Example: Fill in the rest of the table below Table 8.4: Kw is temperature dependent Temp (C) Kw [H+] in pure pH of pure water 0 1.5 x 10-15 0.39 x 10-7 7.47 10 3.0 x 10-15 20 6.8 x 10-15 0.82 x 10-7 7.08 25 1.0 x 10-14 30 1.5 x 10-14 1.22 x 10-7 6.92 40 3.0 x 10-14 1.73 x 10-7 6.77 50 5.5 x 10-14 H+ and OH- are inversely related Because the product [H+] x [OH-] is constant at a given temperature, it follows that as one goes up, the other must go down (since Kw = [H+][OH-]) Table 8.5 Solutions are defined as acidic, basic, or neutral based on the relative concentrations of H+ and OH- Type of sol’n Relative concentrations pH at 25C Acid Neutral Alkaline Example: A sample of blood at 25C has [H+]=4.60 x 10-8 mol dm-3. Calculate the concentration of OH- and state whether the blood is acidic, neutral or basic. How you expect its pH to be altered at body temperature (37C)? pH and pOH scales are pOH= From the relationship: KW = [H+][OH-] -log KW = -log([H+][OH-]) -log KW = (-log[H+]) + (-log[OH-]) pKW = pH + pOH at 25C, KW = 1.0 x10-14, thus 14.00 = pH + pOH at 25C Given any one of the following we can find the other three: [H+],[OH-],pH and pOH Example: juice has a pH of 2.90 at 25C. Calculate its [H+],[OH-], and pOH.
Example: Fill in the rest of the table below Table 8.4: Kw is temperature dependent Temp (C) Kw [H+] in pure pH of pure water 0 1.5 x 10-15 0.39 x 10-7 7.47 10 3.0 x 10-15 20 6.8 x 10-15 0.82 x 10-7 7.08 25 1.0 x 10-14 30 1.5 x 10-14 1.22 x 10-7 6.92 40 3.0 x 10-14 1.73 x 10-7 6.77 50 5.5 x 10-14 H+ and OH- are inversely related Because the product [H+] x [OH-] is constant at a given temperature, it follows that as one goes up, the other must go down (since Kw = [H+][OH-]) Table 8.5 Solutions are defined as acidic, basic, or neutral based on the relative concentrations of H+ and OH- Type of sol’n Relative concentrations pH at 25C Acid Neutral Alkaline Example: A sample of blood at 25C has [H+]=4.60 x 10-8 mol dm-3. Calculate the concentration of OH- and state whether the blood is acidic, neutral or basic. How you expect its pH to be altered at body temperature (37C)? pH and pOH scales are pOH= From the relationship: KW = [H+][OH-] -log KW = -log([H+][OH-]) -log KW = (-log[H+]) + (-log[OH-]) pKW = pH + pOH at 25C, KW = 1.0 x10-14, thus 14.00 = pH + pOH at 25C Given any one of the following we can find the other three: [H+],[OH-],pH and pOH Example: juice has a pH of 2.90 at 25C. Calculate its [H+],[OH-], and pOH.
u08_notetaker_acid_base.docx