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Titration Lab

Titration Lab. Weak Acid/Strong Base Example: 50.0 mL of 0.1000 M Acetic acid titrated against 0.1000 M NaOH. Initial pH (no OH -1 added). Here we will calculate pH using the K a and an I, ∆, E table. We should expect an acidic pH, as no OH -1 has been added yet.

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Titration Lab

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  1. Titration Lab Weak Acid/Strong Base Example: 50.0 mL of 0.1000 M Acetic acid titrated against 0.1000 M NaOH

  2. Initial pH (no OH-1 added) • Here we will calculate pH using the Ka and an I, ∆, E table. • We should expect an acidic pH, as no OH-1 has been added yet. HC2H3O2(aq) ↔ H3O+ + C2H3O2-1 I 0.1000 M ~0 0 ∆ -x +x +x __________________________________ E 0.1000 M –x x x Ka = 1.8 x 10-5 = x2/(0.1000 – x) approx “x” to zero x = 0.013 and pH = 1.87

  3. What are the major and minor species in solution? • In the beaker the major specie is the molecular acetic acid molecule. • Minor species are the H3O+ and C2H3O2- ions.

  4. Add 25.0 mL of OH- At this point pH = pKa. Why? (25.0 mL)(.1000 mmol OH-/mL) = 2.5 mmol of base consumed giving 2.5 mmol C2H3O2- produced and 2.5 mmol of HC2H3O2 Ka= [H3O+][C2H3O2-]= [H3O+][2.5mmol]/[2.5mmol]=1.8 x 10-5 [HC2H3O2] And pH = 4.74 (still acidic)

  5. What are the major and minor species in solution now? • In the beaker the major specie is the molecular acetic acid molecule and C2H3O2- ions. • Minor species are the H3O+ ions.

  6. At Equivalence Point (50.0 mL of OH-sol’n added) • Q. What happens at equivalence pt? • A. All OH- is neutralized! • H3O++ OH- 2H2O • Remember: with a weak acid pH will be slightly basic at this point. (Why?)

  7. What’s Going On At Equivalence? • At equivalence the major specie is the acetate ion, C2H3O2- • Remember, anion of weak acid is strong conjugate base • Now we are out of Ka and into Kb • C2H3O2-(aq)+ H2O(l) ↔ HC2H3O2(aq) + OH-(aq)

  8. What are the major and minor species in solution now? • In the beaker the major specie is the acetate ion, C2H3O2- . • Minor species are the HC2H3O2 and OH- ions.

  9. Calculating pH at Equivalence • Kw = KaKb so Kb = Kw/Ka • Kb = [OH-][HC2H3O2]/[C2H3O2-] • This gives Kb = 10-14/(1.8x10-5)= 5.6 x10-10 • (0.0500 mL)(0.1000 C2H3O2- M)/0.100L =concentration of the acetate ion = 0.0500M • Kb = x2/0.0500M • Giving x = 5.3 x 10-6; and pOH = 5.27 • Giving a pH of 8.72, basic as we expect!

  10. Wrap Up • Keep track of your ions! • Do stoichiometry first - determine ion consumed • Done by writing equilibrium reaction • Solve for pH by using I Δ E tables with changes in terms of x • Plug into Ka or Kb expression and solve • Validate any approximations • pH value should make sense!

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