Acid/Base Indicators Substance that changes color in the

Acid/Base Indicators  Substance that changes color in the

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: Dissociate partially in water Ex: HC2H3O2 or vinegar Episode 1102 Strong bases: Dissociate completely in water Ex: NaOH Weak Bases: Dissociate partially in water Ex: NH4OH or ammonia solution

Episode 1102 Acid/Base Concentration pH = - log [H+] 0 -----ACID-----7-----BASE-----14 Episode 1102 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+ + Cl1 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 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 Calculate the pH for a solution of H2SO4 with a molarity of 1 x 104 M. 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 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 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+ + OH1 mol NaOH = 1 mol OH3.0 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 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 + 2OH1 mol Ca(OH)2 = 2 mol OH1 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 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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