Lesson 19 Temperature and pH

Kw and Temperature Energy + H2O ⇋ H+ + OH- @ 25 oC Kw = 1.0 x 10-14  Higher Temperatures @ 50 oC shifts right Kw > 1.0 x 10-14 pH < 7 [H+] > 1.0 x 10-7 M pOH < 7 [OH-] > 1.0 x 10-7 M @ this temperature pH + pOH = pKw < 14

Kw and Temperature Energy + H2O ⇋ H+ + OH- @ 25 oC Kw = 1.0 x 10-14  Lower Temperatures shifts left Kw < 1.0 x 10-14 pH > 7 [H+] < 1.0 x 10-7 M pOH > 7 [OH-] < 1.0 x 10-7 M @ this temperature pH + pOH = pKw > 14

Important Points If you're not @ 25oC [H+][OH-] ≠ 1.0 x 10-14 pH + POH ≠ 14 All other equations are valid and……. Pure water is always neutral, but pH ≠ 7 and……. [H+] = [OH-] and …… pH = pOH

Is the temperature higher or lower than 25 oC? Calculate the Kw. The pH of pure water is 6.50 Is the temperature higher or lower than 25 oC? Calculate the Kw. Energy + H2O ⇋ H+ + OH- Higher [H+] = 10-6.50 = 3.1623 x 10-7 M [OH-] = 3.1623 x 10-7 M Kw = [H+][OH-] = (3.1623 x 10-7)2 Kw = 1.0 x 10-13 Is water acidic, basic or neutral? neutral

 2. The Kw = 5.0 x 10-13 for pure water. Is the temperature higher or lower than 25 oC? Calculate the pH , POH, and pKw. Energy + H2O ⇋ H+ + OH- Kw > 1.0 x 10-14 shifted right…… temperature is higher Kw = 5.0 x 10-13 = [H+][OH-] = x2 x = [H+] = 7.071 x 10-7 M pH = pOH = 6.15 pKw = 12.30 Is water acidic, basic or ? neutral

3. 25. 0 ml of 0. 100 M NaOH, 10. 0 mL 0. 200 M KOH, and 20. 0 mL of 3. 25.0 ml of 0.100 M NaOH, 10.0 mL 0.200 M KOH, and 20.0 mL of 0.100 M H2SO4 are poured into the same beaker. What is the resulting pH? 0.0250 L x 0.100 mole NaOH = 0.00250 mol L 0.0100 L x 0.200 mole KOH = 0.00200 mol Total = 0.00450 mol 0.0200 L x 0.100 mole H2SO4 = 0.00200 mol

2XOH + H2SO4 → X2SO4 + 2HOH I 0.00450 mol 0.00200 mol C 0.00400 mol 0.00200 mol E 0.00050 mol 0.00000 Total Volume = 25.0 mL + 10.0 mL + 20.0 mL = 55.0 mL Molarity = 0.00050 mol = 0.0090909 M 0.0550 L pOH = 2.04 pH = 11.96