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Chapter 16 – Acid‐Base Equilibria
16.1 Acids & Bases: A Brief Review
‐ Arrhenius acids and bases: ‐‐ acid: an H +^ donor HA(aq) H(aq) ^ A(aq) ‐‐ base: an OH‐^ donor MOH (^) (aq) M(aq) ^ OH(aq) ‐ Brønsted‐Lowry acids and bases: ‐‐ acid: an H+^ donor HA(aq)^ H(aq)^ ^ A(aq) ‐‐ base: an H+^ acceptor H (^) (aq)^ B(aq) BH(aq)
16.2 Brønsted‐Lowry Acids & Bases
‐ The H+^ Ion in Water ‐‐ H 3 O ^ is called the hydronium ion and is what truly happens when H +^ is in water ‐‐ since all aqueous solution are in water we will use H 3 O +^ & H+^ interchangeably ‐ Conjugate Acid‐Base Pairs ‐‐ conjugate acid: the acid that is created after the Brønsted‐Lowry base has accepted the proton, BH+ ‐‐ conjugate base: the base that is created after the Brønsted‐Lowry acid has donated the proton, A‐
‐‐ examples: basic: (^) 3( g ) 2 ( ) l 4( aq ) ( aq ) b 1.76 105 base acid conj acid conj base
NH H O NH ^ OH ^ K
^ ^ ^
acidic: (^) (^) 2 3 a 1 acid (^) base conj base conj acid
HCl H O Cl ^ H O K (^)
‐ Relative Strengths of Acids & Bases
‐‐ a strong acid will completely dissociate/ionize in solution: HNO 3( (^) aq ) H O 2 ( ) (^) l H O 3 ( aq^ ) NO 3( aq ) ‐‐‐ as soon as a strong acid is placed in water it will ionize completely ‐‐‐‐ all the reactant goes to product ‐‐‐‐ product is a very weak conjugate acid and base ‐‐‐ other strong acids: HCl, HBr, H 2 SO 4 , HI, HClO 4 , HClO 3 , HNO (^3) ‐‐ the same is the case for a strong base
‐‐‐ examples of strong bases: MOH (where M is alkali metal), NH 2 ^ , H‐ ‐‐ a weak acid will only partially dissociate HNO 2( ^ aq ) H (^) ( aq^ ) NO 2( ^ aq ) Ka 4.0 10 ^4 ‐‐‐ the eq constant is called Ka “a” for acid ‐‐‐ in this case some of the reactant is present at eq ‐‐ the larger the K (^) a the more an acid dissociates, the stronger the acid ‐‐‐ e.g. for nitric acid Ka >> 1 ‐‐ we talk about K (^) a 's at length a little later
16.3 The Autoionization of Water
‐ as we will soon learn water may act as either an acid or a base ‐ amphoteric ‐‐ water as a base: HA(aq) H O 2 (l) H O 3 (aq)^ A(aq) ‐‐ water as an acid: B(aq) H O 2 (l) BH (^) (aq)^ OH(aq) ‐ Autoionization occurs when a species can ionize itself this occurs for water as shown below: 2 H O 2 ( ) (^) l OH (^) ( aq^ ) H O 3 ( aq^ ) K (^) c [ OH (^) ( aq^ ) ][ H O 3 ( aq )] ‐ since water is in excess (it’s the solvent) the concentration is constant Kw [ OH (^) ( aq^ ) ][ H O 3 ( aq )] ‐ relationship between [ OH (^) ( aq^ )]and [ H O 3 ( aq )] 3 ( ) ( ) 3 ( ) ( ) 3 ( ) ( )
[ ] [ ]
[ ] [ ]
[ ] [ ]
aq aq aq aq aq aq
H O OH acidic H O OH neutral H O OH basic
‐ the value for both [ OH ( aq^ )]and [ H O 3 ( aq^ )]in pure water are 1.0 x 10 ‐^7 at 298K Kw (1.0 10 ^7 )(1.0 10 ^7 ) 1.0 10 ^14 ‐ Example: Determine the hydroxide concentration in a solution with [ H O 3 ( aq^ )] 1.89 10 ^4 M. (^1411) 3 ( ) ( ) ( ) (^3) ( ) 4
[ ][ ] [ ] 1.0^10 5.29 10
[ ] 1.89 10
w aq aq aq^ w aq
K H O OH OH K M
H O
16.4 The pH Scale
‐ pH is a log scale which describes the "power of hydrogen" ‐ the pH is related to the [ H O 3 ( aq^ )]: pH log[ H O 3 ( aq )] ‐‐ for pure water: pH log[ H O 3 ( aq^ )] log(1.0 10 ^7 ) 7 this is neutral pH ‐‐ when a solution has a pH < 7.0 it is acidic ‐‐ when a solution has a pH > 7.0 it is basic ‐ pOH & Other ‘p’ Scales ‐‐ all ‘p’ scales mean ‘‐log’ of ‐‐ pOH is exactly like pH except it is dependent upon [ OH ( aq^ )]: pOH = ‐log [ OH (^) ( aq )] ‐‐ In Chapter 17 we will have pK (^) a and pK (^) b meaning the –log(Ka ) etc. ‐‐ the pH and pOH are related to K (^) w
‐‐ the larger the K (^) a the more strongly the equilibrium will lie toward product ‐‐ the more likely the acid is to dissociate and raise the acidity of the solution ‐‐ the stronger the acid
‐ Calculating K (^) a from pH ‐‐ Similar to what we did last chapter with the %dissociation, we can get the K (^) a from the pH ‐‐ Example: What is the Ka of a 0.050 M solution of HC 7 H 5 O 2 if the pH of this solution is 2.75? HC H O 7 5 2( (^) aq ) C H O 7 5 2( ^ aq ) H ( aq ) We begin with an ICE Table HC H O 7 5 2( (^) aq ) C H O 7 5 2( aq ) H (^) ( aq ) Initial 0.050 0.0 0. Change ‐x +x +x Eq 0.050 – x +x +x Next, we write down symbolic representation of the equilibrium/acid dissociation expression: 7 5 2( ) ( ) 7 5 2( )
[ ][ ]
[ ]
a^ aq^ aq aq
K C H O^ H
HC H O
Now we remember we don’t know K (^) a but need to find it which means we must know x: [ H (^) ( aq^ )] 10 ^ pH 10 ^ 2.75^ 1.77 10 ^3 M Finally we plug in x for our equilibrium expression and solve 7 5 2( ) ( ) 2 2 5 7 5 2( )
[ ][ ] 0.00177 6.5 10
[ ] 0.050 0.050 0.
a^ aq^ aq aq
K C H O^ H^ x HC H O x
(^) (^)
‐ Percent Ionization ‐‐ degree of ionization/dissociation: percentage that an acid ionizes ( ) ( ) ( )
[ ] 100%
aq aq aq [ ]
HA H A H
HA
^ ‐‐ Example: Determine the percent dissociation of 0.050M of benzoic acid. HC H O 7 5 2( (^) aq ) C H O 7 5 2( ^ aq ) H (^) ( aq^ ) Ka 6.5 10 ^5 We already will find [ H ^ ] 1.77 10 ^3 M therefore
[ ] 1.77 10 3
HHA 0.050 MM 100% 3.54%
It should be small since our K (^) a is so small ‐ Using K (^) a to Calculate pH/pOH ‐‐ we use ICE tables
‐‐ Ex: Calculate [H+^ ] and the pOH of 0.050M of benzoic acid. HC H O 7 5 2( (^) aq ) C H O 7 5 2( ^ aq ) H (^) ( aq^ ) Ka 6.5 10 ^5 HC H O 7 5 2( (^) aq ) C H O 7 5 2( aq ) H (^) ( aq ) Initial 0.050 0.0 0. Change ‐x +x +x Eq 0.050 – x +x +x
7 5 2( ) ( ) (^25) 7 5 2( ) 2 6 5 2 5 6 5 5 2 6
(^5 )
3
[ ][ ] 6.5 10
[ ] 0.
[ ] 1.77 10
a^ aq^ aq aq
K C H O^ H^ x HC H O x x x x x
x
x x M H M pH pOH
^ ^ ^
pOH 14 pH 14 log[ H ] 11. ‐ Polyprotoic Acids ‐‐ for acids containing 1 proton we call them monoprotic ‐‐ for two they are diprotic (e.g. H 2 SO 4 ) ‐‐ for three they are called triprotic (H 3 PO 4 ) ‐‐ Example: Calculate the [H+^ ] of 0.050M of sulfuric acid. 2 4( ) ( ) 4( ) 4( ) ( ) 4(^2 )^2
aq aq aq a aq aq aq a
H SO H HSO K
HSO H SO K
Initially all 0.050M of the H 2 SO 4 dissociates completely into 0.050M H (^) ( aq^ ) HSO 4( aq ) HSO 4( aq ) H^ ( aq^ ) SO 4(^2 aq ) Initial 0.050 0.050 0. Change ‐x +x +x Eq 0.050 ‐ x 0.050+x +x
(^)
( ) 4(^2 ) (^2) 4( ) 2 4 2 2 4 (^2 )
(^23)
[ ][ ] (0.050 ) 1.2 10
[ ] (0.050 )
a^ aq^ aq aq
K H^ SO^ x x HSO x x x x x x
x
x x M
^ ^
[H+^ ] = 0.050+0.0085 = 0.059M
‐‐ Why is Ka1 > Ka2? 2 4( ) ( ) 4( ) 4( ) ( ) 4(^2 )^2
aq aq aq a aq aq aq a
H SO H HSO K
HSO H SO K
4 4 4 4 2 3 3 3 5 14 10 (^3 2 4 )
for :? we will not find this in a table BUT we can find the of to it: 1.8 10 10 5.56 10 1.8 10 for we will have to
a a b b a w b
NH CN NH CN
NH NH H O NH H O K
K K NH
NH H O NH OH K K K
K
CN
2
use the of HCN to get its ?
a b b
K K
CN ^ H O HCN OH K
10 14 5 (^2 3 ) 4
the soln is basic
a b^ w a b a
HCN H O CN H O K K K
K
K CN K NH
Example: Calculate the pH of a 0.25M NaC 2 H 3 O 2 , K (^) a = 1.76x10‐^5. (^14 ) 5
b^ w a
K K
K
(^)
C H O 2 3 2( ^ aq ) H O 2 ( ) (^) l HC H O 2 3 ( (^) aq ) OH ( aq ) C H O 2 3 2(^ aq ) HC H O 2 3 (^ aq ) OH aq Initial 0.250 0.0 0. Change ‐x +x +x Eq 0.250 ‐ x +x +x 2 3 ( ) (^210) 2 3 2( )
[ ][ ] 5.68 10
[ ] 0.
b^ aq aq
K HC H O^ OH^ x C H O x
(^) (^) because we have a large concentration of acetate & a small K (^) b we will try and assume 0.250 >> x (^2 210 )
5
0.250 0.250 5.68^10 1.19^10
: 1.19^10 100% 0.005% 5%
x x (^) x M x ck
^ ^ ^ ^
therefore our assumption is valid and [OH‐] = 4.77x10‐^5 M pH = 14 ‐ pOH = 14 + log[OH] = 9.
16. 10 Acid‐Base Behavior & Chemical Structure
‐ electronegativity ‐ draw trend ‐‐ acid strength down a column in the periodic table ‐‐‐ as we go down a column we decrease EN ‐‐‐ a more EN atom will have a stronger bond to H + ‐‐‐‐ if the H‐X bond is strong then it will be hard to dissociate HX ‐‐‐‐ this will lead to a weaker acid in solution ‐‐‐ acid strength: HF < HCl < HBr < HI
‐‐ acid strength going across a row in the periodic table
‐‐‐ as we go across a row from left to right we increase EN ‐‐‐ we also increase polarity of the H‐X bond ‐‐‐‐ this means that if the H‐X bond is broken it does not necessarily lead to H +^ (e.g. CH 4 ) ‐‐‐‐ if we don't have H+^ then we have a weaker acid ‐‐‐ acid strength: CH 4 < NH 3 < H 2 O < HF ‐ oxoacids: HnXO (^) m ‐‐ electronegativity effects ‐‐‐ for oxoacids of the halogens the trend is: HOI < HOBr < HOCl ‐‐‐ this trend occurs because as we increase the EN of the halogen we are pulling electron‐density away from the O‐atom ‐‐‐ if we take electron‐density away from the O‐atom we weaken the O‐H bond ‐‐‐ this will in turn allow H+^ the freedom to break‐away and go into soln ‐‐ number of oxygens in the acid (oxoacids) ‐‐‐ if we increase the number of O‐atoms in an acid then we increase its strength ‐‐‐‐ this because we increase the oxidation number of the central atom ‐‐‐‐ Example: Oxoacids of Chlorine Acid Oxidiation State of Cl K (^) a HClO +1 2.9 x 10 ‐^8 HClO 2 +3 1.1 x 10 ‐^2 HClO 3 +5 1 HClO 4 +7 1 x 10 8
16.11 Lewis Acids & Bases
‐ Definition: ‐‐ a base is a substance that donates pairs of electrons ‐‐ an acid is a substance that accepts pairs of electrons
‐‐ e.g. NH 3 BF 3 H NBF 3 3 where NH 3 donates e‐'s (base) and B is the acceptor (acid)
