 2 H > = ( 0.150 − x )  M = 0.145  M < HCO 2 > = x = 5.2 × 10 − 3  M < H + > = ( 1.00 × 10 − 7 + x )  M = 5.2 × 10 − 3  M

Thus the pH of the solution is –log(5.2 × 10−3) = 2.28. We can likewise use these concentrations to calculation the fraction of the original acid that is ionized. In this case, the percent ionization is the proportion of (or ) come the analysis concentration, multiplied by 100 to give a percentage:

percent ionization = < H + > C HA × 100 = 5.2 × 10 − 3  M 0.150 × 100 = 3.5 %

Always check to make sure that any simplifying assumption was valid. together a general ascendancy of thumb, approximations such together those used right here are valid only if the quantity being neglected is no an ext than about 5% that the quantity to which it is being included or indigenous which the is gift subtracted. If the amount that to be neglected is much greater than around 5%, then the approximation is probably not valid, and also you have to go back and deal with the difficulty using the quadratic formula. In the ahead demonstration, both simplifying assumptions were justified: the percent ionization is only 3.5%, which is well below the about 5% limit, and the 1.00 × 10−7 M because of the autoionization of water is much, much less than the 5.2 × 10−3 M because of the ionization that formic acid.

As a general rule, the contribution because of the autoionization of water deserve to be ignored as long as the product of the acid or the basic ionization constant and the analysis concentration of the acid or the base is at least 10 times greater than the or from the autoionization the water—that is, if

By substituting the appropriate values for the formic acid solution into Equation 16.45, we view that the simplifying presumption is precious in this case:

Doing this basic calculation before solving this kind of difficulty saves time and permits you to write simplified expressions for the final concentrations of the varieties present. In practice, that is vital to include the contribution due to the autoionization the water just for exceptionally dilute solutions of an extremely weak mountain or bases.

Example 8 illustrates exactly how the procedure outlined previously deserve to be supplied to calculate the pH that a equipment of a weak base.

### Example 8

Calculate the pH and percent ionization that a 0.225 M systems of ethylamine (CH3CH2NH2), i beg your pardon is supplied in the synthetic of some dyes and also medicines. The pKb the ethylamine is 3.19 at 20°C.

Given: concentration and also pKb

Asked for: pH and also percent ionization

Strategy:

A create the well balanced equilibrium equation because that the reaction and also the equilibrium continuous expression. Calculation Kb native pKb.

B usage Equation 16.45 to watch whether you deserve to ignore as result of the autoionization the water. Then usage a tabular style to create expressions for the last concentrations the all types in solution. Substitute these values right into the equilibrium equation and also solve for . Usage Equation 16.42 to calculation the percent ionization.

C use the partnership Kw = to achieve . Then calculation the pH that the solution.

Solution:

A We start by creating the balanced equilibrium equation for the reaction:

CH 3 CH 2 NH 2 (aq) + H 2 O(l) ⇌ CH 3 CH 2 NH 3 + (aq) + oh − (aq)

The corresponding equilibrium constant expression is together follows:

K b = < CH 3 CH 2 NH 3 + > < oh − > < CH 3 CH 2 NH 2 >

From the pKb, we have Kb = 10−3.19 = 6.5 × 10−4.

B To calculate the pH, we require to recognize the H+ concentration. Unfortunately, H+ walk not show up in one of two people the chemistry equation or the equilibrium consistent expression. However, and in one aqueous equipment are connected by Kw = . Hence if we have the right to determine , we deserve to calculate and also then the pH. The early stage concentration that CH3CH2NH2 is 0.225 M, and the early is 1.00 × 10−7 M. Since ethylamine is a weak base, the extent of the reaction will be small, and it makes sense to let x same the amount of CH3CH2NH2 that reacts v water. The adjust in is as such −x, and also the adjust in both and is +x. To check out whether the autoionization the water can safely be ignored, us substitute Kb and CB right into Equation 16.46:

KbCB = (6.5 × 10−4)(0.225) = 1.5 × 10−4 > 1.0 × 10−6

Thus the simplifying assumption is valid, and we will certainly not encompass due to the autoionization that water in our calculations.

H 2 O(1) + CH 3 CH 2 NH 2 (aq) ⇌ CH 3 CH 2 NH 3 + (aq) + oh − (aq)
initial 0.225 0 1.00 × 10−7
change x +x +x
final (0.225 − x) x x

Substituting the quantities from the last line of the table into the equilibrium continuous expression,

K b = < CH 3 CH 2 NH 3 + > < oh − > < CH 3 CH 2 NH 2 > = ( x ) ( x ) 0.225 − x = 6.5 × 10 − 4

As before, we assume the amount of CH3CH2NH2 the ionizes is little compared v the initial concentration, for this reason f = 0.225 − x ≈ 0.225. V this assumption, we have the right to simplify the equilibrium equation and solve for x:

K b = x 2 0.225 = 6.5 × 10 − 4 x = 0.012 = < CH 3 CH 2 NH 3 + > f = < five − > f

The percent ionization is therefore

percent ionization = < five − > C B × 100 = 0.012  M 0.225  M × 100 = 5.4 %

which is in ~ the upper limit of the about 5% variety that deserve to be ignored. The last hydroxide concentration is for this reason 0.012 M.

C We have the right to now identify the making use of the expression because that Kw:

K w = < oh − > < H + > 1.01 × 10 − 14 = ( 0.012  M ) < H + > 8.4 × 10 − 13  M = < H + >

The pH that the solution is −log(8.4 × 10−13) = 12.08. Alternatively, we might have calculation pOH as −log(0.012) = 1.92 and also determined the pH together follows:

pH  +  pOH = p K w = 14.00 pH = 14.00 − 1.92 = 12.08

The two approaches are equivalent.

Exercise

Aromatic amines, in i beg your pardon the nitrogen atom is bonded straight to a phenyl ring (−C6H5) tend to be lot weaker bases than basic alkylamines. Because that example, aniline (C6H5NH2) has actually a pKb of 9.13 at 25°C. What is the pH of a 0.050 M solution of aniline?

The previous instances illustrate a crucial difference in between solutions of solid acids and also bases and also solutions that weak acids and also bases. Because strong acids and also bases ionize essentially completely in water, the percent ionization is always approximately 100%, nevertheless of the concentration. In contrast, the percent ionization in solutions of weak acids and also bases is tiny and relies on the analysis concentration the the weak mountain or base. As depicted for benzoic acid in number 16.16 "The Relationship in between the analytical Concentration the a Weak Acid and also Percent Ionization", the percent ionization that a weak mountain or a weak base in reality increases as its analytical concentration decreases. The percent ionization also increases together the size of Ka and Kb increases.

Figure 16.16 The Relationship in between the analysis Concentration the a Weak Acid and also Percent Ionization

As shown here because that benzoic mountain (C6H5CO2H), the percent ionization decreases together the analytical concentration of a weak acid increases.

Unlike the Ka or the Kb, the percent ionization is no a constant for weak acids and also bases however depends top top both the Ka or the Kb and the analysis concentration. Consequently, the procedure in instance 8 need to be offered to calculation the percent ionization and pH for solutions of weak acids and also bases. Example 9 and also its equivalent exercise demonstrate that the combination of a dilute solution and a relatively large Ka or Kb can offer a percent ionization much higher than 5%, making it vital to use the quadratic equation to determine the concentrations of species in solution.

### Note the Pattern

The percent ionization in a equipment of a weak mountain or a weak base increases together the analysis concentration decreases and as the Ka or the Kb increases.

### Example 9

Benzoic mountain (C6H5CO2H) is provided in the food market as a preservative and medically together an antifungal agent. The pKa at 25°C is 4.20, making it a rather stronger acid than acetic acid. Calculate the portion of benzoic acid molecules that room ionized in every solution.

a 0.0500 M equipment a 0.00500 M systems

Given: concentrations and also pKa

Strategy:

A compose both the balanced equilibrium equation for the ionization reaction and the equilibrium equation (Equation 16.15). Usage Equation 16.25 to calculate the Ka from the pKa.

B because that both the focused solution and the dilute solution, usage a tabular format to compose expressions for the last concentrations of all varieties in solution. Substitute this values into the equilibrium equation and solve because that f because that each solution.

C use the worths of f and also Equation 16.41 to calculation the percent ionization.

Solution:

A If we abbreviate benzoic mountain as PhCO2H whereby Ph = –C6H5, the well balanced equilibrium equation because that the ionization reaction and the equilibrium equation can be created as follows:

PhCO 2 H(aq) ⇌ H + (aq) + PhCO 2 − (aq) K a = < H + > < PhCO 2 − > < PhCO 2 H >

From the pKa, we have actually Ka = 10−4.20 = 6.3 × 10−5.

B for the much more concentrated solution, we set up ours table of early stage concentrations, changes in concentrations, and final concentrations:

PhCO 2 H(aq) ⇌ H + (aq) + PhCO 2 − (aq)
initial 0.0500 1.00 × 10−7 0
change x +x +x
final (0.0500 − x) (1.00 × 10−7 + x) x

Inserting the expressions for the final concentrations right into the equilibrium equation and also making our normal assumptions, the and also are negligible as result of the autoionization of water,

K a = < H + > < PhCO 2 − > < PhCO 2 H > = ( x ) ( x ) 0.0500 − x = x 2 0.0500 = 6.3 × 10 − 5 1.8 × 10 − 3 = x

This worth is less than 5% of 0.0500, for this reason our simplifying assumption is justified, and also at equilibrium is 1.8 × 10−3 M. We reach the same conclusion using CHA: KaCHA = (6.3 × 10−5)(0.0500) = 3.2 × 10−6 > 1.0 × 10−6.

C The percent ionized is the proportion of the concentration that PhCO2− to the analysis concentration, multiply by 100:

percent ionized = < PhCO 2 − > C PhCO 2 H × 100 = 1.8 × 10 − 3 0.0500 × 100 = 3.6 %

Because just 3.6% that the benzoic mountain molecules are ionized in a 0.0500 M solution, our simplifying assumptions are confirmed.

B for the much more dilute solution, we proceed in specifically the same manner. Ours table of concentration is thus as follows:

PhCO 2 H(aq) ⇌ H + (aq) + PhCO 2 − (aq)
initial 0.00500 1.00 × 10−7 0
change x +x +x
final (0.00500 − x) (1.00 × 10−7 + x) x

Inserting the expressions for the final concentrations into the equilibrium equation and also making our normal simplifying assumptions,

K a = < H + > < PhCO 2 − > < PhCO 2 H > = ( x ) ( x ) 0.00500 − x = x 2 0.00500 = 6.3 × 10 − 5 5.6 × 10 − 4 = x

Unfortunately, this number is greater than 10% of 0.00500, so our assumption that the fraction of benzoic acid that is ionized in this solution could be neglected and that (0.00500 − x) ≈ x is no valid. Furthermore, we see that KaCHA = (6.3 × 10−5)(0.00500) = 3.2 × 10−7 −6. Hence the pertinent equation is together follows:

x 2 0.00500 − x = 6.3 × 10 − 5

which need to be resolved using the quadratic formula. Multiplying the end the quantities,

x2 = (6.3 × 10−5)(0.00500 − x) = (3.2 × 10−7) − (6.3 × 10−5)x

Rearranging the equation to fit the standard quadratic equation format,

x2 + (6.3 × 10−5)x − (3.2 × 10−7) = 0

This equation deserve to be solved by using the quadratic formula:

x = − b ± b 2 − 4 a c 2 a = − ( 6.3 × 10 − 5 ) ± ( 6.3 × 10 − 5 ) 2 − 4 ( 1 ) ( − 3.2 × 10 − 7 ) 2 ( 1 ) = − ( 6.3 × 10 − 5 ) ± ( 1.1 × 10 − 3 ) 2 = 5.3 × 10 − 4  or − 5.9 × 10 − 4

Because a an unfavorable x value synchronizes to a an unfavorable , which is no physically meaningful, we use the confident solution: x = 5.3 × 10−4. Therefore = 5.3 × 10−4 M.

C The percent ionized is therefore

percent ionized = < PhCO 2 − > C PhCO 2 H × 100 = 5.3 × 10 − 4 0.00500 × 100 = 11 %

In the more dilute systems (C = 0.00500 M), 11% the the benzoic mountain molecules room ionized versus only 3.6% in the more concentrated equipment (C = 0.0500 M). Diminish the analytical concentration through a factor of 10 results in an around threefold boost in the portion of benzoic acid molecules that space ionized.

Exercise

Lactic acid (CH3CH(OH)CO2H) is a weak acid with a pKa the 3.86 in ~ 25°C. What percent of the lactic acid is ionized in each solution?

## Determining Keq native Ka and also Kb

In ar 16.2 "A Qualitative description of Acid–Base Equilibriums", girlfriend learned just how to use Ka and Kb values to qualitatively predict whether reactants or products are favored in one acid–base reaction. Tabulated worths of Ka (or pKa) and also Kb (or pKb), add to the Kw, enable us to quantitatively identify the direction and also extent the reaction for a weak acid and also a weak base by calculating K because that the reaction. To illustrate just how to do this, we start by creating the dissociation equilibriums for a weak acid and a weak base and also then summing them:

Equation 16.48

mountain HA ⇌ H + + A − K a base B  +  H 2 O ⇌ HB + + five − K b sum HA  +  B +  H 2 O ⇌ H + + A − + HB + + oh − K amount = K a K b

The overall reaction has actually H2O ~ above the left and H+ and also OH− top top the right, which way it involves the autoionization the water (H2O⇌H++OH−) in addition to the acid–base equilibrium in i beg your pardon we are interested. Us can obtain an equation that includes only the acid–base equilibrium by simply adding the equation for the reverse of the autoionization of water (H++OH−⇌H2O), because that which K = 1/Kw, come the all at once equilibrium in Equation 16.48 and canceling:

Equation 16.49

HA  +  B +   H 2 O ⇌ H + + A − + HB + + oh − K sum = K a K b H + + oh − ⇌ H 2 O 1 / K w HA  +  B ⇌ A − + HB + K = ( K a K b ) / K w

Thus the equilibrium continuous for the reaction the a weak acid through a weak base is the product the the ionization constants the the acid and also the base divided by Kw. Instance 10 illustrates how to calculate the equilibrium consistent for the reaction the a weak acid through a weak base.

### Example 10

Fish have tendency to damn it rapidly, also when refrigerated. The cause of the result “fishy” odor is a mixture that amines, specifically methylamine (CH3NH2), a volatile weak base (pKb = 3.34). Fish is frequently served v a wedge the lemon due to the fact that lemon juice has citric acid, a triprotic acid v pKa values of 3.13, 4.76, and also 6.40 that can neutralize amines. Calculation the equilibrium consistent for the reaction of overabundance citric acid with methylamine, assuming that just the first dissociation continuous of citric acid is important.

Given: pKb because that base and pKa for acid

Strategy:

A create the well balanced equilibrium equation and the equilibrium constant expression for the reaction.

B convert pKa and pKb come Ka and also Kb and also then use Equation 16.49 to calculate K.

Solution:

A If we abbreviate citric mountain as H3citrate, the equilibrium equation for its reaction v methylamine is together follows:

CH 3 NH 2 (aq) + H 3 citrate(aq) ⇌ CH 3 NH 3 + (aq) + H 2 citrate − (aq)

The equilibrium continuous expression for this reaction is as follows:

K = < CH 3 NH 3 + > < H 2 citrate − > < CH 3 NH 2 > < H 3 citrate >

B Equation 16.49 is K = (KaKb)/Kw. Converting pKa and also pKb to Ka and also Kb gives Ka = 10−3.13 = 7.4 × 10−4 for citric acid and also Kb = 10−3.34 = 4.6 × 10−4 for methylamine. Substituting these values into the equilibrium equation,

K = K a K b K w = ( 7.4 × 10 − 4 ) ( 4.6 × 10 − 4 ) 1.01 × 10 − 14 = 3.4 × 10 7

The worth of pK can additionally be calculated directly by taking the an unfavorable logarithm that both sides of Equation 16.49, i beg your pardon gives

pK = pKa + pKb − pKw = 3.13 + 3.34 − 14.00 = −7.53

Thus K = 10−(−7.53) = 3.4 × 107, in commitment with the previously value. In either case, the K values show that the reaction that citric acid with the volatile, foul-smelling methylamine lies an extremely far come the right, donate the development of a much much less volatile salt through no odor. This is one reason a small lemon juice helps make less-than-fresh fish more appetizing.

Exercise

Dilute aqueous ammonia solution, frequently used together a cleaning agent, is additionally effective together a deodorizing agent. To check out why, calculate the equilibrium constant for the reaction of aqueous ammonia through butyric acid (CH3CH2CH2CO2H), a particularly foul-smelling substance linked with the odor of rancid butter and smelly socks. The pKb the ammonia is 4.75, and the pKa that butyric mountain is 4.83.

### Summary

If the concentration that one or an ext of the varieties in a equipment of an acid or a base is established experimentally, Ka and also Kb can be calculated, and also Ka, pKa, Kb, and also pKb have the right to be supplied to quantitatively explain the ingredient of services of acids and also bases. The concentration of all varieties present in solution can be determined, as can the pH of the solution and the percent of the mountain or base that is ionized. The equilibrium consistent for the reaction that a weak acid through a weak base deserve to be calculated native Ka (or pKa), Kb (or pKb), and Kw.

### Key Takeaway

for a equipment of a weak acid or a weak base, the percent ionization rises as the Ka or the Kb increases and as the analytical concentration decreases.

### Key Equations

Percent ionization of acid

Equation 16.41: CHA×100

Percent ionization of base

Equation 16.42: CB×100

Equilibrium continuous for reaction the a weak acid with a weak base

Equation 16.49: K=KaKbKw

Write one expression because that the analytical concentration (C) of H3PO4 in terms of the concentrations of the varieties actually existing in solution.

For fairly dilute options of a weak mountain such as acetic acid (CH3CO2H), the concentration that undissociated acetic acid in solution is regularly assumed to it is in the very same as the analysis concentration. Define why this is a precious practice.

What is the relationship in between the Ka the a weak acid and its percent ionization? walk a compound through a big pKa value have a greater or a lower percent ionization than a compound v a small pKa value (assuming the same analytical concentration in both cases)? Explain.

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For a dilute systems of a weak mountain (HA), present that the pH that the solution deserve to be approximated utilizing the following equation (where CHA is the analytical concentration the the weak acid):

pH = − log K a ⋅ C HA

Under what conditions is this approximation valid?

The pKa the NH3 is approximated to be 35. The conjugate base, the amide ion (NH2−), have the right to be isolated together an alkali metal salt, together as salt amide (NaNH2). Calculation the pH that a solution ready by including 0.100 mol of salt amide to 1.00 together of water. Walk the pH different appreciably indigenous the pH that a NaOH solution of the exact same concentration? Why or why not?

A 0.200 M solution of diethylamine, a substance supplied in insecticides and also fungicides, is just 3.9% ionized in ~ 25°C. Create an equation showing the equilibrium reaction and then calculate the pKb of diethylamine. What is the pKa that its conjugate acid, the diethylammonium ion? What is the equilibrium constant expression because that the reaction of diethylammonium chloride through water?

A 1.00 M systems of fluoroacetic acid (FCH2CO2H) is 5% dissociated in water. What is the equilibrium consistent expression for the dissociation reaction? calculate the concentration that each species in solution and also then calculate the pKa the FCH2CO2H.

The pKa of 3-chlorobutanoic mountain (CH3CHClCH2CO2H) is 4.05. What portion is dissociated in a 1.0 M solution? do you expect the pKa the butanoic mountain to be greater than or much less than the pKa of 3-chlorobutanoic acid? Why?

The pKa of the ethylammonium ion (C2H5NH3+) is 10.64. What percentage of ethylamine is ionized in a 1.00 M equipment of ethylamine?

The pKa that Cl3CCO2H is 0.64. What is the pH the a 0.580 M solution? What portion of the Cl3CCO2H is dissociated?

The pH the a 0.150 M systems of aniline hydrochloride (C6H5NH3+Cl−) is 2.70. What is the pKb the the conjugate base, aniline (C6H5NH2)? do you intend the pKb the (CH3)2CHNH2 come be better than or less than the pKb the C6H5NH2? Why?