- By
**chase** - Follow User

- 126 Views
- Uploaded on

Download Presentation
## PowerPoint Slideshow about 'Applications of Aqueous Equilibria' - chase

**An Image/Link below is provided (as is) to download presentation**

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

### Applications of Aqueous Equilibria

Chapter 8

E-mail: benzene4president@gmail.com

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

Applications of Aqueous Equilibria– ch. 8

1. What is the pH at the equivalence point for the following titrations?

a. NaOH with HBr

pH = 7 pH > 7 pH < 7

b. HCl with NaCH3COO

pH = 7 pH > 7 pH < 7

c. KOH with HCN

pH = 7 pH > 7 pH < 7

Applications of Aqueous Equilibria– ch. 8

2. Which of the following will result in a buffer solution upon mixing?

a. 0.1 mol HClO3 and 0.1 mol NaClO3 are put into 1L ofwater

b. 0.5 mol H2S and 0.8 molNaHS are put into 1L of water

c. 0.03 mol KHC2O4 and 0.02 mol K2C2O4 are put into 1L of water

d. 0.1 molLiOH and 0.2 mol H3PO4 are put into 1L of water

e. 0.1 mol HNO3 and 0.04 mol NH3 are put into 1 L of water

Applications of Aqueous Equilibria– ch. 8

Buffers⇒ Solutions that can resist change in pH

Composed of a weak acid and it’s conjugate base

3 ways to make a buffer:

1. Mix a weak acid and it’s conjugate base

2. Mix a strong base with a weak acid – the strong base

must be the limiting reagent

3. Mix a strong acid with a weak base – the strong acid

must be the limiting reagent

Applications of Aqueous Equilibria– ch. 8

Buffers are classic examples of Le Chatlier’s Principle

Applications of Aqueous Equilibria– ch. 8

3. Which of the following buffer solutions can “absorb” the most acid without changing the pH? (All solutions have the same volume)

a. 0.4 M HF /0.5 M NaF

b. 0.8 M HF /0.7 M NaF

c. 0.3 M HF/0.7 M NaF

Applications of Aqueous Equilibria– ch. 8

4. Calculate the pH of the following:

a. A 35 mL solution that contains 0.2 M HN3 (Ka = 1.9 x 10-5) and 0.1 M NaN3

b. A solution prepared by mixing 30 mL of 0.2 M HN3 with 80 mL of 0.1 M NaN3

Applications of Aqueous Equilibria– ch. 8

Henderson-Hasselbalch (Buffer) Equation

pH = pKa + log

or

pH = pKa + log

Applications of Aqueous Equilibria– ch. 8

5. Calculate the pH for the following titrations:

a. 25 mL of 0.1 M HCl is mixed with 25 mL of 0.2 M KNO2

(Kb = 2.5 x 10-11)

b. 50 mL of 0.1 M HCl is mixed with 25 mL of 0.2 M KNO2

c. 60 mL of 0.1 M HClis mixed with 25 mL of 0.2 M KNO2

d. Draw a titration curve and label 1) the equivalence point, 2) the region with maximum buffering, 3) where pH=pKa, 4) the buffer region, 5) where the pH only depends on [HA] and 6) where the pH only depends on [A-]

Applications of Aqueous Equilibria– ch. 8

3 possible scenarios when titrating a weak acid with a strong base or titrating a weak base with a strong acid

1. Buffer ⇒nweak > nstrong

use pH = pKa + log

2. Equivalence point ⇒nweak= nstrong

pH only depends on the conjugate of the weak acid or weak base

use Ka = or Kb =

to solve for unknown

3. Beyond the equivalence point ⇒nweak< nstrong

pH depends on the [xs strong acid] or [xs strong base]

Applications of Aqueous Equilibria– ch. 8

Titration Curves

Titration of a weak acid

with a strong base

Titration of a weak base

with a strong acid

>7

<7

pH

pH

Buffer

Zone

Volume of strong base

Volume of strong acid

Applications of Aqueous Equilibria– ch. 8

6. Calculate the pH of the following titrations:

a. 40 mL of 0.20 M KOH is mixed with 40 mL of 0.5 M HClO (Ka = 3 x 10-8)

b. 80 mL of 0.25 M KOH is mixed with 40 mL of 0.5 M HClO

c. 100 mL of 0.25 M KOH is mixed with 40 mL of 0.5 M HClO

d. Draw a titration curve and label 1) the equivalence point, 2) the region with maximum buffering, 3) where pH=pKa, 4) the buffer region, 5) where the pH only depends on [HA] and 6) where the pH only depends on [A-]

Applications of Aqueous Equilibria– ch. 8

7. Calculate the pH for the following:

a. 10 mL of 1 M HCl is added to a 100 mL buffer with 0.5 M CH3CH2COOH (Ka = 1.34 x 10-5) and 0.6 M CH3CH2COONa.

b. 20 mL of 0.2 M KOH is added to a 100 mL buffer with 0.5 M CH3CH2COOH and 0.6 M CH3CH2COONa.

Applications of Aqueous Equilibria– ch. 8

8. How would you prepare 1.0 L of a buffer at pH = 9.0 from 1.0 M HCN (Ka = 6.2 x 10-10) and 1.5 M NaCN?

Applications of Aqueous Equilibria– ch. 8

9. How many grams of NaOH need to be added to a 50 mL of 0.3 M HNO2 (Ka = 4.0 x 10-4) to have a solution with a pH = 4?

Applications of Aqueous Equilibria– ch. 8

10. Determine the solubility in mol/L and g/L for the following compounds:

a. BaCO3 (Ksp = 1.6 x 10-9)

b. Ag2S (Ksp = 2.8 x 10-49)

Applications of Aqueous Equilibria– ch. 8

Solubility ⇒the maximum amount of solute that can dissolve into a given amount of solvent at any one temperature at this point the solution is said to be saturated and in equilibrium

Applications of Aqueous Equilibria– ch. 8

11. Determine the Ksp values for the following compounds:

a. Al(OH)3 (solubility = 5 x 10-9mol/L)

b. MgF2 (solubility = 0.0735 g/L)

Applications of Aqueous Equilibria– ch. 8

12. What is the solubility of Ca3(PO4)2 (Ksp = 1 x 10-54) in 0.02M solution of Na3PO4?

Applications of Aqueous Equilibria– ch. 8

13. What is the solubility of Zn(OH)2 (Ksp = 4.5 x 10-17) in a solution with pH of 11?

Applications of Aqueous Equilibria– ch. 8

14. Will BaCrO4 (Ksp = 8.5 x 10-11) precipitate when 200 mL of 1 x 10-5 M Ba(NO3)2 is mixed with 350 mL of 3 x 10-5 M KCrO4?

Applications of Aqueous Equilibria– ch. 8

You have completed ch. 8

Answer Key – ch. 8

1. What is the pH at the equivalence point for the following titrations?

a. NaOH(strong base) with HBr(strong acid)

pH = 7 pH > 7 pH < 7

b. HCl(strong acid) with NaCH3COO (weak base)

pH = 7 pH > 7 pH < 7

c. KOH (strong base) with HCN (weak acid)

pH = 7 pH > 7 pH < 7

Answer Key – ch.8

2. Which of the following will result in a buffer solution upon mixing?

a. 0.1 mol HClO3 and 0.1 mol NaClO3 are put into 1L ofwater

Strong acid and it’s conjugate base is not a buffer

b. 0.5 mol H2S and 0.8 molNaHS are put into 1L of water

weak acid and it’s conjugate base is a buffer

c. 0.03 mol KHC2O4 and 0.02 mol K2C2O4 are put into 1L of water

weak acid and it’s conjugate base is a buffer

d. 0.1 molLiOH and 0.2 mol H3PO4 are put into 1L of water strong base and weak acid will be a buffer as long as the strong base is the limiting reagent

e. 0.1 mol HNO3 and 0.04 mol NH3 are put into 1 L of water

strong acid and weak base could be a buffer if the strong as is limiting – however in this case the weak base is limiting so no buffer

Answer Key – ch. 8

3. Which of the following 50 mL solutions can absorb the most acid without changing the pH?

a. 0.4 M HF /0.5 M NaF

b. 0.8 M HF /0.7 M NaF

c. 0.3 M HF/0.7 M NaF

In order for a buffer to absorb an acid you want

[base] to be high and the [acid] to be low and vice

versa if absorbing base

Answer Key – ch. 8

4. Calculate the pH of the following.

a. 0.2 M HN3 (Ka = 1.9 x 10–5) /0.1 M NaN3

HN3/N3 – ⇒weak acid/conjugate base ⇒ buffer

pH = pKa + log

pH = - log (1.9 x 10–5) + log

pH = 4.4

b. 30 mL of 0.2 M HN3/ 80 mL of 0.1 M NaN3

mixing solutions causes dilution so we can use

M1V1 = M2V2 to get the new concentrations or we can alter the Henderson Hasselbalcheqn (HH)⇒

pH = pKa + log

… continue to next slide

Answer Key – ch. 8

4. b. …continued

nA- ⇒ (80 mL)(0.1 mol/L) = 8 mmol

nHA⇒ (30 mL)(0.2 mol/L) = 6 mmol

pH = - log(1.9 x 10–5) + log(8/6)

pH = 4.8

Answer Key – ch.8

5. Calculate the pH for the following:

In all 3 scenarios we’re adding a strong acid (HCl) to a salt with a weak base (NO2–) ⇒ since we have a strong substance (HCl) we can assume the neutralization reaction goes to completion ⇒ work stoichiometrically in moles

a. 25 mL of 0.1 M HCl /25 mL of 0.2 M KNO2

Limiting

Reagent

2.5 mmol H+

5 mmol NO2–

After the neutralization rxnis complete

there’sHNO2/NO2–present ⇒ buffer

using the altered HH eqn⇒

pH = pKa + log

pH = - log(4 x 10–4) + log(2.5/2.5)

pH = 3.4

note since the [HA] = [A–] ⇒ pH = pKa

Answer Key – ch. 8

5. b. 50 mL of 0.1 M HCl /25 mL of 0.2 M KNO2

5 mmol H+

5 mmol NO2–

Note ⇒

Equivalence

Point

- After the neutralization rxn
- is complete there’s only a
- weak acid (HNO2) in
- solution ⇒ change mmol to M
- [HNO2] = = 0.067 M
- Use Ka to solve for (x)
- 4x10–4=
- x = 0.00518 = [H3O+]
- pH = -log(0.00518)
- pH = 2.3

Answer Key – ch. 8

c. 60 mL of 0.1 M HCl/25 mL of 0.2 M KNO2

6 mmol H+

5 mmol NO2–

After the neutralization rxn is

complete there is strong acid (H+)

and weak acid (HNO2) present

A weak acid in the presence of a

strong acid becomes insignificant

[H+] = = 0.0118 M

pH = - log(0.0118)

pH = 1.9

Limiting

Reagent

Answer Key – ch. 8

6. Calculate the pH of the following:

In all 3 scenarios we’re adding a strong base (OH–) to a weak acid (HClO)⇒ since we have a strong substance (OH–) we can assume the neutralization reaction goes to completion ⇒ work stoichiometrically in moles

a. 40 mL of 0.2 M KOH/40 mL of 0.5 M HClO

8 mmol OH–

20 mmolHClO

Limiting

Reagent

After the neutralization rxn is complete

there’s HClO/ClO–present ⇒ buffer

using the altered HH eqn⇒

pH = pKa + log

pH = -log(3.5x10–8) + log

pH = 7.3

Answer Key – ch. 8

6. b. 80 mL of 0.25 M KOH/40 mL of 0.5 M HClO

Note ⇒

Equivalence

Point

20 mmol OH–

20 mmolHClO

- After the neutralization rxn
- there’s only weak base (ClO–) in
- solution ⇒ change mmol to M
- [ClO–] = = 0.167M
- Need Kbto solve for (x)
- Kb= = = 2.86x10–7
- 2.86x10–7=
- x = 2.18x10–4= [OH–]
- pOH = -log(2.18x10–4) = 3.7
- pH = 14 – 3.7 = 10.3

Answer Key – ch. 8

6. c. 100 mL of 0.25 M KOH/40 mL of 0.5 M HClO

25 mmol OH–

20 mmolHClO

- After the neutralization rxn is
- complete there’s strong base
- (OH–) and weak base (ClO–)
- present ⇒ A weak base
- is insignificant in the presence
- of a strong base ⇒
- [OH–] = = 0.0357 M
- pOH = - log(0.0357) = 1.45
- pH = 14 – 1.45
- pH = 12.5

Limiting

Reagent

Answer Key – ch. 8

7. Calculate the pH of the following:

a. 10 mL of 1 M HCl is added to a 100 mL buffer with 0.5 M CH3CH2COOH (pKa = 4.87) and 0.6 M CH3CH2COONa. A strong acid (HCl) will react with the base in the buffer (CH3CH2COO–)

After the neutralization rxn

is complete the solution is still

a buffer

using the altered HH eqn⇒

pH = pKa + log

pH = 4.87 + log

pH = 4.79

Answer Key – ch. 8

7 b. 20 mL of 0.2 M KOH is added to a 100 mL buffer with 0.5 M CH3CH2COOH and 0.6 M CH3CH2COONa.The strong base (OH–) will react with the acid in the buffer (CH3CH2COOH)

After the neutralization rxn

is complete the solution is still

a buffer

using the altered HH eqn⇒

pH = pKa + log

pH = 4.87 + log

pH = 4.98

Answer Key – ch. 8

8. How would you prepare 1.0 L of a buffer at pH = 9.0 from 1.0 M HCN (Ka = 6.2 x 10-10) and 1.5 M NaCN?

The total volume of the buffer has to be 1.0 L => so if we designate X as the volume of HCN then the volume of NaCN must be 1-X => so the moles of HCN are (1.0 M)(X L) = X moles and the moles of NaCN are (1.5 M)(1-X) = 1.5-1.5X moles => now we can plug these into pH = pKa + log

9.0 = -log(6.2 x 10-10)+log => X=0.71

Therefore the buffer is made by mixing 0.71 L of 1.0M HCN with 0.29 L of 1.5M NaCN

Answer Key – ch. 8

9. How many grams of NaOH need to be added to a 50 mL of 0.3 M HNO2 (Ka = 4.0 x 10-4) to have a solution with a pH = 4?

pH = pKa + log

4 = -log(4.0 x 10-4)+log

x = 1.2 mmol of NaOH

Molar mass of NaOH = 40 g/mol

(1.2 mmol)(40 g/mol) = 48 mg of NaOH

Answer Key – ch. 8

10. Determine the solubility in mol/L and g/L for the following compounds:

a. BaCO3 (Ksp = 1.6 x 10-9) ⇒ the solubility is defined as the maximum amount of solute that can be dissolved in a particular amount of solvent at any one temperature or the saturation point ⇒ the solute is at equilibrium for saturated solutions

Use Ksp to solve for x

1.6 x 10-9 = (x)(x)

x = 4 x 10–5

since the molar ratio is 1:1

the molar solubility of BaCO3

= 4 x 10–5 mol/L

or

solubility = (4 x 10–5 mol/L)(197.34 g/mol)

= 0.0079 g/L

Answer Key – ch. 8

10. b. Ag2S (Ksp = 2.8 x 10-49)

Use Ksp to solve for x

2.8 x 10-49 = (2x)2(x)

x = 4.14 x 10–17

Molar solubility = 4.14 x 10–17 M

or

solubility = (4.14 x 10–17mol/L)(247.8 g/mol)

= 1 x 10–14 g/L

Answer Key – ch. 8

11. Determine the Ksp values for the following compounds:

a. Al(OH)3 (solubility = 5 x 10-9mol/L) ⇒the solubility tells us how much will dissolve in order to get to equilibrium

Ksp = (5 x 10-9)(1.5 x 10-8)3 = 1.7 x 10–32

Answer Key – ch. 8

11.b. MgF2 (solubility = 0.0735 g/L)⇒ first we need the molar solubility (0.0735 g/L)/(62.31 g/mol) = 0.00118 mol/L

Ksp = (0.00118)(0.0024)2 = 6.6 x 10–9

Answer Key – ch. 8

12. What is the solubility of Ca3(PO4)2 (Ksp = 1 x 10-54) in 0.02M solution of Na3PO4? The solute and the solution have the phosphate ion in common ⇒ this will result in lower solubility than if the solute were to be dissolved in pure water aka the common ion effect

Use Ksp to solve for x

1 x 10-54 = (3x)3(0.02)2

x = 4.5 x 10–18

molar solubility = 4.5 x 10–18M

Insignificantly

small

Answer Key – ch. 8

13. What is the solubility of Zn(OH)2 (Ksp = 4.5 x 10-17) in a solution with pH of 11? The solute and the solution have the hydroxide ion in common ⇒ this will result in lower solubility than if the solute were to be dissolved in pure water aka the common ion effect ⇒ since the pH is 11 the pOH is 3 ⇒ [OH-] = 10–3 M

Use Ksp to solve for x

4.5 x 10-17= (x)(10–3)2

x = 4.5 x 10-11

molar solubility = 4.5 x 10-11M

Insignificantly small

Answer Key – ch. 8

14. Will BaCrO4 (Ksp = 8.5 x 10-11) precipitate when 200 mL of 1 x 10-5 M Ba(NO3)2 is mixed with 350 mL of 3 x 10-5 M KCrO4?

The solution is saturated with BaCrO4 if [Ba2+][CrO42–] = 8.5 x 10-11

however if [Ba2+][CrO42–] > 8.5 x 10-11 the solution is supersaturated

and will precipitate out some BaCrO4or if [Ba2+][CrO42–] < 8.5 x 10-11

the solution is unsaturated and there will be no noticeable change

We can use M1V1 = M2V2 to get the new concentrations

[Ba2+] = (1 x 10–5 M)(200 mL)/(550 mL) = 3.64 x 10–6 M

[CrO42–]= (3 x 10-5 M)(350 mL)/(550 mL) = 1.91 x 10 -5 M

[Ba2+][CrO42–] = (3.64 x 10–6 M )(1.91 x 10 -5 M)= 6.95 x 10 – 11

the solution is unsaturated and no precipitate will form

Download Presentation

Connecting to Server..