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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].

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A.P. 1999 #1

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  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

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