acid-base equilibria the reaction of weak acids with water, or the reaction of weak bases with...
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Acid-Base EquilibriaThe reaction of weak acids with water,
OR
the reaction of weak bases with water,
always results in an equilibrium!!
The equilibrium constant for the reaction of a weak acid with water is Ka
1
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Acid-Base Equilibria
eg. HF(aq) + H2O(l)
Ka =[H3O+] [F-]
[HF]
H3O+(aq) + F-
(aq)
Keq = ?
2
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Acid-Base Equilibria
For any weak acid
Why is H2O(l) omitted from the Ka expression?
Ka =[H3O+] [conjugate base]
[weak acid]
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Acid-Base Equilibria
the equilibrium constant for the reaction of a weak base with water is Kb
HS-(aq) + H2O(l)
Kb =
H2S(aq) + OH-(aq)
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Acid-Base Equilibria
For any weak base
Kb =[OH-] [conjugate acid]
[weak base]
5
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eg.
Write the expression for Kb for S2-(aq)
ANSWER:
S2-(aq) + H2O(l)
Kb =[OH-] [HS-]
[S2-]
HS-(aq) + OH-
(aq)
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5.a) Use Ka to find [H3O+] for 0.100 mol/L HF(aq)
HF(aq) + H2O(l) H3O+(aq) + F-
(aq)
[HF]
][F ]OH[K
-3
a
Ka = 6.6 x 10-4
Mar. 25
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x]- [0.100
[x] [x]10 x 6.6 4-
x 2 = (0.100)(6.6 x 10-4)
x 2 = 6.6 x 10-5
x = 8.1 x 10-3 mol/L
1st try - Ignore xMar. 25
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2nd try– Include x
0.0081] - [0.100
[x] [x]10 x 6.6 4-
x 2 = (0.0919)(6.6 x 10-4)
x 2 = 6.0654 x 10-5
x = 7.8 x 10-3 mol/L
Different than 1st try:CANNOT IGNORE
DISSOCIATION
Mar. 28
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3rd try– Include new x
0.0078] - [0.100
[x] [x]10 x 6.6 4-
x 2 = (0.0922)(6.6 x 10-4)
x 2 = 6.0852 x 10-5
x = 7.8 x 10-3 mol/L
[H3O+] = 7.8 x 10-3 mol/L
Same as 2nd try:
Mar. 28
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5.b) find [H3O+] for 0.250 mol/L CH3COOH(aq
CH3COOH(aq) + H2O(l) H3O+(aq) + CH3COO-
(aq)
COOH][CH
]COO[CH ]OH[K
3
-33
a
Ka = 1.8 x 10-5
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x]- [0.250
[x] [x]10 x 1.8 5-
x 2 = (0.250)(1.8 x 10-5)
x 2 = 4.5 x 10-6
x = 2.1 x 10-3 mol/L
1st try - Ignore xMar. 28
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2nd try– Include x
0.0021] - [0.250
[x] [x]10 x 1.8 5-
x 2 = (0.2479)(1.8 x 10-5)
x 2 = 4.462 x 10-6
x = 2.1 x 10-3 mol/L
[H3O+] = 2.1 x 10-3 mol/L
Same as 1st try:
Mar. 28
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14
To ignore OR not to ignore:
that is the question
Mar. 28
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pH of a weak acid
Step #1: Write a balanced equation
Step #2: ICE table OR assign variables
Step #3: Write the Ka expression
Step #4: Check (can we ignore dissociation)
Step #5: Substitute into Ka expression
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pH of a weak acid
eg. Find pH of 0.100 mol/L HF(aq).
Step #1: Write a balanced equation
HF(aq) + H2O(l) H3O+(aq) + F-
(aq)
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Step #2: Equilibrium Concentrations
Let x = [H3O+] at equilibrium
[F-] = x
[HF] = 0.100 - x
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Step #3: Write the Ka expression
Ka =[H3O+] [F-]
[HF]
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Step #4: Check (can we ignore dissociation)
dissociation (- x) may be IGNORED
= 151 (0.100)
6.6 x 10-4
Acid dissociation CANNOT beIGNORED in this question.
[weak acid]
Ka
If > 500
We have to use the – x
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Step #5: Substitute into Ka expression
x]- [0.100
[x] [x]10 x 6.6 4-
x2 = 6.6 x 10-5 - 6.6 x 10-4 x
x2 + 6.6 x 10-4 x - 6.6 x 10-5 = 0
a = 1 b = 6.6 x 10-4 c = -6.6 x 10-5
Quadratic Formula!!
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2a
4acbbx
2
2(1)
)10x4(1)(-6.6)10x(6.610x6.6x
-52-4-4
2
0.00026410x6.6x
-4
mol/L0.0078x Ignore
negative roots
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a) Find the [H3O+] in 0.250 mol/L HCN(aq)
Check: 4.0 x 108
x = 1.24 x 10-5
[H3O+] = 1.24 x 10-5
b) Calculate the pH of 0.0300 mol/L HCOOH(aq)
Check: 167
x = 2.24 x 10-3
pH = 2.651
Try these:Mar. 31
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HCN + H2O H⇋ 3O+ + CN-
Let x = [H3O+]
x = [CN-]
0.250 – x = [HCN]
Check:
23
Ka = [H3O+] [CN-] [HCN]
= 4.0 x 108 0.250
6.2 x 10-10
Quadratic NOTneeded
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24
x]- [0.250
[x] [x]10 x 6.2 10-
x = 1.25 x 10-5
[H3O+] = 1.25 x 10-5 mol/L
pH = 4.904
x2 = 1.55 x 10-10
Mar. 31
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HCOOH + H2O H⇋ 3O+ + HCOO-
0.0300 0 0
-x +x +x
0.0300 – x x x
Check:
25
Ka = [H3O+] [HCOO-] [HCOOH]
= 167 0.0300
1.8 x 10-4
Quadratic needed
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x]- [0.0300
[x] [x]10 x 1.8 4-
A = 1 B = 1.8 x 10-4 C = -5.4 x 10-6
x = 2.24 x 10-3
[H3O+] = 2.24 x 10-3 mol/L pH = 2.651
x2 = 5.4 x 10-6 - 1.8 x 10-4x
x2 + 1.8 x 10-4x - 5.4 x 10-6 = 0
Mar. 31
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Practice1. Formic acid, HCOOH, is present in the sting of
certain ants. What is the [H3O+] of a 0.025 mol/L solution of formic acid? (0.00203 mol/L)
2. Calculate the pH of a sample of vinegar that contains 0.83 mol/L acetic acid.
( [H3O+] = 3.87 x 10-3 pH = 2.413 )
3. What is the percent dissociation of the vinegar in 2.?
% diss = 0.466 %
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Practice4. A solution of hydrofluoric acid has a molar
concentration of 0.0100 mol/L. What is the pH of this solution?
( [H3O+] = 0.00226 pH = 2.646 )
5. The word “butter” comes from the Greek butyros. Butanoic acid, C3H7COOH, gives rancid butter its distinctive odour. Calculate the [H3O+] of a 1.0 × 10−2 mol/L solution of butanoic acid.
(Ka = 1.51 × 10−5 ) (Ans: 3.89 x 10-4 mol/L)
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pH of a weak base same method as acids ignore dissociation if
to calculate Kb (usually given on the exam)
K x K Ka b w KK
Kbw
a
29
[weak base]
Kb
> 500
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pH of a weak baseCalculate the pH of 0.0100mol/L Na2CO3(aq)
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CO32- + H2O HCO⇋ 3
- + OH-
0.0100 0 0
-x +x +x
0.0100 – x x x
Check:
31
Kb = [OH-] [HCO3-]
[CO32-]
= 47 0.0100
2.13 x 10-4→ Quadratic needed
Kb = 1.00 x 10-14
4.7 x 10-11
= 2.13 x 10-4
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x]- [0.0100
[x] [x]10 x 2.13 4-
x2 = 2.13 x 10-6 - 2.13 x 10-4x
x2 + 2.13 x 10-4x - 2.13 x 10-6 = 0
A = 1 B = 2.13 x 10-4 C = -2.13 x 10-6
x = 1.36 x 10-3
[OH-] = 1.36 x 10-3 mol/LpOH = ?? pH = 11.13
Apr. 4
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pH of a weak baseCalculate the pH of 0.500 mol/L NaNO2(aq)
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Apr. 4
Na+ NO2- H2O
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NO2- + H2O HNO⇋ 2 + OH-
0.500 0 0
-x +x +x
0.500 – x x x
Check:
34
Kb = [OH-] [HCO3-]
[CO32-]
= 3.6 x 1010 0.500
1.39 x 10-11 OK to ignore –x hereie. NO Quadratic
Kb = 1.00 x 10-14
7.2 x 10-4
= 1.39 x 10-11
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x]- [0.500
[x] [x]10 x 1.39 11-
x2 = 6.95 x 10-12
[OH-] = 2.6 x 10-6 mol/L
pOH = ??
pH = 8.42
x = 2.6 x 10-6
Apr. 4
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Calculating Ka from [weak acid] and pH
eg. The pH of a 0.072 mol/L solution of benzoic acid, C6H5COOH, is 2.68. Calculate the numerical value of the Ka for this acid.
- Equation- Find [H3O+] from pH
- Subtract from [weak acid]- Substitute to find Ka See p. 591 #6 & 8
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C6H5COOH(aq) + H2O(l) H3O+(aq) + C6H5COO-
(aq)
[H3O+] = 10-2.68 = 0.00209 mol/L
[C6H5COOH] = 0.072 – 0.00209
= 0.06991 mol/L
Find Ka
Ka =(0.00209)(0.00209)
(0.06991)= 6.2 x 10-5
[C6H5COO-] = 0.00209 mol/L
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Calculating Ka from [weak acid] and pH
eg. The pH of a 0.072 mol/L solution of benzoic acid, C6H5COOH, is 2.68. Calculate the % dissociation for this acid.
See p. 591 #’s 5 & 6[H3O+] = 10-2.68
= 0.00209 mol/L
100%xacid] [weak
]O[Hdiss % 3
= 2.9 %
38
100% x 0.072
0.00209
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a) 0.250 mol/L chlorous acid, HClO2(aq); pH = 1.31
0.012 19.5%b) 0.150 mol/L cyanic acid, HCNO(aq); pH =
2.150.00035 4.7%
c) 0.100 mol/L arsenic acid, H3AsO4(aq); pH = 1.70
0.0050 20%d) 0.500 mol/L iodic acid, HIO3(aq); pH = 0.670
0.160 42.8%
Calculate the acid dissociation constant, Ka , and the percent dissociation for each acid:
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More Practice: Weak Acids:
pp. 591, 592 #’s 6 -8 Weak Bases:
p. 595 #’s 11 - 16 (Kb’s on p. 592)
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Acid-Base Stoichiometry
Solution Stoichiometry (Review)
1. Write a balanced equation
2. Calculate moles given ( OR n = CV)
3. Mole ratios
4. Calculate required quantity
OR OR m = nM
M
mn
C
nV
V
nC
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42
# #required given xcoeffic ien t
coeffic ien trequired
given
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eg. 25.0 mL of 0.100 mol/L H2SO4(aq) was used to neutralize 36.5 mL of NaOH(aq). Calculate the molar concentration of the NaOH solution.
H2SO4(aq) + NaOH(aq) → H2O(l) + Na2SO4(aq)22
nH2SO4 =
nNaOH =
CNaOH =
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Acid-Base Stoichiometry
pp. 600, 601 – Sample Problems
p. 602 #’s 17 - 20
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Dilution Given 3 of the four variables Only one solution CiVi = CfVf
Stoichiometry Given 3 of the four variables Two different solutions 4 step method
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Excess Acid or BaseTo calculate the pH of a solution produced by mixing an acid with a base:
write the B-L equation (NIE) calculate the moles of H3O+ and OH-
subtract to determine the moles of excess H3O+ or OH-
divide by total volume to get concentration calculate pH
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eg. 20.0 mL of 0.0100 M Ca(OH)2(aq) is mixed with 10.0 mL of 0.00500 M HCl(aq).
Determine the pH of the resulting solution.
ANSWER:
Species present:
Ca2+OH- H3O+ Cl- H2O
SB SA
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C = 0.0200 mol/LV = 0.0200 L
C = 0.00500 mol/L V = 0.0100 L
NIE: OH- + H3O+ → 2 H2O
4.00 x 10-4 mol OH- 5.0 x 10-5 mol H3O+
3.5 x 10-4 mol excess OH-
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n = CV n = CV
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= 0.01167 mol/L
[OH-] = 0.01167 mol/L
pOH = 1.933
pH = 12.067
totalV
nC
49
L0.0300
mol10x3.5 4
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Indicators An indicator is a weak acid that
changes color with changes in pH HIn is the general formula for an
indicator To choose an indicator for a titration,
the pH of the endpoint must be within the pH range over which the indicator changes color
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HIn(aq) + H2O(l) H3O+(aq) + In-
(aq)
Colour #1 Colour #2
HIn is the acid form of the indicator.
Adding H3O+
Adding OH-
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causes colour 1 (LCP)
removes the H3O+ & causes colour #2
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methyl orange
HMo(aq) + H2O(l) H3O+(aq) + Mo-
(aq)
red yellow
bromothymol blue
HBb(aq) + H2O(l) H3O+(aq) + Bb-
(aq)
yellow blue
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Indicators: p. 231.a) HMv + H2O(l) H3O+
(aq) + Mv-(aq)
b) HBb + H2O(l) H3O+(aq) + Bb-
(aq)
2. Indicator pH colour
thymol blue 3.0 yellow
methyl red 7.9 yellow
phenolpthalein 7.1 colourless
indigocarmine 13.5 yellow
3.a) pH range: 2.8 – 4.5
b) pH range: 8.0 – 8.2
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Acid-Base Titration (p. 603 → ) A titration is a lab technique used to
determine an unknown solution concentration. A standard solution is added to a known
volume of solution until the endpoint of the titration is reached.
The endpoint occurs when there is a sharp change in colour
The equivalence point occurs when the moles of H3O+ equals the moles of OH-
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Acid-Base Titration The colour change is caused by the indicator
added to the titration flask. An indicator is a chemical that changes color
over a given pH range (See indicator table) A buret is used to add the standard solution standard solution - solution of known
concentration
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Acid-Base Titration
primary standard - a standard solution which can be made by direct weighing of a stable chemical.
Data from titrations allows us to calculate an unknown solution concentration.
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Titration Calculationeg. Clem Student performed a titration by adding
0.250 mol/L HCl(aq) to 10.0 mL samples of Ca(OH)2(aq). Use the data below to determine the molar concentration of Ca(OH)2(aq).
Trial 1 2 3 4
Final volume 8.48 mL 15.70 mL 22.91 mL 30.14 mL
Initial volume 1.05 mL 8.48 mL 15.70 mL 22.91 mL
Volume HCl(aq)
used57
7.43 mL 7.23 mL7.21 mL7.22 mL
Omit first trialOVERSHOT the endpoint
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Equation:
2 HCl(aq) + Ca(OH)2(aq) → CaCl2(aq) + 2 H2O(l)
nHCl =
nCa(OH)2 =
C = 58
C = 0.250 mol/LVave = 0.00722 L
C = ? mol/LVave = 0.0100 L
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Acid-Base Titration
Titration Lab – pp. 606, 607
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Multi-Step Titrations (p. 609 - 611)
Polyprotic acids donate their protons one at a time when reacted with a base.
eg. Write the equations for the steps that occur when H3PO4(aq) is titrated with NaOH(aq)
H3PO4(aq) + OH-(aq)
H2PO4-(aq) + OH-
(aq)
HPO42-
(aq) + OH-(aq) 60
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Multi-Step TitrationsH3PO4(aq) + OH-
(aq) → H2PO4-(aq) + H2O(l)
H2PO4-(aq) + OH-
(aq) → HPO42-
(aq) + H2O(l)
HPO42-
(aq) + OH-(aq) PO4
3-(aq) + H2O(l)
H3PO4(aq) + 3 OH-(aq) PO4
3-(aq) + 3 H2O(l)
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Multi-Step Titrations
Write the balanced net ionic equations, and the overall equation, for the titration of Na2S(aq) with HCl(aq).
p. 611 #’s 21.b), 22, & 23
LAST TOPIC!! Titration Curves
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Properties / Operational Definitions Acid-Base Theories and Limitations
Arrhenius – H-X and X-OHModified – react with water → hydroniumBLT – proton donor/acceptor (CA and CB)
Writing Net Ionic Equations (BLT) Strong vs. Weak pH & pOH calculations Equilibria (Kw, Ka, Kb) Titrations/Indicators/Titration Curves Dilutions and Excess Reagent questions
Acids and Bases
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Step #2: ICE table
[HF] [H3O+] [F-]
I
C
E
0.100 mol/L 0 0
-x +x +x
0.100 - x x x
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CO32-
(aq) + H2O(l) HCO3-(aq) + OH-
(aq)
0.0100mol/L CO23-
(aq)
[PO43-] [HPO4
2-] [OH-]
I
C
E
0.0100 mol/L 0 0
-x + x + x
0.0100 - x x x65
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Acid-Base Stoichiometry Solution stoichiometry (4 question sheet) Excess reagent problems (use NIE) Titrations Titration curves Indicators STSE: Acids Around Us
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A primary standard is a pure substance that is stable enough to be stored indefinitely without decomposition, can be weighed accurately without special precautions when exposed to air, and will undergo an accurate stoichiometric reaction in a titration.
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15. pH of 0.297 mol/L HOCl
HOCl(aq) + H2O(l) H3O+(aq) + OCl-(aq)
Let x = [H3O+] at equilibrium
[OCl-] = x
[HOCl] = 0.297 - x
Ka =[H3O+] [OCl-]
[HOCl]
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Check:
dissociation (- x) may be IGNORED
= 1.02 x 107 (0.297)
2.9 x 10-8
[0.297]
[x] [x]10 x 2.9 8-
X = 9.28 x 10-5 pH = 4.03
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0.484 mol/L0.07000 L
0.125 mol/L0.02500 L
16. NIE: OH- + H3O+ → 2 H2O
0.03388 mol OH- 0.003125 mol H3O+
0.030755 mol excess OH-
[OH-] = 0.3237 mol/L pOH = 0.490
pH = 13.51070
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17. Ignore dissociation
[OH-] = 0.0146 mol/L
% diss = 2.92 %
18. Vave = 10.975 mL
nNaOH = 0.001262 mol
nH2SO4 = 0.000631 mol
C = 0.0252 mol/L
19. Kb = 3.93 x 10-4
% diss = 6.27
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c) 2.50 mol/L NaCN(aq) Kb = 1.61 x 10-5
Check: 1.5 x 105
x = 6.34 x 10-3
pOH = 2.20 pH = 11.80
d) 0.100 mol/L K2S(aq) Kb = 0.0769
Check: 1.3
x = 0.0573
pOH = 1.24 pH = 12.7672