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Intersection 11 11/12/05 Reading: 16.8 (p794-800) 17.2 (p 828-836)

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Presentation on theme: "Intersection 11 11/12/05 Reading: 16.8 (p794-800) 17.2 (p 828-836)"— Presentation transcript:

1 Intersection 11 11/12/05 Reading: 16.8 (p794-800) 17.2 (p 828-836)

2 Equilibrium Representations

3 Water Projects…Now What?

4 Outline Strong vs. Weak Acids and Bases Which H’s are acidic? Which groups are basic? Periodic Trends and Acid Strength What affects the pH of a solution? Polyprotic acids Concept Questions Acid/Base Titration: A Closer Look

5 Strong vs. Weak A

6 Vocabulary Strong (16 definitions): Weak (10 definitions): Favorable reaction.. ionizing freely in solution ionizing only slightly in solution Strong Reaction exothermic spontaneous product favored goes to completion A

7 Strong Acids An acid that dissociates completely (the equilibrium is shifted all of the way to its conjugate base and hydronium ion) is said to be a strong acid. HCl (aq) + H 2 O (l) → H 3 O + (aq) + Cl - (aq) acid conj. base An acid that does not dissociate completely (an equilibrium is established in solution between the acid, its conjugate base, and hydronium ion) is said to be a weak acid. HClO 2(aq) + H 2 O (l) ↔ H 3 O + (aq) + ClO 2 - ( aq) acid conj base K a = ([H 3 O + ][ClO 2 - ]) / [HClO 2 ] A

8 Strong Bases A base that dissociates completely (the equilibrium is shifted all of the way to its conjugate acid and hydroxide) is said to be a strong base. NaOH (aq) + H 2 O (l) → OH - (aq) + Na + (aq) + H 2 O (l) base conj. acid A base that does not dissociate completely (an equilibrium is established in solution between the base, its conjugate acid, and hydroxide) is said to be a weak base. (CH 3 ) 3 N (aq) + H 2 O (l) ↔ (CH 3 ) 3 NH + (aq) + OH - (aq) base conj. acid K b = ([(CH 3 ) 3 NH + ][OH - ]) / [(CH 3 ) 3 N] A

9 Strong Acid Weak Acid A

10 There are six strongly dissociating acids: HCl HNO 3 HBr HClO 4 HI H 2 SO 4 There are also five bases that dissociate completely in solution (strong): LiOH Ca(OH) 2 NaOH Ba(OH) 2 KOH You should commit the strong acids and bases to memory. Appendix F in your text book lists K a and K b values for many weakly dissociating acids and bases. A

11 Problem 1 Trimethylamine (CH 3 ) 3 N has a K b of 6.5 x10 -5. Write out its chemical reaction with water: What is the [OH - ] of a 0.010 M solution of triethylamine? What is the pOH? What is the pH? M

12 Which H’s are acidic? What groups are basic? M

13 What kinds of hydrogen atoms (protons) are acidic? H-halogen (HF, HCl, HBr, HI) H2OH2O H 2 S (K a1 = 8.9x10 -8 ) Oxoacids (H-polyatomic ions) (H 2 CO 3, HNO 3, etc.) HCN M

14 Organic Acids RCOOH M

15 Bases OH R 3 N M

16 Periodic Trends and Acid Strength A

17 Groups 7 and Period 2 Acid strength: HI > HBr > HCl> HF HF>H 2 O >> H 3 N >>> CH 4 A

18 Oxoacids I HOY + H 2 O H 3 O + + OY - Since a negatively charged ion must be formed, it will be most stable when Y is an atom that is most effective at attracting electron density to itself, thereby stabilizing the negative charge. So the more electronegative atom as Y, yields a stronger acid Acid strength: HOI < HOBr < HOCl Y = I, Br, Cl A

19 Oxoacids II What trend do you see and why? NameHypochlorousChlorousChloricPerchloric FormulaHOClHOClOHOClO 2 HOClO 3 Experimental K a 3.2 x10 -8 1.3 x10 -2 1x10 2 2 x10 7 A

20 What affects the pH? M

21 Question 1 How could you change the pH of a solution of acetic acid (CH 3 COOH)? M

22 Question 2 Apatite, Ca 5 (PO 4 ) 3 OH is the mineral in teeth. Ca 5 (PO 4 ) 3 OH(s) 5 Ca +2 (aq) + 3PO 4 -4 (aq) + OH - (aq) Sour milk contains lactic acid. Not removing sour milk from the teeth of young children can lead to tooth decay. Use chemical principles to explain why. M

23 Can salts affect the pH of a solution? Ca(OH) 2 Na(CH 3 COO) NH 4 Cl NaNO 3 K sp = 7.9x10 -6 K b NH3 = 1.8 x10 -5 K a CH3COOH = 1.8 x10 -5 NH 4 + + H 2 O NH 3 + H 3 O + NaH 2 PO 4 Na 2 HPO 4 M

24 Name ions found in salts that would not affect the pH: M

25 Question 3 A carbonated beverage is left open to the atmosphere. Will the pH change? CO 2 (aq) + H 2 O(l) ↔ H 2 CO 3 (aq) H 2 CO 3 + H 2 O(l) ↔ H 3 O + + HCO 3 ¯ K a1 = 4.2 × 10 -7 HCO 3 ¯ + H 2 O(l) ↔ H + + CO 3 2 ¯ K a2 = 4.8 × 10 -11 M


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