1. Polyprotic acids – like sulfurous acid, H 2 SO 3 – have more than one ionizable H. H 2 SO 3 (aq) H + (aq) + HSO 3 – (aq) (K a1 = 1.7 x 10 –2 ) HSO 3 – (aq) H + (aq) + SO 3 2– (aq) (K a2 = 6.4 x 10 –8 ) -- It is harder to remove additional H + s, so K a values w /each H + removed. -- Usually, K a2 is at least 1000X smaller than K a1. In such cases, one can calculate [H + ] and pH based only on K a1 (i.e., ignore K a2 and pretend you have a monoprotic acid).

2. pH = 4.40 Find the pH of a 0.0037 M carbonic acid solution. (K a1 = 4.3 x 10 –7, K a2 = 5.6 x 10 –11 ) H 2 CO 3 (aq) H + (aq) + HCO 3 – (aq) at eq. [ ]0.0037 – xxx 4.3 x 10 –7 = x2x2 0.0037 4.3 x 10 –7 = x2x2 x = [H + ] = 3.97 x 10 –5 Mx = [H + ] = 3.99 x 10 –5 M pH = 4.40 shortcut

3. ([H + ] after 1 st ionization = 3.97 x 10 –5 M) at eq. [ ]3.97 x 10 –5 – y3.97 x 10 –5 + yy HCO 3 – (aq) H + (aq) + CO 3 2– (aq) 3.97 x 10 –5 – y 5.6 x 10 –11 = (3.97 x 10 –5 + y) (y) 3.97 x 10 –5 – y 5.6 x 10 –11 = (3.97 x 10 –5 y + y 2 ) y = add’l [H + ] = 5.6 x 10 –11 M totally negligible pH = 4.40

4. Weak Bases weak base + H 2 O conjugate acid + OH – K b = [conj. acid] [OH – ] [weak base] Weak bases are often nitrogen- containing molecules (“amines”) or anions. (all of these have lone e – pairs) Example: OH – (aq)+H–N–CH 3 (aq) H H + : H H 2 O(l) [ ] + dimethylamine (i.e., proton acceptors)

5. What is the conc. of ammonia in a solution of pH 9.35? NH 3 (aq) + H 2 O(l) NH 4 + (aq) + OH – (aq) Since pOH = 4.65, [OH – ] = 10 –4.65 = 2.24 x 10 –5 M. w.c?2.24 x 10 –5 x – 2.24 x 10 –5 Ammonia’s K b = 1.8 x 10 –5. 1.8 x 10 –5 = (2.24 x 10 –5 ) 2 x – 2.24 x 10 –5 x = [NH 3 ] init. = 5.0 x 10 –5 M

6. Anions related to Weak Acids NaClO(aq) Na + (aq) + ClO – (aq) This is the conj. base of HClO, so it acts as a weak base in H 2 O (i.e., it accepts H + ). ClO – (aq) + H 2 O(l) HClO(aq) + OH – (aq) Sodium hypochlorite is the active ingredient in bleach, which is used as a laundry additive and to disinfect water for swimming pools.

7. 0.6058 mol NaClO What mass of NaClO is required to make 2.0 L of a pH 10.50 solution? The K b for ClO – is 3.3 x 10 –7. Since pOH = 3.50, [OH – ] = 10 –3.50 = 3.16 x 10 –4 M. ClO – (aq) + H 2 O(l) HClO(aq) + OH – (aq) x = [ClO – ] init. = [NaClO] init. = 0.3029 M w.c? K b = [HClO] [OH – ] [ClO – ] 3.16 x 10 –4 x – 3.16 x 10 –4 3.3 x 10 –7 = (3.16 x 10 –4 ) 2 x – 3.16 x 10 –4 M NaClO = mol NaClO L soln 45 g NaClO 0.3029 M 2.0 L