ph = - log [h + ] or ph = - log [h 3 o + ] example: if [h + ] = 1 x 10 -10 ph = - log 1 x 10 -10 ph...
TRANSCRIPT
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pH = - log [H+]
or
pH = - log [H3O+]
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Example: If [H+] = 1 X 10-10
pH = - log 1 X 10-10
pH = - (- 10)pH = 10
What would be the pH of a 0.000018 M HNO3 solution?
Example: If [H+] = 1.8 X 10-5
pH = - log 1.8 X 10-5
pH = - (- 4.74)pH = 4.74
pH Problems
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If the pH of Coke is 3.12, [H+] = ???
[H+] = 10-3.12 = 7.6 x 10-4 M
More pH Problems
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Identify as SA, SB, WA, WB
Big or Small K value?
H2O can function as both
an ACID and a BASE.In pure water there can
be AUTOIONIZATION
Pure Water – Acid or Base?
HH22O + HO + H22O O H H33OO++ + OH + OH--
WB SAWA SB
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Equilibrium constant for water = KwKw = [H3O+] [OH-] = 1.00 x 10-14 at 25 oC
Kw = [H3O+] [OH-]
In pure water, [H3O+] = [OH-]
so Kw = [x][x] = [x]2
and so, [H3O+] = [OH-] = 1.00 x 10-7
THEREFORE, the pH is…
pH of Pure Water
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•What would be the pH of a 1.0 M HCl solution? Of a 0.01 M HCl solution?
[H+] and [OH-] have an inverse relationship in aqueous solution
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pOH is from the base perspectivepOH = - log [OH-]
Since we are dealing with aqueous solution…
[H+] [OH-] = 1.00 x 10-14
pH + pOH always equals14
pOH
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pH pOH [H+] [OH-] Acid or Base
2.45
4.75
3.5 x 10-10
0.00084
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Why is the pH of a 0.1 or 10-1 M acetic acid not 1?
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A strong acid ionize 100% but a weak acid does not!
Weak acid has Ka < 1
Leads to small [H3O+]
Strong vs Weak Acids
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Weak base has Kb < 1
Leads to small [OH-]
Strong vs Weak Bases
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Complete ionization!
Virtually no reactants left
No equilibrium
The equivalence point is the point where the number of moles of base equal the number of moles of acid.
http://www.youtube.com/watch?v=ILn79QpYwPc
http://www.youtube.com/watch?v=HnGy8Um6ibM
http://www.youtube.com/watch?v=H63dHo-T1TM
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At equivalence point, the pH > 7
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A polyprotic acid has two or more hydrogen which can ionize in multiple steps.
H2CO3 (aq) H+ + HCO3- K1 = 4.5 x10-7
HCO3 (aq) H+ + CO3-2 K2 = 4.7 x10-11
H2CO3 (aq) 2 H+ + CO3-2 Ka = ?
(Overall Ka = K1 x K2)
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What is the Ka of the equation below?
H3PO4 (aq) 3 H+ + PO4
-3
H3PO4 H+ + H2PO4
- K1 = 7.1 x 10-3
H2PO4- H+
+ HPO4-2
K2 = 6.3 x 10-8
HPO4-2
H+ + PO4
-3 K3 = 4.5 x 10-
13
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When dissolved, the salt created from a strong acid and a weak base will be acidic.
NH4Cl (s) NH4+ + Cl-
Why? NH4
+ + H2O NH4OH + H3O+
The ammonium ion acts as an acid. It will have a Ka value.
conjugate of NH3
conjugate of HCl
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When dissolved, the salt created from a strong base and a weak acid will be basic.
NaC2H3O2 (s) Na+ + C2H3O2-
So?C2H3O2
- + H2O HC2H3O2 + OH-
The acetate ion acts as a base. It will have a Kb value.
How will the products react with H+ and OH-?
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For an acid-base conjugate pair:
The conjugate of a strong is weak, and the conjugate of a weak is strong. Why?
(Ka)(Kb) = Kw = 1.4 x 10-14
Ka is the acid ionization constantKb is the base ionization constantKw is the ionization constant of water
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Acetic acid has a Ka value of 1.7 x 10-
5.
• What is the conjugate base?
• What is the Kb value of the conjugate base?
Carboxyl or
organic acid group
(-COOH)
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Methylamine has a Kb value of 4.38 x 10-4.
• What is the conjugate acid?
• What is the Ka value of the conjugate base?
Amine group (-NH2)
Methyl group
(-CH3)
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Information Given Concentration of salt Ka or Kb value
Additional Information Needed Identify as Acidic, Basic , or Neutral Equation of ions with water Keq expression
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Find the pH of a 0.500 M solution of KCN. The Ka value of HCN is 5.8 x 10-
10.
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Dissociation of Salt
KCN + H2O K+ + CN-
Identify base and acid
KOH (strong base) + HCN (weak acid)
The salt must be basic. CN- is a strong conjugate base
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Reaction between CN- and water
CN- + H2O OH-
+ HCN
Write Kb expression
Kb = [HCN][OH-] [CN-]
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Use Ka of HCN to find Kb of CN-
Ka•Kb = Kw(5.8 x 10-10)•(Kb) = (1.0 x 10-14)Kb = 1.7 x 10-5
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Use Kb and Kb expression to solve for [OH-] (set up ice table first)
CN- + H2O OH-
+ HCN
I .500M n/a 0 0
C -x +x +x
E .500 - x x x
1.7 x 10-5 = [x][x] [.500 - x]
X = 2.9 x 10-3
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Using [OH-], find pOH and pH
pOH = -log[2.9 x 10-3]pOH = 2.54
pOH + pH = 142.54 + pH = 14pH = 11.46
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What would be the pH of a 0.200 M ammonium chloride solution? Kb of ammonia is 1.7 x 10-5.• Write dissociation of ammonium chloride• Write reaction of ammonium and water• Write Ka expression for ammonium• Calculate Ka value using Kb and Kw• Solve for [H+]• Find pH
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What would be the pH of a 0.200 M ammonium chloride solution? Kb of ammonia is 1.7 x 10-5.
Ka of NH4+ =
[H3O+] [NH3]
[NH4+]
5.88 x 10-10 =__x2__(.200)
x = 1.08 x 10-5 pH = 4.96
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A buffer solution has the ability to resist changes in pH upon the addition of small amounts of either acid or base.
A buffer contains:• A weak acid or weak base, AND• The salt of the weak acid or base
NH3 and NH4Cl (Weak Base and Acidic Salt)
HC2H3O2 and NaC2H3O2 (Weak Acid and Basic Salt)
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The NH3 (base) will neutralize any acid i.e. HCl (extra H+ ions) by combining with the extra H+ ions to form NH4
+ ions.
The NH4+ (conjugate acid) will
neutralize any base i.e. KOH (extra OH- ions) by donating its H+ ion to form HOH.
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HCl + CH3COO- CH3COOH + Cl-
BUFFER: some H+,some C2H3O2-, HC2H3O2 and
Na+,C2H3O2- ADD: HCl
ADD: KOH
KOH + CH3COOH
CH3COOK + HOH
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Consider mixture of HC2H3O2 and NaC2H3O2
The common ion (C2H3O2-) suppresses the
ionization of the weak acid. This is called the common ion effect.
NaC2H3O2 (aq) Na+ + C2H3O2-
HC2H3O2 (aq) H+ + C2H3O2
-
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Which of the following are buffer systems?
(a) KCl/HF (b) NH4NO3/NH3
(c) KCl/HCl(d) NaHCO3/H2CO3
(e) Ca(OH)2/CaSO4
(f) NH3/HNO2
Which buffers have a pH above 7?
-------------- No Common Ion
--------------- Strong Acid (not weak)
----------------------------- Strong Base (not weak)-------------------- Weak Base and Acid (no salt)
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What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK?Ka for HCOOH = 1.8 x 10 -4
Mixture of weak acid and conjugate base!
Initial (M)
Change (M)
Equilibrium (M)
HCOOH (aq) H+ (aq) + HCOO- (aq)
-x +x
0.30 - x
+x
x 0.52 + x
0.30 0.00 0.52
x = 1.04 X 10 -4 pH = 4.0
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Henderson-Hasselbach equation (on reference sheet)
For weak acid and its salt
For weak base and its salt
pH = pKa + log [A-][HA]
pOH = pKb + log [HB+][B]
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What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK?Ka for HCOOH = 1.8 x 10 -4
pH = 3.77 + log[0.52][0.30] = 4.0
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Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system.
The Kb of NH3 is 1.8 x 10-5
NH3 + HOH NH4+ + OH-
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What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution?
NH3 + HOH NH4+ + OH-
(.30)(.08) n/a (.36)(.08) (.050)(.020)
I .024 moles .0288 .0010 moles C + .001 - .0010 - .0010 E .025 .0278 0
E . 25 M .278 M
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At equivalence point, the pH > 7
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Halfway to equivalence, [H+] = Kathis means that the pH = pKa
Prior to equivalence, [H+] > Kathis means that the pH < pKa
Between halfway to equivalence, [H+] < Kathis means that the pH >pKa
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Which indicator would you use for a titration of HNO3 with NH3 ?
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Strong Acid vs. Strong Base• 100 % ionized! pH = 7 No equilibrium!
Weak Acid vs. Strong Base• Acid is neutralized; Need Kb for conjugate base equilibrium
Strong Acid vs. Weak Base• Base is neutralized; Need Ka for conjugate acid equilibrium
Weak Acid vs. Weak Base• Depends on the strength of each; could be conjugate acid,
conjugate base, or pH 7
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These are usually formed from a transition metal surrounded by ligands (polar molecules or negative ions).
As a "rule of thumb" you place twice the number of ligands around an ion as the charge on the ion... example: the dark blue Cu(NH3)4
2+ (ammonia is used as a test for Cu2+ ions), and Ag(NH3)2
+. Memorize the common ligands.
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Ligands Names used in the ion
H2O aqua
NH3 ammine
OH- hydroxy
Cl- chloro
Br- bromo
CN- cyano
SCN- thiocyanato (bonded through sulphur) isothiocyanato (bonded through nitrogen)
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Names: ligand first, then cationExamples:• tetraamminecopper(II) ion: Cu(NH3)4
2+
• diamminesilver(I) ion: Ag(NH3)2+.
• tetrahydroxyzinc(II) ion: Zn(OH)4 2-
The charge is the sum of the parts (2+) + 4(-1)= -2.
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Aluminum also forms complex ions as do some post transitions metals. Ex: Al(H2O)6
3+
Transitional metals, such as Iron, Zinc and Chromium, can form complex ions.
The odd complex ion, FeSCN2+, shows up once in a while
Acid-base reactions may change NH3 into NH4+ (or vice
versa) which will alter its ability to act as a ligand. Visually, a precipitate may go back into solution as a
complex ion is formed. For example, Cu2+ + a little NH4OH will form the light blue precipitate, Cu(OH)2. With excess ammonia, the complex, Cu(NH3)4
2+, forms. Keywords such as "excess" and "concentrated" of
any solution may indicate complex ions. AgNO3 + HCl forms the white precipitate, AgCl. With excess, concentrated HCl, the complex ion, AgCl2-, forms and the solution clears.
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Total number of bonds from the ligands to the metal atom.
Coordination numbers generally range between 2 and 12, with 4 (tetracoordinate) and 6 (hexacoordinate) being the most common.
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molecular formula
Lewis base/ligand
Lewis acid donor atom
coordination number
Ag(NH3)2+ NH3 Ag+ N 2
[Zn(CN)4]2- CN- Zn2+ C 4
[Ni(CN)4]2- CN- Ni2+ C 4
[PtCl6] 2- Cl- Pt4+ Cl 6
[Ni(NH3)6]2+ NH3 Ni2+ N 6