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Published byMarianna Hart Modified over 8 years ago
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Titration Curves
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Problem 50.0 mL of 0.10 M acetic acid (K a = 1.8 x 10 -5 ) are titrated with 0.10 M NaOH. Calculate the pH after the additions of 0, 10, 25, 40, 50, 60, and 75 mL samples of NaOH. Then, construct a titration curve and label it properly.
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Problem 50.0 mL of 0.20 M carbonic acid (K a = 4.4 x 10 -7 ) are titrated with 0.20 M NaOH. Calculate the pH after the additions of 0, 10, 25, 40, 50, 60, and 75 mL samples of NaOH. Then, construct a titration curve and label it properly.
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Weak Base-Strong Acid Titration Curves Before the addition. Construct an “equilibrium” reaction table ONLY! K b = [OH - ] [BH + ] and obtain [OH - ]. [B] Calculate the pOH, then the pH.
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Additions before the equivalence point. Construct a “stoichiometry” reaction table. Determine MOLES of base in excess (not neutralized) and MOLES of conjugate acid formed. Divide MOLES by the TOTAL VOLUME to obtain [BH + ] and [B]. Construct an “equilibrium” reaction table. K b = [OH - ] [BH + ] and obtain [OH - ]. [B] Calculate the pOH, then the pH.
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Additions at the equivalence point Construct a “stoichiometry” reaction table. Determine MOLES of conjugate acid formed. Divide MOLES by the TOTAL VOLUME to obtain [BH + ]. Calculate K a (K a x K b = K w ). Construct an “equilibrium” reaction table, reacting the conjugate base with water. K a = [A - ] [H 3 O + ] and obtain [H 3 O + ]. [HA] Calculate the pH.
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Additions beyond the equivalence point Construct a “stoichiometry” reaction table. Determine MOLES of acid in excess (not neutralized) and the MOLES of conjugate acid. Divide MOLES by the TOTAL VOLUME, Because [H 3 O + ] excess >> [H 3 O + ] conj. acid, use [OH - ] excess to calculate the pH.
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Problem 20.0 mL of 0.10 M triethylamine, (CH 3 CH 2 ) 3 N, (K b = 5.2 x 10 -4 ) are titrated with 0.100 M HCl. Calculate the pH after the additions of 0.0, 10, 15, 19, 19.95, 20, 20.05, and 25 mL samples of HCl. Then, construct a titration curve and label it properly.
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