810 1033 911 1232 1113 Problem Prepare 3L of a pH 49 buffer in which the total

810 1033 911 1232 1113 problem prepare 3l of a ph 49

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8–10                              10.33 9–11                             12.32     11–13
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Problem Prepare 3L of a pH 4.9 buffer in which the total concentration of buffering species is 2 M.
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Acid-Base Titration • A - + H 3 O + HA + H 2 O K eq >>1 The reaction of any base (strong or weak) with a strong acid is quantitative. HA + OH - A - + H 2 O K eq >>1 The reaction of any acid (strong or weak) with a strong base is quantitative. Assumes that the H + is ionizable. e.g. the third proton of H 3 PO 4 does not react with OH - .
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Titration Performed by adding a strong acid to a base or a strong base to an acid in a controlled way that allows us to easily see the equivalence point. The equivalence point is the point where we have added exactly as many moles of acid as there are moles of base. The equivalence point is the point where we have added exactly as many moles of base as there are moles of acid.
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Titration of a Strong Acid with a Strong Base HCl(aq) + NaOH(aq) H 2 O(l) + NaCl(aq) H 3 O + + OH - 2 H 2 O K eq = 1/K w = 10 14 We can predict what the titration curve (pH vs amount of base added) will look like: Before any base is added: pH = – log [Acid]. After base has been added that exceeds the amount of acid initally: pH = - log [(n base added – n acid initially )/total volume of solution in liters] At the equivalence point: pH = 7.
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Demonstration Titrate a strong acid with a strong base using phenolphthalein as an indicator of end point. Titrate a strong acid with a strong base using a pH electrode to draw the curve and determine the equivalence point.
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Titration of a Weak Acid with a Strong Base HAc(aq) + NaOH(aq) H 2 O(l) + NaAc(aq) We can predict qualitatively what the titration curve (pH vs amount of base added) will look like: Before any base is added: HAc + H 2 O H 3 O + + Ac - pH = – log [(K a * [HAc] init ) 1/2 ]. After base has been added that exceeds the amount of acid initally: pH = - log [(n base added – n acid initially )/total volume of solution in liters] At the equivalence (stoichiometric) point: Ac - + H 2 O HAc + OH - pH = 14 + log [(K b * [Ac - ]) 1/2 ]. (Note: K b = K w /K a ) At the half-equivalence point: [Ac - ] = [HAc] pH = pK a + log ([A - ] eq /[HA] eq ) pH = pK a Note the buffer region that extends from 10>([Ac-]/[HAc])>0.1
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Titration of a Weak Acid with a Strong Base
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Demonstration Titrate a weak acid with a strong base using a pH electrode to draw the curve and determine the equivalence point.
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Titration of a Weak Polyprotic Acid with a Strong Base H 2 A (aq) + 2 OH - (aq) 2 H 2 O(l) + A - (aq) We can predict qualitatively what the titration curve (pH vs amount of base added) will look like: Before any base is added: H 2 A + H 2 O H 3 O + + HA - K eq =K a1 pH = – log [(K a * [H 2 A] init ) 1/2 ]. Be careful here!! At the first stoichiometric point: 2HA - H 2 A + A -2 K eq =K a2 /K a1 pH = - log (K a1 K a2 ) 1/2 At the second stoichiometric point: A -2 + H 2 O HA - + OH - K b =K W /K A2 pOH = - log (K b *[A -2 ] eq ) 1/2 ) pH = 14 - pOH After base has been added that exceeds the amount of acid initally: pH = - log [(n base added – n acid initially )/total volume of solution in liters]
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Titration of a Polyprotic Acid with a Strong Base
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  • Spring '08
  • SHUMAN
  • Chemistry, pH, Buffer solutions, buffer solution

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