1. CH. 16 Problems

2. © 2012 Pearson Education, Inc. A.Na + and OH – B.H + and OH – CNa + and Cl – D.H + and Cl –

3. © 2012 Pearson Education, Inc. A.Na + and OH – B.H + and OH – CNa + and Cl – D.H + and Cl –

4. © 2012 Pearson Education, Inc. A.Both substances act as Brønsted-Lowry bases. B.Neither substance acts as a Brønsted-Lowry base. C.H 2 S(aq) D.CH 3 NH 2 (aq)

5. © 2012 Pearson Education, Inc. A.Both substances act as Brønsted-Lowry bases. B.Neither substance acts as a Brønsted-Lowry base. C.H 2 S(aq) D.CH 3 NH 2 (aq)

6. © 2012 Pearson Education, Inc. 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.

7. © 2012 Pearson Education, Inc. 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.

8. © 2012 Pearson Education, Inc. A.pH = 17.00; basic because pH > 7 B.pH = 11.00; basic because pH > 7 C.pH = 3.00; acidic because pH < 7 D.The pH cannot be determined without [H + ] information. Solution is basic because pOH < 7.

9. © 2012 Pearson Education, Inc. A.pH = 17.00; basic because pH > 7 B.pH = 11.00; basic because pH > 7 C.pH = 3.00; acidic because pH < 7 D.The pH cannot be determined without [H + ] information. Solution is basic because pOH < 7.

10. © 2012 Pearson Education, Inc. pK a (HF)pK b (F – ) A.3.1710.83 B.10.833.17 C.9.604.40 D.4.409.60 K a = 6.8*10 -4

11. © 2012 Pearson Education, Inc. pK a (HF)pK b (F – ) A.3.1710.83 B.10.833.17 C.9.604.40 D.4.409.60 K a = 6.8*10 -4

12. © 2012 Pearson Education, Inc. NO 3 – CO 3 2– A.IncreaseIncrease B.No changeDecrease C.DecreaseDecrease D.No changeIncrease

13. © 2012 Pearson Education, Inc. NO 3 – CO 3 2– A.IncreaseIncrease B.No changeDecrease C.DecreaseDecrease D.No changeIncrease

15. © 2012 Pearson Education, Inc. Chemistry, The Central Science, 12th Edition Theodore 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 Figure to predict whether the equilibrium lies to the left (K c < 1) or to the right (K c > 1): Solution Analyze 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 CO 3 2-, the base in the forward reaction, and SO 4 2 , the conjugate base of HSO 4 . We can find the relative positions of these two bases in the figure to determine which is the stronger base.

16. © 2012 Pearson Education, Inc. Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward Continued Solve The CO 3 2  ion appears lower in the right-hand column in the figure and is a stronger base than SO 4 2 . CO 3 2  will get the proton to become HCO 3 , while SO 4 2  will remain mostly unprotonated. The resulting equilibrium lies to the right, favoring products (that is, K c > 1): Comment Of the two acids HSO 4  and HCO 3 , the stronger one (HSO 4  ) gives up a proton more readily, and the weaker one (HCO 3  ) 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 Exercise For each reaction, use the figure to predict whether the equilibrium lies to the left or to the right: (a) (b) (a) left, (b) right

17. © 2012 Pearson Education, Inc. Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward Calculating K a 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 K a. What percentage of the acid is ionized? Solution Analyze Given the molar concentration of an aqueous solution of weak acid and the pH of the solution, and asked to determine the K a 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 equilibrium reaction. The ionization of formic acid can be written as The equilibrium-constant expression is From the measured pH, we can calculate [H + ] :

18. © 2012 Pearson Education, Inc. Chemistry, The Central Science, 12th Edition Theodore 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  10  3 M at equilibrium, we can construct the following table: Neglect the very small concentration of H + (aq) due to H 2 O 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 K a : Check The magnitude of our answer is reasonable because K a for a weak acid is usually between 10  2 and 10  10. IONIZED

19. © 2012 Pearson Education, Inc. Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward Using K a 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 Solution Analyze Given the molarity of a weak acid and find the pH. From the table, K a for HCN is 4.9  10  10.

20. © 2012 Pearson Education, Inc. Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward Using K b to Calculate OH  Calculate the concentration of OH  and the pH in a 0.15 M solution of NH 3. Plan Use same procedure as used in solving problems of weak acids, write equation, construct table, insert equilibrium concentrations into expression Solution Analyze Given the concentration of a weak base, determine the concentration of OH . K b = 1.8*10 -5 Because K b is small, the amount of NH 3 that reacts with water is much smaller than the NH 3 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

21. © 2012 Pearson Education, Inc. Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward Using Percent Dissociation to Calculate Calculate the concentration of OH  and the K a in a 0.15 M solution of NH 3 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 K a * K b = K w to convert K b to K a. Solution Analyze 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”.