Strong Acids Strong acids fully dissociate so: They are fully separated into their ions They are good conductors of electricity Are not in a state of equilibrium.

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Presentation transcript:

Strong Acids Strong acids fully dissociate so: They are fully separated into their ions They are good conductors of electricity Are not in a state of equilibrium A 10x dilution results in pH increase of 1 React quickly as [H 3 O + ] is high Examples include: H 2 SO 4, HNO 3, HCl

Strong acid pH HCl (aq) + H 2 O (l)  Cl - (aq) +H 3 O + (aq) pH = -log[H 3 O + ] Because strong acid fully dissociate, [HCl] is equal to [H 3 O + ]. pH = -log[HCl]

Water dissociation constant, K w K w = [H 3 O + ] x [OH - ] Increases with increased temperature At 25 deg K w = 1X10 -14

Strong Base Completely dissociates so [OH - ] = [Base] [H 3 O + ]= So pH = -log KwKw [OH - ] 1X [OH - ]

Weak Acids Do not fully dissociate so: Are in equilibrium biased to left Have a low concentration of ions Are poor conductors of electricity React slowly as [H 3 O + ] is low CH3COOH + H2O CH3COO- + H3O+ As they are in equilibrium, when they are diluted water concentration is increased and there is a shift to the right to compensate. This means predicting pH increase is harder.

Acidity Constant, K a Equilibrium constant K c = But water is a solvent so K c = SoK a = [H 3 O + ] [A - ] [HA] [H2O] [H 3 O + ] [A - ] [HA] [H 3 O + ] [A - ] [HA]

Weak acid pH Because So K a = Rearrange and pH = -log[H 3 O + ] Now try Qs 1 & 2 pg 253 study guide [H 3 O + ] = [A - ] [H 3 O + ] 2 [HA] [H 3 O + ]=K a x [HA]

pK a pK a = -logK a As the strength of acid decreases K a gets smaller pK a gets larger Conjugate base gets stronger

Weak Base Dissociates only partially Use K a of conjugate acid to calculate pH H 3 O + = pH = -log K w x K a c(B) K w x K a c(B)

Using K b K b = 1x /K a [OH - ] = √K b x c(B) Then H 3 O + = K w /[OH - ] pH = -logH 3 O +