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+ + 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

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 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

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+ + 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

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

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