2 Ionic product of water Kw

Ionic Product for water [H+ (aq)][OH- (aq)] [H2O(l)] Kc= This equilibrium has the following equilibrium expression Kc x [H2O (l)] = [H+ (aq) ][OH- (aq)] Rearrange to Because [H2O (l)] is much bigger than the concentrations of the ions, we assume its value is constant and make a new constant Kw Kw = [H+ (aq) ][OH- (aq) ] Learn this expression At 25oC the value of Kw for all aqueous solutions is 1×10-14 mol2dm-6 The Kw expression can be used to calculate [H+ (aq)] ions if we know the [OH- (aq)] ions and vice versa. Finding pH of pure water Pure water/ neutral solutions are neutral because the [H+ (aq) ] = [OH- (aq)] Using Kw = [H+ (aq) ][OH- (aq) ] then when neutral Kw = [H+ (aq) ]2 and [H+ (aq) ] = √ Kw At 25oC [H+ (aq) ] = √ 1×10-14 = 1×10-7 so pH = 7 At different temperatures to 25oC the pH of pure water changes. Le Chatelier’s principle can predict the change. The dissociation of water is endothermic so increasing the temperature would push the equilibrium to the right giving a bigger concentration of H+ ions and a lower pH. Example 2 : Calculate the pH of water at 50ºC given that Kw = 5.476 x 10-14 mol2 dm-6 at 50ºC [H+ (aq) ] = √ Kw = √ 5.476 x 10-14 =2.34 x 10-7 mol dm-3 pH = – log 2.34 x 10-7 = 6.6 It is still neutral though as [H+ (aq) ] = [OH- (aq)]. In all aqueous solutions and pure water the following equilibrium occurs: H2O (l) H+ (aq) + OH- (aq). Calculating pH of Strong Base For bases we are normally given the concentration of the hydroxide ion. To work out the pH we need to work out [H+ (aq)] using the kw expression. Strong bases completely dissociate into their ions. NaOH  Na+ + OHExample 3: What is the pH of the strong base 0.1M NaOH Assume complete dissociation. Kw = [H+ (aq)][OH- (aq)] = 1×10-14 [H+ (aq)] = kw/ [OH- (aq)] = 1×10-14 / 0.1 = 1×10-13 M pH = – log[1×10-13 ] =13.00