lec23 - HA = n A – A – further accepts a proton from...

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1 Last Time • Titrations – analyte, titrant – Stoichiometric point • Neutralization reactions – Strong acid/strong base – S point at pH = 7 – Strong acid/weak base – S point at pH < 7 – Weak acid/strong base – S point at pH > 7 Strong Acid/Weak Base • Analyte – weak base : Titrant – strong acid B (aq) + H 3 O + (aq) HB(aq) + H 2 O(l) • pH < 7 at the stoichiometric point – n H3O = initial n B = n HB – HB further donates a proton to water – calculate pH like the pH of a salt solution • Above stoichiometric point pH calculation is like a weak base with a salt, below, [H 3 O + ]

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2 Weak Acid/Strong Base • Analyte – weak acid : Titrant – strong base HA(aq) + OH (aq) A (aq) + H 2 O(l) • pH > 7 at the stoichiometric point – n OH = initial n
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Unformatted text preview: HA = n A – A – further accepts a proton from water – calculate pH like the pH of a salt solution • Below stoichiometric point pH calculation is like a weak acid with a salt, above, ∝ [OH – ] Example Calculate the pH at the following three points during the titration of 25 mL of a 0.1 M NH 3 solution (K b = 1.8 x 10 –5 ) with 0.2 M HCl. (1) When V added = 11 mL (2) When V added = 13 mL (3) At the stoichiometric point 3 Determination of K a HA(aq) + OH – (aq) → A – (aq) + H 2 O(l) • At stoichiometric point, n OH = initial n HA = n A • Halfway to the stoichiometric point – n OH = n A = n HA = initial n HA /2 – using total volume : [HA] = [A – ] Ka = [H 3 O + ] or pK a = pH ←...
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lec23 - HA = n A – A – further accepts a proton from...

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