Applications of Aqueous Equilibria Chapter 8 E-mail:...

Applications of Aqueous Equilibria

Chapter 8E-mail: [email protected]

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

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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 combinations will result in a buffer solution

upon dissolving in 1 L of water? Calculate the pH of the solutions that are buffers?

a. 0.1 mol HClO3 and 0.1 mol NaClO3

b. 0.5 mol H2S and 0.8 mol NaHS

c. 0.03 mol NH4I and 0.02 mol NH3

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 limiting amount of strong base with a weak acid

3. Mix a limiting amount of strong acid with a weak base

pH = pKa + log or pH = pKa + log


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. A 65.0-mL sample of 0.12 M HNO2 (Ka = 4.0 x 10–4 ) is titrated with

0.11 M NaOH. What is the pH after 28.4 mL of NaOH has been added?

A) 10.43

B) 7.00

C) 3.57

D) 3.00

E) 3.22

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]


Volume of strong base

pH

>7

Titration Curves

Titration of a weak acid with a strong base

Titration of a weak base with a strong acid

Volume of strong acid

pH

<7

Buffer Zone

Applications of Aqueous Equilibria– ch. 8 5. A 84.0-mL sample of 0.04 M KNO2 (Ka for HNO2 = 4.0 x 10–4 ) is

titrated with 0.2 M HI. What is the pH after 13.6 mL of HI has been added?

Applications of Aqueous Equilibria– ch. 8 6. A 63.0 mL sample of 0.11 M HCN (Ka = 6.2 x 10–10) is titrated with

0.06 M KOH. What is the pH at the equivalence point?

Applications of Aqueous Equilibria– ch. 8 7. A 75.0 mL of 0.22M NH3 (Kb = 1.8 x 10–5) is titrated with 0.50M

HCl. What is the pH at the equivalence point?

Applications of Aqueous Equilibria– ch. 8 8. A 15.0 mL of 0.40M HF (Ka = 7.2 x 10–4) is titrated with 0.85M

LiOH. What is the pH after 8 mL of LiOH has been added?

Applications of Aqueous Equilibria– ch. 8 9. A 120 mL of 0.14M NaCH3COO is titrated with 0.36 M HCl. What

is the pH after 50 mL of HCl has been added?

Applications of Aqueous Equilibria– ch. 8 10. Draw a titration curve for a weak acid being titrated by a strong

base and label the following points:

a. the equivalence point

b. the region with maximum buffering

c. where pH = pKa

d. the buffer region

e. where the pH only depends on [HA]

f. where the pH only depends on [A-]

Applications of Aqueous Equilibria– ch. 8 11. Draw a titration curve for a weak base being titrated by a strong

acid and label the following points:

a. the equivalence point

b. the region with maximum buffering

c. where pH = pKa

d. the buffer region

e. where the pH only depends on [HA]

f. where the pH only depends on [A-]

Applications of Aqueous Equilibria– ch. 8 12. A solution contains 0.36 M HA (Ka = 2.0 x 10-7) and 0.24 M NaA.

Calculate the pH after 0.04 mol of NaOH is added to 1.00 L of this solution.

a. 6.76

b. 6.52

c. 6.64

d. 6.40

e. 6.70

Applications of Aqueous Equilibria– ch. 8 13. 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)


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 14. Determine the Ksp values for the following compounds:

a. Ag3PO4 (solubility = 1.8 x 10-18 mol/L)

b. MgF2 (solubility = 0.0735 g/L)

Applications of Aqueous Equilibria– ch. 8 15. 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 16. What is the solubility of Zn(OH)2 (Ksp = 4.5 x 10-17) in a solution

with pH of 11? Is Zn(OH)2 more soluble in acidic or basic solutions?

Applications of Aqueous Equilibria– ch. 8 17. 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?


You have completed 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 of water Strong acid and it’s conjugate base is not a buffer

b. 0.5 mol H2S and 0.8 mol NaHS 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 mol LiOH 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


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

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 Hasselbalch eqn (HH)⇒pH = pKa + log


… continue to next slide

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. 85. 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

H+ NO2– HNO2

I 2.5 5 0

∆ -2.5 -2.5 +2.5

F 0 2.5 2.5

2.5 mmol H+ 5 mmol NO2–Limiting

Reagent After the neutralization rxn is complete there’s HNO2/NO2

– present ⇒ bufferusing 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

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


H+ NO2– HNO2

I 5 5 0

∆ -5 -5 +5

F 0 0 5

5 mmol H+ 5 mmol NO2–

Note ⇒Equivalence

Point After the neutralization rxnis 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

HNO2 H2O ⇌ H3O+ NO2–

I 0.067 N/A 0 0

∆ -x N/A +x +x

E 0.067 N/A x x

Answer Key – ch. 8c. 60 mL of 0.1 M HCl/25 mL of 0.2 M KNO2

H+ NO2– HNO2

I 6 5 0

∆ -5 -5 +5

F 1 0 5

6 mmol H+ 5 mmol NO2–

LimitingReagent

After the neutralization rxn is complete there is strong acid (H+)

and weak acid (HNO2) presentA weak acid in the presence of a strong acid becomes insignificant

[H+] = = 0.0118 MpH = - log(0.0118)

pH = 1.9

Answer Key – ch. 86. 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 mmol HClO

OH– HClO H2O ClO–

I 8 20 N/A 0

∆ - 8 -8 N/A +8

F 0 12 N/A 8

LimitingReagent

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



I 20 20 N/A 0

∆ - 20 -20 N/A 20

F 0 0 N/A 20

Note ⇒Equivalence

PointAfter the neutralization rxn

there’s only weak base (ClO–) insolution ⇒ change mmol to M

[ClO–] = = 0.167M

Need Kb to solve for (x)Kb = = = 2.86x10–7

2.86x10–7= x = 2.18x10–4= [OH–]

pOH = -log(2.18x10–4) = 3.7pH = 14 – 3.7 = 10.3

ClO– H2O ⇌ OH– HClO

I 0.167 N/A 0 0

∆ - x N/A +x +x

F 0.167 N/A x x

Answer Key – ch. 86. c. 100 mL of 0.25 M KOH/40 mL of 0.5 M HClO



I 25 20 N/A 0

∆ - 20 -20 N/A 20

F 5 0 N/A 20

LimitingReagent

After the neutralization rxn iscomplete 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) = 1.45pH = 14 – 1.45

pH = 12.5

Answer Key – ch. 87. 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–)

CH3CH2COO– H+ CH3CH2COOH

I 60 10 50 mmol

∆ - 10 - 10 + 10

F 50 0 60

After the neutralization rxn is complete the solution is still

a bufferusing the altered HH eqn ⇒

pH = pKa + log

pH = 4.87 + log pH = 4.79

Answer Key – ch. 87 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)

CH3CH2COOH OH- CH3CH2COO–

I 50 mmol 4 mmol 60 mmol

∆ - 4 - 4 + 4

F 46 0 64

After the neutralization rxn is complete the solution is still

a bufferusing the altered HH eqn ⇒

pH = pKa + log

pH = 4.87 + log pH = 4.98

Answer Key – ch. 88. 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. 89. 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?

HNO2 OH- H2O NO2-

I 15mmol x mmol N/A 0

Δ -x -x N/A +x

F 15-x 0 N/A x

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. 810. 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

BaCO3 ⇌ Ba2+ CO32–

I N/A 0 0

∆ N/A +x +x

Eq N/A x x

Use Ksp to solve for x1.6 x 10-9 = (x)(x)

x = 4 x 10–5

since the molar ratio is 1:1the molar solubility of BaCO3

= 4 x 10–5 mol/Lor

solubility = (4 x 10–5 mol/L)(197.34 g/mol) = 0.0079 g/L

Answer Key – ch. 810. b. Ag2S (Ksp = 2.8 x 10-49)

Ag2S ⇌ 2 Ag+ S2–

I N/A 0 0

∆ N/A +2x +x

Eq N/A 2x x

Use Ksp to solve for x2.8 x 10-49 = (2x)2(x)

x = 4.14 x 10–17

Molar solubility = 4.14 x 10–17 Mor

solubility = (4.14 x 10–17 mol/L)(247.8 g/mol)= 1 x 10–14 g/L

Answer Key – ch. 811. Determine the Ksp values for the following compounds:

a. Al(OH)3 (solubility = 5 x 10-9 mol/L) ⇒ the solubility tells us how

much will dissolve in order to get to equilibrium

Al(OH)3 ⇌ Al3+ 3 OH–

I N/A 0 0

∆ N/A + 5 x 10-9 +3(5 x 10-9 )

Eq N/A 5 x 10-9 1.5 x 10-8

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

Answer Key – ch. 811. 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

MgF2 ⇌ Mg2+ 2 F–

I N/A 0 0

∆ N/A + 0.00118 +2(0.00118 )

Eq N/A 0.00118 0.0024

Ksp = (0.00118)(0.0024)2 = 6.6 x 10–9

Answer Key – ch. 812. 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

Ca3(PO4)2 ⇌ 3 Ca2+ 2 PO43-

I N/A 0 0.02

∆ N/A + 3x +2x

Eq N/A 3x 0.02 + 2x

Insignificantlysmall

Use Ksp to solve for x1 x 10-54 = (3x)3(0.02)2

x = 4.5 x 10–18

molar solubility = 4.5 x 10–18 M

Answer Key – ch. 813. 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

Zn(OH)2 ⇌ Zn2+ 2 OH-

I N/A 0 10–3 M

∆ N/A + x +2x

Eq N/A x 10–3 + 2x

Insignificantly small

Use Ksp to solve for x4.5 x 10-17= (x)(10–3)2

x = 4.5 x 10-11

molar solubility = 4.5 x 10-11M

Answer Key – ch. 814. 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 BaCrO4 or 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

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