4.12:Mixtures of Strong Acids and Bases Chemistry 12

4.12 Mixing Strong Acids and Bases 100% dissociation The following neutralizing reaction occurs: H3O+ +OH-  2 H2O If H3O+ = OH- then neither acid or base is in excess and pH = pOH = 7 (at 25C)

H3O+ is larger and pH is less 2 H2O(l) + 59J ↔ H3O+(aq) + OH−(aq) A Little Review How does H3O+ and pH of pure water at 30 C compare to that of 25C? H3O+ is larger and pH is less 2 H2O(l) + 59J ↔ H3O+(aq) + OH−(aq) And how does pH and pOH of pure water at 30 C compare to each other? They are equal Why ???

H3O+ or OH- in excess If H3O+ or OH- is in excess we do not have to worry about the following equilibrium: 2 H2O ↔ H3O+ + OH- Reason 1: KA (aka as Kw here) is very small thus product concentrations are very small and  insignificant ([H3O+] = 0.0000001 M) Reason 2: excess H3O+ (or OH-) would force eqb left  further reducing the concentration of the products from above. Therefore calculations are very much like our last section (4.11).

Ba(OH)2  Ba2+ + 2OH- HCl  H+ + Cl- Recall M1V1 = M2V2  M2 = M1V1/V2 E.g.) 25.0 mL of 0.0420 M Ba(OH)2 is added to 125.0 mL of 0.0120 M HCl. Calculate the pH. Ba(OH)2  Ba2+ + 2OH- HCl  H+ + Cl- Recall M1V1 = M2V2  M2 = M1V1/V2 [OH-] = (2 x 0.0420) x 25.0 mL = 0.0140 M 150.0 mL [H3O+] = (0.0120 x 125.0 mL) = 0.0100 M

[OH-] [H3O+] Excess [OH-] 0.0140 -0.0100 = 0.0040 M OH- pOH = -log [0.0040] = 2.3979 pH = 14 – 2.3979 = 11.60 Or you could convert to moles [H3O+] Excess [OH-] 2 sig figs

Same Question mol OH- = ML = (2 x 0.0420) x 0.0250 = 0.00210 mol mol H3O+ = ML = 0.0120 x 0.125 = 0.00150 mol 0.00210 - 0.00150 =0.00060 mol OH- M = mol/L = 0.00060/0.150 = 0.0040M OH- pOH = -log [0.0040] = 2.3979 pH = 14 – 2.3979 = 11.60

Complete Calculations Assignment (p. 20) Hebden # 58-67