Strong Bases There are six strong acids. HCl, HNO3, HClO4, HI, HBr, and H2SO4 When given any of the above acids always assume 100% ionization. 100% ionization is shown by using a single arrow in the ionization equation. The ionization equation for H2SO4 is not intuitively obvious to the most casual observer.

Strong Base Problem What is the pH of a solution prepared by dissolving 20.0 g Ba(OH)2/L? m = 20.0 g Ba(OH)2/L Ba(OH)2(aq) → Ba2+(aq) + 2OH-(aq) [Ba(OH)2] = n/V

1 mol Ba(OH)2 20.0 g Ba(OH)2 x 171.32 g Ba(OH)2 [Ba(OH)2] = 1.00 L 2 mol OH- [OH-] = x 1 mol Ba(OH)2 1.00 L [OH-] = 0.234 M

pOH = -log[OH-] = -log(0.234) = 0.631 pH + pOH = 14.00 pH = 14.00 – pOH = 14.00 – 0.631 = 13.37

A Weak Base Problem Calculate the pH of an ammonia solution with a molarity of 0.25 M. (b) Determine the pKb of the solution. [NH3] = 0.25 M Kb = 1.8 x 10-5 Take special note of the equilibria equation. Unlike weak acid problems where H2O as a reactant is optional, it is not the case with a weak base problem.

NH3(aq) + H2O(l) NH4+(aq) + OH-(aq) [ ]I 0.25 0 0 [ ]c -x +x +x [ ]e 0.25-x x x Remember that there no entries for liquids and solids. [NH4+] [OH-] Ka = [NH3] x 0.25 - x x × 1.8 x 10-5 =

Assume 0.25 – x ≈ 0.25 and check later. = x = [NH4+] = [OH-] = 2.1 x 10-3 M [OH-] [NH3] × 100% %ion = 2.1 x 10-3 M %ion = × 100% = 0.84 % 2.5 x 10-1 M

The %ion = 0.84% is well within the 5%, therefore the assumption is valid. pOH = -log[OH-] = -log(2.1 x 10-3) = 2.68 pH + pOH = 14.00 pH = 14.00 – 2.68 = 11.32 (b) pKb = -logKb = -log(1.8 x 10-5) = 4.74