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Acid/Base Titrations

Acid/Base Titrations. Titrations . Titration Curve – always calculate equivalent point first Strong Acid/Strong Base Regions that require “different” calculations B/F any base is added Half-way point region At the equivalence point After the equivalence point. Strong Acid/Strong Base.

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Acid/Base Titrations

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  1. Acid/Base Titrations

  2. Titrations • Titration Curve – always calculate equivalent point first • Strong Acid/Strong Base • Regions that require “different” calculations • B/F any base is added • Half-way point region • At the equivalence point • After the equivalence point

  3. Strong Acid/Strong Base • 50 mL of 0.02000 M KOH • Titrated with 0.1000 M HBr • First -find Volume at equivalence • M1V1 = M2V2 • (0.050 L)(0.02000M) = 0.1000 V • V = 10.0 mL

  4. Strong Acid/Strong Base • 50.00 mL of 0.02000 M KOH • Titrated with 0.1000 M HBr • Second – find initial pH pH = - logAH ~ -log [H+] pOH = -logAOH ~ -log [OH-] pH = 12.30

  5. (~6 ml) Limiting Reactant Strong Acid/Strong Base • 50 mL of 0.02000 M KOH • Titrated with 0.1000 M HBr • Third– find pH at mid-way volume • KOH (aq) + HBr (aq) -> H2O (l) + KBr(aq) 0.001000 mol 0.0006000 mol Before After 0.000400 mol 0 mol 0.0006000 mol 0.0006000 mol pH = 11.8

  6. Strong Acid/Strong Base • 50 mL of 0.02000 M KOH • Titrated with 0.1000 M HBr • Fourth – find pH at equivalence point • KOH (aq) + HBr (aq) -> H2O (l) + KBr(aq) 0.001000 mol 0.0010000 mol Before After 0 mol 0 mol 0.0010000 mol 0.0010000 mol pH = 7.0

  7. Limiting Reactant Strong Acid/Strong Base • 50 mL of 0.02000 M KOH • Titrated with 0.1000 M HBr • Finally – find pH after equivalence point 12 ml • KOH (aq) + HBr (aq) -> H2O (l) + KBr(aq) 0.001000 mol 0.001200 mol Before After 0 mol 0.0002000 mol 0.0010000 mol pH = 2.5

  8. Titration of WEAK acid with a strong base

  9. Titration of a weak acid solution with a strong base. • 25.0 mL of 0.1000M acetic acid • Ka = 1.8 x 10-5 • Titrant = 0.100 M NaOH • First, calculate the volume at the equivalence-point • M1V1 = M2V2 • (0.0250 L) 0.1000 M = 0.1000 M (V2) • V2 = 0.0250 L or 25.0 mL

  10. Titration of a weak acid solution with a strong base. • 25.0 mL of 0.1000M acetic acid • Ka = 1.8 x 10-5 • Titrant = 0.100 M NaOH • Second, Calculate the initial pH of the acetic acid solution

  11. Titration of a weak acid solution with a strong base. • 25.0 mL of 0.1000M acetic acid • Ka = 1.8 x 10-5 • Titrant = 0.100 M NaOH • Third, Calculate the pH at some intermediate volume

  12. Titration of a weak acid solution with a strong base. • 25.0 mL of 0.1000M acetic acid • Ka = 1.8 x 10-5 • Titrant = 0.100 M NaOH • Fourth, Calculate the pH at equivalence

  13. Titration of a weak acid solution with a strong base. • 25.0 mL of 0.1000M acetic acid • Ka = 1.8 x 10-5 • Titrant = 0.100 M NaOH • Finally calculate the pH after the addition 26.0 mL of NaOH

  14. pH after equivalence? Buffer Region pH @ equivalence? Equivalence point M1V1=M2V2 Initial pH

  15. pH after equivalence? Buffer Region pH @ equivalence Equivalence point M1V1=M2V2 Initial pH

  16. pH after equivalence Dominated by remaining [OH-] Buffer Region pH @ equivalence Equivalence point M1V1=M2V2 Initial pH

  17. Weak Base titrated with strong acid • Consider a 100 ml of a 0.0100 M base with 0.0500 M HCl • Kb = 1 x 10-5

  18. Initial pH Buffer Region pH after equivalence Dominated by remaining [H+] pH @ equivalence

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