Acids and
Bases
Unit 9
Properties of Acids and Bases
• Taste (stupid idea in the • Reactions with metals
lab, eh?) – Acids – bubbles! H2
– Acids – sour – Bases – nada
– Bases – bitter
• Electrical
• Touch
Conductivity
– Acids – nothing – unless
you have a cut or booboo – Both – yep
– Bases – feel slippery • Indicators
– Acids – turns blue litmus
paper red
– Bases – turns red litmus
paper blue (brb)
Definitions
• Arrhenius Definition • Lewis Definition
– Acids – produce H+ – Acid – electron pair
– Bases – produce OH- acceptor
• Brønsted-Lowery – Base – electron pair
donator
– Acids
• Donates H+
• Donates a proton
– Bases
• Accepts a H+
• Accepts a proton
Hydronium Ion
• H+ does not actually exist in solution
• H+ „sees‟ water‟s two non-bonding pairs of
e-
• Forms H3O+
pH : potential Hydronium
• Common scale for strength
– Reason why?
– Actual [H3O+] is very low in most solutions
• So, Swedish dude named Søren Sørenson
devised the pH scale…
• “0 – 14”
• Closer to 0: increasing
acidity (decreasing
alkalinity)
• Closer to 14:
increasing alkalinity
(decreasing acidity)
• 7: neutral (equal
acidity and alkalinity)
pOH: potential Hydroxide
• Opposite of pH !!!
• pH + pOH = 14
Calculating pH and pOH
• pH = -log [H3O+]
• pOH = -log [OH-]
pH to conc.
• Conc. = antilog of negative p
• H+ conc. = 10(-pH)
• OH- conc. = 10(-pOH)
Finding one from the other
• pH + pOH = 14
– duh, 14 – one of them = the other
• [H3O+] [OH-] = KW
– KW = 1 x 10-14
– 1 x 10-14 / one of them = the other
Conjugate Acid – Base Pairs
• Conjugate Acid
– Has received hydrogen ion
• Conjugate Base
– Has lost hydrogen ion
• HCl (aq) + NaOH (aq) HOH (l) + NaCl (aq)
• NH3(g) + H2O(l) NH4+(aq) + OH-(aq)-
Determining Strengths of Acids
or Bases and pH
• Dissociation
– Strong acids or bases dissociate completely
– Weak acids or bases do not
Acid Dissociation Constant
• HA(g) + H2O(l) H3O+(aq) + A-(aq)
• Ka = [H3O+][A-] / [HA]
• [H+] = Ka times ([acid] /[conjugate base])
• The largerer the strongerer
Base Dissociation Constant
• B(aq) + H2O(l) HB+(aq) + OH-(aq)
• Kb = [HB+][OH-] / [B]
• The largerer the strongerer
Water Dissociation Constant
• Kw = Ka x Kb
• Kw = 1 x 10-14
Acid – Base Properties of Salts
• Acid + Base water + salt
– Salt: any cation with any anion
• Salts
– Dissociate in water
– Are either neutral, slightly acidic, or slightly
basic
– Hydrolysis of the salt !!!
• Strong (or weak) acid plus strong (or weak)
base
– Neutral salt
– NaOH + HCl NaCl + HOH
• Strong acid + weak base
– Slightly acidic salt
– NH4OH + HCl NH4Cl + HOH
• Strong base + weak acid
– Slightly basic salt
– NaOH + HC2H3O2 NaC2H3O2 + HOH
Buffers
• Resist changes in pH
• Weak acid and conj. Base
• HC2H3O2 and NaC2H3O2
– Weak acid reacts with added hydroxide ions
• H+ + OH- HOH then more acid dissociates!
– Conjugate base reacts with added hydronium
ions
• C2H3O2- + H+ HC2H3O2
Titrations !!!
• A quantitative determination of adding an
acid to a base (duh, or vice versa)
Strong-Strong Titration Problems
• In a titration of 25.00 mL of 0.220 M
NaOH with 0.252 M HCl:
a) What is the initial pH?
b) What is the volume required to reach
equivalence?
c) What is the pH after 10.0 mL is added?
d) What is the pH after 21.8 mL is added?
e) What is the pH after 22.5 mL is added?
NaOH + HCl NaCl + H2 O
.220 M .252 M
a) pOH = -log .220 = .658 pH = 13.34
b) Equivalence Vol?
.0218 L HCl
.025 L x .220 mol NaOH 1 mol HCl L
x x = or
L 1 mol NaOH .252 mol HCl 21.8 mL
c) pH after 10.00 mL HCl added?
.252 mol HCl 1 mol NaOH
.010 L x x = .00252 mol NaOH used
L 1 mol HCl
.025 L x .220 mol NaOH
= .00550 mol NaOH originally
L
.00550 mol - .00252 mol = .00298 mol NaOH left
.00298 mol
=.0851 M NaOH = [OH-]
. 035 L
pOH = -log .0851 = 1.07
pH = 12.9
NaOH + HCl NaCl + H2 O
.220 M .252 M
d) pH = 7 (21.8 mL = equivalence volume)
e) pH after 22.5 mL HCl added?
22.5 mL - 21.8 mL = .67 mL excess HCl
mol 1.69 x 10-4 mol HCl
.252 M HCl = = = .00356 M HCl
.00067 L .0475 L
pH = -log .00356 = 2.45
You Try!
You begin a titration with 20.0 mL of 0.500 M HCl. Your are titrating
with 0.500 M NaOH. HCl + NaOH H2O + NaCl
a) What is the initial pH?
b) What is the volume needed to neutralize?
c) What is the pH after 10.0 mL of NaOH is added?
d) What is the pH after 20.0 mL of NaOH is added?
e) What is the pH after 20.20 mL of NaOH is added?
a) Initial pH? pH = -log .500 = .301
b) Equivalence pt. volume?
.500 mol HCl 1 mol NaOH 1000 mL
20.0 mL HCl x x x = 20.0 mL
1000 mL 1 mol HCl .500 mol NaOH
c) pH after 10.0 mL NaOH added?
.500 mol HCl
20.0 mL x = .0100 mol HCl
1000 mL
10.0 mL x .500 mol NaOH = .00500 mol NaOH added
1000 mL
.0100 - .005 = .005 mol HCl left = .167 M HCl = [H+]
.030 L
pH = -log .167 pH = .778
d) pH after 20.0 mL NaOH added?
20.0 mL is the neutralization volume so pH = 7
e) pH after 20.2 mL NaOH added? .20 mL excess NaOH
.2 mL x .500 mol NaOH = 1.0 x 10-4 mol excess NaOH
1000 mL
.0402 L
= .00249 M NaOH = [OH-]
pOH = -log .00249 pOH = 2.60 pH = 11.40
Compare pH changes when strong acid is
added with and without buffer.
• Strong acid with no buffer:
– What is the pH change if you add 6.00 mL of
0.500 M HCl to 500 mL water?
.500 M = mole HCl =.00300 mol HCl
.00600 L
.00300 mol = 5.93 x 10-3 M H+
.506 L
pH = -log 5.93 x 10-3 = 2.23
Initial pH = 7.0 pH = 4.77
Strong Acid with Buffer:
What is the pH change if you add 6.00 mL of 0.500 M HCl to a 500.
mL of a buffer where [NH4+] = [NH3] = 1.0 M? (Kb = 1.8 x 10-5)
First: Calculate starting pH in the buffer using ICE
The Reaction NH3 + H2O NH4+ + OH-
Initial (M) 1.0 1.0 ≈0
Change (M) -x +x +x
Equilibrium (M) 1.0 - x 1.0 + x x
[NH4+][OH-]
Kb = pOH = -log 1.8 x 10-5
[NH3]
(1.0 + x)(x)
1.8 x 10 -5 = pOH = 4.74
(1.0 - x)
1.0x Starting pH = 9.26
1.8 x 10 -5 =
1.0
x = 1.8 x 10-5 M
Strong Acid with Buffer: Cont‟d.
What is the pH change if you add 6.00 mL of 0.500 M HCl to a 500. mL
of a buffer where [NH4+] = [NH3] = 1.0 M? (Kb = 1.8 x 10-5)
Second: Calculate the neutralizing effect of adding HCl. Use moles and
an ICE type of strategy. The buffer contains .500 mol each of NH3 and
NH4+. The .0030 mol of HCl added will react with the basic NH3.
Buffer Reaction: NH3 + H3O + NH4+ + H2O
Initial buffer (mol) .500 ≈0 .500
Add (mol) +.0030
Change (mol) -.0030 -.0030 +.0030
After neutralization (mol) .497 ≈0 .503
M (divide by .506 L .982 M .994 M
Using the Kb expression & neutralization concentrations, calculate
the [OH-]:
[NH4+][OH-] Kb[NH3] (1.8 x 10-5)(.982)
Kb = [OH-] = = = 1.78 x 10-5 M
[NH3] [NH4+] .994
The Enddd!!! Yaaay!!
Your welcome for the wonderful Easter-ish
pretty colors!
Sincerely,
Maggie & Katie!
This is a platypus….
That is Mac. Macapus