1. 4.12:Mixtures of Strong Acids and Bases
Chemistry 12

2. 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)

3. 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 ???

4. 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+] = 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).

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

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

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

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