8.3 Bases Similar to weak acids, weak bases react with water to a solution of ions at equilibrium. The general equation is: B(aq) + H2O(l)  HB+(aq) + OH-(aq) where, B represents any base The equilibrium expression for this reaction is: Kc = [HB+][OH-] [B][H2O]

Base Dissociation Constant, Kb Just as with weak acids, the concentration of water is almost constant in a solution of weak base. Therefore, we can group the constant terms in the equation. [H2O] Kc = [HB+][OH-] = Kb [B] Table 8.3 (Pg. 404 lists some common base dissociation constants, Kb)

Calculating pH and pOH An aqueous solution of ammonia has a concentration of 0.105 mol/L. Calculate the pH of the solution. 1. Write the balanced chemical equation. NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq) Given: [NH3] = 0.105 mol/L 2. Set up an ICE table

P.P. #29, Continued Concentration (mol/L) NH3(aq) H2O(l) NH4+(aq) OH- (aq) Initial 0.105 ~0 Change -x +x Equilibrium 0.105 - x Write the equation for the base dissociation constant, Kb Kb =[NH4+][OH-] [NH3] From the table, Kb(NH3) = 1.8 x 10-5

P.P. #29, Continued Substitute Kb and the equilibrium values from the ICE table into the equation. Solve for x. 1.8 x 10-5 = (x) (x) (0.105 – x) Look at [NH3]/Kb = 0.105/1.8x10-5 = 5833 > 100, so the amount of NH3 that dissociates is negligible compared to the initial [NH3] Therefore, (0.105 – x), becomes (0.105)  solve for x

P.P. #29, Continued Solve for x 1.8 x 10-5 = (x) (x) (0.105) 6. Solve for pOH pOH = -log[OH-] = -log[1.374 x 10-3] = 2.832

P.P. #29, Continued Solve for pH pH = 14 – pOH = 14 – 2.832 = 11.168 Therefore, the pH of the ammonia solution is 11.14

The relationship b/n Ka and Kb For any acid and its conjugate base Ka(Kb) = [H3O+][OH-] = Kw What does this mean? The stronger an acid is, the weaker its conjugate base is. The conjugate of a strong acid is always a weak base. The conjugate of a strong base is always a weak acid.

Buffer Solutions Contain a mixture of: A weak acid and its conjugate base OR A weak base and its conjugate acid Buffer solutions resist a change in pH when a moderate amount of acid or base is added. Buffer solutions are made 2 ways: Use a weak acid and one of its salts Eg. Mix acetic acid and sodium acetate Use a weak base and one of its salts Eg. Mix ammonia and ammonium chloride

Making a Buffer http://www.chembio.uoguelph.ca/educmat/chm19104/chemtoons/chemtoons5.htm

How do buffers work? Consider a buffer made with acetic acid and sodium acetate. Acetic acid is weak, so [CH3COOH] is high Sodium acetate is very soluble, so [CH3COO-] is high Adding and acid or base has only a slight effect on pH because the H3O+ or OH- ions are removed by one of the components of the buffer solution. http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/buffer12.swf

Importance of Buffers Very important in biological systems For example: pH of arterial blood is ~7.4 It must stay b/n 7.0 and 7.5 or the organism may die. Blood is buffered by the equilibrium between CO3 ions and HCO3 ions. (formed by the reaction of dissolved CO2 and H2O)

8.4 Acid-Base Titration Curves = Graph of the pH of an acid (or base) against the volume of and added base (or acid) Titration reactions are used to find the equivalence point Equivalence point – point where the acid and base completely react with one another. If, V1, V2 and C1 are known C2 can be found (C1V1 = C2V2)

How pH indicators show equivalence points pH changes rapidly near the equivalence point A single drop of titrant can change the pH by 2 pH units. The colour change indicates equivalence even if the colour change happens at pH of 8 As long as it is in the steep section of the curve Strong acid titrated with strong base

Selecting a pH Indicator Weak acid titrated with strong base.

Section Review Which indicator should be used for each curve? Is this the titration curve for a weak acid titrated with a strong base? Explain.