Weak Acid – Strong Base Titrations pH curve for CH 3 COOH titrated with NaOH pH 7 mL of NaOH added The equivalence point is when, say, 50.0 mL of 0.10 M NaOH have been added to 50.0 mL of 0.10 M CH 3 COOH, but pH is > 7 at that point because... pH at equiv. pt. the CH 3 COO – generated by the rxn acts as a weak base. A B

Calculate the pH at the equivalence point if 40.0 mL of M benzoic acid (C 6 H 5 COOH, K a = 6.3 x 10 –5 ) are titrated with M sodium hydroxide. At equivalence pt., mol H + = mol OH – (0.04) = 0.05 (X L OH – ) X = L NaOH So V tot = ( ) L = 0.06 L We are at the eq. pt. from the rxn. btwn. a weak acid and a strong base. Strong wins. Use the K b equation. C 6 H 5 COO – + H 2 O C 6 H 5 COOH + OH –

(0.04) – x – x w.c?0 + x w.c?xx Need “conjugatized” K b for C 6 H 5 COO – … 6.3 x 10 –5 1 x 10 –14 = 1.59 x 10 –10 x2x – x = x = [OH – ] =1.63 x 10 –6 M pH = 8.21pOH = 5.79 X

BEFORE: Calculate the pH when 10.0 mL of M sodium hydroxide are added to 40.0 mL of M benzoic acid (C 6 H 5 COOH, K a = 6.3 x 10 –5 ). available OH – =0.080 M (0.01 L)= mol OH – available H + =0.025 M (0.04 L)= mol H + So… mol OH – rips mol H + off of C 6 H 5 COOH, leaving mol C 6 H 5 COO – and mol of unreacted C 6 H 5 COOH. 100 kids w /Ipods AFTER: 20 kids w /Ipods 80 kids crying 80 bullies w /Ipods 80 bullies w /nothing mol C 6 H 5 COOH mol C 6 H 5 COOH mol C 6 H 5 COO – mol H 2 O mol OH –

Now find pH: Before eq., we have… [C 6 H 5 COOH] = mol / 0.05 L = M [C 6 H 5 COO – ] = mol / 0.05 L = M – x x x (H + > OH –, so use K a eq.) 6.3 x 10 –5 = 0.016x + x – x x = [H + ] = x 10 –5 MpH = 4.80 X X C 6 H 5 COOH C 6 H 5 COO – + H + ? C 6 H 5 COO – + H 2 O C 6 H 5 COOH + OH –

Titration curves for polyprotic acids (e.g., H 2 CO 3 ) look something like pH curve for H 2 CO 3 pH mL of base added -- they have... two equivalence points