1 / 9

What is the equilibrium chemical equation?

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).

zia-walton
Download Presentation

What is the equilibrium chemical equation?

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. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 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]}

  2. 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.

  3. 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.

  4. 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.

  5. 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.

  6. 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:

  7. 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

  8. 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.

  9. The NaOH will increase the moles of base and decrease the moles of conj. acid. pH = 8.83

More Related