Question

What is the hydroxide concentration, [OH-], in a 3.85 M NaC2H3O2 solution? Consider the following information to help you answer this question: Sodium acetate is a water-soluble salt. Like all water-soluble ionic compounds, it dissociates completely when dissolved in water: H2O NaC2H3O2(s) ---------------> Na+(aq) + C2H3O2-(aq) Since the Na+ ion is the conjugate acid of a strong base (NaOH) it is too weak of an acid to react with water: Na+(aq) + H2O(l) ---------> N.R. Note: N.R. means “No Reaction”. However, the C2H3O2- ion is the conjugate base of a weak acid (HC2H3O2). Therefore, it is a strong enough base to react with water: C2H3O2-(aq) + H2O(l) <----------> HC2H3O2(aq) + OH-(aq) The above equilibrium reaction is the one for which you need to set up and solve an ICE table. However, the Kb value for C2H3O2- is normally not published in tables. It doesn’t need to be, because you can get it from two other values that are published: the Ka for HC2H3O2 and KW for the ionization of water. HC2H3O2(aq) + H2O(l) <----------> H3O+(aq) + C2H3O2-(aq) Ka = 1.8 x 10-5 2H2O(l) <----------> H3O+(aq) + OH-(aq) KW = 1.0 x 10-14 By combining the above two equations in the appropriate way, you can produce the reaction for which the equilibrium constant is desired. Once you have the required equilibrium constant, you can set up the ICE table in the usual way. Note that the initial concentration of C2H3O2- is the concentration given for NaC2H3O2. This is because the reaction proceeds to 100% completion, and the reaction has one-to-one stoichiometry. For every mole of NaC2H3O2, you get a mole of C2H3O2- (and a mole of Na+).