Post on 26-Mar-2015
transcript
ACID AND BASE
By Dr. Taing N. You 2009 ~2010
Water• Water behave both like acid and base• The dissociation of water is the most fundamental of acid-
base reaction
• Water break apart to form H+ and hydroxyl OH- • In pure water, the equilibrium constant is defined by
• The equilibrium constant less then 1 suggest that the reaction prefers to stay on the side of the reactants.
2H O H OH
14wK H . OH 1.00x10
Definition of ACID and BASE Arrhenius
Acid: generate [H+] in solution Base: generate [OH-] in solution Normal Arrhenius equation: acid + base < --> salt + water
Example: HCl + NaOH <----> NaCl + H2O
Lewis Acid: Accept an electron pair Base: Donate an electron pair
The advantage of this theory is that many more reaction can be considered acid-base reaction as they do not have to occur in solution
Definition of ACID and BASE
Bronted –Lowery Acid: anything that donates a [H+] Base: anything that accepts a [H+] Normal Arrhenius equation: acid + base < --> acid + base
Example: HNO2 + H2O < -- > NO2- + H3O+
Each acid has a conjugate base and each base has a conjugate acid.
pH and pOH• The acidity or basicity of a substance is defined most typically by the
pH, defined as follow:
• pOH gives another way to measure the acidity of a solution. It is just the opposite of pH
pH + pOH = 14 pH = 7 neutral (pH of pure water) pH < 7 acid pH > 7 base
pH log[H ]
pOH log OH
Salt• A salt is formed when acid reacts with base• Acid releases H+ while base releases OH- “Hydrolysis”• pH of the salt depends on the strengths of the original acids
and bases• Conjugate of a strong acid is very weak and cannot undergo
hydrolysis and vice versa for conjugate of strong base
acid Base Salt pH
Strong Strong pH = 7
Weak Strong pH > 7
Strong Weak pH < 7
Weak Weak Depends on which is stronger
Acid – Base character
H-X is to be acid: Hydrogen must have positive number Hydrogen ionizes to form a positive +1
H-X is base when H have a negative charge
NaH is not an acid because H has a -1 charge, it’s a base CH4 is not an acid because H does not ionize
H-O-X to be base:
X-O-H must break to form OH-
H-O-X is acid if H ionizes to form H+
Strong acid
• They are completely ionized in solution• There are 7 strong acid:
• Hydrochloric acid HCl• Bromic acid HBr• Iodic acid HI• Sulfuric acid H2SO4 • Nitric acid HNO3 • Chloric acid HClO3 • Perchloric acid HClO4
Calculate pH of a strong acid
• It is easiest to follow the standard “Start, Change, Equilibrium” process
• Ex: Determine the pH of a 0.25 M sol of HBr
HBr (aq) H+ (aq) + Br-
Start 0.25 M 0 M 0 M
Change - 0.25 + 0.25 + 0.25
Equilibrium 0 0.25 0.25
pH log H log 0.25 0.6
Weak acid
• They are the most common type of acid
• The equilibrium for dissociation of acid is know as Ka •
• The larger the value of Ka the stronger the acid• Example: Determine the pH of 0.3 M acetic acid with the Ka of
1.8 x 10-5 . • Write the equilibrium equation for acid:•
(aq) (aq) (aq)HA H A
a
[H ].[A ]K
[HA]
2 3 2 2 3 2HC H O H C H O
Weak acid
• Write the equilibrium expression
• Start, Change, Equilibrium
52 3 2a
2 3 2
[H ].[C H O ]K 1.8x10
[HC H O ]
HC2H3O2(aq) H+ (aq) + HC2H3O2-
Start 0.3 M 0 M 0 M
Change - x + x +x
Equilibrium 0 x x
Week acid
• Substitute the variable and solve for [H+]
25
a
3
(x).(x) xK 1.8x10
0.3 x 0.3
x [H ] 2.3x10
pH log([H ]) 2.64
Strong base
• Like strong acid they completely ionize in solution• There are 8 strong bases
– Lithium hydroxide LiOH– Sodium hydroxide NaOH– Potassium hydroxide KOH– Rubidium(I) cation hydroxide RbOH– Cesium(I) cation hydroxide CsOH– Calcium hydroxide Ca(OH)2
– Strontium dihydroxide Sr(OH)2
– Barium dihydroxide Ba(OH)2
Calculate pH of a strong base
• It is easiest to follow the standard “Start, Change, Equilibrium” process
• Ex: Determine the pH of a 0.o10 M sol of Ba(OH)2
Ba(OH)2 (aq) Ba2+ (aq) + 2OH-(aq)
Start 0.10 M 0 M 0 M
Change - 0.10 + 0.10 + 0.10
Equilibrium 0 0.10 0.10
pOH log OH log 0.10 1
pH 14 1.0 13.0
Weak base
• This group follow the following equation
• Kb is base dissociation constant:
2weak base H O conjugateacid OH
b2
4b
3 2
14a b w
[conjugateacid].[OH ]K
[weak base].[H O]
[NH ].[OH ]Example: K
[NH ].[H O]
K xK K 1.00x10
Weak base
• The similar “Start, Change, Equilibrium” process is used to
calculate the pH of weak base, however a few step is added
• Ex: Determine the pH of 0.15 M NH3 with Kb = 1.8 x 10-5
• Write the equilibrium expression and the Kb value
3 2 4NH H O NH OH
54a
3 2
[NH ].[OH ]K 1.8x10
[NH ][H O]
Weak base• Start, Change, Equilibrium process
• Substitute de variable and solve for [OH-]
NH3 (aq) + H2O NH4+ + OH-
Start 0.15 M - - 0 M 0 M
Change - x - - + x + x
Equilibrium 0.15 - x - - x x
25
b
(x).(x) xK 1.8x10
0.15 x 0.15
3x [OH ] 1.6 x10 M
pOH log[OH ] 2.80
pH 14 2.80 11.20
Common ion effect
• When adding a salt to a weak acid or base that contains one of the ions present in the acid or base. Molarity of the salt must be added in the calculation of pH
• The same process is use to calculate Ka and Kb as in weak acid
or weak base.
• Example: Find the pH of a solution formed by dissolving 0.100 mol of HC2H3O2 with a Ka of 1.8x10-8 and 0.200 mol of NaC2H3O2
in a total volume of 1.00 L.
Common ion effect
HC2H3O2(aq) H+ (aq) + HC2H3O2-
Start 0.10 M 0 M 0.20 M
Change - x + x +x
Equilibrium 0.10 - x x 0.2 + x
8a
6
6
(x)(0.2 x)) (x)(0.2)K 1.8x10
(0.1 x) (0.1)
x [H ] 9.0x10
pH log(9.0 x10 ) 5.05
Acid Base titration: Strong acid/strong Base
An acid-base titration is when you add a base to an acid until the equivalence point is reached which is where the moles of acid equal the moles of base. For the titration of a strong base and a strong acid, this equivalence point is reached when the pH of the solution is seven (7) as seen on the following titration curve
Acid Base titration: Strong acid/strong Base
The equivalence point is reached when the pH is greater than seven (7). The half equivalence point is when half of the total amount of base needed to neutralize the acid has been added. It is at this point where the pH = pKa of the weak acid.
Acid Base titration In an acid-base titration, the base will react with the weak acid and form a solution that contains the weak acid and its conjugate base until the acid is completely gone. To solve these types of problems, we will use the weak acid's Ka value and the molarities in a similar way as we have before. Before demonstrating this way, let us first examine a short cut, called the Henderson-Hasselbalch Equation. This can only be used when you have some acid and some conjugate base in your solution.
Acid Base titration
Where:
pH: the log of the molar concentration of the hydrogenpKa: the equilibrium dissociation constant for an acid[base]: the molar concentration of a basic solution [acid]: the molar concentration of an acidic solution
If you only have acid, then you must do a pure Ka problem and if you only have base (like when the titration is complete) then you must do a Kb problem.
Example Problem
25.0 mL of .400 M KOH is added to 100. mL of .150 M benzoic acid, HC7H5O2 (Ka=6.3x10-5). Determine the pH of the solution