patino_chm2046_chapter16

Any of the four b bh oh or kb may be unknown adding

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Unformatted text preview: you is ), to work with the following equilibrium: to B + H2O ⇄ BH+ + OH− Kb, K of base B. Any of the four, [B], [BH+], [OH−], or Kb may be unknown Adding an Acid to a Buffer Adding Q: What is the pH when 0.020 mol HCl is added to a solution solution with 0.12 mol of NH3 and 0.095 mol NH4Cl. Kb = 1.8 × 10-5 with Cl. Solution H3O+ from HCl will react with NH3 to produce NH4Cl. [NH3] will dec and [NH4Cl] will increase. The new concns (& K) will Cl] will determine the pH. will NH3 + H3O+ → NH4+ + H2O NH + mol NH3 + mol H3O+ → mol NH4+ mol + 0.12 0.02 0.095 0.12 0.095 -0.02 -0.02 +0.02 -0.02 0.10 0 0.115 K >>1 before rxn before changes after rxn The buffer remains ‘alive’. Now we’ll substitute The concentrations in the Equil. of NH3/NH4+ concentrations Adding an Acid to a Buffer Adding Q: What is the pH when 0.020 mol HCl is added to What 0.12 mol of NH3 and 0.095 mol NH4Cl. Kb = 1.8 × 10-5 0.12 Cl. Solution NH3 + H2O ⇄ NH4+ + OH− NH after rxn 0.10 mol 0.115 mol ? Kb 0.10 mol ———— V [OH−] = 1.8× 10−5 × ————— = 1.57 × 10−5 M [OH ————— 0.115 mol ———— pOH = 4.81 pOH V pH = 9.19 pH Important: pH doesn’t depend on V of buffer solxn Adding an Acid to a Buffer REMOVED!!! REMOVED!!! Q: What is the pH when 0.020 mol HCl is added to Q: What 0.12 mol of NH3 and 0.095 mol NH4Cl. Kb = 1.8 × 10-5 0.12 Cl. Solution Ka = 5.6 × 10-10 ⇒ pKa = 9.25 [base] pH = pKa + log [acid] 0.10 pH = 9.25 + log 0.115 pH = 9.25 +( - 0.061) pH = 9.19 pH decreased from 9.36 to 9.19 (∆ = 0.17) when 0.020 mol HCl was added to buffer containing 0.12 mol NH3 and HCl was 0.095 mol NH4Cl!! 0.095 Preparing a Buffer Preparing We want to buffer a solution at pH = 4.30. We This means [H3O+] = 10-pH = 5.0 × 10-5 M This It is best to choose an acid such that [H3O+] is is about equal to Ka (or pH ≈ pKa). about —then get the exact [H3O+] by adjusting the ratio of by acid to conjugate base. acid [Acid] +] = [H3O • Ka [Conj. base] For a buffer of ba...
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