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What is the equilibrium chemical equation? - PowerPoint PPT Presentation


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Calculate the pH of a 1.00 L solution that is 0.150 M in Benzoic acid (HC 7 H 5 O 2 ) and 0.050 M in potassium benzoate (KC 7 H 5 O 2 ). The pK a of benzoic acid is 4.20. This is a buffer solution since it contains a weak acid and its conjugate base (a salt of the acid).

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slide1

Calculate the pH of a 1.00 L solution that is 0.150 M in Benzoic acid (HC7H5O2) and 0.050 M in potassium benzoate (KC7H5O2). The pKa of benzoic acid is 4.20

This is a buffer solution since it contains a weak acid and its conjugate base (a salt of the acid).

What is the equilibrium chemical equation?

KC7H5O2dissociates in water to K+andC7H5O2-

The acid establishes an equilibrium.

HC7H5O2 + H2O ↔ H3O+ + C7H5O2-

Since this is a buffer solution, the Henderson-Hasselbalch equation can be used: pH = pKa+ Log{[conj. Base])/[Acid]}

slide2

Calculate the pH of a 1.00 L solution that is 0.150 M in Benzoic acid (HC7H5O2) and 0.050 M in potassium benzoate (KC7H5O2). The pKa of benzoic acid is 4.20)

Since this is a buffer solution, the Henderson-Hasselbalch equation can be used: pH = pKa+ Log{[conj. Base])/[Acid]}

This is the problem on slide 1

a. What will the pH be after the addition of 10.0 mL of 0.10 M HCl?

Approach: Determine the moles of each species and then decide which one

Increase in and which one decrease.

slide3

Calculate the pH of a 1.00 L solution that is 0.150 M in Benzoic acid (HC7H5O2) and 0.050 M in potassium benzoate (KC7H5O2). The pKa of benzoic acid is 4.20)

Since this is a buffer solution, the Henderson-Hasselbalch equation can be used: pH = pKa+ Log{[conj. Base])/[Acid]}

This is the problem on slide 1

a. What will the pH be after the addition of 100.0 mL of 0.10 M HCl?

Approach: Determine the moles of each species and then decide which one

Increase in and which one decrease.

The HCl will increase the moles of acid and decrease the moles of base.

slide4

Calculate the pH of a 1.00 L solution that is 0.150 M in Benzoic acid (HC7H5O2) and 0.050 M in potassium benzoate (KC7H5O2). The pKa of benzoic acid is 4.20)

Since this is a buffer solution, the Henderson-Hasselbalch equation can be used: pH = pKa+ Log{[conj. Base])/[Acid]}

This is the problem on slide 1

b. What will the pH be after the addition of 100.0 mL of 0.10 M HCl?

Approach: Determine the moles of each species and then decide which one

Increase in and which one decrease.

The HCl will increase the moles of acid and decrease the moles of base.

slide5

Calculate the pH of a 1.00 L solution that is 0.150 M in Benzoic acid (HC7H5O2) and 0.050 M in potassium benzoate (KC7H5O2). The pKa of benzoic acid is 4.20)

Since this is a buffer solution, the Henderson-Hasselbalch equation can be used: pH = pKa+ Log{[conj. Base])/[Acid]}

This is the problem on slide 1

c. What will the pH be after the addition of 100.0 mL of 0.10 M NaOH?

Approach: Determine the moles of each species and then decide which one

Increase in and which one decrease.

The NaOH will increase the moles of base and decrease the moles of acid.

slide6

Calculate the pH of a solution that is 0.150 M in hydrazine (N2H4) and 0.050 M in its conjugate acid (N2H5Cl ). The Kb of hydrazine is 1.7 x 10-6.

What the chemistry?

Dissociate the salt:

N2H5Cl → N2H5++ Cl-

Write equilibrium expression for the weak base:

N2H4 + H2O ↔ N2H5+ + OH-

I. 0.150 0.050 0

C. -x +x +x

E. 0.150 – x 0.050 + x +x

pOH = 5.29 and pH = 8.71

X = [OH1-] = 5.1 x 10-6

Or with the Henderson Hasselbach equation:

slide7

Calculate the pH of a 1.00 L solution that is 0.150 M in hydrazine (N2H4) and 0.050 M in its conjugate acid (N2H5Cl ). The Kb of hydrazine is 1.7 x 10-6.

What the chemistry?

Dissociate the salt:

N2H5Cl → N2H5++ Cl-

Write equilibrium expression for the weak base:

N2H4 + H2O ↔ N2H5+ + OH-

Or with the Henderson Hasselbach equation: (from slide number 6)

Determine the pH after the addition of 100.0 mL of 0.10 M HCl.

The HCl will increase the moles of conj. acid and decrease the moles of base.

pH = 8.59

slide8

Calculate the pH of a 1.00 L solution that is 0.150 M in hydrazine (N2H4) and 0.050 M in its conjugate acid (N2H5Cl ). The Kb of hydrazine is 1.7 x 10-6.

What the chemistry?

Dissociate the salt:

N2H5Cl → N2H5++ Cl-

Write equilibrium expression for the weak base:

N2H4 + H2O ↔ N2H5+ + OH-

Or with the Henderson Hasselbach equation: (from slide number 6)

Determine the pH after the addition of 100.0 mL of 0.10 M NaOH.

The answer is on the next slide. Try working the problem first.