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Applications of Aqueous Equilibria

Applications of Aqueous Equilibria. Solutions of Acids or Bases Containing a Common Ion. Common Ion Effect: shift in equilibrium caused by addition of a compound having an ion in common with a dissolved acid or base. Calculate pH of a solution of 0.500 M HCN and 0.200 M NaCN

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Applications of Aqueous Equilibria

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  1. Applications of Aqueous Equilibria

  2. Solutions of Acids or Bases Containing a Common Ion • Common Ion Effect: shift in equilibrium caused by addition of a compound having an ion in common with a dissolved acid or base. • Calculate pH of a solution of 0.500 M HCN and 0.200 M NaCN • Addition of the common ion CN- drives equilibrium to the left (Le Chatelier’s Principle) • NaCN is a soluble salt and will dissociate completely

  3. Solutions of Acids or Bases Containing a Common Ion • Ka = 6.2 x 10-10 • HCN + H2O ⇌H3O+ + CN- • init 0.500 0 0.200 • change -x +x +x • eq 0.500-x x 0.200 + x

  4. Solutions of Acids or Bases Containing a Common Ion • Ka = 6.2 x 10-10 = (x)(0.200 + x) / (0.500-x) • we can assume x is << 0.200 • x = 6.2 x 10-10 (0.500)/0.200 = 1.6 x 10-9 • pH = -log(1.6 x 10-9) = 8.81 • CN- is a stronger base than HCN is an acid

  5. Solutions of Acids or Bases Containing a Common Ion • for HA + H2O ⇌H3O+ + A- and Ka = [H3O+][A-] / [HA] • [H3O+] = Ka[HA] / [A-] • and pH = pKa + log[A-] • [HA] where pKa = -log Ka

  6. Solutions of Acids or Bases Containing a Common Ion • Henderson-Hasselbalch equation: • pH = pKa + log([conjugate base] / [acid]) • or pH = pKa + log ([base] / [conjugate acid]) • Calculate pH of a solution of 0.200 M NH3 and 0.300 M NH4Cl • Kb = 1.8 x 10-5, Ka =10-14 /(1.8 x 10-5)= 5.6 x 10-10 • pH = -log(5.6 x 10-10) + log(0.200 / 0.300) = 9.08 • We could also derive pOH = pKb + log([conjugate acid] / [base])

  7. Buffered Solutions • A buffer is a solution of a weak acid or base and its conjugate which is resistant to changes in pH. • HA + OH-⇌H2O + A- • strong base converted to weak base • A- + H3O+⇌HA + H2O • strong acid converted to weak acid • pH of a buffer can be calculated with the Henderson-Hasselbalch equation

  8. Buffered Solutions • Calculate the pH of a solution which is 1.500 M HOAc and 1.250 M NaOAc • Ka = 1.8 x 10-5 pH = -log(1.8 x 10-5) + log(1.250/1.500) = 4.67 • Add 10 mL 0.100 M HCl to 100 mL water and 100 mL buffer • To water, [HCl] =(10 mL)(0.100 M)/(110 mL) = 0.0091 M, pH = 2.04 • To buffer, OAc- + H3O+ → HOAc + H2O • .125 mol .001 mol .150 mol before reaction • .124 mol 0 mol .151 mol after reaction • pH = -log(1.8 x 10-5) + log [(.124)/(.151)] = 4.66 • pH changes 0.01 pH unit • Buffers work best when pH = pKa ± 1.00 and when there is enough acid/base to neutralize appreciable amounts of added acid or base.

  9. Buffering Capacity Henderson-Hasselbalch equation: pH = pKa + log([base]/[acid]) • For the pH to change by ± 1 pH unit, the ratio [base]/[acid] must change by a factor of 10. • Preparing a Buffer • Select an acid (or base) with a pKa (or pKb) as close as possible to the desired buffered pH • Want a acid/base ratio of close to 1

  10. Titrations and pH Curves • Strong Acid-Strong Base Titrations • HCl + NaOH → NaCl + H2O 100% • Calculate the pH in the titration of 100.0 mL of 0.200 M HCl with 0.400 M NaOH after 0, 25.00, 50.00, and 75.00 mL base are added. • (100.0 mL)(0.200 M HCl) = 20.0 mmol HCl • at 0 ml, [H3O+] = 0.200 M, pH = -log(0.200) = 0.70 • at 25.00 mL, mmol OH- = (25.00 mL)(0.400mmol OH- /mL soln) = 10.0 mmol OH-

  11. Titrations and pH Curves

  12. Titrations and pH Curves • pH = -log(0.0800) = 1.10 • at 50.00 mL, mmol OH- = (50.00 mL)0.400 mmol OH- / mL soln = 20.0 mmol OH-

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