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General Form of Dissociation

Acids and Bases Base Equations • Acid • HA  H+ + A- or + - • H2O + A  H + AOH Chapter 14 • Base • BOH  B+ + OH- or + - • H2O + B  BH + OH • Strong /bases will have a , weak acids/bases will have a ⇌

Strong Bases Some weak acids and bases Strong acids these make a lightning bolt on the periodic table! • Acids Bases Acid formula Acid Formula Name Formula Name Formula

• Carbonic acid Carbonate Hydrochloric HCl Sulfuric H2SO4 Sodium NaOH Calcium Ca(OH) • Acetate acid Acid 2 Hydrobromic HBr Nitric Acid HNO Hydroxide Hydroxide • Ammonium Ammonia 3 acid Potassium KOH Strontium Sr(OH)2 • triethylamine Hydriodic HI Perchloric HClO4 Hydroxide Hydroxide + acid Acid • (CH3CH2)3NH (CH3CH2)3N All group Barium Ba(OH)2 1 metals Hydroxide Chloric HClO3 Acid

Strong acids and bases Weak + Strong reaction Examples • Strong acids and bases are not at equilibrium, • Reacting an acid with a base will produce water. there is no reverse reaction. • HCl + NaOH  H O + NaCl 2 • Calcium hydroxide reacts with chlorous acid • Strong acids and bases will never be formed in • Reacting a weak acid with a strong base will • Acetic acid reacts with potassium hydroxide a net ionic equation. produce water and conjugate base. • reacts with calcium nitrite • Adding strong acid or base normally means • Sodium hydroxide and carbonic acid + - - - • Nitric acid reacts with sodium chlorite you will have H or OH as a reactant, the rest • H2CO3 + OH  H2O + HCO3 • Hydroiodic acid reacts with sodium is a spectator ion. • Reacting a weak base with a strong acid will hydroxide • All other acids/bases can be formed by produce conjugate acid. reacting the conjugate ion with a strong • Nitric acid and ammonia acid/base. + + • NH3 + H  NH4

1 Acids and Bases Brønsted-Lowery Follow the proton

• Water is the product of all neutralization + - • The Brønsted-Lowery definitions, the • HF + H2O ⇌ H3O + F reactions between an acid and a base proton transfer: + • H O ⇌ H+ + OH- H H+ 2 (l) (aq) (aq) • An acid is a proton (H+) donor. • The Arrhenius Definitions. • A base is a proton (H+) acceptor. • NH + H O ⇌ NH + + OH- • An acid contains H and forms H+ (proton) 3 2 4 • which is an acid/base? + + • When that H is donated to H O it forms H O H+ H+ 2 3 • HF + H O ⇌ H O+ + F- (hydronium). 2 3 • NH + H O ⇌ NH + + OH- • What about the reverse reaction? • A base contains OH and forms OH- (hydroxide). 3 2 4 10 11

Label Acid, Base, Conjugate Acid, Conjugate acids and bases Conjugate Base Ammonia Smell

- + • NH + H O ⇌ NH ++OH- • When you run the reverse reaction you find • HClO3 + H2O ⇌ ClO3 +H3O 3 2 4 the products are also acids and bases. The • A B CB CA • Ammonia is a gas with a distinct odor - - • Ammonium and hydroxide are both odorless. acids and bases that are formed are called • ClO + H2O ⇌ HClO +OH conjugate acids or bases • B A CA CB • What will happen if you add an acid or a base to + - - 2- + this equilibrium? • H2O + HF ⇌ H3O + F • HSO4 + H2O ⇌ SO4 +H3O • base acid conjugate acid conjugate base • A B CB CA • If base is added to the solution the equilibrium will shift to the left so you will smell ammonia, if + - • AgOH + H O ⇌ Ag+ + H O+OH- • NH3 + H2O ⇌ NH4 + OH 2 2 hydroxide is removed (acid is added), it will shift • base acid CA CB • B A CA CB to the right to so you won’t smell anything.

Pet “Stain” Problem Conjugate acids and bases … Acid Equilibrium constant • Urine has ammonia in it. • For some acid “A” • Most cleansers are basic. • Conjugate acids and bases determine if an acid or base is strong or weak. • HA + H O ⇌ H O+ + A- • After cleaning, we still leave small amounts (aq) 2 (l) 3 (aq) (aq) • If the conjugate acid/base readily reacts to + - behind. • Ka = [H3O ] [ A ] run the reverse reaction it is a weak • If it is small amount of ammonia and a basic acid/base. • [HA] cleanser the equilibrium will be shifted to the • If it does not react in the reverse reaction the • It is also written ammonia side so some thing with a great sense of acid or base is strong. • H A ⇌ H+ + A- smell (dog) could pick it up. + - • Ka = [H ] [ A ] • A slightly acidic cleanser shifts the equilibrium to • [HA] the ammonium side to solve this problem

2 Problems

• Acid Dissociation (Ionization ) Reactions. • Strong acids dissociate completely in water. • Write the simple dissociation (ionization) K values of • At equilibrium, Q = K >> 1 because [HA] a reaction for each of the following acids. Then a is approx. 0. some weak write the K expression a • Weak acids are mostly undissociated. acids • Hydrochloric acid • At equilibrium, Q = Ka << 1 because • Acetic acid + - [H3O ] and [ A ] are very small. • The ammonium ion • The smaller the Ka value, the weaker the + • The annilium ion (C6H5NH3 ) acid. 3+ 21 • The hydrated aluminum(III) ion [Al(H2O)6]

The Autoionization of Water and Strong acids Water the Ion-Product (Dissociation)

• Strong acids are not an equilibrium reaction. • Water is a stronger base than the conjugate • Kw is the constant for Water + - • HCl + H2O  H3O + Cl acid of a strong base. + - • H2O(l) + H2O(l) ⇌ H3O (aq) + OH (aq) • Ka cannot be accurately determined in water • Water would be weaker than the conjugate • At 25o C because the reaction lies so far to the right that acid of a weak base • K = K = [H O+] [OH-] = 1.0 x 10-14 [HCl] is too small to measure. c w 3 • Remember water can act as an acid or a • if [H O+] = [OH-] = 1.0 x 10-7M • That is true for all strong acids, so K is base. 3 a + - normally only used for weak acids. • if [H3O ] > [OH ] = Acidic. • Amphoteric substances, like water, act as + - either an acid or a base • if [H3O ] < [OH ] = Basic. + - • if [H3O ] = [OH ] = Neutral.

+ - Calculating [H3O ] and [OH ] Problem pH + -pH • [H3O ] = 10 + - o -13 • Equation sheet • Calculate [H3O ] or [OH ] as required for • At 60 C, the value of Kw is 1.0 x10 . each of the following solutions at 25o C, and • Using LeChatelier’s principle, predict state whether the solution is neutral, acidic, whether the reaction or basic. + - • H2O(l) + H2O(l) ⇌ H3O (aq) + OH (aq) -5 - • 1.0 x10 M OH • is exothermic or endothermic. -7 - • 2.3 x10 M OH + - • Calculate [H3O ] and [OH ] in a neutral + • 10.0 M H3O solution at 60o C.

3 Sig Figs and pH pH problem

• The number of decimal places in the log value, • The pH of a sample of human blood was pH value, is equal to the number of significant measured to be 7.41 at 25o C. figures in the number that we took the logarithm • Calculate pOH, [H O+], and [OH-] for the of, concentration. 3 sample. • Calculate pH and pOH for each of the following solutions. • 2.7 x 10-3 M OH- -5 + • 3.4 x 10 M H3O • 1.54 x 10-10 M OH-

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