Ch.16 Test Bank Acid-Base Equilibria In Aqueous Solutions - Solution Bank | Chemistry Molecular Nature 8e by Neil D. Jespersen. DOCX document preview.

Ch.16 Test Bank Acid-Base Equilibria In Aqueous Solutions

Chemistry: Molecular Nature of Matter, 8e (Jespersen)

Chapter 16 Acid-Base Equilibria in Aqueous Solutions

1) If the OH− ion concentration in an aqueous solution at 25.0 °C is measured as 3.4 × 10−3 M, what is the pH?

A) 2.47

B) 7.22

C) 8.24

D) 11.53

E) 16.47

Diff: 1

Section: 16.1

2) The pH of a solution is measured to be 6.30. What are the values of [H3O+] and [OH−] for this solution?

A) [H3O+] = 5.0 × 10−7, [OH−] = 2.0 × 10−8

B) [H3O+] = 6.3 × 10−7, [OH−] = 7.7 × 10−8

C) [H3O+] = 2.0 × 106, [OH−] = 5.0 × 107

D) [H3O+] = 2.0 × 10−8, [OH−] = 5.0 × 10−7

E) [H3O+] = 1.6 × 10−1, [OH−] = 1.3 × 10−1

Diff: 1

Section: 16.1

3) The pH of a solution is measured to be 10.40. What are the values of [H3O+] and [OH−] for this solution?

A) [H3O+] = 4.0 × 10−11, [OH−] = 4.0 × 103

B) [H3O+] = 2.5 × 10−4, [OH−] = 4.0 × 10−11

C) [H3O+] = 1.0 × 10−10, [OH−] = 3.6 × 10−4

D) [H3O+] = 4.0 × 10−11, [OH−] = 2.5 × 10−4

E) [H3O+] = 9.6 × 10−2, [OH−] = 2.8 × 10−1

Diff: 1

Section: 16.1

4) If the OH− ion concentration in an aqueous solution at 25.0 °C is 6.6 × 10−4 M, what is the molarity of the H+ ion?

A) 1.5 × 10−1 M

B) 1.5 × 10−4 M

C) 6.6 × 10−10 M

D) 1.5 × 10−11 M

E) 6.6 × 10−11 M

Diff: 1

Section: 16.1

5) An aqueous solution at 25.0 °C has an H+ concentration of 4.0 × 10−2 molar. What is the molarity of the OH− ion ?

A) 4.0 × 10−2 M

B) 4.0 × 10−9 M

C) 4.0 × 10−12 M

D) 2.5 × 10−13 M

E) 25.0 M

Diff: 2

Section: 16.1

6) What is the molarity of OH− ions if the H+ ion concentration in an aqueous solution at 25.0 °C is 0.100 M?

A) 0.100 M

B) 1.00 × 10−7 M

C) 10 × 10−12 M

D) 1.00 × 10−13 M

E) 0 M

Diff: 2

Section: 16.1

7) If the H+ ion concentration in an aqueous solution at 25.0 °C has a value of 0.10 M, what is the pOH of the solution?

A) 1.00

B) 7.00

C) 12.00

D) 13.00

E) 11.40

Diff: 2

Section: 16.1

8) If the OH− ion concentration in an aqueous solution at 25.0 °C has a value of 0.100 M, what is the pH of the solution?

A) 1.00

B) 7.00

C) 12.00

D) 13.00

E) 11.40

Diff: 2

Section: 16.1

9) If the H+ ion concentration in an aqueous solution at 25.0 °C has a value of 0.10 M, then what is the pH of the solution?

A) −1.00

B) 0.100

C) 1.00

D) 6.90

E) 13.00

Diff: 2

Section: 16.1

10) If the OH− ion concentration in an aqueous solution at 25.0 °C has a value of 0.10 M, then what is the pOH of the solution?

A) −1.00

B) 0.10

C) 1.00

D) 6.90

E) 13.00

Diff: 2

Section: 16.1

11) If the OH− ion concentration in an aqueous solution at 25.0 °C is 3.4 × 10−3 M, then what is the molarity of the H+ ion?

A) 2.9 × 10−3 M

B) 2.9 × 10−12 M

C) 3.4 × 10−17 M

D) 5.8 × 10−9 M

E) 6.6 × 10−10 M

Diff: 2

Section: 16.1

12) If the H+ ion concentration in an aqueous solution at 25.0 °C is measured as 6.6 × 10−4 M, what is the pH?

A) 3.00

B) 3.18

C) 6.60

D) 9.55

E) 10.82

Diff: 2

Section: 16.1

13) Calculate the pH of a beer in which the hydrogen ion concentration is 3.9 × 10−5 M.

A) 4.41

B) 3.90

C) 10.11

D) 5.03

E) 9.61

Diff: 2

Section: 16.1

14) A scientist is testing a new cleaning chemical and measures the pH of the cleaner to be 10.70. What is the [H3O+] for this cleaner?

A) 2.0 × 10−11 M

B) 10.7 M

C) 9.3 × 10−10 M

D) 1.03 M

E) 5.0 × 10−10 M

Diff: 2

Section: 16.1

15) The pH of urine is approximately 3.0. How many times greater is the [H3O+] in urine than in orange juice, which has an approximate pH of 6.0?

A) 100

B) 3.0

C) 1000

D) 1.8

E) 2

Diff: 1

Section: 16.1

16) The pH of a specific household cleaner is approximately 11.0. How many times greater is the [OH−] in this cleaner than in an ammonia solution, which has an approximate pH of 8.0?

A) 100

B) 3.0

C) 1000

D) 1.8

E) 8

Diff: 1

Section: 16.1

17) The pH of coffee is approximately 5.0. How many times greater is the [H+] in coffee than in neutral water?

A) 5.0

B) 1.4

C) 200

D) 2

E) 100

Diff: 2

Section: 16.1

18) Calculate the pH of a mixture made by adding 50.0 mL of 0.20 M HCl(aq) to 150.0 mL of water at 25.0 °C.

A) 0.70

B) 1.00

C) 1.18

D) 1.30

E) 13.00

Diff: 1

Section: 16.2

19) Calculate the pH of a mixture made by adding 50.0 mL of 0.016 M HNO3(aq) to 150.0 mL of water at 25.0 °C.

A) 1.80

B) 2.40

C) 1.40

D) 2.10

E) 11.6

Diff: 1

Section: 16.2

20) A mixture is made by adding 50.0 mL of 0.20 M NaOH(aq) to 50.0 mL of water. At 25.0 °C, what is its pH?

A) 1.00

B) 4.55

C) 7.00

D) 13.00

E) 13.30

Diff: 1

Section: 16.2

21) A mixture is made by adding 25.0 mL of 0.10 M KOH(aq) to 25.0 mL of water. At 25.0 °C, what is its pH?

A) 1.30

B) 12.70

C) 13.00

D) 1.00

E) 13.30

Diff: 1

Section: 16.2

22) Calculate the pH of a 0.020 M solution of Ca(OH)2 whose temperature is 25.0 °C.

A) 1.40

B) 0.040

C) 1.69

D) 12.60

E) 12.30

Diff: 2

Section: 16.2

23) Calculate the pH of a 0.0018 M solution of Mg(OH)2 whose temperature is 25.0 °C.

A) 2.44

B) 2.74

C) 11.00

D) 11.25

E) 11.55

Diff: 2

Section: 16.2

24) Calculate the pH of a 0.25 M solution of KOH whose temperature is 25.0 °C.

A) 12.61

B) −0.60

C) 0.60

D) 0.25

E) 13.40

Diff: 1

Section: 16.2

25) Calculate the pH of a 0.025 M solution of NaOH whose temperature is 25.0 °C.

A) 0.60

B) 1.60

C) 1.30

D) 12.39

E) 13.39

Diff: 1

Section: 16.2

26) Given 0.01 M solutions of each of the following acids, which solution would have the lowest pH?

A) Hypoiodous acid (HOI), Ka = 2.3 × 10−11

B) Hypobromous acid (HOBr), Ka = 2.5 × 10−9

C) Lactic acid (HC3H5O3), Ka = 1.3 × 10−4

D) Chlorous acid (HClO2), Ka = 1.1 × 10−2

E) Boric acid (H3BO3), Ka = 5.9 × 10−10

Diff: 2

Section: 16.3

27) Given 0.01 M solutions of each of the following bases, which solution would have the highest pH?

A) Aniline (C6H5NH2), Kb = 3.9 × 10−10

B) Dimethylamine ((CH3)2NH), Kb = 5.1 × 10−4

C) Hydrazine (N2H4), Kb = 1.3 × 10−6

D) Methylamine (CH3NH2), Kb = 4.4 × 10−4

E) Pyridine (C5H5N), Kb = 1.7 × 10−9

Diff: 1

Section: 16.3

28) Given 0.01 M solutions of each of the following acids, which solution would have the lowest pH?

A) Nitrous acid (HNO2), pKa = 3.34

B) Phenol (HC6H5O), pKa = 9.89

C) Acetic acid (HC2H3O2), pKa = 4.74

D) Hydrofluoric acid (HF), pKa = 3.46

E) Formic acid (HCHO2), pKa = 3.74

Diff: 1

Section: 16.3

29) Given 0.01 M solutions of each of the following bases, which solution would have the highest pH?

A) Aniline (C6H5NH2), pKb = 9.39

B) Hydrazine (N2H4), pKb = 5.89

C) Hydroxylamine (HONH2), pKb = 7.96

D) Methylamine (CH3NH2), pKb = 3.36

E) Butylamine (C4H9NH2), pKb = 3.23

Diff: 1

Section: 16.3

30) The ionization constant, Ka, for macnic acid is 5.0 × 10−5. What is the pKa of this acid?

A) 2.00 × 104

B) 4.30

C) 5.70

D) 1.75 × 10−1

E) 10.70

Diff: 1

Section: 16.3

31) The ionization constant, Ka, for lactic acid is 1.38 × 10−4. What is the pKa of this acid?

A) 7.25 × 103

B) 2.59

C) 3.86

D) 5.38

E) 3.59

Diff: 1

Section: 16.3

32) A certain base has a pKb value of 4.74. What is the Kb value for this base?

A) 1.8 × 10−5

B) 5.5 × 104

C) 5.5 × 10−10

D) 6.7 × 10−1

E) 9.7 × 10−1

Diff: 1

Section: 16.3

33) The ionization constant, Ka, for benzoic acid is 6.28 × 10−5. What is the pKa of this acid?

A) 1.59 × 104

B) 4.20

C) 4.45

D) 5.64

E) 9.80

Diff: 1

Section: 16.3

34) The ionization constant, Kb, for ammonia has a value of 1.76 × 10−5. What is the pKb of this base?

A) +3.24

B) −4.75

C) +4.75

D) −9.25

E) +9.25

Diff: 1

Section: 16.3

35) Hydrazine is a base, and the value of its constant, Kb, is 1.70 × 10−6. What is the value of pKb for this base?

A) +4.30

B) +5.77

C) −5.77

D) +8.23

E) −8.23

Diff: 1

Section: 16.3

36) Butanoic acid, HC4H7O2, has a pKa value = 4.82. What is the value of the ionization constant for butanoic acid?

A) 8.08 × 10−3

B) 1.51 × 10−5

C) 4.82 × 10−7

D) 6.61 × 10−10

E) 1.24 × 10−12

Diff: 1

Section: 16.3

37) Hypochlorous acid, HClO, has a pKa value = 7.82. What is the value of the ionization constant for hypochlorous acid?

A) 5.42 × 10−4

B) 3.31 × 10−7

C) 7.52 × 10−7

D) 1.51 × 10−8

E) 1.84 × 10−11

Diff: 1

Section: 16.3

38) A 0.100 M solution of an acid, HA, has a pH = 2.00. What is the value of the ionization constant, Ka for this acid?

A) 1.1 × 10−2

B) 1.1 × 10−3

C) 1.1 × 10−4

D) 1.0 × 10−3

E) 1.0 × 10−4

Diff: 2

Section: 16.3

39) A 0.200 M solution of an acid, HA, has a pH = 3.50. What is the value of the ionization constant Ka for this acid?

A) 3.2 × 10−4

B) 1.6 × 10−3

C) 6.1 × 101

D) 5.0 × 10−7

E) 7.9 × 10−4

Diff: 2

Section: 16.3

40) Formic acid, HCO2H, has an ionization constant with the value: Ka = 1.76 × 10−4. Calculate the value of pKb for the formate ion, CO2H−, the conjugate base of formic acid.

A) 3.75

B) 5.35

C) 8.65

D) 10.25

E) 12.24

Diff: 1

Section: 16.3

41) Benzoic acid, HC7H5O2, has an ionization constant with the value: Ka = 6.4 × 10−5. Calculate the value of pKb for the benzoate ion, C7H5O2−, the conjugate base of benzoic acid.

A) 4.19

B) 4.35

C) 9.65

D) 9.81

E) 13.46

Diff: 2

Section: 16.3

42) Toluic acid, HC8H7O2, has an ionization constant with the value: Ka = 1.23 × 10−5. Calculate the value of pKb for the conjugate base of toluic acid.

A) 3.91

B) 5.00

C) 7.00

D) 8.23

E) 9.09

Diff: 2

Section: 16.3

43) The Ka of hydrocyanic acid (HCN) is 4.9 × 10−10. What is the Kb of CN−?

A) 4.9 × 104

B) 4.9 × 10−10

C) 9.1 × 10−10

D) 9.3

E) 2.0 × 10−5

Diff: 2

Section: 16.3

44) A 0.400 M solution of an acid, HQ, has a pH = 1.301. What is the value of the ionization constant, Ka, for this acid?

A) 5.00 × 10−2

B) 1.25 × 10−3

C) 5.56 × 10−3

D) 6.25 × 10−3

E) 7.14 × 10−3

Diff: 2

Section: 16.4

45) A 0.200 M solution of an acid has a pH = 1.000. What is the value of the ionization constant, Ka, for this acid?

A) 0.200

B) 0.0400

C) 1.00 × 10−3

D) 1.00

E) 1.00 × 10−1

Diff: 2

Section: 16.4

46) A 0.300 M solution of an acid, HZ, has a pH = 1.301. What is the value of the ionization constant, Ka, for this acid?

A) 1.67 × 10−1

B) 2.00 × 10−1

C) 1.00 × 10−2

D) 7.14 × 10−3

E) 8.33 × 10−3

Diff: 2

Section: 16.4

47) A 0.200 M solution of a weak base in water has a pH = 10.40 at 25°C. Calculate the value of Kb for this base.

A) 1.0 × 10−5

B) 3.2 × 10−7

C) 2.2 × 10−5

D) 4.0 × 10−11

E) 5.0 × 10−5

Diff: 2

Section: 16.4

48) A 0.075 M solution of a weak base in water has a pH = 11.21 at 25°C. Calculate the value of Kb for this base.

A) 1.6 × 10−3

B) 5.1 × 10−22

C) 2.2 × 10−2

D) 3.6 × 10−5

E) 6.2 × 10−12

Diff: 2

Section: 16.4

49) Calculate the pH of a 0.14 M HNO2 solution that is 5.7% ionized.

A) 11.90

B) 0.85

C) 1.70

D) 13.10

E) 2.10

Diff: 2

Section: 16.4

50) Calculate the pH of a 0.80 M HOCN solution that is 5.00% ionized.

A) 1.40

B) 1.30

C) 0.0969

D) 0.523

E) 12.7

Diff: 2

Section: 16.4

51) Calculate the pH of a 0.095 M weak acid solution that is 1.9% ionized.

A) 3.74

B) 0.744

C) 2.74

D) 1.72

E) 1.02

Diff: 2

Section: 16.4

52) The pH of a 0.035 M weak acid solution is 3.30. What is the percent ionization for this acid?

A) 1.4

B) 3.3

C) 1.1

D) 0.05

E) 3.5

Diff: 2

Section: 16.4

53) A 0.100 M solution of an acid, HA, has a pH = 2.00. What is the value of the ionization constant, Ka, for this acid?

A) 1.1 × 10−2

B) 1.1 × 10−3

C) 1.1 × 10−4

D) 1.0 × 10−3

E) 1.0 × 10−4

Diff: 2

Section: 16.4

54) The ionization constant, Ka, for benzoic acid, HC7H5O2, is 6.28 × 10−5. What is the pH of a 0.15 molar solution of this acid?

A) 0.82

B) 2.52

C) 4.20

D) 5.03

E) 5.79

Diff: 2

Section: 16.5

55) The ionization constant, Ka, for acetic acid, HC2H3O2, is 1.76 × 10−5. What is the pH of a 0.0800 molar solution of this acid?

A) 1.097

B) 2.013

C) 2.926

D) 4.754

E) 5.852

Diff: 2

Section: 16.5

56) What is the pH of a 1.00 molar solution of nitrous acid? The Ka for nitrous acid is 7.1 × 10−4.

A) 1.57

B) 1.93

C) 2.04

D) 2.67

E) 3.15

Diff: 2

Section: 16.5

57) The ionization constant, Kb, for the weak base trimethylamine, (CH3)3N, is 7.4 × 10−5. What is the pH of a 0.040 molar solution of this base?

A) 8.37

B) 9.87

C) 10.70

D) 11.23

E) 12.60

Diff: 2

Section: 16.5

58) The ionization constant, Ka, for dichloroacetic acid, HC2HO2Cl2 , is 5.0 × 10−2. What is the pH of a 0.15 molar solution of this acid?

A) 1.06

B) 1.19

C) 1.30

D) 1.56

E) 1.82

Diff: 2

Section: 16.5

59) Trihydroxybenzoic acid (THB) is a not-so-weak acid with an ionization constant:

Ka = 2.1 × 10−2. Calculate the [H+] and the pH of a 0.050 molar solution of THB.

Hint: Simplified assumptions do not work in this case.

A) 1.9 × 10−2, 1.72

B) 2.4 × 10−2, 1.63

C) 2.5 × 10−2, 1.59

D) 3.2 × 10−2, 1.49

E) 1.1 × 10−3, 2.98

Diff: 3

Section: 16.5

60) Dichloroacetic acid HC2HCl2O2, is a not-so-weak acid with an ionization constant:

Ka = 5.5 × 10−2. Calculate the [H+] and the pH of a 0.100 molar solution of this acid.

Hint: Simplified assumptions do not work in this case.

A) 1.9 × 10−2, 1.72

B) 5.2 × 10−2, 1.29

C) 2.5 × 10−2, 1.59

D) 3.2 × 10−2, 1.49

E) 1.1 × 10−3, 2.98

Diff: 3

Section: 16.5

61) The ionization constant, Ka, for HCN(aq) is 6.2 × 10−10. What is the pH of a 0.10 molar solution of sodium cyanide, which contains the cyanide ion?

A) 5.10

B) 8.90

C) 9.21

D) 11.10

E) 11.30

Diff: 2

Section: 16.6

62) The ionization constant, Ka, for HOI(aq), hypoiodous acid, is 2.3 × 10−11. What is the pH of a 0.050 molar solution of sodium hypoiodite, which contains the hypoiodite ion, OI−?

A) 5.97

B) 8.03

C) 10.64

D) 11.65

E) 12.70

Diff: 2

Section: 16.6

63) What is the pH of a 1.00 molar solution of KCl?

A) 0.00

B) 6.90

C) 7.00

D) 7.10

E) 13.00

Diff: 1

Section: 16.6

64) What is the pH of a 1.00 molar solution of NH4Cl(aq)? The Kb for NH3 is 1.8 × 10−5.

A) 2.37

B) 4.63

C) 9.26

D) 9.37

E) 11.63

Diff: 2

Section: 16.6

65) What is the pH of a 1.00 × 10−9 molar solution of NaCl?

A) 5.00

B) 6.00

C) 6.80

D) 7.00

E) 9.00

Diff: 1

Section: 16.6

66) Which one of the following salts will form a basic solution upon dissolving in water?

A) NaCl

B) NaNO2

C) NH4NO3

D) KBr

E) AlCl3

Diff: 1

Section: 16.6

67) What is the pH of a 1.00 molar solution of NaCN(aq)? The Ka for HCN = 6.2 × 10−10.

A) 2.40

B) 4.61

C) 9.40

D) 11.60

E) 13.00

Diff: 2

Section: 16.6

68) What is the pH of a 1.00 molar solution of NaNO2(aq)? The Ka for HNO2 is 7.0 × 10−4.

A) 1.57

B) 3.15

C) 5.42

D) 8.58

E) 10.85

Diff: 2

Section: 16.6

69) What is the pH of a 1.00 molar solution of NaNO3(aq)?

A) 6.00

B) 6.90

C) 7.00

D) 7.10

E) 13.00

Diff: 1

Section: 16.6

70) What is the pH of a 1.00 molar solution of NaC2H3O2(aq)?

The Ka for acetic acid (HC2H3O2) is 1.8 × 10−5.

A) 2.37

B) 4.63

C) 4.75

D) 9.37

E) 11.63

Diff: 2

Section: 16.6

71) What is the pH of a 1.00 molar solution of NaBr(aq)?

A) 0.00

B) 6.90

C) 7.00

D) 7.10

E) 13.00

Diff: 1

Section: 16.6

72) A buffer solution is prepared by taking 0.400 moles of acetic acid (pKa = 4.76) and 0.250 moles of sodium acetate in sufficient water to make 1.400 liters of solution. Calculate the pH of this solution.

A) 4.46

B) 4.56

C) 4.66

D) 4.86

E) 4.96

Diff: 2

Section: 16.7

73) For H3PO3, Ka1 = 1.0 × 10−2 and Ka2 = 2.6 × 10−7. Calculate a value for the [H2PO3−] in a 0.500 molar solution of H3PO3.

A) 1.0 × 10−2 mol L−1

B) 6.6 × 10−2 mol L−1

C) 7.1 × 10−2 mol L−1

D) 8.5 × 10−3 mol L−1

E) 3.3 × 10−3 mol L−1

Diff: 2

Section: 16.7

74) A buffer solution is prepared by taking 0.250 moles of acetic acid (pKa = 4.76) and 0.400 moles of sodium acetate in sufficient water to make 1.800 liters of solution. Calculate the pH of this solution.

A) 4.56

B) 4.66

C) 4.86

D) 4.96

E) 5.06

Diff: 2

Section: 16.7

75) A buffer solution is prepared by taking 0.400 moles of acetic acid (pKa = 4.76) and 0.250 moles of calcium acetate in sufficient water to make 1.400 liters of solution. Calculate the pH of this solution.

A) 4.46

B) 4.56

C) 4.66

D) 4.86

E) 4.96

Diff: 2

Section: 16.7

76) A buffer solution is prepared by taking 0.250 moles of acetic acid (pKa = 4.76) and 0.400 moles of barium acetate in sufficient water to make 1.400 liters of solution. Calculate the pH of this solution.

A) 4.25

B) 4.45

C) 5.00

D) 5.27

E) 5.35

Diff: 2

Section: 16.7

77) The pKa of acetic acid, HC2H3O2, is 4.76. A buffer solution was made using an unspecified amount of acetic acid and 0.30 moles of NaC2H3O2 in enough water to make 2.00 liters of solution. Its pH was measured as 4.40. How many moles of HC2H3O2 were used?

Hint: The opposite operation of log(x) is 10x.

A) 0.13 mol

B) 0.18 mol

C) 0.60 mol

D) 0.69 mol

E) 1.37 mol

Diff: 3

Section: 16.7

78) The pKa of acetic acid, HC2H3O2, is 4.76. A buffer solution was made using an unspecified amount of acetic acid and 0.30 moles of NaC2H3O2 in enough water to make 1.50 liters of solution. Its pH was measured as 4.50 on a meter. How many moles of HC2H3O2 were used?

Hint: The opposite operation of log(x) is 10x.

A) 0.31 mol

B) 0.55 mol

C) 0.61 mol

D) 1.09 mol

E) 1.21 mol

Diff: 3

Section: 16.7

79) The pKa of acetic acid, HC2H3O2, is 4.76. A buffer solution was made using an unspecified amount of NaC2H3O2 and 0.30 moles of acetic acid in enough water to make 2.00 liters of solution. Its pH was measured as 4.60. How many moles of NaC2H3O2 were used?

Hint: The opposite operation of log(x) is 10x.

A) 0.10 mol

B) 0.21 mol

C) 0.30 mol

D) 0.42 mol

E) 0.44 mol

Diff: 3

Section: 16.7

80) The pKa of acetic acid, HC2H3O2, is 4.76. A buffer solution was made using an unspecified amount of NaC2H3O2 and 0.30 moles of acetic acid in enough water to make 1.50 liters of solution. Its pH was measured as 4.55. How many moles of NaC2H3O2 were used?

Hint: The opposite operation of log(x) is 10x.

A) 0.10 mol

B) 0.18 mol

C) 0.30 mol

D) 0.37 mol

E) 0.49 mol

Diff: 3

Section: 16.7

81) You have 500.0 mL of a buffer solution containing 0.30 M acetic acid (CH3COOH) and 0.20 M sodium acetate (CH3COONa). What will the pH of this solution be after the addition of 20.0 mL of 1.00 M NaOH solution? [Ka = 1.8 × 10−5]

Hint: Adding NaOH will change the amounts of both the acid and the base. Calculate the pH based on the new amounts.

A) 4.74

B) 5.07

C) 4.71

D) 4.56

E) 4.42

Diff: 3

Section: 16.7

82) You have 500.0 mL of a buffer solution containing 0.30 M acetic acid (CH3COOH) and 0.20 M sodium acetate (CH3COONa). What will the pH of this solution be after the addition of 20.0 mL of 1.00 M HCl solution? [Ka = 1.8 × 10−5]

Hint: Adding HCl will change the amounts of both the acid and the base. Calculate the pH based on the new amounts.

A) 4.74

B) 5.07

C) 4.71

D) 4.56

E) 4.42

Diff: 3

Section: 16.7

83) You have 500.0 mL of a buffer solution containing 0.30 M ammonia (NH3) and 0.20 M ammonium chloride (NH4Cl). What will the pH of this solution be after the addition of 20.0 mL of 1.00 M HCl solution? [Kb = 1.8 × 10−5]

Hint: Adding HCl will change the amounts of both the acid and the base. Calculate the pH based on the new amounts.

A) 4.74

B) 9.26

C) 9.58

D) 9.20

E) 9.29

Diff: 3

Section: 16.7

84) You have 500.0 mL of a buffer solution containing 0.30 M ammonia (NH3) and 0.20 M ammonium chloride (NH4Cl). What will the pH of this solution be after the addition of 20.0 mL of 1.00 M NaOH solution? [Kb = 1.8 × 10−5]

Hint: Adding NaOH will change the amounts of both the acid and the base. Calculate the pH based on the new amounts.

A) 4.74

B) 9.26

C) 9.58

D) 9.20

E) 9.29

Diff: 3

Section: 16.7

85) Which one of the following is a buffer solution?

A) 0.10 M NaCl

B) 0.40 M HNO3 and 0.10 M NaNO3

C) 0.40 M KF and 0.10 M HF

D) 0.40 M CH3COONa

E) 0.40 M HBr and 0.10 M NaBr

Diff: 1

Section: 16.7

86) Which one of the following cannot act as a buffer solution?

A) 0.40 M HNO3 and 0.10 NaNO3

B) 0.20 M NH3 and 0.25 M NH4Cl

C) 0.50 M HCN and 0.10 NaCN

D) 1.0 M KF and 1.0 M HF

E) 0.10 M HC2H3O2 and 0.10 M NaC2H3O2

Diff: 1

Section: 16.7

87) Which one of the following cannot act as a buffer solution?

A) 0.40 M HNO2 and 0.10 NaNO2

B) 0.20 M NH3 and 0.25 M NH4Cl

C) 0.50 M HCN and 0.10 NaCN

D) 1.0 M KCl and 1.0 M HCl

E) 0.10 M HC2H3O2 and 0.10 M NaC2H3O2

Diff: 1

Section: 16.7

88) Which one of the following is a buffer solution?

A) 0.50 M CH3NH3Cl

B) 0.10 M H2SO4 and 0.10 NaHSO4

C) 0.50 M Ca(OH)2 and 0.10 KCl

D) 1.0 M NaOCN and 0.2 M HOCN

E) 0.10 M NH4NO2

Diff: 1

Section: 16.7

89) Assuming equal concentrations of conjugate base and acid, which of the following mixtures is suitable for making a buffer solution with an optimum pH in the range of 9.1-9.5?

A) CH3NH2/CH3NH3Cl (Ka = 2.3 × 10−11)

B) NaNO3/HNO3

C) NaOCN/HOCN (Ka = 2.0 × 10−4)

D) NaNO2/HNO2 (Ka = 4.6 × 10−4)

E) KCN/HCN (Ka = 4.9 × 10−10)

Diff: 1

Section: 16.7

90) For H3PO3, which is actually a diprotic acid, Ka1 = 1.0 × 10−2 and Ka2 = 2.6 × 10−7. Calculate a value for the [H+] in a 0.500 molar solution of H3PO3.

A) 1.0 × 10−2 mol L−1

B) 6.6 × 10−2 mol L−1

C) 7.1 × 10−2 mol L−1

D) 8.5 × 10−3 mol L−1

E) 3.3 × 10−3 mol L−1

Diff: 2

Section: 16.8

91) For H3PO3, which is actually a diprotic acid, Ka1 = 1.0 × 10−2 and Ka2 = 2.6 × 10−7. Calculate a value for the [HPO32−] in a 0.500 molar solution of H3PO3.

A) 2.6 × 10−7 mol L−1

B) 6.6 × 10−2 mol L−1

C) 7.1 × 10−2 mol L−1

D) 8.5 × 10−5 mol L−1

E) 3.3 × 10−3 mol L−1

Diff: 2

Section: 16.8

92) Selenous acid (H2SeO3) is a diprotic acid, for which Ka1 = 4.5 × 10−3 and Ka2 = 5.0 × 10−8. Determine the concentration of selenite (SeO3−2) in a 0.21 M selenous acid solution.

A) 4.5 × 10−3 M

B) 2.1 × 10−2 M

C) 5.0 × 10−8 M

D) 0.10 M

E) 9.5 × 10−11 M

Diff: 2

Section: 16.8

93) For H3PO4, a tripotic acid, Ka1 = 7.1 × 10−3, Ka2 = 6.3 × 10−8, and Ka3 = 4.5 × 10−13. What is the pH of a solution that contains 0.200 moles of K2HPO4 and 0.300 moles of Na3PO4 per liter of solution?

Hint: Treat this as a buffer and use the Ka that matches the given solutions.

A) 7.20

B) 7.34

C) 12.17

D) 12.35

E) 12.52

Diff: 3

Section: 16.8

94) For H3PO4, a tripotic acid, Ka1 = 7.1 × 10−3, Ka2 = 6.3 × 10−8, and Ka3 = 4.5 × 10−13. What is the pH of a solution that contains 0.300 moles of Na2HPO4 and 0.200 moles of Na3PO4 per liter of solution?

Hint: Treat this as a buffer and use the Ka that matches the given solutions.

A) 7.20

B) 7.34

C) 12.17

D) 12.35

E) 12.52

Diff: 3

Section: 16.8

95) For H3PO4, a tripotic acid, Ka1 = 7.1 × 10−3, Ka2 = 6.3 × 10−8, and Ka3 = 4.5 × 10−13. What is the pH of a solution which contains 0.400 moles of Na2HPO4 and 0.300 moles of K3PO4 per liter of solution?

Hint: Treat this as a buffer and use the Ka that matches the given solutions.

A) 7.20

B) 12.22

C) 12.47

D) 13.09

E) 13.68

Diff: 3

Section: 16.8

96) For H3PO4, a tripotic acid, Ka1 = 7.1 × 10−3, Ka2 = 6.3 × 10−8, and Ka3 = 4.5 × 10−13. What is the pH of a solution that contains 0.300 moles of Na2HPO4 and 0.400 moles of KH2PO4 per liter of solution?

Hint: Treat this as a buffer and use the Ka that matches the given solutions.

A) 7.08

B) 7.20

C) 7.33

D) 10.20

E) 12.22

Diff: 3

Section: 16.8

97) For H3PO4, a tripotic acid, Ka1 = 7.1 × 10−3, Ka2 = 6.3 × 10−8, and Ka3 = 4.5 × 10−13. What is the pH of a solution that contains 0.300 moles of Na2HPO4 and 0.200 moles of NaH2PO4 per liter of solution?

Hint: Treat this as a buffer and use the Ka that matches the given solutions.

A) 7.02

B) 7.38

C) 7.87

D) 8.70

E) 12.52

Diff: 3

Section: 16.8

98) For H3PO4, a tripotic acid, Ka1 = 7.1 × 10−3, Ka2 = 6.3 × 10−8, and Ka3 = 4.5 × 10−13. What is the pH of a solution that contains 0.400 moles of K2HPO4 and 0.300 moles of NaH2PO4 per liter of

solution?

Hint: treat this as a buffer and use the Ka that matches the given solutions.

A) 7.08

B) 7.20

C) 7.33

D) 10.20

E) 12.22

Diff: 3

Section: 16.8

99) Which is the best appropriate acid-base indicator to use in the titration of 0.100 molar HCl(aq) by 0.100 molar NaOH(aq)? The NaOH(aq) is the titrant.

A) thymol blue, for which pKind = 8.8, also HInd is yellow and Ind− is blue

B) bromothymol blue, for which pKind = 6.8, also HInd is yellow and Ind− is blue

C) phenolphthalein, for which pKind = 9.1, also HInd is colorless and Ind− is pink

D) cresol red, for which pKind = 7.9, also HInd is yellow and Ind− is red

E) Any of these indicators will suffice for this titration of a strong acid by a strong base.

Diff: 1

Section: 16.9

100) A mixture contains 0.060 moles of NaH2PO4 and 0.080 moles of K2HPO4. It is titrated with 0.500 molar NaOH(aq) to neutralize it completely. How many mL of the NaOH solution are required?

A) 280 mL

B) 313.3 mL

C) 372 mL

D) 400 mL

E) 440 mL

Diff: 2

Section: 16.9

101) Which is the most appropriate acid-base indicator to use in the titration of 0.100 molar acetic acid(aq) by 0.100 molar NaOH(aq)? This is the titration of a weak acid by a strong base.

A) phenol red, for which pKIn = 7.3; HInd is yellow and Ind− is red

B) bromothymol blue, for which pKIn = 6.8; HInd is yellow and Ind− is blue

C) phenolphthalein, for which pKIn = 9.1; HInd is colorless and Ind− is pink

D) cresol red, for which pKIn = 7.9; HInd is yellow and Ind− is red

E) All of the indicators listed will work equally well for the titration of a weak acid by a strong base.

Diff: 1

Section: 16.9

102) Which is the most appropriate acid-base indicator to use in the titration of 0.100 molar ammonia solution by 0.100 molar HCl(aq)?

A) thymol blue, for which pKIn = 8.8; HInd is yellow and Ind− is blue

B) methyl orange, for which pKIn = 3.8; HInd is red and Ind− is yellow

C) thymolphthalein, for which pKIn = 9.9; HInd is colorless and Ind− is blue

D) cresol red, pKIn = for which 7.9; HInd is yellow and Ind− is red

E) All of the indicators listed will work equally well for the titration of a weak acid by a strong base.

Diff: 1

Section: 16.9

103) A solution is made by taking 100.0 mL of 4.0 × 10−3 molar NaOH(aq) and mixing in 10.0 mL of 4.0 × 10−2 molar HCl(aq). A small portion was tested using thymol blue indicator, for which the pKIn = 8.8, and the colors are yellow for HIn, and blue for In−. What color would be observed?

A) blue

B) green

C) yellow

D) orange

E) red

Diff: 1

Section: 16.9

104) The pH at the equivalence point of the titration of perchloric acid solution by sodium hydroxide solution is

A) < 2.00.

B) 4.00 < pH < 7.00.

C) 7.00.

D) 7.00 < pH < 11.00.

E) > 11.0.

Diff: 1

Section: 16.9

105) The pH at the equivalence point of the titration of nitric acid solution by sodium hydroxide solution is

A) < 2.00.

B) 4.00 < pH < 7.00.

C) 7.00.

D) 7.00 < pH < 11.00.

E) > 11.0.

Diff: 1

Section: 16.9

106) The pH at the equivalence point of the titration of ammonia solution by hydrochloric acid solution is

A) < 2.00.

B) 4.00 < pH < 7.00.

C) 7.00.

D) 7.00 < pH < 11.00.

E) > 11.0.

Diff: 1

Section: 16.9

107) Given the following two solutions:

0.200 M HC2H3O2(aq) with pKa = 4.74

0.200 M NaOH(aq)

A mixture is made using 50.0 mL of the HC2H3O2 solution and 25.0 mL of the NaOH solution. What is the solution's pH?

Hint: After NaOH is added, treat the solution as a buffer.

A) 2.81

B) 4.46

C) 4.74

D) 5.06

E) 12.82

Diff: 3

Section: 16.9

108) Given the following two solutions:

0.200 M HCl(aq)

0.200 M NH3 (aq) with pKb = 4.74

A mixture is made using 50.0 mL of the NH3 solution and 25.0 mL of the HCl solution. What is the solution's pH?

Hint: After HCl is added, treat the solution as a buffer.

A) 8,96

B) 9.56

C) 4.74

D) 9.26

E) 4.44

Diff: 3

Section: 16.9

109) A mixture is made by mixing 50.0 mL of 0.200 M HCl and 50.0 mL of water. What is its pH?

A) 0.70

B) 1.00

C) 1.30

D) 1.61

E) 2.00

Diff: 2

Section: 16.9

110) A solution is made by taking 100.0 mL of 2.0 × 10−2 molar NaOH(aq) and mixing in 50.0 mL of 3.0 × 10−2 molar HCl(aq). A small portion was tested using tyrolian blue indicator, for which the pKIn = 12.5, and the colors are blue for HIn, and yellow for In−. What color would be observed as a result of the testing?

A) blue

B) green

C) yellow

D) orange

E) red

Diff: 2

Section: 16.9

111) A solution is made by taking 2.42 grams of NaCl, 6.44 grams of NaClO4, 3.81 grams of KNO3, and 4.64 grams of KBr in enough water to make 500 mL of solution. A portion was treated with thymol blue indicator, for which the pKIn = 8.8, and the colors are yellow for HIn, and blue for In−. What color should be obtained with the test portion?

A) blue

B) green

C) yellow

D) orange

E) red

Diff: 2

Section: 16.9

112) The pH at the equivalence point of the titration of acetic acid solution by sodium hydroxide solution is

A) < 2.00.

B) 3.50 < pH < 7.00.

C) 7.00.

D) 7.00 < pH < 11.00.

E) > 11.0.

Diff: 2

Section: 16.9

113) 50.0 mL of 0.10 M HCl(aq) was added to 40.0 mL of 0.10 M NaOH(aq) and the mixture was stirred, then tested with a pH meter. What is its pH at 25.0 °C?

A) 1.95

B) 2.35

C) 7.00

D) 12.00

E) 12.05

Diff: 2

Section: 16.9

114) 20.0 mL of 0.10 M H2SO4(aq) was added to 50.0 mL of 0.10 M NaOH(aq) and the mixture was stirred, then tested with a pH meter. What is its pH at 25.0 °C?

A) 1.85

B) 2.00

C) 7.00

D) 11.00

E) 12.15

Diff: 2

Section: 16.9

115) 40.0 mL of 0.20 M HCl(aq) was added to 50.0 mL of 0.20 M NaOH(aq) and the mixture was stirred, then tested with a pH meter. What is its pH at 25.0 °C?

A) 1.65

B) 2.00

C) 8.00

D) 11.00

E) 12.35

Diff: 2

Section: 16.9

116) 40.0 mL of 0.20 M HCl(aq) was added to 50.0 mL of 0.10 M NaOH(aq) and the mixture was stirred, then tested with a pH meter. What is its pH at 25.0 °C?

A) 0.70

B) 1.48

C) 1.65

D) 12.00

E) 12.52

Diff: 2

Section: 16.9

117) Given the following:

0.20 M NaOH(aq)

0.20 M HCl(aq)

A mixture is made using 50.0 mL of the HCl and 50.0 mL of the NaOH.

What is its pH at 25.0 °C?

A) 3.62

B) 4.57

C) 6.30

D) 7.00

E) 9.46

Diff: 2

Section: 16.9

118) Given the following solutions:

0.20 M NaOH(aq)

0.20 M HCl(aq)

A mixture is made using 50.0 mL of the NaOH solution and 25.0 mL of the HCl solution.

What is its pH at 25.0 °C?

A) 7.00

B) 12.52

C) 12.82

D) 13.00

E) 13.30

Diff: 2

Section: 16.9

119) Given the following two solutions:

0.200 M NH3(aq) with pKb = 4.76

0.200 M HCl(aq)

A mixture is made using 25.0 mL of the HCl solution and 50.0 mL of the NH3 solution). What is its pH?

Hint: After mixing the HCl and NH3, treat the resulting solution like a buffer.

A) 1.18

B) 8.87

C) 8.94

D) 9.24

E) 9.54

Diff: 3

Section: 16.9

120) 40.0 mL of 0.10 M HCl(aq) was added to 50.0 mL of 0.10 M NaOH(aq) and the mixture was stirred, then tested with a pH meter. What is its pH at 25.0 °C?

A) 1.95

B) 2.00

C) 7.00

D) 11.90

E) 12.05

Diff: 2

Section: 16.9

121) At a temperature of 50.0 °C, the value of the ion product constant for H2O is 5.5 × 10−14. What is the [H+] in pure water at this temperature?

Diff: 2

Section: 16.1

122) Calculate the H+ ion concentration in lemon juice having a pH of 3.3.

Diff: 1

Section: 16.1

123) In a sample of blood at 25° C, [H+] = 4.6 × 10−8 M. Find the molar concentration of OH−, and

indicate whether the sample is acidic, basic, or neutral.

Diff: 2

Section: 16.1

124) Because rain washes pollutants out of the air and minerals out of the soil, the lakes in many parts of the world have undergone pH changes. In a New England state, the water in one lake was found To have a hydrogen ion concentration of 3.2 × 10−5 mol L−1. What are the calculated pH and pOH values of the lake's water?

Diff: 2

Section: 16.1

125) What is the pH of a sodium hydroxide solution at 25°C in which the hydroxide ion concentration equals 0.0026 M?

Diff: 2

Section: 16.1

126) A food scientist is looking at the acid content of a certain fruit juice. If the [H+] is 6.9 × 10−5, the pH of the juice is ________.

Diff: 2

Section: 16.1

127) What is the pH of a 4.1 × 10−3 M solution of HNO3?

Diff: 2

Section: 16.2

128) What is the pH of a 0.19 M solution of KOH?

Diff: 1

Section: 16.2

129) The pH of a solution made by mixing 30.00 mL of 0.0600 M NaOH with 30.00 mL of water at 25°C is ________.

Diff: 1

Section: 16.2

130) Calculate the pH of a 0.00035 M solution of Ba(OH)2.

Diff: 1

Section: 16.2

131) The Ka of nitrous acid (HNO2) is 7.1 × 10−4. What is the Kb of NO2−?

Diff: 1

Section: 16.3

132) The pKa of a weak acid is 6.50. What is the value of Ka for this acid?

Diff: 1

Section: 16.3

133) Hypochlorous acid, HClO, has an ionization constant with the value: Ka = 3.0 × 10−-8. The value of pKb for the hypochlorite ion, OCl−, which is the conjugate base of hypochlorous acid would be ________.

Diff: 1

Section: 16.3

134) A certain acid has a pKa equal to 4.88. Is this acid stronger or weaker than acetic acid, which has a pKa of 4.74? What is the value of Ka for this acid?

Diff: 2

Section: 16.3

135) Solutions of chlorine bleach contain the hypochlorite ion, OCl−, which is a weak

base. Write the appropriate equation for Kb for this anion.

OCl- + H2O An equilibrim arrow, which has a forward single barbed arrow and a backward single barbed arrow one below the other.
HOCl + OH-

Therefore, the Kb equation is

Kb = ([HOCl] [OH-]/OCl-)

Diff: 1

Section: 16.3

136) A 0.200 M solution of an acid, HA, has a pH = 3.10. What is the value of the ionization constant, Ka, for this acid?

Diff: 2

Section: 16.4

137) A 0.350 M solution of an unknown base has a pH = 10.3. What is the value of the ionization constant, Kb, for the unknown base?

Diff: 2

Section: 16.4

138) A 0.150 M solution of an unknown base has a pH = 11.1. The value of the ionization constant, Kb, for the unknown base is ________.

Diff: 2

Section: 16.4

139) A 0.100 molar solution of a moderately strong monoprotic acid in water has a pH = 1.88.

Calculate the value of Ka, the acid ionization constant for this acid.

Diff: 2

Section: 16.4

140) What is the pH of a 0.0045 M HA (weak acid) solution that is 7.5% ionized?

Diff: 2

Section: 16.4

141) What is the pH of a 0.0115 M HA (weak acid) solution that is 3.4% ionized?

Diff: 2

Section: 16.4

142) The pH of a 0.0085 M HA (weak acid) solution is 4.70. What is the percent ionization of the acid in this solution?

Diff: 2

Section: 16.4

143) What is the pH of a 0.0100 M CH3NH2 solution? The Kb of methylamine is 4.4 × 10−4.

CH3NH2(aq) + H2O(l) An equilibrim arrow, which has a forward single barbed arrow and a backward single barbed arrow one below the other.
CH3NH3+ + OH-

CH3NH2(aq) + H2O(l) An equilibrim arrow, which has a forward single barbed arrow and a backward single barbed arrow one below the other.
CH3NH3+ + OH-

Initial 0.0100 — 0 0

Change -x — +x +x

Equilibrium 0.0100 - x — +x +x

We can make an approximation, and assume x is small. This leaves us with the following relationship:

Kb = ((x) with superscript (2)/0.0100); solving for x, we get 0.0021 M. However, this number is significant with respect to the starting amount of 0.0100. Comparing the two, we see that about 21% of the methylamine solution has ionized. Since this is the case, we need to use the quadratic equation to help us. We use the equation:

Kb = ((x) with superscript (2)/0.0100 - x), which leaves us with two roots: 0.001889 and -0.002329. Since we can't have a negative concentration, the answer is 0.001889. We get the pOH by taking the negative log of this concentration, giving us 2.7. We find the pH by subtracting 2.7 from 14, giving us pH = 11.27.

Diff: 2

Section: 16.5

144) What is the pH of a 0.231 M solution of hydrazine, N2H4? The pKb of hydrazine is 6.02.

N2H4(aq) + H2O(l) An equilibrim arrow, which has a forward single barbed arrow and a backward single barbed arrow one below the other.
N2H5+(aq) + OH-(aq)

Since we're given the pKb of hydrazine, we need to find the Kb. This can be found by taking the antilog of the −pKb, giving us 9.55 × 10−7. Setting up a table:

N2H4(aq) + H2O(l) An equilibrim arrow, which has a forward single barbed arrow and a backward single barbed arrow one below the other.
N2H5+(aq) + OH-(aq)

Initial 0.231 — 0 0

Change -x — +x +x

Equilibrium 0.231 - x — +x +x

Because Kb is very small, we'll neglect x. This gives us: Kb = ((x) with superscript (2)/0.231). Solving for x, we find [OH−] = 4.70 × 10−4. This shows us the hydrazine has only ionized by 0.2%; our approximation is valid. We can find the pOH by taking the negative log of [OH−], giving us 3.33. Subtracting the pOH from 14 tells us the pH is 10.67.

Diff: 2

Section: 16.5

145) A 0.125 M solution of an acid, HA, which has a Ka value of 4.2 × 10−7, has a pH of ________.

Diff: 2

Section: 16.5

146) An aqueous solution of Ca(NO3)2 would result in a solution that would be ________.

Diff: 2

Section: 16.6

147) An aqueous solution of NaOCl would result in a solution that would be ________.

Diff: 2

Section: 16.6

148) What is the pH of a 0.210 M solution of protonated pyridine, C5H5NH+?

The Kb of pyridine is 1.5 × 10−9.

Hint: Use Ka for weak acids and Kb for weak bases.

C5H5NH+(aq) + H2O(l) An equilibrim arrow, which has a forward single barbed arrow and a backward single barbed arrow one below the other.
C5H5N(aq) + H3O+(aq)

Initial 0.210 — 0 0

Change -x — +x +x

Equilibrium 0.210 - x — +x +x

Setting up the Ka equation, we find Ka = ((x) with superscript (2)/0.210 - x). We make the approximation that x is much smaller than 0.210. We solve for x and find x = 1.18 × 10−3. Compared to the starting acid concentration of 0.210, the solution is only 0.2% ionized, so our approximation is valid. The pH of the solution is thus 2.93.

Diff: 3

Section: 16.6

149) What is the pH of a 2.50 molar solution of NH4Cl(aq)? The Kb for NH3 is 1.8 × 10−5.

Hint: Use Ka for weak acids and Kb for weak bases.

Diff: 3

Section: 16.6

150) Consider the following conjugate bases: Cl−, Br−, SO42−, NO2−, ClO4−, ClO−, and C2O42−. Which are weak bases and which are negligible bases?

Diff: 1

Section: 16.6

151) A solution is made by dissolving 15.00 grams of acetic acid and 3.25 grams of sodium acetate in 500.00 mL of water. What is the pH of the solution? The Ka of acetic acid is

1.8 × 10−5.

pH = pKa + log ([A-]/[HA]) . In our example, A− is acetate, and HA is acetic acid. We convert both acetic acid and sodium acetate from grams to moles:

mol acetic acid = 15.00 g acetic acid × = 0.2498 mol acetic acid

mol sodium acetate = 3.25 g sodium acetate × (1 mol sodium acetate/82.03 g sodium acetate)

= 0.0396 mol sodium acetate

The volume of the sample doesn't matter because it will cancel out in the Henderson-Hasselbalch equation.

We are thus left with: pH = 4.74 + log(0.0396 mol/0.2498 mol), and a pH of 3.94.

Diff: 2

Section: 16.7

152) A solution of a molecular base B (0.50 M) and its salt BH+Cl− (0.80 M) has a pH = 9.40. What is the value of Kb for this base, B?

Diff: 2

Section: 16.7

153) A solution of a weak monoprotic acid, HA (0.50 molar), and its potassium salt, KA

(0.75 molar), has a measured pH = 4.88. What is the value of Ka for this acid?

Diff: 2

Section: 16.7

154) Calculate the concentration of oxalate ion (C2O4−2) in a 0.175 M solution of oxalic acid (C2H2O4).

[For oxalic acid, Ka1 = 6.5 × 10−2, Ka2 = 6.1 × 10−5.]

Diff: 2

Section: 16.7

155) How many grams of sodium formate must be added to 100.00 mL of a 0.125 M formic acid solution to achieve a pH of 4.900? The pKa of formic acid is 3.740.

Hint: Use the Henderson-Hasselbalch equation and then dimensional analysis to find the grams of HCOONa.

(10) with superscript (pH - pKa) × [HA] = [A-].

Using this equation, we find [A−] = 1.81 M. Since there is only 100.00 mL of solution, we can now find the mass of sodium formate necessary to achieve the desired pH.

mass sodium formate

= 1.81 (mol sodium formate/L sodium formate) × 0.10000 L solution × (68.01 g sodium formate/1 mol sodium formate)

= 12.3 g sodium formate

Diff: 3

Section: 16.7

156) Hydrosulfuric acid (H2SO3) is a diprotic acid with ionization constants of Ka1 = 5.7 × 10−8 and Ka2 = 1.0 × 10−19.

Determine the concentration of sulfide ions (SO3−-2) in a 0.10 M hydrosulfuric solution.

Diff: 2

Section: 16.8

157) Carbonic acid, H2CO3, is a diprotic acid with ionization constants:

Ka1 = 4.5 × 10−7, Ka2 = 4.7 × 10−11.

Calculate the pH of a 0.100 molar solution of Na2CO3.

Diff: 2

Section: 16.8

158) For H3PO3, which is actually a diprotic acid, Ka1 = 1.0 × 10−2 and Ka2 = 2.6 × 10−7. Calculate the pH of a 0.100 molar solution of Na2HPO3.

Diff: 2

Section: 16.8

159) Selenous acid (H2SeO3) is a diprotic acid, for which Ka1 = 4.5 × 10−3 and Ka2 = 5.0 × 10−8. Determine the concentration of selenite (SeO3−2) in a 0.27 M selenous acid solution.

Diff: 2

Section: 16.8

160) 0.20 M HNO3 is titrated with 0.20 M NH3. At the equivalence point, is the solution acidic, basic, or neutral? (Kb for NH3 = 1.8 × 10−5)

Diff: 2

Section: 16.9

161) 50.00 mL of 0.10 M HNO2 (nitrous acid, Ka = 4.5 × 10−4) is titrated with a 0.10 M KOH solution. What would be the pH of the solution after 25.00 mL of the KOH solution is added?

Hint: After the partial neutralization, treat the solution as a buffer.

Diff: 3

Section: 16.9

162) Potassium hydrogen phthalate, abbreviated KHP, is used to standardize NaOH solutions. A standard KHP solution is made by dissolving 2.12 grams of KHP in 100.00 mL of water. The KHP solution is then titrated with NaOH solution. It takes 23.12 mL of NaOH to reach the endpoint. What is the concentration of the NaOH?

2.12 g KHP ×(1 mol KHP/204.22 g KHP) = 0.0104 mol KHP

To find the concentration of the NaOH solution:

0.104 mol KHP × (1 mol NaOH/1 mol KHP) × (1/0.02312 L) = 4.50 M NaOH

Diff: 2

Section: 16.9

163) The ion product constant for water, Kw, varies with the temperature of the water.

Diff: 2

Section: 16.1

164) A relation involving acids and bases is, Ka + Kb = Kw.

Diff: 2

Section: 16.1

165) The pKa is a reliable measure of the strength of an acid. Strong acids have larger pKa values than weak acids.

Diff: 1

Section: 16.3

166) A solution of sodium cyanide, NaCN, in water should be basic, not neutral or acidic.

Diff: 1

Section: 16.6

167) A solution of ammonium nitrate, NH4NO3, in water should be basic, not neutral or acidic.

Diff: 2

Section: 16.6

168) The titration of a strong acid by a strong base gives sharper and more well-defined end points than titration of a weak acid by a weak base.

Diff: 2

Section: 16.9

169) At 60 °C the value of Kw is 9.5 × 10−14. Considering this, what is the pOH of pure water this temperature?

A) 6.02

B) 6.51

C) 7.00

D) 7.49

E) 9.50

Diff: 2

Section: 16.1

170) At 60 °C the value of Kw is 9.5 × 10−14. Considering this, what is the pOH of a 2.00 × 10−3 M HCl(aq) solution at this temperature?

A) 2.70

B) 6.51

C) 10.32

D) 11.30

E) 13.02

Diff: 2

Section: 16.1

171) At 60 °C the value of Kw is 9.5 × 10−14. Considering this, what is the pH of a 5.00 × 10−2 M Ba(OH)2(aq) solution at this temperature?

Hint: Be sure to start with the balanced equation.

A) 1.00

B) 1.30

C) 12.02

D) 12.70

E) 13.00

Diff: 3

Section: 16.2

172) At 60 °C the value of Kw is 9.5 × 10−14. Considering this, what is the calculated value for the pH of a solution made by dissolving 1.00 g of sodium hydroxide in enough water to make 500 ml of solution at this temperature?

Diff: 2

Section: 16.2

173) The pH of a solution made by mixing 30.00 mL of 0.060 M NaOH with 165.00 mL of water at 25°C is ________.

Diff: 1

Section: 16.2

174) A solution of nitrous acid has a pH of 2.10. What is the initial concentration of nitrous acid in this solution? The Ka of nitrous acid is 7.1 × 10−4.

HNO2(aq) + H2O(l) An equilibrim arrow, which has a forward single barbed arrow and a backward single barbed arrow one below the other.
NO2-(aq) + H3O+(aq)

Initial x — 0 0

Change -0.0079 — +0.0079 +0.0079

Equilibrium x - 0.0079 — +0.0079 +0.0079

We then use the relationship: 7.1 × 10−4 = ((0.0079) with superscript (2)/x - 0.0079). Solving for x, we find the concentration of nitrous acid is 0.097 M.

Diff: 2

Section: 16.5

175) A solution of barbituric acid has a pH of 4.20. What is the concentration of barbituric acid in the solution? The pKa of barbituric acid, HC4H3N2O3, is 4.01.

HC4H3N2O3(aq) + H2O(l) An equilibrim arrow, which has a forward single barbed arrow and a backward single barbed arrow one below the other.
C4H3N2O3-(aq) + H3O+(aq)

Initial x — 0 0

Change -6.30 × 10-5 — +6.30 × 10-5 +6.30 × 10-5

Equilibrium x - 6.30 × 10-5 — +6.30 × 10-5 +6.30 × 10-5

We need the Ka, but we're given the pKa. We can convert between them by:

Ka = (10) with superscript ((-pK) with subscript (a)). This gives us a Ka of 9.77 × 10−5. We then use our standard Ka relationship to solve for the initial acid concentration:

9.77 × (10) with superscript (-5) = (((6.30 × (10) with superscript (-5))) with superscript (2)/x - 6.30 × (10) with superscript (-5)).

Solving for x gives an acid concentration of 1.04 × 10−4 M.)

Diff: 2

Section: 16.5

176) What is the pH of a 0.300 M solution of sodium phenolate, NaC6H5O?

The Ka of phenol is 1.3 × 10−10.

Kb for phenolate is thus 7.7 × (10) with superscript (−5) .

We can now set up a balanced equation and use a table to find what we're looking for.

C6H5O-(aq) + H2O(l) An equilibrim arrow, which has a forward single barbed arrow and a backward single barbed arrow one below the other.
C6H5OH(aq) + OH-(aq)

Initial 0.300 — 0 0

Change -x — +x +x

Equilibrium 0.300 - x — +x +x

Setting up the Kb equation, Kb = ((x) with superscript (2)/0.300 - x), we'll make an approximation that x is smaller than 0.300. This leaves us with Kb = ((x) with superscript (2)/0.300). Solving for x gives 0.00481 as the concentration of OH−. Comparing this concentration to the initial base concentration, we find the solution is only 1.6% ionized, so our approximation is good. The pOH of the solution is thus 2.32, and the pH is 11.68.

Diff: 2

Section: 16.5

177) Assuming equal concentrations of conjugate acid and base, which one of the following mixtures is suitable for making a buffer solution with an optimum pH of 9.2-9.3?

A) CH3COONa/CH3COOH (Ka = 1.8 × 10−5)

B) NaOCl/HOCl (Ka = 3.2 × 10−8)

C) NaCl/HCl

D) NH3/NH4Cl (Ka = 5.6 × 10−10)

E) NaNO2/HNO2 (Ka = 4.5 × 10−4)

Diff: 2

Section: 16.7

178) Assuming equal concentrations of conjugate acid and base, which one of the following mixtures is suitable for making a buffer solution with an optimum pH of 4.6-4.8?

A) CH3COONa/CH3COOH (Ka = 1.8 × 10−5)

B) NH3/NH4Cl (Ka = 5.6 × 10−10)

C) NaOCl/HOCl (Ka = 3.2 × 10−8)

D) NaNO2/HNO2 (Ka = 4.5 × 10−4)

E) NaCl/HCl

Diff: 2

Section: 16.7

179) You are asked to go into the lab and prepare an acetic acid [Ka = 1.8 × 10−5]/sodium acetate buffer solution with a pH of 4.00 ± 0.02. What molar ratio of CH3COOH to CH3COONa should be used?

Diff: 2

Section: 16.7

180) The pKa for HC2H3O2(aq) is 4.76. What should be the pH of a mixture containing 100.0 mL of

0.300 M NaC2H3O2(aq), 100 mL of 0.300 M HC2H3O2(aq), and 100 mL of 0.100 M NaOH?

Hint: After neutralizing the acid and base, treat this solution as a buffer. Organizing your information will be just as important as applying chemistry concepts.

Diff: 3

Section: 16.7

181) The pKb for NH3(aq) is 4.76. What should be the pH of a mixture containing 100.0 mL of 0.300 M NH4Cl(aq), 100 mL of 0.300 M NH3(aq), and 100 mL of 0.100 M HCl(aq)?

Hint: After neutralizing the acid and base, treat this solution as a buffer. Organizing your information will be just as important as applying chemistry concepts.

Diff: 3

Section: 16.7

182) The pKa for HC2H3O2(aq) is 4.76. What should be the pH of a mixture containing

100.0 mL of 0.300 M NaC2H3O2(aq), 100 mL of 0.300 M HC2H3O2(aq), and 100 mL

of 0.100 M HCl(aq)?

Hint: After neutralizing the acid and base, treat this solution as a buffer. Organizing your information will be just as important as applying chemistry concepts.

Diff: 3

Section: 16.7

183) A mixture contains 400.0 mL of 1.00 M NH3(aq), whose pKb is 4.76, and 400.0 mL

of 1.00 M NH4Cl(aq). 200.0 mL of 0.0500 M NaOH(aq) were added to the mixture. What was the pH of the mixture before addition of the base and after addition of the base?

Hint: Calculate the pH of this buffer solution once before the acid and base are neutralized, and again once after. Organizing your information will be just as important as applying chemistry concepts.

Diff: 3

Section: 16.7

184) A mixture contains 400.0 mL of 1.00 M NH3(aq), whose pKb is = 4.76, and 400.0 mL of 1.00 M NH4Cl(aq). 200.0 mL of 0.0500 M HCl(aq) were added to the mixture. What is the pH of the buffer before addition of the acid and after addition of the acid?

Hint: Calculate the pH of this buffer solution once before the acid and base are neutralized, and again once after. Organizing your information will be just as important as applying chemistry concepts.

Diff: 3

Section: 16.7

185) A mixture contains benzoic acid, a monoprotic acid (Ka = 6.28 × 10−5, 0.250 M), and sodium benzoate (0.400 moles/liter). How many mL of a HCl(aq) solution whose pH is 0.523 should be added to 500 mL of this buffer to change its pH from what it is to pH = 4.20?

Hint: Determine the pH of the original buffer, the original [base]/[acid] ratio, and then the target [base]/[acid] ratio.

Diff: 3

Section: 16.7

186) Carbonic acid, H2CO3, is a diprotic acid with ionization constants:

Ka1 = 4.5 × 10−7, Ka2 = 4.7 × 10−11.

What is the value of the ratio, [H2CO3]/[HCO3−], in a solution containing both species that is maintained at pH = 7.00 by means of a buffer?

A) 1.00

B) 0.33

C) 0.25

D) 0.22

E) 1.33

Diff: 2

Section: 16.7

187) A student does a titration of NH3(aq)(pKb = 4.76) with HCl(aq). When 100.0 mL of the HCl solution had been added to 500.0 mL of 0.300 M NH3(aq), he measured the pH of the mixture and found it was 8.94. What was the molarity of the HCl solution?

Hint: Treat this solution like a buffer and remember that adding HCl simultaneously increases NH4+ and decreases NH3.

Diff: 3

Section: 16.8

188) A student does a titration of NH3(aq) (pKb = 4.76) with HCl(aq). When 100.0 mL of the HCl solution had been added to 500.0 mL of 0.300 M NH3(aq), she measured the pH of the mixture and found it was 9.54. What was the molarity of the HCl solution?

Hint: Treat this solution like a buffer and remember that adding HCl simultaneously increases NH4+ and decreases NH3.

Diff: 3

Section: 16.9

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Document Type:
DOCX
Chapter Number:
16
Created Date:
Aug 21, 2025
Chapter Name:
Chapter 16 Acid-Base Equilibria In Aqueous Solutions
Author:
Neil D. Jespersen

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