CH. 16Problems

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A. Na+ and OH–

B. H+ and OH–

C Na+ and Cl–

D. H+ and Cl–

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A. Na+ and OH–

B. H+ and OH–

C Na+ and Cl–

D. H+ and Cl–

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A. Both substances act as Brønsted-Lowry bases.B. Neither substance acts as a Brønsted-Lowry

base.

C. H2S(aq)

D. CH3NH2(aq)

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A. Both substances act as Brønsted-Lowry bases.B. Neither substance acts as a Brønsted-Lowry

base.

C. H2S(aq)

D. CH3NH2(aq)

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A. No, the pH range is 1-14.B. No, the definition of pH does not permit it to

have a negative value.C. Yes, for any solution with a concentration of

base greater than 1 M.D. Yes, for any solution with a concentration of

acid greater than 1 M.

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A. No, the pH range is 1-14.B. No, the definition of pH does not permit it to

have a negative value.C. Yes, for any solution with a concentration of

base greater than 1 M.D. Yes, for any solution with a concentration of

acid greater than 1 M.

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A. pH = 17.00; basic because pH > 7B. pH = 11.00; basic because pH > 7C. pH = 3.00; acidic because pH < 7D. The pH cannot be determined without [H+]

information. Solution is basic because pOH < 7.

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A. pH = 17.00; basic because pH > 7B. pH = 11.00; basic because pH > 7C. pH = 3.00; acidic because pH < 7D. The pH cannot be determined without [H+]

information. Solution is basic because pOH < 7.

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pKa(HF) pKb(F–)A. 3.17 10.83B. 10.83 3.17C. 9.60 4.40D. 4.40 9.60

Ka = 6.8*10-4

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pKa(HF) pKb(F–)A. 3.17 10.83B. 10.83 3.17C. 9.60 4.40D. 4.40 9.60

Ka = 6.8*10-4

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NO3– CO3

2–

A. Increase IncreaseB. No change DecreaseC. Decrease DecreaseD. No change Increase

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NO3– CO3

2–

A. Increase IncreaseB. No change DecreaseC. Decrease DecreaseD. No change Increase

© 2012 Pearson Education, Inc.Chemistry, The Central Science, 12th EditionTheodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward

Predicting the Position of a

Proton-Transfer Equilibrium

For the following proton-transfer reaction use Figureto predict whether the equilibrium lies to the left (Kc < 1)or to the right (Kc > 1):

SolutionAnalyze We are asked to predict whether an equilibrium lies to the right, favoring products, or to the left, favoring reactants.

Plan This is a proton-transfer reaction, and the position of the equilibrium will favor the proton going to the stronger of two bases. The two bases in the equation are CO3

2-, the base in the forward reaction, and SO42 , the

conjugate base of HSO4. We can find the relative positions of these two bases in the figure to determine which

is the stronger base.

© 2012 Pearson Education, Inc.Chemistry, The Central Science, 12th EditionTheodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward

Continued

Solve The CO32 ion appears lower in the right-hand column in the figure and is a stronger base than SO4

2. CO3

2 will get the proton to become HCO3, while SO4

2 will remain mostly unprotonated. The resulting equilibrium lies to the right, favoring products (that is, Kc > 1):

Comment Of the two acids HSO4 and HCO3

, the stronger one (HSO4) gives up a proton more readily, and the

weaker one (HCO3) tends to retain its proton. Thus, the equilibrium favors the direction in which the proton moves

from the stronger acid and becomes bonded to the stronger base.

Pratice ExerciseFor each reaction, use the figure to predict whether theequilibrium lies to the left or to the right:

(a)

(b)

(a) left, (b) right

© 2012 Pearson Education, Inc.Chemistry, The Central Science, 12th EditionTheodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward

Calculating Ka from Measured pH

A prepared solution of formic acid (HCOOH) is 0.10 M and found its pH at 25 C to be 2.38. Calculate Ka.What percentage of the acid is ionized?

SolutionAnalyze Given the molar concentration of an aqueous solution of weak acid and the pH of the solution, and asked to determine the Ka for the acid.

Plan We are dealing specifically with the ionization of a weak acid, this problem is very similar to the equilibrium problems encountered in Chapter 15. We can solve this problem starting with the chemical reaction and a tabulation of initial and equilibrium concentrations.

Solve The first step in solving any equilibrium problem is to write the equation for the equilibriumreaction. The ionization of formic acid can be written as

The equilibrium-constant expression is

From the measured pH, we can calculate [H+] :

© 2012 Pearson Education, Inc.Chemistry, The Central Science, 12th EditionTheodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward

Continued

Determine the concentrations involved in the equilibrium. The solution is initially 0.10 M in HCOOH molecules. The ionization of the acid into H+ and HCOO. For each HCOOH molecule that ionizes, one H+ ion and one HCOO ion are produced. Because the pH measurement indicates that [H+] = 4.2 103 M at equilibrium, we can construct the following table:

Neglect the very small concentration of H+ (aq) due to H2O autoionization. The amount of HCOOH that ionizes is very small compared with the initial concentration of the acid. To the number of significant figures we are using, the subtraction yields 0.10 M:

Insert the equilibrium concentrations into the expression for Ka:

Check The magnitude of our answer is reasonable because Ka for a weak acid is usually between 102 and 1010.

IONIZED

© 2012 Pearson Education, Inc.Chemistry, The Central Science, 12th EditionTheodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward

Using Ka to Calculate pH

Calculate the pH of a 0.20 M solution of HCN.

Plan Proceed as before: write equation, construct table, H+ is unknown

SolutionAnalyze Given the molarity of a weak acid and find the pH. From the table, Ka for HCN is 4.9 1010.

© 2012 Pearson Education, Inc.Chemistry, The Central Science, 12th EditionTheodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward

Using Kb to Calculate OH

Calculate the concentration of OH and the pH in a 0.15 M solution of NH3.

Plan Use same procedure as used in solving problems of weak acids, write equation, construct table, insert equilibrium concentrations into expression

SolutionAnalyze Given the concentration of a weak base, determine the concentration of OH . Kb = 1.8*10-5

Because Kb is small, the amount of NH3 that reacts with water is much smaller thanthe NH3 concentration, and so we can neglect x relative to 0.15 M. Then we have

pOH = -Log 1.8*10-3 = 2.80 pH = 14 – 2.8 = 11.20

© 2012 Pearson Education, Inc.Chemistry, The Central Science, 12th EditionTheodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward

Using Percent Dissociation to CalculateCalculate the concentration of OH and the Ka in a 0.15 M solution of NH3 if there is 2.0% dissociation.

Plan Use same procedure as used in solving problems of weak acids, write equation, construct table, find dissociated concentration, insert equilibrium concentrations into expression.Use Ka * Kb = Kw to convert Kb to Ka.

SolutionAnalyze Given concentration and % dissociated of a weak base, determine the concentration of OH .

2% dissociation of initial concentration will be the equilibrium concentrations for “x”. 0.003 0.15 * 0.02 ]X[

5-

3

-4

b 10*6.0 ]15.0[

003][0.003][0.

][NH

]][OH[NH K

10-

5

-14

b

wa 10*1.67

10*0.6

10*1

K

K K