1. Acids and Bases Chapter 14 – One step closer to the AP Test

2. Arrhenius Definition Acids produce hydrogen ions in aqueous solution. Acids produce hydrogen ions in aqueous solution. Bases produce hydroxide ions when dissolved in water. Bases produce hydroxide ions when dissolved in water. Limits to aqueous solutions. Limits to aqueous solutions. Only one kind of base. Only one kind of base. NH 3 ammonia could not be an Arrhenius base. NH 3 ammonia could not be an Arrhenius base.

3. Bronsted-Lowry Definitions  And acid is an proton (H + ) donor and a base is a proton acceptor.  Acids and bases always come in pairs.  HCl is an acid.  When it dissolves in water it gives its proton to water.  HCl(g) H + + Cl -  Water is a base -makes hydronium ion

4. What do all of these have in common? NH 3 H 2 O CO 3 2-

5. Dissociate the Following 1. Hydrochloric acid 2. Acetic acid 3. Ammonium ion 4. Anilinium ion (C 6 H 5 NH 3 + ) 5. Hydrate Aluminum (III) ion [Al(H 2 O) 6 ] 3+ 1. HCl H + + Cl - 2. HC 2 H 3 O 2 H + + C 2 H 3 O 2 - 3. NH 4 + H + + NH 3 4. C 6 H 5 NH 3 + H + + C 6 H 5 NH 2 5. [Al(H 2 O) 6 ] 3+ H + + Al(H 2 O) 5 OH 2+

6. Polyprotic Acid Be careful dissociating polyprotic acids. H 2 SO 4 H + + HSO 4 - SA HSO 4 - H + + SO 4 2- WA H 3 PO 4 H + + H 2 PO 4 - WA

7. Pairs  General equation  HA(aq) + H 2 O(l) H 3 O + (aq) + A - (aq)  Acid + Base Conjugate acid + Conjugate base  This is an equilibrium.  Competition for H + between H 2 O and A -  The stronger base controls direction.  If H 2 O is a stronger base it takes the H +  Equilibrium moves to right.

8. Writing Net Ionic A-B Reactions 1- Determine if the species are an acid, base or salt 2 – Determine if the acid or base are strong or weak If SA or SB dissociate completely (Show with an ) If WA or WB it reaches equilibrium (Show with an ) 3. If a salt, determine if either ion is reactive (Comes from a WA or WB) Determine whether the reactive ion acts as an acid or base (What type of solution is it in) Write the reaction of the ion. 4. Write the neutralizations reaction 5. Cross out SA, SB and Salt reactions 6. Add the remaining reactions together

9. Let ’ s try one together Equimolar amounts of hydrochloric acid and potassium hydroxide are mixed. HCl H + + Cl - SA KOHK + + OH - SB H + + OH - H 2 O

10. Another example 0.1M nitrous acid solution is added to the same volume of a 0.1M sodium hydroxide solution. (Equimolar amounts = H 2 O MUST be a product) HNO 2 H + + NO 2 - wa NaOHNa + + OH - sb H + + OH - H 2 O HNO 2 + OH - H 2 O + NO 2-

11. Example #2  A solution of potassium hydroxide is added to a solution of ammonium chloride. KOHK + + OH - SB NH 4 ClNH 4 + + Cl - salt NH 4 + NH 3 + H + H + + OH - H 2 O NH 4 + + OH - NH 3 + H 2 O

12. Example #3 Dilute acetic acid solution is added to solid magnesium carbonate. HC 2 H 3 O 2 H + + C 2 H 3 O 2 - wa MgCO 3 Mg 2+ + CO 3 2- salt CO 3 2- + H 2 O HCO 3 - + OH - H + + OH - H 2 O HC 2 H 3 O 2 + CO 3 2- HCO 3 - + C 2 H 3 O 2 -

13. And yet another one … from the WS Solutions of ammonia and hydrofluoric acid are mixed. NH 3 + H 2 O NH 4 + + OH - wb HFH + + F - wa H + + OH - H 2 O NH 3 + HFNH 4 + + F -

14. Problem #2 Hydrogen sulfide has is bubbled through a solution of potassium hydroxide. H 2 SH + + HS - wa KOHK + + OH - sb H + + OH - H 2 O H 2 S + OH - H 2 O + HS -

15. Problem #3 – A twist A solution of sulfuric acid is added to a solution of barium hydroxide until the same number of moles of each compound has been added. H 2 SO 4 H + + HSO 4 - sa Ba(OH) 2 Ba 2+ + 2OH - sb H + + OH - H 2 O Ba 2+ + SO 4 2- + H + + OH - H 2 O + BaSO 4

16. Problem #4 A solution of sodium hydroxide is added to a solution of sodium dihydrogen phosphate until the same number of moles of each compound has been added. NaOHNa + + OH - sb NaH 2 PO 4 Na + + H 2 PO 4 - salt H 2 PO 4 - H + + HPO 4 2- H + + OH - H 2 O H 2 PO 4 - + OH - H 2 O + HPO 4 2-

17. Problem #5 Dilute nitric acid is added to crystals of pure calcium oxide. HNO 3 H + + NO 3 - sa CaO + H 2 O Ca 2+ + 2OH - 2H + + 2OH - 2H 2 O CaO + 2H + Ca 2+ + H 2 O

18. Jump to Problem #8 Excess sulfur dioxide gas is bubbles through a dilute solution of potassium hydroxide. KOHK + + OH - sb SO 2 (NMO + H 2 O) HSO 3 - HSO 3 - H + + SO 3 2- H + + OH - H 2 O SO 2 + OH - H 2 O + SO 3 2-