Acids and Bases Chapter 7 E-mail: benzene4president@gmail.combenzene4president@gmail.com Web-site:

Acids and BasesChapter 7 E-mail: benzene4president@gmail.com

Web-site: http://clas.sa.ucsb.edu/staff/terri/

Acids and Bases – ch. 7 1. Arrange the following solutions in order of most acidic to most

basic.

I. [OH– ] = 0.5 M

II. [H+ ] = 0.3 M

III. pOH = 5.9

IV. pH = 1.2

V. [H+ ] = 1.0x10–4M

Acids and Bases – ch. 7

[H3O+]

[OH-]

pH

pOH

Kw = [H3O+][OH-] pKw = pH + pOH

pH = -log[H3O+]

[H3O+] = 10-pH

[OH-] = 10-pOH

pOH = -log[OH-]

Acids and Bases – ch. 7 2. At 25°C the water ionization constant is 1.0 x 10–14 and at 98°C the

water ionization constant is 6.8 x 10–14. What is the pH of neutral water at both temperatures?

Acids and Bases – ch. 7 3. a. Which of the following is a stronger acid?

HNO2 (Ka = 4.0 x 10–4) or HCN (pKa = 9.21)

b. Which is a stronger base?

NO2– or CN–

Acids and Bases – ch. 7

As acid strength ↑ % ionization ↑ Ka ↑ pKa ↓

As base strength ↑ % ionization ↑ Kb ↑ pKb ↓Conjugate pairs are inversely related

As acid strength ↑ conjugate base strength ↓

Kw = (Ka)(Kb)

Acids and Bases – ch. 7 4. Calculate the pH of the following solutions:

a. 0.004 M HBr

b. 0.25 moles of Ba(OH)2 in 7.50 L of solution

Acids and Bases – ch. 7 5. Calculate the pH and the % ionization of the following solutions

a. 0.05 M HClO (Ka = 3.0 x 10–8)

b. 0.33 M CH3COOH (Ka = 1.8 x 10–5) mixed with 0.85 M HCN (Ka = 6.2 x 10–10)

Acids and Bases – ch. 7

6. What concentration of HCOOH (Ka = 1.77x10-4) solution will have a pH of 2.2?

Acids and Bases – ch. 7 7. A solution with 0.5 M of weak mono-protic acid ionizes 11% . What

is the Ka for this acid?

Acids and Bases – ch. 7 8. What concentration of a weak acid that ionizes 0.1% will have a pH

of 3.8?

Acids and Bases – ch. 7 9. What concentration of HNO2 (Ka = 4.0 x 10–4) will have the same

pH as 0.03M HNO3?

Acids and Bases – ch. 7 10. If 0.1M solution of a weak acid has a pH of 3 what will be the pH

of a 0.001M solution of the weak acid?

Acids and Bases – ch. 7 11. Calculate the pH and % ionization for the following:

a. 0.01M NH3 (Kb = 1.8 x 10-5)

b. 0.5 M C6H5NH2 (Kb = 3.8 x 10-10)

Acids and Bases – ch. 7 12. If a 0.35 M solution of weak base has a pH of 11.2 what is the pH

of a 0.08 M solution of the weak base?

Acids and Bases – ch. 7 13. Rank the following 0.1 M salt solutions in order of increasing pH.

i. NaClO4

ii. LiF

iii. C5H5NHI

iv. NH4Cl

v. KCN

HF Ka = 7.2 x 10-4

C5H5N Kb = 1.7 x 10-9

NH3 Kb = 1.8 x 10-5

HCN Ka = 6.2 x 10-10

Acids and Bases – ch. 7

Salts ⇒ soluble ionic compounds that are the result of an acid base reaction ⇒ when a salt is dissolved in water each ion can potentially affect the pH

of the solution ⇒ you need to analyze the ions individually

CationsGroup 1 and 2

metal ions ⇒ neutralAll other cations ⇒ acids

Anions1st 7 ⇒ neutral

All other anions ⇒ bases

Salts

Acids and Bases – ch. 714. Will a solution of NH4F be acidic, basic or neutral?

NH3 Kb = 1.8 x 10-5

HF Ka = 7.2 x 10-4

Acids and Bases – ch. 7 15. How many grams of KF must be dissolved in 500 mL of water in

order to get a pH of 8.4? (Ka of HF = 7.2 x 10-4)

Acids and Bases – ch. 7 16. When 0.100 mole of an unknown soluble salt is dissolved in 1.00 L

of water, the pH of the solution is 8.07. Assume the volume of the solid salt is negligible. What is the identity of the salt?

A) NaCN

B) NaC2H3O2

C) NaCl

D) NaF

E) NaOCl

Acids and Bases – ch. 7

You have completed ch. 7

Answer Key – ch. 71. Arrange the following solutions in order of most acidic to most basic.

A) [OH– ] = 0.5 M pOH = -log(0.5) = 0.3 pH = 14 - 0.3 = 13.7⇒ ⇒

B) [H+ ] = 0.3 M pH = -log(0.3) = 0.52⇒C) pOH = 5.9 pH = 14-5.9 = 8.1⇒D) pH = 1.2

E) [H+ ] = 1.0x10–4M pH = -log(1.0x10⇒ –4) = 4

As a solution gets more acidic…

[H+ ]↑, pH↓, [OH– ]↓, pOH↑

B > D > E > C > A

Answer Key – ch. 72. At 25°C the water ionization constant is 1.0 x 10–14 and at 98°C the

water ionization constant is 6.8 x 10–14. What is the pH of neutral water at these temperatures?

If a solution is neutral ⇒ [H3O+] = [OH-]

For all aqueous solutions ⇒ [H3O+]x[OH-] = Kw

at 25°C ⇒ x2 = 1.0 x 10–14 ⇒ x = 1.0 x 10-7M = [H3O+] = [OH-]

pH = -log(1.0 x 10-7M)

pH = 7

at 98°C ⇒ x2 = 6.8 x 10–14 => x = 2.61x10-7M = [H3O+] = [OH-]

pH = -log(2.61x10-7M)

pH = 6.58

3. a. Which of the following is a stronger acid?

HNO2 (Ka = 4.0 x 10–4) or HCN (pKa = 9.21)

as Ka ↑ % ionization ↑ or acid strength ↑ pKa ↓

Ka =10-pKa ⇒ Ka for HCN =10-9.21 = 6.2x 10-10

Since Ka for HNO2 > Ka for HCN ⇒ HNO2 is a stronger acid

b. Which is a stronger base?

NO2– or CN–

as acid strength ↑ conjugate base strength ↓

since HNO2 is the stronger acid ⇒ NO2– is the weaker base ⇒

CN– is the stronger base

Answer Key – ch. 7

4. Calculate the pH of the following solutions a. 0.004 M HBr ⇒ Strong acid

[H3O+] = 0.004 M ⇒ pH = - log(0.004) ⇒ pH = 2.4

b. 0.25 moles of Ba(OH)2 in 7.50 L of solution ⇒ Strong base

[OH-]= = 0.067M

pOH = -log(0.067M) = 1.18

pH = 14-1.18 = 12.82

Answer Key – ch. 7

Answer Key – ch. 75. Calculate the pH and the % ionization of the following solutions

a. 0.05 M HClO (Ka = 3.0 x 10–8) ⇒ weak acid

HClO H2O ⇌ H3O+ ClO2–

I 0.05 N/A 0 0

∆ - x N/A + x + x

Eq 0.05- x N/A x x

Use Ka to solve for x3 x 10–8 =

x = 3.87 x 10–5 M[H3O+] = 3.87 x 10–5 M

pH = - log(3.87 x 10–5 M) = 4.41% ionized = x100

% ionized = x100 = 0.077%

Insignificantlysmall

Answer Key – ch. 75. …continued

b. 0.33 M CH3COOH (Ka = 1.8 x 10–5) mixed with 0.85 M HCN (Ka = 6.2 x 10–10)

CH3COOH H2O ⇌ H3O+ CH3COO-

I 0.33 N/A 0 0

∆ - x N/A + x + x

Eq 0.33 - x N/A x x

If you have 2 or more acids only the strongest acid will contribute

significantly to the pHSince acetic acid is stronger (higher Ka)

we can ignore the HCN entirely

1.8 x 10–5 = x = 0.00244M = []

pH = -log(0.00244) = 2.61% ionized = = 0.74%Insignificantly

small

Answer Key – ch. 76. What concentration of HCOOH (Ka = 1.77x10-4) solution will have a

pH of 2.2? Since the pH = 2.2 ⇒ [H3O+] = 10–2.2 M or 0.0063 M ⇒ since the molar ratio is 1:1 ⇒ [H3O+] = [HCOO–]

1.77 x 10-4 = x = 0.231M

HCOOH H2O ⇌ H3O+ HCOOH

I x N/A 0 0

∆ -0.0063 N/A +0.0063 +0.0063

Eq x-0.0063 N/A 0.0063 0.0063

Answer Key – ch. 77. A solution with 0.5 M of weak mono-protic acid ionizes 11% . What

is the Ka for this acid?

Ka = = 6.8x10–3

HA H2O ⇌ H3O+ A–

I 0.5 N/A 0 0

∆ -0.11(0.5) N/A +0.11(0.5) +0.11(0.5)

Eq 0.445 N/A 0.055 0.055

11%

Answer Key – ch. 78. What concentration of a weak acid that ionizes 0.1% will have a pH

of 3.8?

Since the pH is 3.8 ⇒ [H3O+] = 10–3.8 or 1.58x10–4 M

1.58x10–4 = 0.0001x

x = 1.58M

HA H2O ⇌ H3O+ A–

I x N/A 0 0

∆ -0.001x N/A +0.001x +0.001x

Eq x -0.001x N/A 0.001x 0.001x

0.1%

9. What concentration of HNO2 (Ka = 4.0 x 10–4) will have the same pH as 0.03M HNO3?

Same pH ⇒ same [H3O+]

Since HNO3 is a strong acid ⇒ [H3O+] = 0.03M

4.0 x 10–4 =

x = 2.28M

Answer Key – ch. 7

HNO2 H2O ⇌ H3O+ NO2–

I x N/A 0 0

∆ -0.03 N/A +0.03 +0.03

Eq x – 0.03 N/A 0.03 0.03

Answer Key – ch. 710. If 0.1M of a weak acid has a pH of 3 what will be the pH of a

0.001M solution of the weak acid?

Same acid with different concentration will have the same Ka so you can set one Ka equation equal to another

=

If the pH = 3 ⇒ [] = 10–3

=

x = 10–4 M

pH = -log(10–4) = 4

Answer Key – ch. 711. Calculate the pH and % ionization for the following:

a. 0.01M NH3 (Ka of NH4+ = 5.6 x 10-10) NH3 is a weak base so

you need Kb ⇒ Kb = = = 1.79x10–5

1.79x10–5 =

x = 4.23x10–4 = [OH–]

pOH = -log(4.23x10–4)

pOH = 3.37

pH = 14 – 3.37

pH = 10.63

NH3 H2O ⇌ OH- NH4+

I 0.01 N/A 0 0

∆ -x N/A +x +x

Eq 0.01-x N/A x x

% ionized =

Answer Key – ch. 711. …continued

b. 0.5 M C6H5NH2 (Kb = 3.8 x 10-10)

3.8 x 10-10 =

x = 1.38x10–5 = [OH–]

pOH = -log(1.38x10–5)

pOH = 4.86

pH = 14 – 4.86

pH = 9.14

C6H5NH2 H2O ⇌ OH- C6H5NH3+

I 0.5 N/A 0 0

∆ – x N/A +x +x

Eq 0.5 – x N/A x x

% ionized =

Answer Key – ch. 712. If a 0.35 M solution of weak base has a pH of 11.2 what is the pH

of a 0.08 M solution of the weak base?

Same weak base with different concentrations will have the same Kb so you can set one Kb equation equal to another

= If the pH = 11.2 pOH = 2.8 [OH⇒ ⇒ –] = 10–2.8 = 1.58 x 10–3

=

x = 7.58 x 10–4 M

pOH = -log(7.58 x 10–4 ) = 3.12

pH = 10.88

Answer Key – ch. 713. Rank the following 0.1 M salt solutions in order of increasing pH.

i. NaClO4 only contains spectator or neutral ions⇒

ii. LiF F⇒ – is a weak base

iii. C5H5NHI C⇒ 5H5NH+ is a weak acid

iv. NH4Cl ⇒ NH4+

is a weak acid

v. KCN CN⇒ – is a weak base

Since HF is a stronger acid than HCN F⇒ – is a weaker base than CN–

Since NH3 is a stronger base than C5H5N NH⇒ 4+

is a weaker acid than C5H5NH+

C5H5NH+ < NH4+

< NaClO4 < F– < CN–

Answer Key – ch. 714. Will a solution of NH4F be acidic, basic or neutral?

NH4F is a salt that contains both a weak acid (NH4+)

and a weak base (F–)

Since HF is a stronger acid than NH3 as a base F⇒ – is a weaker base

than NH4+ as an acid

pH of the salt will be acidic because NH4+ is the stronger component

Answer Key – ch. 715. How many grams of KF must be dissolved in 500 mL of water in

order to get a pH of 8.4? (Ka of HF = 7.2 x 10-4)

F– is a weak base ⇒ since the pH = 8.4 ⇒ pOH = 5.6 ⇒ [OH–] = 10–5.6 M or 2.51 x 10–6 M

We need Kb to solve for x ⇒

Kb = =

Kb = 1.4 x 10–11

1.4 x 10–11 = ⇒ x = 0.454

[F–] = 0.45 M

(0.454 mol/L)(0.5 L) = 0.227 mol KF ⇒ (0.227 mol KF)(58.1 g/mol) = 13.2 g of KF

F– H2O ⇌OH– HF

I x N/A 0 0

∆ -2.51x10–6 N/A +2.51x10–6 +2.51x10–6

Eq x-2.51x10–6 N/A 2.51x10–6 2.51x10–6

Answer Key – ch. 716. When 0.100 mole of an unknown soluble salt is dissolved in 1.00 L

of water, the pH of the solution is 8.07. Assume the volume of the solid salt is negligible. What is the identity of the salt?

A) NaCN B) NaC2H3O2 C) NaCl D) NaF E) NaOCl

You can eliminate NaCl because it only contains spectator or neutral ions your table has a list of K⇒ a values ⇒ In order to get Ka for the

conjugate acid you need Kb for the weak base ⇒ since the pH = 8.07 ⇒ pOH = 5.93 ⇒ [OH–] = 1.17 x 10–6

A– H2O ⇌ OH– HA

I 0.1 N/A 0 0

∆ N/A +1.17 x 10–6 +1.17 x 10–6

Eq 0.1 N/A 1.17 x 10–6 1.17 x 10–6

continue to next slide…

Answer Key – ch. 7Kb = = 1.37 x 10–7

Ka = = 7.3 x 10–4 HF so the salt must be NaF⇒ ⇒