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Lecture May 1
1
Goals for today: Chapter 15Part II
!Hour Exam III !
Thursday May 8
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pH and % Dissociation
2
What is the pH of a 1.5 M solution of HF?
Ka= 3.5 x 10-4
Answers: [H3O+] = 0.023 M; pH = 1.64
% dissociation = ([H3O+]/1.5) x 100 =1.5 %
% dissociation or ionization =(equilibrium/original) x 100%
Can also go backwards to find a Ka
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Kafrom pH and % Dissoc.
3
An unknown weak acid solution has a pH of 2.25for a 0.055 M
solution. What is the Kaof the acid?
[H3O+] = 5.62 x 10-3
Ka = x2/(0.055-x)
Ka= 6.3 x 10-4
% Diss = [(5.62 x 10-3)/0.055] x 100 =10.2 % or 10. %
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Anions as Weak Bases
4
Weak acid:HF(aq) + H2O(l) F-(aq) + H3O+(aq)
Ka= 3.5 x 10-4
What is the pH of a 1.50 M solution of NaF?
1) Write the relevant chemical equilibrium
F-(aq) + H2O(l) HF(aq) + OH-(aq)
2) Set up the ICE table (on board)
3) Write the equilibrium expression andcalculate the Kbfrom Kaof
the conjugate acid
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Example continued
5
Kw= KaKbKb= (1.0 x 10-14)/(3.5 x 10-4) = 2.86x 10-11
Kb= [HF][OH-]/[F-]
4) Use these and the ICE table to solve for x
2.86x10-11= x2/(1.50-x)
x = 6.55x10-6 = [OH-]
5) Calculate the pH (from the pOH)
pOH = 5.18 pH = 8.82 BASIC
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pH of a Salt Solution
6
Calculate the pH of a 0.100 M ammoniumchloride solution.
Ka = 5.6 x 10-10= [NH3] [H3O+] / [NH4+]
x ~ 7.48 x 10-6 = [H3O+]
pH = -log (7.48 x 10-6)= 5.13
I 0.100 0 0
C -x +x +x
E 0.100-x x x
NH4+(aq) NH3(aq) +H3O+(aq)
NH4+(aq) + H2O(l) NH3(aq) +H3O+(aq)
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Polyprotic Acids
7
More than one acidic proton
Structural similarities: the O-H bond H2SO4, H2SO3, H3PO4,
H2CO3Sulfuric acid: Ka1= very large Ka2= 1.2 x 10-2Sulfurous
acid:
Ka1= 1.5 x 10-2
Ka2= 6.3 x 10-8
Carbonic Acid H2CO3 Ka1= 4.2 x 10-7
Ka2= 4.8 x 10-11
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Carbonic Acid H2CO3
8
H2CO3(aq) + H2O(l) HCO3-(aq) + H3O+(aq)
(actually CO2dissolved in H2O)
Ka1 = 4.3 x 10-7
HCO3-(aq) + H2O(l) CO32-(aq) + H3O+(aq)Ka2= 4.7 x 10-11
BUT this can also happen if put NaHCO3in water:
HCO3-(aq) + H2O(l) H2CO3(aq) + OH-(aq)
Kb(HCO3-)= Kw/Ka1= (1.00 x 10-14)/(4.3 x 10-7)= 2.3 x 10-8
dominates
BASIC!!
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Other Polyprotics
9
H3PO4phosphoric acid - triprotic
Ka1= 7.11 x 10-3H3PO4(aq) + H2O(l) H2PO4-(aq) + H3O+(aq)
H2PO4-(aq) + H2O(l) HPO42-(aq) + H3O+(aq)
Ka2= 6.32 x 10-8
HPO42-(aq) + H2O(l) PO43-(aq) + H3O+(aq)
Ka3= 4.5 x 10-11
What if put Na3PO4in water?
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Conjugate Acid-Base Pairs
of Polyprotic Acids
10
Na3PO4
Soluble salt in water gives 3Na+(aq) + PO43-(aq)
PO43-(aq) + H2O(l) HPO42-(aq) + OH-(aq)
Kb= Kw/Ka3 = (1.00x10-14)/(4.5x10-13) =
2.2x10
-2
VERY STRONG BASE!!!
What happens if dissolve Na2HPO4in water??
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Quiz of the Day
11
What is the pH of a 0.28 M solution of NaClO?
Ka(HClO) = 2.9 x 10-8
Kb(ClO-) = (1.0 x 10-14)/(2.9 x 10-8)=3.45 x 10-7
3.45 x 10-7= x2/(0.28-x) x = [OH-]
ClO-(aq) + H2O(l) HClO(aq) + OH-(aq)
x = 3.11 x 10-4pOH = 3.51 pH = 10.49
