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Acid/Base Indicators

Acid/Base Indicators. Substance that changes color in the presence of an acid or a base Red or Blue Litmus Phenolphthalein ( phth ) Bromothymol blue (blue food coloring) Red cabbage juice. Episode 1102. Strong acids: Dissociate completely in water Ex: HCl Weak Acids:

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Acid/Base Indicators

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  1. Acid/Base Indicators • Substance that changes color in the presence of an acid or a base • Red or Blue Litmus • Phenolphthalein (phth) • Bromothymol blue (blue food coloring) • Red cabbage juice Episode 1102

  2. Strong acids: • Dissociate completely in water • Ex: HCl • Weak Acids: • Dissociate partially in water • Ex: HC2H3O2 or vinegar Episode 1102

  3. Strong bases: • Dissociate completely in water • Ex: NaOH • Weak Bases: • Dissociate partially in water • Ex: NH4OH or ammonia solution Episode 1102

  4. Acid/Base Concentration • pH = - log [H+] • 0 -----ACID-----7-----BASE-----14 Episode 1102

  5. Determine the pH of a solution of HCl that has a molarity of 1 x 10-4 M. • HCl is a strong acid- know it completely dissociates. • HCl H+ + Cl- • 1 mole HCl = 1 mol H+ • 1 x 10-4 M HCl = 1 x10-4 M H+ • pH = -log (1x 10-4) • pH = 4 Episode 1102

  6. Calculate the pH for a solution of HNO3 with a molarity of 1 x 10-3 M. • Strong acid – completely dissociates • HNO3 H+ + NO3- • 1 mol HNO3 = 1 mol H+ • 1 x 10-3 M HNO3 = 1 x 10-3 M H+ • pH = -log (1 x 10-3) • pH = 3 Episode 1102

  7. Calculate the pH for a solution of H2SO4 with a molarity of 1 x 10-4M. • Strong acid – completely dissociates • H2SO4 2H+ + SO4-2 • 1 mol H2SO4 = 2 moles H+ • 1 x 10-4 M H2SO4 = 2(1 x 10-4 M H+) • 2(1 x 10-4M H+) = 2 x 10-4 M H+ • pH = -log (2 x 10-4) • pH = 3.7 Episode 1102

  8. Self Ionization of Water • [H+][OH-] = 1 x 10-14 • Origin of pH scale • (-log[H+]) + (-log[OH-]) = -log (1 x 10-14) • pH + pOH = 14 Episode 1102

  9. Calculate the pH of a solution of NaOH with a molarity of 3.0 x 10-2 M. • Notice NaOH – acid or base? • Strong bases – completely dissociates • NaOH Na+ + OH- • 1 mol NaOH = 1 mol OH- • 3.0 x 10-2 M NaOH = 3.0 x 10-2 M OH- • [H+][OH-] = 1 x 10-14 • [H+](3.0 x 10-2) = 1 x 10-14 • [H+] = 3.3 x 10-13 • pH = -log [H+] • pH = -log (3.3 x 10-13) • pH = 12.5 Episode 1102

  10. Find the pH for a solution of Ca(OH)2 with a molarity of 1 x 10-4 M. • Notice Ca(OH)2 – acid or base? • Strong bases – completely dissociates • Ca(OH)2 Ca+2 + 2OH- • 1 mol Ca(OH)2 = 2 mol OH- • 1 x 10-4 M Ca(OH)2 = 2(1 x 10-4 M OH-) • [OH-] = 2 x 10-4 M OH- • [H+][OH-] = 1 x 10-14 • [H+](2 x 10-4) = 1 x 10-14 • [H+] = 5 x 10-11 • pH = -log [H+] • pH = -log (5 x 10-11) • pH = 10.3 Episode 1102

  11. Calculate both the hydrogen ion concentration and the hydroxide concentration for an aqueous solution that has a pH of 4.0. • pH = -log [H+] • 4.0 = -log [H+] • Log is a function of the number 10, so … • -4.0 = log [H+] • 10-4 = [H+] or 1 x 10-4 M [H+] • [H+][OH-] = 1 x 10-14 • (1 x 10-4)[OH-] = 1 x 10-14 • [OH-] = 1 x 10-10 M Episode 1102

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