Title: Acid and Base Equilibrium
1Acid and Base Equilibrium
2Basic Concepts
- Acids sour or tart taste, electrolytes,
described by Arrhenius as sub that inc. the H
conc. - Bases bitter to the taste, slippery,
electrolytes, describes by Arrhenius as sub that
inc. the OH- conc.
3Dissociation of Water
- Pure water exists almost entirely of water
molecules. It is essentially a non-electrolyte. - Water ionizes to a small extent auto-ionization
- The equilibrium expression is
- H20(l) lt-gt H(aq) OH-(aq)
- Kw H OH-
4- Since the H ion does not exist alone in water Kw
is often expressed - Kw H3O OH-
- because water conc is constant it does not
appear in the expression
5Values for Kw and H and OH-
- Kw H3O OH- 1.0 x 10-14
- 1.0 x 10-14 X X
- 1.0 x 10-14 X2
- X 1.0 X 10-7 H3O OH-
6The Bronsted Lowry definition
- holds true for situations not involving water
- Acids donate protons
- Bases accept protons
7- HCl(g) NH3(g) ?? NH4(g) Cl-(aq)\
- donates H accepts H conj acid conj
base - BL-Acid BL-Base
- notice that the reaction doesnt happen in water
and that the OH- concentration has not increased
8Conjugate Acids Bases and Amphoteric Substances
- HNO2(aq) H2O(l) ?? NO2-(aq)H3O(aq)
-
conjugate - acid base base acid
- donates - accepts protons
- note HNO2 is considered a Arrenius acid H inc in
H20 also a BL because donates a Proton - all Arrenius acids and bases are also BL acids
and bases however all BL acids and bases may or
may not be arrenius acids or bases water is not
an Arrenius base in this example
9- every acid has a conjugate base formed from the
removal of a proton from that acid - every base has a conjugate acid formed from the
addition of a proton to that base
10Amphotheric Substances
- NH3(aq)H20(l) ?? NH4(aq) OH-(aq)
- Base Acid Conj Acid Conj Base
- p-acc p-donar
- note water is acting as an acid in this reaction
and a base in the previous one that makes it a
amphortic substance
11Strengths of Acid, Bases
- The stronger the acid is the weaker its conjugate
base (weaker acid ? stronger conj base) - The stronger the base the weaker its conjugate
acid (weaker base ? stronger conj acid) - Stronger acids and bases ionize to a greater
extent than do weak acids and bases. -
12Strong Acids dissociate completely into ions
- HNO3(aq) ? H(aq) NO3-(aq)
- the production of H ions from the acid
dominates ignore the H donated from the water
it is insignificant - The ? equilibrium lies so far to the right
because HNO3 doesnt reform - The neg log of the H from the acid determines
pH. - Strong bases also dissociate completely and the
conc of OH- from the base is the only factor
considered when calculating pH.
13Acids and Bases
- Strong Acids
- HClO4 prechloric
- H2SO4 sulfuric
- HI hydroiodic
- HCl hydochloric
- HBr hydrobromic
- HNO3 nitric
- Strong Bases
- GI hydroxides ex. NaOH,KOH
- G2 Hydroxides Sr(OH)2
- GI Oxides ex.
- Na2O, K2O
- GI,II Amides ex. KNH2,Ca(NH2)2
14The pH scale
- pH is defined as the neg log (base-10) of the
H ion concentration - pH -log H
- What is the pH of a neutral solution
- H 1.0 x 10-7
- pH -log 1.0 x 10-7
- pH 7
15Strong acids and the pH scale
- An acidic solution must have a H conc greater
than 1.0 X 10-7 ex 1.0 X 10-6 - -log 1.0 X 10-6 pH 6
- What is the pH of a basic solution?
- A basic solution is one in which the OH-
- is greater than 1X10-7.
16Calc. pH of strong basic solutions
- Calculate the pH of a sol that has a OH- con.
Of 1.0 X10-5 - Kw 1x10-14 H OH-
- Kw 1x10-14 H 1.0X10-5
- H 1x10-14 1.0 X 10-9
- 1.0X10-5
- pH -log 1.0 x 10-9 pH 9
17The p Scale
- The negative log of a quantity is labeled p
(quantity) - Ex we could reference the quantity of
- OH- directly pOH -logOH-
- From the definition of Kw
- -log Kw (-log H) (-log OH-) -log 1x10-4
- Kw pH pOH 14
18Calc. pH using the p scale
- Ex. OH- conc 1.0X10-5
- -log 1.0X10-5 5 pOH
- pH pOH 14
- pH 5 14
- pH 9
19Weak Acids
- partially ionize in aqueous solution
- mixture of ions and un-ionized acid in sol.
- WA are in equilibrium (H20 is left out because it
a pure liquid) - HA(aq) H20(l) ?? H30(aq) A-(aq)
- Ka is the acid dissociation constant
- Ka H30 A- H A-
- HA HA
20Acid Dissociation Constant
- The larger the Ka value the stronger the acid is
more product is in solution
21Weak Bases
- Weak bases in water react to release a hydroxide
(OH-) ion and their conjugate acid - Weak Base(aq) H2O(l) Conjugate Acid(aq)
OH-(aq)
22- A common weak base is ammonia
- NH3(aq) H2O(l) ?? NH4(aq) OH-(aq)
- Since H2O is a pure liquid it is not expressed in
the equilibrium Kb expression - Kb NH4OH- (base dissociation
- NH3 constant)
- Kb always refers to the equilibrium in which a
base reacts with H2O to form the conjugate acid
and OH-
23Lewis Acids and Bases
- Review - An Arrhenius acid reacts in water to
release a proton - base reacts in water to
release a hydroxide ion - In the Bronstead-Lowry description of acids and
bases acid reacts to donate a proton - a base
accepts a proton
24- G.N.Lewis defination -
- Lewis acid is defined as an electron-pair
acceptor - Lewis base is defined as an electron-pair donor
25- In the example with ammonia, the ammonia is
acting as a Lewis base (donates a pair of
electrons), and the proton is a Lewis acid
(accepts a pair of electrons)
26- Lewis is consistent with the description by
Arrhenius, and with the definition by
Bronstead-Lowry. However, the Lewis description,
a base is not restricted in donating its
electrons to a proton, it can donate them to any
molecule that can accept them.
27Calculating the pH of a Weak Acid
- What is the pH of an aq sol that is 0.0030M
pyruvic acid HC2H3P3? Ka 1.4x10-4 at 25oC - HC2H3P3 ?? H C2H3P3-
- I 0.0030 0 0
- C -X X X
- E 0.0030-X X X
28- Ka H C2H3P3- pluggin in the values
- HC2H3P3 from the
table - 1.4x10-4 X2
- (0.0030-X)
- 1.4x10-4 (0.0030) X2
- 4.2x10-7-1.4x10-4x X2
- X2 1.4x10-4x-4.2x10-7 0 a quadratic
- ignore the neg sol x 5.82 x10-4
- pH -log 5.82 x10-4 pH 3.24
29Learning Check
- What is the pH at 25oC of a solution made by
dissolving a 5.00 grain tablet of aspirin
(acetylsalicylic acid) in 0.500 liters of water?
The tablet contains 0.325g of the acid HC9H7O4.
Ka 3.3x10-4 mm 180.2g/l - H 9.4x10-4 pH 3.03
30Buffers
- A solution that resists changes in pH when a
limited amount of an acid or base is added to it. - Buffers contain either a weak acid and its conj.
base or a weak base and its conj. acid.
31Examples
- Ex. Weak acid and conj. base equal molar
- amounts
- Strong Acid added
- H A- ? HA
- conj. base weak acid
- the conj base interacts with the H ions
- from the strong acid changing them to a
- weak acid
32- Strong base added
- OH HA ? HOH A-
- weak acid conj base
- the weak acid interacts with the OH- ion from
the base to form water and the conj. base - If the concentration of A- and HA are large
and the amount of H or OH- is small the - solution will be buffered or the change in pH
will be minimized. -
33- Buffering capacity the amount of acid or base a
buffer can react with before a significant change
in pH occurs - Ratio of acid to conj base unless the ratio is
close to 1 ( between 110 and 101) will be too
low to be useful.
34Calculating the pH of a buffer
- Note a solution of 0.10 M acedic acid and its
conj base 0.20 M acetate from sodium acetate is a
buffer solution pH 5.07 - Ex. Calc. the pH of a buffer by mixing 60.0 ml of
0.100 M NH3 with 40.0ml of 0.100 M NH4Cl.
35- 1St cal the conc. of each species
- M moles/liters
- mol of NH3 0.10M X/0.060 l 0.0060mol
- mol of NH4 0.10M X/0.040 l 0.0040mol
- NH3 0.0060 mol/ 0.100 l 0.060 M
- NH4 0.0040 mol/ 0.100 l 0.040 M
- NH3 H2O ?? NH4 OH-
- I 0.060M O.040 0
- C -X X X
- E 0.060-x 0.040x X
36- Kb NH4 OH- ( 0.040X)X
- 1.8X10-5 NH3 (0.060-X)
- Ignore X 1.8X10-5 0.040X X 2.7X10-5
- 0.060
- -log (2.7x10-5) 4.6 pOH pH 9.4
- Or using Henderson - Hasselbalch equation
- pOH pKb log conj acid 4.74 log (
0.04 - B
0.06) - 4.6 pOH pH 9.4
37What is the pH of a buffer prepared by adding
30.0ml of .15M HC2H3O2 to 70ml of .2M NaC2H3O2?
- HC2H3O2 .15M x/.03 .0045/.01 .045
- C2H3O2 .20M x/.07 .0140/.01 .140
- ka 1.7x10-5
- HHeq pH pKa log conj base
- acid
- pH 4.77 log(.140 5.3
- .045)
38Adding an acid or a base to a buffer
- Calc the ph of 75ml of the buffer solution
of(0.1M HC2H3O2 and 0.2M NaC2H3O2) - to which 9.5 ml of 0.10M HCl has been added.
Compare the change to that of adding HCl to pure
water. - H C2H3O2 - ?? HC2H3O2
- H 0.10M n/.0095l 0.00095 moles
- C2H3O2 - 0.2M n/.075 0.0150 moles
- HC2H3O2 0.10M n/.075 0.0075 moles
39- Neutralization Reaction
- C2H3O2 C2H3O2 moles H moles
- 0.0150n - .00095n 0.014
- HC2H3O2 Orginial Conc. Conc Contributed by
reaction - 0.075 moles 0.00095 0.0085mol
- C2H3O2 0.014mol/0.085l 0.16M
- HC2H3O2 0.0085/0.085 0.10M
- pH 4.76 log (.16/.10) 4.96
-