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Unit 12: Acids and Bases

Unit 12: Acids and Bases. Section 3: Calculating [H 3 O + ] and [OH - ] from pH. Formula Manipulation. pH = -log [H 3 O + ] log [H 3 O + ] = -pH [H 3 O + ] = antilog (-pH) [H 3 O + ] = 10 -pH  formula to find [H 3 O + ] concentration. Sample Problems.

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Unit 12: Acids and Bases

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  1. Unit 12: Acids and Bases Section 3: Calculating [H3O+] and [OH-] from pH

  2. Formula Manipulation • pH = -log [H3O+] • log [H3O+] = -pH • [H3O+] = antilog (-pH) • [H3O+] = 10-pH formula to find [H3O+] concentration

  3. Sample Problems • Determine the [H3O+] concentration of an aqueous solution that has a pH of 4.0. Is the solution acidic, basic, or neutral? • Since the pH is less than 7.0, the solution is acidic • [H3O+] = 10-pH [H3O+] = 10-4 1 x 10-4 M

  4. Sample Problems • Determine the [H3O+] and [OH-] concentration for a solution that has a pH of 7.52. Is the solution acidic, basic, or neutral? • Since pH is greater than 7.0, the solution is slightly basic • [H3O+] = 10-pH [H3O+] = 10-7.52 3.0 x 10-8M • pOH = 14 – 7.52  6.48 • [OH-] = 10-pOH  [OH-] = 10-6.48 3.3 x 10-7 M

  5. Sample Problems • Determine the [H3O+] and [OH-] concentration for a solution that has a pOH of 2.35. Is the solution acidic, basic, or neutral? • Since the pOH is less than 7.0, the solution is basic • [OH-] = 10-pOH  [OH-] = 10-2.35 4.5 x 10-3 M • pH = 14 – 2.35  11.65 • [H3O+] = 10-pH [H3O+] = 10-11.65 2.2 x 10-12 M

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