Lecture Notes Week 07

Lecture Notes Week 07 - Week 7 Weak Acid (WA) + Strong Base...

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1 – Week 7 – Weak Acid (WA) + Strong Base (SB) e.g., HAc(aq) + OH (aq) Ac (aq) + H 2 O(l) K = “Total Reaction” Consider 50.00 mL 1.0 M HAc and titrate this with 1.0 M NaOH. Calculate the pH as the titration proceeds. Pt. Base added (ml) pH Comments 1. 0 2.37 2. 10 4.14 3. 25 4.74 4. 40 5.35 5. 49 6.43 6. 50 9.22 7. 51 12.0 Detailed Calculations: 1 . Initially we have 1.0 M HAc (K a = 1.8×10 –5 ) Recognise what type of problem this is: HAc(aq) + H 2 O(l) Ac (aq) + H 3 O + (aq) [H 3 O + ] = = = pH = 2.37 . 2. Add 10.00 mL of 1.0 M NaOH (10 mmol OH ) to initial 50.00 mL of 1.0 M HAc (50 mmol HAc) new volume =
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2 Rxn: HAc(aq) + OH (aq) Ac (aq) + H 2 O(l) I 50 10 0 (mmol) C E present in non - trivial amounts. What is the ratio of HAc to Ac - ? What do we call this kind of solution? pH = pK a + log(base/acid) = 3. After 25.00 mL of 1.0 M base added to 50.00 mL of 1.0 M HAc: new volume = What point in the titration have we reached? Rxn: HAc(aq) + OH (aq) Ac (aq) + H 2 O(l) I 50 25 0 (mmol) C E What is the ratio of HAc to Ac - ? pH = pK a + log(base/acid) = 4. After 40.00 mL of 1.0 M base added to 50.00 mL of 1.0 M HAc: new volume = Rxn: HAc(aq) + OH (aq) Ac (aq) + H 2 O(l) I 50 40 0 (mmol) C E pH = pK a + log(base/acid) =
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3 5. After 49.00 mL of base added to 50.00 mL of 1.0 M HAc: new volume = Rxn: HAc(aq) + OH (aq) Ac (aq) + H 2 O(l) I 50 10 0 (mmol) C E Not an effective buffer at all, but we can still use the H-H equation,
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Lecture Notes Week 07 - Week 7 Weak Acid (WA) + Strong Base...

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