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Weak Acid + Strong Base Titration

Weak Acid + Strong Base Titration. There are two key points in a titration process. Equivalence Point – the point in the titration when the moles of added base (OH - ) are equal to the moles of acid in the original sample. Equimolar acid and base species.

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Weak Acid + Strong Base Titration

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  1. Weak Acid + Strong Base Titration There are two key points in a titration process. • Equivalence Point – the point in the titration when the • moles of added base (OH -) are equal to the moles of • acid in the original sample. Equimolar acid and base • species. 2. Halfway Point – the point in the titration when the moles of added base (OH -) are equal to one-half the moles of acid in the original sample.

  2. Mathematical Consideration of the Equivalence Point Example: 0.10 M NaOH added to 25 mL of 0.20 M HC2H3O2 25 ml 0.20 mol = 0.005 mol HC2H3O2 1000 ml 0.005 mol OH -1000 ml = 50.0 ml NaOH to equivalence point 0.10 mol 25 (0.20 M HC2H3O2 )= 0.0667 M HC2H3O2 = 0.0667 M OH - 75 molarities of each species at the equivalence point

  3. HC2H3O2 + OH - HOH + C2H3O2- Initial 0.0667 0.0667 0 Change - 0.0667 - 0.0667 + 0.0667 Final 0.0 0.0 0.0667 Remember -Reactions with strong base or strong acid always go to completion. At the equivalence point, no acid or strong base remains, but the reaction has formed conjugate base. The pH at the equivalence point is not 7.0 but a value greater than 7.0 due to the presence of the conjugate base

  4. C2H3O2 - + HOH - HC2H3O2- + OH - Initial 0.0667 0 0 Change - x x x Equil 0.0667 – x = 0.0667 x x Given: Ka of HC2H3O2 = 1.8 x 10 -5 Kb = 1 x 10 -14 = 5.6 x 10 -10 1.8 x 10 -5 5.6 x 10 –10 = (x) (x) x = OH - = 6.112 x 10 -6 0.0667 pOH = 5.214 pH = 8.79

  5. Titration Curve - Weak Acid and Strong Base The halfway point in the above titration is reached with 25.0 mL of NaOH added, the middle of the flat portion of the titration curve.

  6. Titration Curve - Weak Acid and Strong Base At the halfway point, the solution contains equimolar concentrations of the weak acid and its conjugate base. The pH of the solution is therefore equal to the pKa of the weak acid

  7. A 10.0 mL sample of 0.50 M nitrous acid (HNO2), a weak acid, is titrated with 0.20 M KOH. The Ka of nitrous acidis 4.0 x 10 –4. What volume of base is needed to reach the halfway point of the titration? 10 ml 0.50 mol = 0.005 mol HNO2 1000 ml 0.0025 mol OH –1000 ml = 12.5 ml KOH 0.20 mol

  8. A 10.0 mL sample of 0.50 M nitrous acid (HNO2), a weak acid, is titrated with 0.20 M KOH. The Ka of nitrous acidis 4.0 x 10 –4. What is the pH of the solution at the halfway point of the titration? pH = pKa = - log (4.0 x 10 –4) = 3.40 or

  9. HNO2 + OH -  HOH+ NO2 – I 0.005 0.0025 0 C - 0.0025 - 0.0025 0.0025 F 0.0025 0 0.0025 HNO2 H + + NO2 – I 0.0025 0 0.0025 C - x x x E 0.0025 – x x 0.0025 + x  0.0025  0.0025 4.0 x 10 –4 = (x) (0025) x = H + = 4.0 x 10 -4 0.0025 pH = 3.40

  10. A 10.0 mL sample of 0.50 M nitrous acid (HNO2), a weak acid, is titrated with 0.20 M KOH. The Ka of nitrous acidis 4.0 x 10 –4. What is the pH of the solution at the equivalence point of the titration? 10 ml 0.50 mol = 0.005 mol HNO2 1000 ml 0.0050 mol OH –1000 ml = 25.0 ml KOH 0.20 mol 10 (0.50 M HNO2) = 0.1429 M HNO2 35

  11. A 10.0 mL sample of 0.50 M nitrous acid (HNO2), a weak acid, is titrated with 0.20 M KOH. The Ka of nitrous acidis 4.0 x 10 –4. What is the pH of the solution at the equivalence point of the titration? HNO2 + OH - HOH + NO2 - I 0.1429 0.1429 0 C - 0.1429 - 0.1429 0.1429 F 0 0 0.1429 NO2 - + HOH  OH - + HNO2 I 0.1429 0 0 C - x x x F 0.1429-x x x  x Kb = 1 x 10 –14 = (x) (x) x = OH - = 1.89 x 10 -6 4.0 x 10 –4 0.1429 pOH = 5.7235 pH = 8.28

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