Chemistry, The Central Science, 10th edition

Theodore L. Brown; H. Eugene LeMay, Jr.;and Bruce E. Bursten

Chapter 17Additional Aspects of Aqueous Equilibria

Troy WoodUniversity of BuffaloBuffalo, NY 2006, Prentice Hall

What is [H+] of a 0.050 M HF solution dissolved in 0.20 M NaF? Ka (HF) = 6.8 104.

1. 1.7 104 M2. 3.4 104 M3. 6.8 104 M

4. 1.4 103 M5. 2.7 103 M

Correct Answer:

The solution to problem is on the following slide.

1. 1.7 104 M2. 3.4 104 M3. 6.8 104 M

4. 1.4 103 M5. 2.7 103 M

Correct Answer:

[HF]

]][F[H

aK

[HF] [H+] [F]

Initial 0.050 ~0 0.20

Change -x +x +x

Eq. 0.050 - x x 0.20 + x

(0.20)

)(0.050)10(6.8

][F

[HF] 4

][H

aK

4101.7][H

Calculate the pH of a 0.50 M solution of sodium acetate in 0.050 M acetic acid (HOAc).

pKa (HOAc) = 4.74.

1. 3.742. 4.243. 4.74

4. 5.245. 5.74

Correct Answer:

[acid]

[base]logppH aK

[0.050]

[0.50]log4.74pH

5.7410log4.74pH

1. 3.742. 4.243. 4.74

4. 5.245. 5.74

Calculate the pH of a buffer solution containing 0.25 moles sodium acetate and 0.30 moles acetic acid (HOAc) to which 0.20 moles HCl are added.

pKa (HOAc) = 4.74.

1. 3.742. 4.243. 4.74

4. 5.245. 5.74

Correct Answer:

[acid]

[base]log4.74pH

0.20][0.30

0.20][0.25log4.74pH

3.74[0.50]

[0.05]log4.74pH

1. 3.742. 4.243. 4.74

4. 5.245. 5.74

What is the pH of an aqueous solution to which 51.0 mL of 0.10 M NaOH have been added to 50.0 mL 0.10 M HCl?

1. 2.002. 2.703. 3.004. 3.305. 4.00

Correct Answer:

1. 2.002. 2.703. 3.004. 3.305. 4.00

The solution to problem is on the following slide.

Correct Answer:

Moles H+ = (0.10 M)(0.0500 L) = 0.0050

Moles OH = (0.10 M)(0.0490 L) = 0.0049

10 1.0 0.049)(0.050

0.0049)(0.0050 ][H 3

pH = log(1.0 103) = 3.00

What is the pH of an aqueous solution to which 26.0 mL of 0.10 M NaOH have been added to 50.0 mL 0.050 M HCl?

1. 10.002. 10.703. 11.004. 11.305. 12.00

Correct Answer:

1. 10.002. 10.703. 11.004. 11.305. 12.00

The solution to problem is on the following slide.

Correct Answer:

Moles H+ = (0.050 M)(0.050 L) = 0.0025

Moles OH = (0.10 M)(0.0260 L) = 0.0026

10 2.0 0.025)(0.026

0.0025)(0.0026 ][OH 3

pOH = log(2.0 103) = 2.70pH = 14.00 2.70 = 11.30

What is the molar solubility of BaF2 if its Ksp = 1.0 106?

1. 3.1 103 M

2. 1.0 102 M

3. 6.3 103 M

4. 8.0 103 M

Correct Answer:

The solution to problem is on the following slide.

1. 3.1 103 M

2. 1.0 102 M

3. 6.3 103 M

4. 8.0 103 M

Correct Answer:

Ksp = [Ba2+][F]2

Let x = [Ba2+]; [F] = 2x.

Ksp = (x)(2x)2 = 4x3

x = (Ksp /4)1/3

x = 6.3 103

What is the molar solubility of CaF2 in

0.010 M NaF? Ksp = 3.9 1011?

1. 9.5 106 M

2. 1.9 107 M

3. 3.9 107M

4. 3.9 109 M

Correct Answer:

The solution to problem is on the following slide.

1. 9.5 106 M

2. 1.9 107 M

3. 3.9 107M

4. 3.9 109 M

Correct Answer:

Ksp = [Ca2+][F]2

Let x = [Ca2+]; [F] = 0.010 + 2x.

Ksp = (x)(0.010)2

x = Ksp /(0.010)2

x = 3.9 107

Which of the following substances will not be more soluble in acidic solution than basic solution? The sinkhole below is a dire result of such “solubility.”

1. AgCl2. Ni(OH)2

3. CaCO3

4. BaF2

Correct Answer:

Of these, only AgCl does not produce a basic anion that will react with H+.

1. AgCl2. Ni(OH)2

3. CaCO3

4. BaF2

Calculate [Ag+] when a NaCN solution is added to 0.10 M AgNO3 (and the equilibrium [CN] = 0.10 M. Kf (Ag(CN)2

) = 1 1021?

1. 1 1017 M

2. 1 1018 M

3. 1 1019 M

4. 1 1020 M

5. 1 1021 M

Correct Answer:

The solution to problem is on the following slide.

1. 1 1017 M

2. 1 1018 M

3. 1 1019 M

4. 1 1020 M

5. 1 1021 M

Correct Answer:

Kf = [Ag(CN)2]/[Ag+][CN]2

Assume: Ag+ initially is converted to Ag(CN)2

[Ag+] = [Ag(CN)2]/Kf [CN]2

[Ag+] = (0.10)/[1 1021](0.1)2

[Ag+] = 1 1018

Upon addition of 6 M HCl to an aqueous solution, a precipitate forms. Which of the following cannot be in the precipitate?

1. AgCl2. Hg2Cl2

3. PbCl2

4. CuCl2

Correct Answer:

Chlorides of Ag+, Hg22+,

and Pb2+ are insoluble.

1. AgCl2. Hg2Cl2

3. PbCl2

4. CuCl2