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Strong Acids

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.

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Strong Acids

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  1. 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 [H3O+] is high Examples include: H2SO4, HNO3, HCl

  2. Strong acid pH HCl(aq) + H2O(l) Cl-(aq) + H3O+(aq) pH = -log[H3O+] Because strong acid fully dissociate, [HCl] is equal to [H3O+]. pH = -log[HCl]

  3. Water dissociation constant, Kw Kw = [H3O+] x [OH-] Increases with increased temperature At 25 deg Kw = 1X10-14

  4. Strong Base • Completely dissociates so [OH-] = [Base] [H3O+]= So pH = -log Kw [OH-] 1X10-14 [OH-]

  5. 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 [H3O+] is low CH3COOH + H2O n 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.

  6. Acidity Constant, Ka Equilibrium constant Kc = But water is a solvent so Kc = So Ka = [H3O+] [A-] [HA] [H2O] [H3O+] [A-] [HA] [H3O+] [A-] [HA]

  7. Weak acid pH Because So Ka = Rearrange and pH = -log[H3O+] Now try Qs 1 & 2 pg 253 study guide [H3O+] = [A-] [H3O+]2 [HA] [H3O+]= Ka x [HA]

  8. pKa pKa= -logKa As the strength of acid decreases Ka gets smaller pKa gets larger Conjugate base gets stronger

  9. Weak Base • Dissociates only partially • Use Ka of conjugate acid to calculate pH H3O+ = pH = -log Kw x Ka c(B) Kw x Ka c(B)

  10. Using Kb • Kb = 1x10-14/Ka • [OH-] = √Kb x c(B) • Then H3O+ = Kw/[OH-] • pH = -logH3O+

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