1 x 10 -14

Report
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.
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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 10-4M.
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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.
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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.
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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 …
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-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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