Addition of Strong Acids or Bases to Buffers -- Reactions between strong acids/bases and weak bases/acids proceed to completion. -- We assume that the strong acid/base is completely consumed. -- When adding a “strong” to buffered solutions… (1) (2) Calc. [ ]s after rxn, using stoichiometry Calc. eq. [ ]s using K a or K b

rxn.: 2.00-L of a buffered solution of pH 4.74 contains 0.30 mol of acetic acid (K a = 1.8 x 10 –5 ) and 0.30 mol of sodium acetate. Calculate the pH after mol of sodium hydroxide is added. Ignore volume changes. CH 3 COOH + OH – CH 3 COO – + H 2 O w.c? – Init.  0 mol0.26 mol mol After rxn. (1) Calc. [ ]s after rxn, using stoichiometry (Use mol for rxn, M for ICE!)

CH 3 COOH CH 3 COO – + H + (2) Calc. eq. [ ]s using K a or K b (Since K a > K b, use K a equation.) 0.13 – x x 0 + x 1.8 x 10 –5 = 0.17x + x – x x = [H + ] = x 10 –5 MpH = 4.86 X X CH 3 COO – + H 2 O CH 3 COOH + OH – ? ICE

the pt. at which equivalent amounts of acid and base have been mixed Acid-Base Titrations equivalence point: mol H + = mol OH –

pH curve for HCl titrated with NaOH pH 7 mL of NaOH added Strong Acid – Strong Base Titrations Any indicator whose color change begins and ends along the vertical line is okay. pH at equiv. pt. AABB

Strong Acid – Strong Base Titrations (cont.) -- phenolphthalein (pH ) pink colorless -- methyl red (pH ) yellow red l l

Find pH when mL of 0.10 M nitric acid are mixed with mL of 0.10 M potassium hydroxide. For strong-strong… Rxn. goes to completion; excess determines pH. mol H + =0.10 M ( L)= mol H + mol OH – =0.10 M ( L)= mol OH – (HNO 3 and KOH) = mol OH – excess [OH – ] = 1 x 10 –5 mol ~ L = x 10 –4 M OH – pOH =3.70 pH = 10.30