1 / 5

A.P. 1999 #1

A.P. 1999 #1. a)Write the equilibrium expression (Kb mass action ). RECOGNISE NH 3 AS A WEAK BASE NOT A CONJUGATE BASE) REACTION IS NH 3  NH 4 + + OH -. K b = [NH 4 + ][OH - ] or K b = [BH + ][OH - ] [NH 3 ] [B].

jenn
Download Presentation

A.P. 1999 #1

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. A.P. 1999 #1 a)Write the equilibrium expression (Kb mass action).RECOGNISE NH3 AS A WEAK BASE NOT A CONJUGATE BASE) REACTION IS NH3 NH4+ + OH- Kb = [NH4+][OH-] or Kb = [BH+][OH-] [NH3] [B] b) Calculate the pH of 0.0180 NH3. You were given that the [OH-] is 5.6 x 10-4 M resulting from the dissociation of the 0.0180 M ammonia. -LOG (5.6 x 10-4) = pOH = 3.252, pH = 14 – 3.252 = 10.745

  2. c)Write the equilibrium expression (Kb mass action). RECOGNISE NH3 AS A WEAK BASE NOT A CONJUGATE BASE) REACTION IS NH3 NH4+ + OH- Kb = [NH4+][OH-] or Kb = [BH+][OH-] [NH3] [B] THE NH4+ AND OH- MUST BE EQUAL AS THEY BOTH COME FROM THIS Kb SYSTEM! Kb = [NH4+][OH-] = Kb = [5.6 X 10-4][5.6 X 10-4] = 1.80 x 10-5 = Kb [NH3] [0.0180]

  3. D) Calculate the % IONIZATION OF NH3 % IONIZATION = OH- = 5.60 x 10-4X 100 = 3.11% NH3 0.0180 E-i) CALCULATE VOLUME OF HCl AT EQUIVALENCE. At equivalence, H+ added = B neutralized = BH+ generated. Neutralization goes to completion, stoichiometry Moles H+ added must = moles NH3 neutralized and = NH4+ generatedat equivalence

  4. E - i)continued: NEED 3.6 x 10-4 mol H+ : M = moles/#L : 0.0120M = 3.6 x 10-4 /X X = volume of solution to give 3.6 x 10-4 mole H+ = 30.0 mL E - ii) Determine pH after only 15.0 mL of H= added. Mole H+ added = V x M = 0.0120M x 0.0150L = 1.80 x 10-4 mol H+ added 1.80 x 10-4 mol NH4+ produced [NH4+ ] =1.80 x 10-4 mol NH4+ /0.0350 L = 0.00514 M, At pKb point pH = pKb=9.25 pKb point, weak electrolyte = conjugate

  5. E - iii) At 40.0 ml H+ added, you are 10.0ml past equivalence (30.0 ml from part (i)), and the excess H+ will have nothing to neutralize it, it determines the pH. Mol H+ = 0.0100 L x 0.0120M = 1.20 x 10-4 mol H+ [H+] = 0.000120 mol/ 0.0600L = 0.00200 M pH = - LOG (0.00200) = 2.700

More Related