1. Acids-Bases Reactions

2. Acids & Bases What causes acid rain? And how can we prevent the damage? Why do Perrier drinking chickens give better eggs than chickens who drink regular water? These are the types of questions we will be able to answer after this unit.

3. First Proposed Theory: Arrhenius Theory of Acids & Bases An acid is any substance which releases H + (aq) in water - it is any ionic species that start with “H” - it tastes sour, & turns blue litmus paper red eg. HCl, HNO 3, H 2 SO 4 A base is any substance which releases OH - (aq) in water - it is any ionic species that ends with “OH” - it tastes bitter, feels slippery, & turns red litmus paper blue eg. NaOH, KOH, Ca(OH) 2

4. A sAlt results from the neutralization rxn. of an acid & base Neutralization rxn: Acid + Base → salt + H 2 O Net Ionic Equation: H + + OH - → H 2 O Eg. HCl (aq) + NaOH (aq) → NaCl (aq) + H 2 O (l)

5. Q: Predict the products & balance the following Neutralization Reactions H 2 SO 4(aq) + NaOH (aq) → H 2 SO 4(aq) + 2 NaOH (aq) → Na 2 SO 4(aq) + 2 H 2 O (l) H 3 PO 4(aq) + KOH (aq) → H 3 PO 4(aq) + 3 KOH (aq) → K 3 PO 4(aq) + 3 H 2 O (l) H 4 P 2 O 7(aq) + NaOH (aq) → H 4 P 2 O 7(aq) + 4 NaOH (aq) → Na 4 P 2 O 7(aq) + 4 H 2 O (l)

6. Second Proposed Theory: Bronsted-Lowry Theory of Acids & Bases -more general & includes equilibrium rxns An acid is a substance that DONATES a PROTON ( H + ) An base is a substance that Accepts a PROTON ( H + ) Eg. NH 3 + H 2 O NH 4 + + OH - NH 3 gained a proton → NH 4 + Thus, NH 3 = BASE H 2 O lost a proton → OH - Thus, H 2 O = ACID

7. Your Turn! CH 3 COOH + H 2 O CH 3 COO - + H 3 O + (lost/donated H + ) (gained H + ) ACID BASE Which reactants acts as an acid & which as a base? 1.HNO 3 + H 2 O NO 3 - + H 3 O + 2.HCO 3 - + SO 3 2- CO 3 2- + HSO 3 - 3.HS - + H 2 PO 4 - H 2 S + HPO 4 2- 4.H 3 PO 4 + CH 3 COO - H 2 PO 4 - + CH 3 COOH 5.CO 3 2- + HF HCO 3 - + F -

8. Classifying Acids A monoprotic acid can donate 1 proton. A DIprotic acid can donate up to 2 proton. A triprotic acid can donate up to 3 proton. All acids that can donate more than 1 proton are known as polyprotic acids.

9. Amphiprotic Substances Amphiprotic substances can act as an acid or a base. eg. Water NH 3 + H 2 O NH 4 + + OH - Base acid CH 3 COOH + H 2 O CH 3 COO - + H 3 O + Acid base

10. Apart from water, amphiprotic substances start with H and have a (  ) charge. eg. H 2 PO 4 -, HS -, HCO 3 - Base acid H 3 PO 4  H 2 PO 4 -  HPO 4 2- Accepts H + Donates H +

11. NOTE: In every Bronsted-Lowry rxn. there is an acid and a base on both sides of the eqn. CH 3 COOH + H 2 O CH 3 COO - + H 3 O + acid base base acid

12. Q: Which do you think exhibit amphiprotic behavior? Se 2- HSe - H 2 Se H 3 PO 4 HPO 4 2- HSO 3 -

13. Q: Identify each species as an acid or base. 1.HF + SO 3 2- F - + HSO 3 - 2.H 2 O + HCO 3 - H 3 O + + CO 3 2- 3.NO 2 - + H 2 O OH - + HNO 2 4.H 2 PO - + S 2- HS - + HPO 4 2- 5.N 2 H 5 + + SO 4 2- N 2 H 4 + HSO 4 -

14. Conjugate Acid & Bases A “conjugate” acid-base pair is a pair of chemical species that only differ by ONE proton. Eg. NH 4 + + H 2 O NH 3 + H 3 O + * Acid has the extra proton (ability to donate it) Conjugate Pair Conjugate Acid Conjugate Base NH 4 +, NH 3 NH 4 + NH 3 H 2 O, H 3 O + H3O+H3O+ H2OH2O

15. Assignment: Acid & Base Handout  Questions 1  4

16. Buffer Solutions Buffers are solutions with the ability to resist the addition of strong acids or strong bases, within limits. eg. CH 3 COOH and its salt NaCH 3 COO Most buffer solutions are made up using a WEAK ACID and its sodium salt !

17. When a strong base such as NaOH is added to the buffer CH 3 COOH … CH 3 COOH reacts with and consumes the excess OH - OH - reacts with the H 3 O + ion from the acid in the following reaction: H 2 O + CH 3 COOH H 3 O + + CH 3 COO - H 3 O + + OH - H 2 O

18. When a strong acid such as HCl is added to the buffer CH 3 COOH … H 3 O + reacts with the CH 3 COOH - ions of the salt and form more undissociated CH 3 COOH H 3 O + + CH 3 COO - H 2 O + CH 3 COOH There is a limit to the quantity of H + or OH - that a buffer can absorb without undergoing a significant change in pH.

19. Buffer Components A buffer has two components. HA NaA ---> Na + + A - (weak acid) (a soluble salt of the weak acid) Therefore any extra H 3 O + will be neutralized by the A - in the buffer H 3 O + + A - HA + H 2 O And any extra OH - that is added will be neutralized by the acid HA + OH - A - + H 2 O

20. Sample Problem: CH 3 COOH + H 2 O  CH 3 COO  + H 3 O + 1.0 M 1.0M Original Ratio: Acid = 1.0 = 1.0 Base 1.0 New Ratio: 1.1 = 1.22 0.9 ACID BASE 1.1 0.9 Add 0.1 mol H 3 O +

21. Using K a Original [ H 3 O + ] = (1.8 x 10 -5 )(1.0) = 1.8 x 10 -5 pH = 4.74 After [ H 3 O + ] = (1.8 x 10 -5 )(1.22) = 2.5 x 10 -5 pH = 4.66 pH change of 0.08

22. With no buffer present… Adding 0.1 mol of acid to H 2 O would change the pH by 6.00! pH water = 7 pH water with 0.1M H 3 O + = 1 Thus a difference of 6.