Reactions with water do NOT go to completion, so to find ion concentrations, need to know K eq and solve an equilibrium problem!

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1 Strong Acid and Base Solutions Easy to find ion concentrations! 0.1 M HCl = [H 3 O + ] = 0.1 M [OH ] = 1 x M 0.1 M NaOH = [OH ] = 0.1 M [H 3 O + ] = 1 x M Weak Acid and Base Solutions Reactions with water do NOT go to completion, so to find ion concentrations, need to know K eq and solve an equilibrium problem! Acid Ionization Equilibria Acid ionization (dissociation): acid reacts with water to produce H 3 O + and conjugate base ion HA (aq) + H 2 O (l) weak, monoprotic acid K c = H 3 O + (aq) + A (aq) [H 3 O + ] [A ] [HA] [H 2 O] K c [H 2 O]= [H 3O + ] [A ] [HA] conjugate base K a K a determined experimentally one of two ways: 1. Measure conductivity to get degree of ionization 2. Find ph

2 Lactic acid, HC 3 H 5 O 3, is found in sour milk. A M solution of lactic acid has a ph of What is K a for this acid, and what is the degree of ionization? I C E K a = HLac (aq) + H 2 O (l).025 M x.025 x.025 x H 3 O + (aq) + Lac (aq) ~0 0 +x +x x x x = [H 3 O + ] = 10 ph = K a = (.00178) 2 = ( ) Degree of ionization = 1.37 x x = =.0712 = 7.12% Weakest Acids have the smallest K a values! Use Table 16.1 pg. 656 Given to you on AP Exam

3 Base Ionization Equilibria Base ionization (dissociation): base reacts with water to produce OH and conjugate acid ion B (aq) + H 2 O (l) weak base K c = OH (aq) + HB + (aq) [OH ] [HB + ] [B] [H 2 O] conjugate acid K c [H 2 O]= [OH ] [HB + ] [B] K b Base ionization equilibria treated similarly to acids! Weakest Bases have the smallest K b values! Use Table 16.2 pg. 665 Given to you on AP Exam

4 Calculations with K a and K b When you know K a or K b, you can find [H 3 O + ], [OH ], [HA], [A ], [B], [HB + ], degree of ionization and ph!!! They are equilibrium problems (ICE Table!), and you may need to use the Quadratic Equation! BUT, since dissociations are so small, often times you can make an assumption to avoid the Quadratic... IF [ ] > 100, no Quadratic! K If < 100, Quadratic! Calculations with K a and K b What are the concentrations of nicotinic acid (niacin), H 3 O +, and the nicotinate ion in a solution of 0.10 M nicotinic acid at 25 C? Also, what is the ph of the solution and the degree of ionization? K a = 1.4 x 10 5 HNic (aq) + H 2 O (l) H 3 O + (aq) + Nic (aq) I.010 M ~0 0 C x +x +x E.010 x x x K a =.010 x Degree of ionization = x (714) No Quadratic! K a = = 1.4 x x = 1.2 x 10 3 M [H 3 O + ] and [Nic] [HNic] = 0.1 M ph = log (1.2 x 10 3 ) = 2.92 =.012 = 1.2%

5 Calculations with K a and K b Morphine, C 17 H 19 NO 3, is administered medically to relieve pain. What is the ph of a M solution of morphine at 25 C? K b = 1.6 x 10 6 Also, what is the [H 3 O + ]? Mor (aq) + H 2 O (l) HMor (aq) + OH (aq) I.0075 M ~0 0 C x +x +x E.0075 x x x K b =.0075 x x (4688) No Quadratic! K b = = 1.6 x x = 1.1 x 10 4 M [OH ] poh = log (1.1 x 10 4 ) = 3.96 ph = [H 3 O + ] = 10 ph or K w / [OH ] = 9.12 x M Weak Polyprotic Acids H 2 CO 3 (aq) + H 2 O (l) H 3 O + (aq) + HCO 3 (aq) HCO 3 (aq) + H 2 O (l) K a1 = [H 3 O + ] [HCO 3 ] [H 2 CO 3 ] H 3 O + (aq) + CO 3 2 (aq) [H 3 O + ] [CO 2 3 ] K a2 = [HCO 3 ] 4.3 x x st H + lost much more easily! H 3 PO 4 K a1 > K a2 > K a3

6 Acid Base Properties of Salt Solutions Salt: an ionic compound obtained by a neutralization reaction in aqueous solution *salt solution may be neutral, but often acidic or basic Neutralization of strong acidstrong base Neutralization of weak acids or bases How do you predict if a salt solution is acidic/basic/neutral? Rules pg A salt of a strong base and a strong acid gives a neutral aqueous solution NaOH (aq) + HCl (aq) NaCl (aq) + H 2 O (l) Na + (aq) + Cl (aq) Complete Ionization Both ions unreactive with water No hydrolyzable ions Acid Base Properties of Salt Solutions 2. A salt of a strong base and a weak acid gives a basic aqueous solution NaOH (aq) + HCN (aq) NaCN (aq) + H 2 O (l) Na + (aq) + CN (aq) Hydrolysis of CN CN (aq) + H 2 O (l) HCN (aq) + OH (aq) Basic! *Hydrolysis = ion + water conjugate acid + OH ion + water conjugate base + H 3 O +

7 Acid Base Properties of Salt Solutions 3. A salt of a weak base and a strong acid gives an acidic aqueous solution NH 3 (aq) + HCl (aq) NH 4 Cl (aq) Hydrolysis of NH 4 + NH 4 + (aq) + H 2 O (l) NH 4 + (aq) + Cl (aq) NH 3 (aq) + H 3 O + (aq) Acidic! Acid Base Properties of Salt Solutions 4. A salt of a weak base and a weak acid...both ions hydrolyze, so acidic/basic depends on relative strengths of the ions *Compare K a of the cation with the K b of the anion: K a > K b = acidic solution K b > K a = basic solution NH 3 (aq) + HCN (aq) NH 4 CN (aq) NH 4 + (aq) + CN (aq) K a for NH 4 + = 5.6 x K b for CN = 2.0 x 10 5 Basic!

8 Predict if a salt solution is acidic/basic/neutral GOT TO KNOW THE IONS THAT COME FROM OUR STRONG ACIDS AND BASES!!! STRONG ACIDS ClO 4 Cl NO 3 SO 4 2 Br I STRONG BASES Li + Ca +2 Na + K + Sr +2 Ba +2 a) KCl b) NaF K + Cl Na + F c) Zn(NO 3 ) 2 d) NH 4 NO 2 Zn +2 NO 3 NH + 4 NO 2 K a for NH 4 + = 5.6 x K b for NO 2 = 2.2 x Typically, the K a and K b values for the conjugate acids and bases are not found listed in tables, so... K a x K b = K w Relationship between K a and K b for conjugate acid base pairs: HCN (aq) + H 2 O (l) CN (aq) + H 3 O + (aq) K a CN (aq) + H 2 O (l) HCN (aq) + OH (aq) K b 2 H 2 O (l) H 3 O + (aq) + OH (aq) K w When two reactions are added, their equilibrium constants are multiplied!

9 Find the missing K! K x a K b = K w K b for CN =? K a for HCN = 4.9 x x x = 2.0 x 10 5 K a for NH 4 + =? K b for NH 3 = 1.8 x x x 10 5 = 5.6 x Another reason to know this... find the ph of a salt solution! ph of a Salt Solution What is the ph of 0.10 M solution of sodium nicotinate at 25 C? K a for HNic = 1.4 x 10 5 Nic (aq) + H 2 O (l) HNic (aq) + OH (aq) E.10 x x x 1.0 x 10 K b = K.10 x b = 14 = 7.1 x x x (huge!) No Quadratic! x = x = 8.4 x 10 6 M = [OH ] poh = 5.1 ph = 8.9