1 / 13

The common ion effect Ch 3 COOH + H 2 O ↔CHCOO - + H 3 O + If we add crystalls of CH 3 COONa

The common ion effect Ch 3 COOH + H 2 O ↔CHCOO - + H 3 O + If we add crystalls of CH 3 COONa The concentration of CH 3 COO _ Will increase the equilibrium will shift to the left , thus the concentration H 3 O + will decrease and pH will increase. Example:

brencis-casimir
Download Presentation

The common ion effect Ch 3 COOH + H 2 O ↔CHCOO - + H 3 O + If we add crystalls of CH 3 COONa

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. The common ion effect Ch3COOH + H2O ↔CHCOO- + H3O+ If we add crystalls of CH3COONa The concentration of CH3COO_ Will increase the equilibrium will shift to the left , thus the concentration H3O+ will decrease and pH will increase Dr.abbas alakhras

  2. Example: a) what is the pH of 0.03 Ch3COOH Ka Ch3COOH = 1.8x10-5 b) What will be the pH if 0.03 M CH3COOH is added . To the above Solution : Ch3COOH + H2O ↔CHCOO- + H3O+ 0.3-x مهمله)) x x Ka = x2/ 0.3 = 1.8x10-5 X= 2.3x 10-3 M = [H+] then then PH=2.63 Dr.abbas alakhras

  3. b)Ch3COOH + H2O ↔CHCOO- + H3O+ 0.3 البدايه) (0.3 0.3-x مهمله)) 0.3+ x x مهمله Ka = 0.3 (x) / 0.3 = 1.8x10-5= (H+) pH= 4.74 Dr.abbas alakhras

  4. Example :(important ) page 663 17-2 Calculate the florid ion concentration and PH of a solution that is 0.2M in HF and 0.1 M in HCl Dr.abbas alakhras

  5. Solution [F-] === ??? pH=== ???? HCl= 0.1 mol HF = 0.2 mole V= 1.0 L Ka HF= 6.8x10-4 [HCl] = 0.1/ 1= 0.1 M [HF] = 0.2/ 1= 0.2 M HF + H2O ↔ H+ + F- 0.2 البدايه)) 0.1 zero HCl مصدرها 0.2 – x 0.1+x x Dr.abbas alakhras

  6. مصدرها HCL Ka= (0.1)(x)/ 0.2 = 6.8x10-4 X = [F-] = 1.4x10-3M PH= - log H+ = -log 0.1 = 1 Practice page: 664 Dr.abbas alakhras

  7. NH3+ H2O ↔ NH4+ + OH- If we add acrystalls of NH4Cl the equilibrium will shift to left , [OH-] ↓ [H+]↑ PH ↓ Dr.abbas alakhras

  8. Buffer solution A buffer consist of weak acid its conjugate base or weak base and its conjugate acid • A buffer resist a small charge in PH when strong acid or base is added to the buffer Dr.abbas alakhras

  9. Which of the following pairs is not a buffer: HCN/ NaCN ▓ HCl/ NaCl متفرج ( NO) HF/ NaF ▓ HNO3/NO3- متفرج ( NO) CH3COOH/CH3COONa ▓ NH3/NH4Cl ▓ H2CO3/CO3-2ليست مترافقه H2CO3/HCO3- ▓ Dr.abbas alakhras

  10. {CH3COOH 0.3M / 0.15 M NaOH} CH3COOH + NaOH→ CH3COONa+ H2O 0.3 0.15 zero Limiting reactant 0.3 – 0.15 0.15 – 0.150.15 Conjugate pair buffer Dr.abbas alakhras

  11. Homework Is the following pairs is a buffer: NH3 0.3 M / 0.3 M HCl What is the PH of the a above solution K b NH3= 1.8x10-5 Dr.abbas alakhras

  12. Henderson – Hasse lalch equation PH = PKa + log [A-] / [HA] Weak acid Conjugate base PH = PK b + log [HB+] / [B] Weak base Conjugate acid PK b = - log [Kb] Dr.abbas alakhras

  13. Dr.abbas alakhras

More Related