ACIDS, BASES & ACID/BASE EQUILIBRIA

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1 Chapter 5 ACIDS, BASES & ACID/BASE EQUILIBRIA ill, Petrucci, McCreary & Perry 4 th Ed. ACIDS & BASES The Arrhenius Definition: In Water: Acid Substance which increases the concentration of ydrogen Ion, [ + ]. Base Substance which increases the concentration of ydroxide Ion, [ - ]. Strong Acids: 2 S 4, I, Br, Cl, N 3, Cl 4. Strong Bases: ydroxides of Li +, Na +, K +, Rb +, Cs +, Ba 2+, Sr 2+, Ca 2+. ACIDS & BASES The Bronsted-Lowry Definition: The B-L definition is independent of solvent. Acid Substance which donates a ydrogen Ion, +. Base Substance which accepts a ydrogen Ion, +. Conjugate Acid Base Pairs: Species which differ by +. Examples (Acid/Base): Cl/Cl -, N 4+ /N 3, 3 + / 2, 2 / -

2 ACIDS & BASES The Lewis Definition: The Lewis definition focuses on electrons and is independent of solvent. Acid Substance which accepts an electron pair. Base Substance which donates an electron pair. N F + B F F N F B F F ther Lewis Acids Many Metal Ions: 6 + Al 3+ Al 6 3+ Some Acids are not what they seem: 3 B 3 Boric Acid is not an xoacid, but a Lewis Acid: B B Bronsted Acid/Base Reactions A competition for ydrogen Ion (Protons) Cl + Cl + Conjugates Conjugates - + Stronger Acid than 3 + Stronger Base than Cl - Weaker Base than 2 Weaker Acid than Cl

3 B/L Acids/Bases are Named by Function - N + N + Strong Base Very Weak Acid Conjugates Conjugates Very Weak Acid - Strong Base (but weaker than N 2 - ) Strong Base Very Weak Acid Conjugates Conjugates Very Weak Acid Strong Base (but weaker than -2 ) B/L Acid/Base Strength Relationships Strongest Acids <=> Weakest Conjugate Bases Strongest Bases <=> Weakest Conjugate Acids Weak Acids <=> Weak Conjugate Bases Weak Acids/Bases have pk a/b between 2 & 2 Weak Acids & Weak Bases <0% C C <0% 3 N N 4 + pka & pkb of Conjugate Acid/Base Pairs I I Br Br Cl Cl 3 2 F F C CN C CN N 3 N 4 2 Strong Weak Very Weak

4 Competition for Protons Determining the Direction of Reaction: Direction 2 S + C Favored C weaker acid than acetic weaker base than S - ion. stronger acid than 2 S S stronger base than acetate ion See Table 5., p 620, for Relative Acid/Base Strengths The ydrogen Sulfide Ion is a stronger base than the acetate ion and will successfully compete for the ydrogen Ion. Factors Affecting Acid Strength ) Polarity of the -X Bond. The more electronegative X is, the more polar is, more charge separation in, the bond. δ X δ Where δ means not a full positive charge. 2) Strength of the -X Bond. The stronger the bond the harder it is to ionize. δ + X δ X This Bond must break to form ions! Trends in Binary Acid Strength F < Cl < Br < I Most Greatest Polarity Bond Strength Least Least 2 < 2 S < 2 Se < 2 Te In Water F is a weak acid. Cl, Br and I are strong acids.

5 Acidity Trends in xo-acids All xo-acids contain --X Bonds of near equal strength. The - Bond is strong! Electronegativity Trends --Cl > --Br > --I xygen Substitution is more important! Cl < Cl 2 < Cl 3 < Cl 4 Acidity Trends in xo-acids Why xygen Substitution is more important! Cl < Cl 2 < Cl 3 < Cl 4 Cl < Cl < Cl < Cl The key is the Conjugate Base Strength! δ δ Cl > Cl > δ δ Cl δ > δ δ δ Cl Each added xygen reduces the average charge! δ PLYPRTIC ACIDS More Than ne Proton Ionizes Binary Types: 2 S + S 2 S > S S + S 2 xo-acid Types: 2 S 4 + S 4 2 S 4 > S 4 S S 4 3 P 4 > > P P 4

6 2 The Auto-Ionization of Water Kc = [ 3 ] [ ] [ 2 ] 2 = 3.2 x 0-8 In dilute aqueous solution: [ 2 ] ~ constant 998 g/l [ 2 ] = = 55.4 mol/l 8.02 g/mol Kc [ 2 ] 2 = [ 3 ] [ ] = 3.2 x 0-8 (55.4 mol/l) 2 Kc [ 2 ] 2 Kw = [ 3 ] [ ] =.0 x 0-4 Ion Product of Water For Pure Water at 25 o C: [ 3 = [-] =.0 x 0-7 M For Acid Solution: [ 3 [ >.0 x 0-7 M [-] For Basic (Alkaline) Solution: In all cases the [ - ] X [ + ] = K w [-] [ <.0 x 0-7 M >.0 x 0-7 M <.0 x 0-7 M p: A Convenient Measure of [ + ] p -log [ p [ [-] Basic (Alkaline) Solution: p Neutral Solution: Acidic Solution: -log [-] pkw -log(kw) pkw = Some others we will use later pka pkb -log(ka) -log(kb)

7 Strong Acid Solution A solution which is M Cl: 00% Cl (aq) + (aq) + Cl- (aq) -x +x +x M M M p = -log(0.080) = -(-.0) =.0 [-] = Kw M =.0 x M =.25 x 0-3 M p = -log [ - ]= -log (.25 x 0-3)= 2.90 A Strong Base Solution A solution which is M Ca() 2 : Ca() 2(aq) 00% Ca2+ (aq) (aq) -x +x +2x M M M p = -log(0.080) = -(-.0) =.0 [ = Kw M =.0 x 0-4 =.25 x 0-3 M M p = -log(.25 x 0-3 M) = -(-2.90) p = 2.90 Calculation of [ + ] or [ - ] from p range juice has a p of Calculate the [ & [-]. Note: p + p = 4.00 (pkw). p = 3.54 = -log[ [ = [ = 2.88 x 0-4 M Always change the sign of p or p before taking the antilog. p = p = 0.46 [-] = [-] = 3.47 x 0 - M

8 Weak Acid Equilibria A (aq) + 2 (aq) 3 + (aq) + A- (aq) Kc = [ 3 [A-] [A] [ 2 ] Ka Kc [ 2 ] Ka = but [ 2 ] ~ [ 3 [A-] [A] [[A-] [A] constant! Weak Acid Equilibria: 0.0 M A in water. [initial] [equil] A (aq) + (aq) + A- (aq) 0.0 M 0 M 0 M -x +x +x x +x +x Ka = [ [A-] (x)(x) = [A] (0.0 - x) [[A-] = Ka [A] but [ = [A-] 2 [ = Ka [A] eq but [A] >> x, a weak acid [ ~ = Ka [A] init 2 Calculation of the Ionization Constant, Ka, of a weak acid from a p measurement. A M solution of Lactic Acid, Lac, has a p = 2.75, calculate the Ka. Lac aq aq + Lac aq -x +x +x Ka = [ ][Lac ] [Lac]

9 Percent Ionization & Acid Concentration Any weak acid (or base) dissociates to a greater extent as it becomes more dilute. LeChatelier s Principle & the Common Ion Effect A (aq) + 2 (aq) 3 + (aq) + A- (aq) Kc = [ 3 [A-] [A] [ 2 ] as [ 2 ] increases and K c remains constant the values of [ 3 + ] & [A - ] must increase and [A] must decrease! Example: Weak acid Equilibrium Calculate the Equilibrium Concentrations for M propionic acid, Ka =.3 x 0-5 [init] [eq] Pr aq aq + Pr aq x +x +x x x x Ka = [ ][Pr ] [Pr] Solution: Ka = [ ][Pr ] [Pr] = (x) (x) ( x) =.3 x 0-5 If Ka/[A] < 0-2 This equation will produce a x << [A] quadratic unless we approximate. eq and Errors less than 5% are acceptable. [A] eq ~ [A] initial (x) (x) (x) (x) Ka = =.3 x 0-5 ( x) ~ (0.020) a quadratic not a quadratic x = 5.0 x 0-4 x = 5. x 0-4 p = 3.30 ~ p = 3.29

10 Ionization Equilibria Weak Bases N 3(aq) + 2 (aq) N 4 + (aq) + - (aq) Kc = [N 4 [-] [N 3 ] [ 2 ] Kb Kc [ 2 ] but [ 2 ] ~ constant! = [N 4 [N 3 ] [-] Kb [N 3 ] = [N 4 [-] but: [N 4 = [-] [-] = Kb [N 3 ] eq 2 ~ Kb [N 3 ] init 2 Calculation of K b from p Measurements A M solution of Quinine, Qn, has a p = Calculate the K b for quinine. Qn aq Qn aq + aq [ ] eq x x x [Qn ] = [ ] = 0- ( ) = p [Qn ] = [ ] = 6.92 x 0-5 M Kb = [Qn ] [ ] [Qn] = (6.92 x 0-5)(6.92 x 0-5) ( x 0-5) Kb = (6.92 x 0-5) 2 = 3.2 x 0-6 ( ) Acid/Base Properties of Salts ydrolysis of Salts Salts of SA/SB are neutral. Both the conjugate base/acid are very weak neither the conjugate acid or base alter p = Salts of SA/WB are acidic. The conjugate acid of the weak base makes p < Salts of WA/SB are basic. The conjugate base of the weak acid makes the p > 7.00

11 pka & pkb of Conjugate Acid/Base Pairs I I Br Br Cl Cl 3 2 F F C CN C CN N 3 N 4 2 Strong Weak Very Weak Salts of a Strong Acid / Weak Base Calculate the p of 0.0 M Ammonium Chloride N 4(aq) + (Cl ) (aq) + N 3(aq) + (Cl ) x x x Ka = [N 3 ][ [N 4 x [ -] [-] rearranging Kw [N 3 ] [[-] Ka = [N4 [-] Kb Ka =.0 x x 0-5 => K w Kb For conjugate pairs Ka x Kb = Kw Calculate the p of 0.0 M Ammonium Chloride N 4(aq) + (Cl ) (aq) + N 3(aq) + (Cl ) x x x Ka = [N 3 ][ [N 4 = Ka = 5.6 x 0-0 = Kw Kb (x) (x) x [ = x = (5.6 x 0-0)(0.0) =.0 x x 0-5 ~ x = 7.5 x 0-6 M p = 5.3 an acidic solution!

12 Salts of a Weak Acid / Strong Base Calculate the p of 0.0 M Sodium Cyanide (Na (aq) ) + CN (aq) CN (aq) + (aq) + (Na (aq) ) [ ] eq x x x Kb = [CN] [ -] [CN-] Kb = x [ [ Kw [CN] [[-] [CN-][ Ka Kb =.6 x 0-5 = K w Ka =.0 x x 0-0 Calculate the p of 0.0 M Sodium Cyanide (Na (aq) ) + CN (aq) CN (aq) + (aq) + (Na (aq) ) [ ] eq x x x Kb = [CN] [ -] [CN-] Kb =.6 x 0-5 =.6 x 0-5 = (x) (x) x [-] = x = (.6 x 0-5) (0.0) ~ x =.3 x 0-3 M p = 2.9 p = =. a basic solution! The Common Ion Effect: Calculate the p of a 0.0 M acetic acid solution, then add enough dry sodium acetate to make the solution 0.0 M acetate. C 2 3 2(aq) (aq) + C 2 3 2(aq) [ ] eq x x x [ = (.8 x 0-5)(0.0) The new dissociation z is a lot smaller than x. 2 =.3 x 0-3 M we dropped -x here! p = 2.89 When we add the base and common ion acetate we will suppress the ionization of acetic acid! [ ] eq z z z

13 Now calculate the p of a 0.0 M acetic acid solution after adding the dry sodium acetate to make the solution 0.0 M in acetate. C 2 3 2(aq) (aq) + C 2 3 2(aq) [ ] eq z z z [C ] Ka = [ ] [C ] Solving for [: we can drop +z here! = (z)(0.0 + z) (0.0 - z) we can drop -z here! [ = Ka [C ] = (.8 x 0-5)(0.0 - z) [C ] (0.0 + z) Calculate the p of a 0.0 M acetic acid solution, then add enough dry sodium acetate to make the solution 0.0 M acetate. 2 C 2 3 2(aq) (aq) + C 2 3 2(aq) [ ] eq z z z [ we dropped -z here! (.8 x 0-5)(0.0 - z) = ~ (.8 x 0-5)(0.0) (0.0 + z) (0.0) we dropped +z here! [ =.8 x 0-5 M. p = 4.75 Compare this to just the weak acid! p = 2.89 Buffers the ow to Control p C 2 3 2(aq) (aq) + C 2 3 2(aq) [ ] eq x x x this amount of acid will absorb a lot of hydroxide ion. this amount of base will not absorb a lot of hydrogen ion. If we add the conjugate base the solution will absorb both! [ ] eq z z z [ = these amounts of both conjugates will absorb comparable amounts of hydrogen or hydroxide ion. Ka (0.0) (0.0) = Ka =.8 x 0-5 M. p = 4.75 = pka

14 The enderson asselbalch Equation A (aq) + (aq) + A- (aq) [ = Ka [A] [A-] or p = pka - log [A] [A-] [A-] p = pka + log [A] This equation will work for an acid or base buffer! or [CB] p = pka + log [CA] But it will only work when [CB] ~ [CA]!! Buffers the ow to Control p N 3(aq) + 2 (aq) N 4 + (aq) + - (aq) Kb = [-][N 4 [N 3 ] [ = Ka [N 4 [N 3 ] Ka = K w [N 3 ] = [ -][ [N 3 ] = [ Kb [-][N 4 [N 4 The enderson asselbalch Equation the pka of the conjugate acid p = pka + log [N 3] [N 4 [CB] = pka + log [CA] This equation will work for an acid or base buffer! Buffers Change p only to the Extent that the Conjugate Ratios Change. the pka of the conjugate acid [A-] p = pka + log [A] [CB] = pka + log [CA] This equation will work for an acid or base buffer! the pka of the conjugate acid p = pka + log [N 3 ] [N 4 [CB] = pka + log [CA]

15 Buffers Change p only to the Extent that the Conjugate Ratios Change. N 3(aq) N 4 + (aq) + - (aq) Adding acid will cause N 3 to react to form more N 4 Adding base will cause N 4 to react to form more N 3 [ = Ka [N 4 [N 3 ] => p = pka + log [N 3] [N 4 Determining the p of a buffer to which acid or base is added is a problem in stoichiometry and calculating the new conjugate ratio! Challenging a Buffer 5-8: A solution M in C and M in C - is prepared. What is its p? C (aq) (aq) + C (aq) + Na (aq) [ ] eq M ~ 0 M M M 5-85: To 50.0 ml of the above buffer is added.00 ml of M Cl. What is the new p? Challenging a Buffer 5-8: A solution M in C and M in C - is prepared. What is its p? C (aq) (aq) + C (aq) + Na (aq) [ ] eq M ~ 0 M M M

16 5-85: To 50.0 ml of the above buffer is added.00 ml of M Cl. What is the new p? Titration of Acids and Bases The quantitative addition of an acid solution to a base solution until the equivalence point is reached. At the equivalence point the following equation applies: Moles Acid = Moles Base Vol Acid x M Acid = Vol Base x M Base M Acid = [ ] M Base = [ ] Titrating a Weak Base with Strong Acid Calculate the volume of M Cl required to titrate 32.0 ml of M ammonia. Calculate p at endpoint. N 3(aq) + (aq) + Cl (aq) N 4(aq) + Cl (aq) moles acid = moles base (at equivalence point)

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