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Strong and Weak Acids. The strength of an acid is determined by the extent to which it ionizes, its percent ionization, not the concentration of the acid,

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Presentation on theme: "Strong and Weak Acids. The strength of an acid is determined by the extent to which it ionizes, its percent ionization, not the concentration of the acid,"— Presentation transcript:

1 Strong and Weak Acids

2 The strength of an acid is determined by the extent to which it ionizes, its percent ionization, not the concentration of the acid, the concentration of its hydronium ions, or its ability to react with a metal

3 Strong Acid An acid that nearly completely dissociates All molecules of the acid break up to form the ions soluble in water If more than one proton is being removed, not all steps need to be complete dissociation.

4 Weak Acid An acid that only slightly dissociates in a water solution Only a small percent of acid molecules donate their hydrogen, and most remain the same. Example: CH 3 COOH

5 A strong acid essentially ionizes 100%. An example of a strong acid is hydrochloric acid, HCl (aq) HCl (g) + H 2 O (l)  H 3 O + (aq) + Cl - (aq) 0.10 mol 0.10 mol0.10 mol 100% ionization few moleculesmany ions

6 An example of a weak acid is acetic acid, CH 3 COOH. CH 3 COOH (l) + H 2 O (l)  H 3 O + (aq) + CH 3 COO - (aq) 0.10 mol << 0.10 mol << 0.10 mol 5% ionization at 25  C many moleculesfew ions

7 Strong Base A base that dissociates almost completely into its ions. All oxides and hydroxides of group 1 and 2 are strong bases. Ex: NaOH

8 Weak Base Most bases are weak They dissociate only slightly in a water solution Example: NH 3

9 Strong acids are strong electrolytes and weak acids are weak electrolytes

10 A strong base dissociates 100%. An example of a strong base is sodium hydroxide, NaOH. NaOH (s) + H 2 O (l)  Na + (aq) + OH - (aq) 0.10 mol 0.10 mol 0.10 mol 100% dissociation few formula units (NaOH) many ions

11 A weak base ionizes to a small extent. An example of a weak base is NH3(g). NH 3(g) + H 2 O (l)  NH 4 + (aq) + OH - (aq) 0.10 mol << 0.10 mol << 0.10 mol 5% ionization at 25  C many molecules few ions

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13 Strong bases are strong electrolytes and weak bases are weak electrolytes.

14 Examples of Strong Acids and Bases Strong AcidsStrong Bases HClO4 perchloric acidLiOH lithium hydroxide HCl hydrochloric acidNaOH sodium hydroxide HNO3 nitric acidKOH potassium hydroxide H2SO4 sulfuric acidRbOH rubidium hydroxide HBr hydrobromic acidCsOH cesium hydroxide HI hydriodic acidCa(OH)2 calcium hydroxide Sr(OH)2 strontium hydroxide Ba(OH)2 barium hydroxide

15 Ionization of Water

16 16 In water, H + is transferred from one H 2 O molecule to another. one water molecule acts as an acid, while another acts as a base. H 2 O + H 2 O H 3 O + + OH −........ :O: H + H:O: H:O:H + + :O:H −........ H H H water water hydronium hydroxide ion (+) ion (-) Ionization of Water

17 17 Pure Water is Neutral In pure water, the ionization of water molecules produces small, but equal quantities of H 3 O + and OH − ions. molar concentrations are indicated in brackets as [H 3 O + ] and [OH − ]. [H 3 O + ] = 1.0 x 10 − 7 M [OH − ] = 1.0 x 10 − 7 M Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

18 18 Acidic Solutions Adding an acid to pure water increases the [H 3 O + ]. cause the [H 3 O + ] to exceed 1.0 x 10 -7 M. decreases the [OH − ].

19 19 Basic Solutions Adding a base to pure water increases the [OH − ]. causes the [OH − ] to exceed 1.0 x 10 − 7 M. decreases the [H 3 O + ]. Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

20 20 The ion product constant, K w, for water is the product of the concentrations of the hydronium and hydroxide ions. K w = [ H 3 O + ] [ OH − ] can be obtained from the concentrations in pure water. K w = [ H 3 O + ] [ OH − ] K w = [1.0 x 10 − 7 M] x [ 1.0 x 10 − 7 M] = 1.0 x 10 − 14 Ion Product of Water, K w

21 21 [H 3 O + ] and [OH − ] in Solutions In neutral, acidic, or basic solutions, the K w is always 1.0 x 10 −14.

22 22 Guide to Calculating [H 3 O + ]

23 23 Calculating [H 3 O + ] What is the [H 3 O + ] of a solution if [OH − ] is 5.0 x 10 -8 M? STEP 1: Write the K w for water. K w = [H 3 O + ][OH − ] = 1.0 x 10 −14 STEP 2: Rearrange the K w expression. [H 3 O + ] = 1.0 x 10 -14 [OH − ] STEP 3: Substitute [OH − ]. [H 3 O + ] = 1.0 x 10 -14 = 2.0 x 10 -7 M 5.0 x 10 - 8


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