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Understanding the Equilibrium and pH Calculations for Ammonia as a Weak Base

This document outlines the equilibrium expression for ammonia (NH3) as a weak base, including its dissociation into ammonium (NH4+) and hydroxide (OH-) ions. We derive the data needed to calculate the pH of a 0.0180 M NH3 solution, given an [OH-] concentration of 5.6 x 10^-4 M. We also explore the percent ionization of NH3, the volume of HCl required for neutralization, and the resulting pH values after varying amounts of HCl are added, including scenarios before and after the equivalence point.

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Understanding the Equilibrium and pH Calculations for Ammonia as a Weak Base

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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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