chapter 15 acids and bases chemistry ii. stomach acid & heartburn the cells that line your...
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Chapter 15Acids and Bases
Chemistry II
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Properties of Acids
• sour taste
• react with “active” metals i.e., Al, Zn, Fe, but not Cu, Ag, or Au
2 Al + 6 HCl → 2 AlCl3 + 3 H2
Corrosive
• react with carbonates, producing CO2
marble, baking soda, chalk, limestone
CaCO3 + 2 HCl → CaCl2 + CO2 + H2O
• change color of vegetable dyes blue litmus turns red
• react with bases to form ionic salts
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Common Acids
Chemical Name Formula Uses Strength
Nitric Acid HNO3 explosive, fertilizer, dye, glue Strong
Sulfuric Acid H2SO4 explosive, fertilizer, dye, glue,
batteries Strong
Hydrochloric Acid HCl metal cleaning, food prep, ore
refining, stomach acid Strong
Phosphoric Acid H3PO4 fertilizer, plastics & rubber,
food preservation Moderate
Acetic Acid HC2H3O2 plastics & rubber, food preservation, Vinegar
Weak
Hydrofluoric Acid HF metal cleaning, glass etching Weak
Carbonic Acid H2CO3 soda water Weak
Boric Acid H3BO3 eye wash Weak
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Structure of Acids
• binary acids have H-atoms attached to a nonmetal atom
e.g. HCl, HF
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Structure of Acids
• oxy acids have H-atoms attached to an O-atom:
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Structure of Acids
• carboxylic acids have COOH group:
e.g. Acetic acid
CH3COOH
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Properties of Bases
• also known as alkalis
• taste bitter alkaloids = plant product that is alkaline
often poisonous
• solutions feel slippery
• change color of vegetable dyes different color than acid red litmus turns blue
• react with acids to form ionic salts neutralization
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Common BasesChemical
Name Formula
Common Name
Uses Strength
sodium hydroxide
NaOH lye,
caustic soda soap, plastic,
petrol refining Strong
potassium hydroxide
KOH caustic potash soap, cotton, electroplating
Strong
calcium hydroxide
Ca(OH)2 slaked lime cement Strong
sodium bicarbonate
NaHCO3 baking soda cooking, antacid Weak
magnesium hydroxide
Mg(OH)2 milk of
magnesia antacid Weak
ammonium hydroxide
NH4OH, {NH3(aq)}
ammonia water
detergent, fertilizer,
explosives, fibers Weak
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Structure of Bases
• most ionic bases contain OH- ionsNaOH, Ca(OH)2
• some contain CO32- ions
CaCO3 , NaHCO3
CO32− + 2H2O → HCO3
− + H2O + OH−
HCO3− + H2O + OH− → H2CO3 + 2OH−
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Definitions of Acids and Bases
• What are the main characteristics of molecules and ions that exhibit acid and base behavior?
• 3 different definitions:ArrheniusBrønsted-LowryLewis
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Arrhenius Theory (1880s)
• acids ionize in water to produce H+ ions and anions because molecular acids are not made of ions, they cannot
dissociate they must be pulled apart, or ionized, by the water
HCl(aq) → H+(aq) + Cl–(aq)
CH3COOH(aq) → H+(aq) + CH3COO–(aq)
• bases dissociate in water to produce OH- ions and cations ionic substances dissociate in water
NaOH(aq) → Na+(aq) + OH–(aq)
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Hydronium Ion
• the H+ ions produced by the acid are so reactive they cannot exist in water H+ ions are protons!!
• instead, they react with a water molecule(s) to produce complex ions, mainly hydronium ion, H3O+
H+ + H2O H3O+
Chemists use H+(aq) and H3O+(aq) interchangeably
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Arrhenius Acid-Base Reactions
• the H+ from the acid combines with the OH- from the base to make a molecule of H2O
H+ + OH- → H2O
• the cation from the base combines with the anion from the acid to make a salt
Na+ + Cl- → NaCl
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
acid + base → salt + water
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Problems with Arrhenius Theory
• does not explain why molecular substances, like NH3, dissolve in water to form basic solutions – even though they do not contain OH– ions
• does not explain how some ionic compounds, like Na2CO3 or Na2O, dissolve in water to form basic solutions – even though they do not contain OH– ions
• does not explain why molecular substances, like CO2, dissolve in water to form acidic solutions – even though they do not contain H+ ions
• does not explain acid-base reactions that take place outside aqueous solution
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Brønsted-Lowry Theory (1923)
• in a Brønsted-Lowry Acid-Base reaction, an H+ is transferred
Acid: proton (H+) donorBase: proton (H+) acceptor
• base structure must contain an atom with unshared pair of e-
• in an acid-base reaction, the acid molecule gives an H+ to the base molecule
H–A + :B :A⇌ – + H–B+
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Brønsted-Lowry Acids
• Brønsted-Lowry acids are H+ donorsAnything that has H+ can potentially be Brønsted-Lowry
acid
• HCl(aq) is acidic because HCl transfers an H+ to H2O (base or proton acceptor), forming H3O+ ions
HCl(aq) + H2O(l) → Cl–(aq) + H3O+(aq)
acid base
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Brønsted-Lowry Bases
• Brønsted-Lowry bases are H+ acceptorsany material that has atoms with lone pairs can
potentially be a Brønsted-Lowry base
• NH3(aq) is basic because NH3 accepts an H+ from H2O (acid, proton donor), forming OH–(aq)
NH3(aq) + H2O(l) NH⇌ 4+(aq) + OH–(aq)
base acid
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Amphoteric Substances
• amphoteric substances can act as either an acid or a base have both transferable H and atom with lone pair
• water acts as base, accepting H+ from HCl
HCl(aq) + H2O(l) → Cl–(aq) + H3O+(aq)
• water acts as acid, donating H+ to NH3
NH3(aq) + H2O(l) NH⇌ 4+(aq) + OH–(aq)
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Brønsted-Lowry Acid-Base Reactions
• one of the advantages of Brønsted-Lowry theory is that it allows reactions to be reversible
H–A + :B :A⇌ – + H–B+
• the original base has an extra H+ after the reaction – so it will act as an acid in the reverse process
• and the original acid has a lone pair of e- after the reaction – so it will act as a base in the reverse process
:A– + H–B+ ⇌ H–A + :B
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Conjugate Pairs
• In a Brønsted-Lowry Acid-Base reaction, the original base becomes an acid in the reverse reaction, and the original acid becomes a base in the reverse process
• each reactant and the product it becomes is called a conjugate pair
• the original base becomes the conjugate acid; and the original acid becomes the conjugate base
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Brønsted-Lowry Acid-Base Reactions
H–A + :B ⇌ :A– + H–B+
acid base conjugate conjugate base acid
HCHO2 + H2O ⇌ CHO2– + H3O+
acid base conjugate conjugate base acid
H2O + NH3 ⇌ HO– + NH4+
acid base conjugate conjugate base acid
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Conjugate (joined) Pairs
In the reaction H2O + NH3 OH⇌ – + NH4+
H2O and OH– constitute an Acid/Conjugate Base pair
NH3 and NH4+ constitute a
Base/Conjugate Acid pair
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Ex 15.1a – Identify the Brønsted-Lowry Acids and Bases and Their Conjugates in the Reaction
H2SO4 + H2O ⇌ HSO4– + H3O+
acid base conjugate conjugate base acid
H2SO4 + H2O ⇌ HSO4– + H3O+
When the H2SO4 becomes HSO4, it lost an H+ so
H2SO4 must be the acid and HSO4 its conjugate base
When the H2O becomes H3O+, it accepted an H+ so H2O must be the base and H3O+ its conjugate acid
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Practice – Write the formula for the conjugate acid of the following bases
H2O
NH3
CO32−
H2PO4−
H3O+
NH4+
HCO3−
H3PO4
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Practice – Write the formula for the conjugate base of the following acids
H2O
NH3
CO32−
H2PO4−
OH−
NH2−
No H+, it cannot be an acid
HPO42−
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Strong or Weak• a strong acid is a strong electrolyte
practically all the acid molecules ionize, →
• a strong base is a strong electrolytepractically all the base molecules form OH– ions, either
through dissociation or reaction with water, →
• a weak acid is a weak electrolyteonly a small percentage of the molecules ionize, ⇌
• a weak base is a weak electrolyteonly a small percentage of the base molecules form OH–
ions, either through dissociation or reaction with water, ⇌
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Strong Acids• The stronger the acid, the more
willing it is to donate H+
use water as the standard base
• strong acids donate practically all their H+’s 100% ionized in water strong electrolyte
• [H3O+] = [strong acid]
HCl ® H+ + Cl-
HCl + H2O® H3O+ + Cl-
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Weak Acids
• weak acids donate a small fraction of their H’smost of the weak acid
molecules do not donate H to water
much less than 1% ionized in water
• [H3O+] << [weak acid]
HF ⇌ H+ + F-
HF + H2O ⇌ H3O+ + F-
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Polyprotic Acids
• often acid molecules have more than one ionizable H+ – these are called polyprotic acids the ionizable H+’s may have different acid strengths or be equal 1 H+ = monoprotic, 2 H + = diprotic, 3 H + = triprotic
HCl = monoprotic, H2SO4 = diprotic, H3PO4 = triprotic
• polyprotic acids ionize in steps each ionizable H + removed sequentially
• removing of the first H + automatically makes removal of the second H + harder H2SO4 is a stronger acid than HSO4
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Acids Conjugate Bases HClO4 ClO4
-1 H2SO4 HSO4
-1
HI I-1
HBr Br-1
HCl Cl-1
HNO3 NO3-1
H3O+1 H2O
HSO4-1 SO4
-2
H2SO3 HSO3-1
H3PO4 H2PO4-1
HNO2 NO2-1
HF F-1
HC2H3O2 C2H3O2-1
H2CO3 HCO3-1
H2S HS-1
NH4+1 NH3
HCN CN-1
HCO3-1 CO3
-2
HS-1 S-2
H2O OH-1
CH3-C(O)-CH3 CH3-C(O)-CH2-1
NH3 NH2-1
CH4 CH3-1
OH-1 O-2
Incr
easi
ng A
cidi
ty
Increasing Basicity
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Strengths of Acids & Bases
• commonly, acid or base strength is measured by determining the equilibrium constant of a substance’s reaction with water
HA + H2O A⇌ - + H3O+
B: + H2O HB⇌ + + OH-
• the farther the equilibrium position lies to the products, the stronger the acid or base
• the position of equilibrium depends on the strength of attraction between the base form and the H+
stronger attraction means stronger base or weaker acid
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General Trends in Acidity
• the stronger an acid is at donating H+, the weaker the conjugate base is at accepting H+
• higher oxidation number = stronger oxyacid H2SO4 > H2SO3; HNO3 > HNO2
• cation stronger acid than neutral molecule; neutral stronger acid than anion H3O+ > H2O > OH-; NH4
+ > NH3 > NH2-
base trend opposite
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Acid Ionization Constant, Ka
• acid strength measured by the size of the equilibrium constant when react with H2O
HA + H2O A⇌ - + H3O+
• acid ionization constant, Ka
larger Ka = stronger acid
[HA]
]O[H][A 3a
K
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40
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Autoionization of Water
• Water is actually an extremely weak electrolyte therefore there must be a few ions present
• about 1 out of every 10 million water molecules form ions through a process called autoionization
H2O H⇌ + + OH–
H2O + H2O H⇌ 3O+ + OH–
• all aqueous solutions contain both H3O+ and OH–
the concentration of H3O+ and OH– are equal in water [H3O+] = [OH–] = 10-7M @ 25°C
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Ion Product of Water
• the product of H3O+ and OH– concentrations is always the same n
• the number is called the ion product of water - symbol, Kw
Kw = [H3O+] [OH–] = 1 x 10-14 @ 25°C
if you measure one of the concentrations, you can calculate the other
• as [H3O+] increases the [OH–] must decrease so the product stays constant inversely proportional
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Acidic and Basic Solutions
• all aqueous solutions contain both H3O+ and OH– ions
• neutral solutions have equal [H3O+] and [OH–] [H3O+] = [OH–] = 1 x 10-7
• acidic solutions have a larger [H3O+] than [OH–] [H3O+] > 1 x 10-7; [OH–] < 1 x 10-7
• basic solutions have a larger [OH–] than [H3O+] [H3O+] < 1 x 10-7; [OH–] > 1 x 10-7
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Example 15.2b – Calculate the [OH] at 25°C when the [H3O+] = 1.5 x 10-9 M, and determine if the solution is acidic, basic, or neutral
The units are correct. The fact that the [H3O+] < [OH] means the solution is basic
[H3O+] = 1.5 x 10-9 M
[OH]
Check:
Solution:
Concept Plan:
Relationships:
Given:
Find:
[H3O+] [OH]
]-][OHOH[ 3wK
]OH[]-OH[
]-OH][OH[
3
w
3w
K
K
M 107.6105.1
100.1]-[OH 6
9
14
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Complete the Table[H+] vs. [OH-]
OH-H+ H+ H+ H+ H+
OH-OH-OH-OH-
[OH-]
[H+] 100 10-1 10-3 10-5 10-7 10-9 10-11 10-13 10-14
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Complete the Table[H+] vs. [OH-]
OH-H+ H+ H+ H+ H+
OH-OH-OH-OH-
[OH-]10-14 10-13 10-11 10-9 10-7 10-5 10-3 10-1 100
[H+] 100 10-1 10-3 10-5 10-7 10-9 10-11 10-13 10-14
even though it may look like it, neither H+ nor OH- will ever be 0the sizes of the H+ and OH- are not to scale
because the divisions are powers of 10 rather than units
Acid Base
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pH
• the acidity/basicity of a solution is often expressed as pH
• pH = -log[H3O+], [H3O+] = 10-pH
exponent on 10 with a positive signpHwater = -log[10-7] = 7need to know the [H+] concentration to find pH
• pH < 7 is acidic; pH > 7 is basic, pH = 7 is neutral
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Sig. Figs. & Logs
• when you take the log of a number written in scientific notation, the digit(s) before the decimal point come from the exponent on 10, and the digits after the decimal point come from the decimal part of the number
log(2.0 x 106) = log(106) + log(2.0)
= 6 + 0.30303… = 6.30303...
• since the part of the scientific notation number that determines the significant figures is the decimal part, the sig figs are the digits after the decimal point in the log
log(2.0 x 106) = 6.30
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Question
Complete the following table of pH values
[H+] pH Significant Figures
1 x 10-7 7.0 1
1.0 x 10-7
6.80
4.39
1.78 x 10-11
1.6 x 10-7 2
3
2
4.1 x 10-52
7.00
10.750
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pH
• the lower the pH, the more acidic the solution; the higher the pH, the more basic the solution 1 pH unit corresponds to a factor of 10 difference in acidity
• normal range 0 to 14 pH can be negative (very acidic) or larger than 14 (very
alkaline)
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52
pH of Common SubstancesSubstance pH
1.0 M HCl 0.0
0.1 M HCl 1.0
stomach acid 1.0 to 3.0
lemons 2.2 to 2.4
soft drinks 2.0 to 4.0
plums 2.8 to 3.0
apples 2.9 to 3.3
cherries 3.2 to 4.0
unpolluted rainwater 5.6
human blood 7.3 to 7.4
egg whites 7.6 to 8.0
milk of magnesia (sat’d Mg(OH)2) 10.5
household ammonia 10.5 to 11.5
1.0 M NaOH 14
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Example 15.3b – Calculate the pH at 25°C when the [OH] = 1.3 x 10-2 M, and determine if the solution is
acidic, basic, or neutral
pH is unitless. The fact that the pH > 7 means the solution is basic
[OH] = 1.3 x 10-2 M
pH
Check:
Solution:
Concept Plan:
Relationships:
Given:
Find:
]-][OHOH[ 3wK
2
14
3
3w
103.1
100.1]OH[
]-OH][OH[
K M 107.7]O[H 133
[H3O+][OH] pH
]OH[log pH 3-
12.11 pH
107.7log- pH 13
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pOH
• another way of expressing acidity/basicity of a solution is pOH
• pOH = -log[OH], [OH] = 10-pOH
pOHwater = -log[10-7] = 7need to know the [OH] concentration to find pOH
• pOH < 7 is basic; pOH > 7 is acidic, pOH = 7 is neutral
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pH and pOH Complete the Table
OH-H+ H+ H+ H+ H+
OH-OH-OH-OH-
[OH-]10-14 10-13 10-11 10-9 10-7 10-5 10-3 10-1 100
[H+] 100 10-1 10-3 10-5 10-7 10-9 10-11 10-13 10-14
pH
pOH
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pH and pOHComplete the Table
OH-H+ H+ H+ H+ H+
OH-OH-OH-OH-
[OH-]10-14 10-13 10-11 10-9 10-7 10-5 10-3 10-1 100
[H+] 100 10-1 10-3 10-5 10-7 10-9 10-11 10-13 10-14
pH 0 1 3 5 7 9 11 13 14
pOH 14 13 11 9 7 5 3 1 0
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Relationship between pH and pOH
• the sum of the pH and pOH of a solution = 14.00at 25°Ccan use pOH to find pH of a solution
14.00pOHpH
00.14]-[OHlog]OH[log
100.1log]-][OHOH[log
100.1]-][OHOH[
3
143
14w3
K
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pK• a way of expressing the strength of an acid or base is pK
pKa = -log(Ka), Ka = 10-pKa
pKb = -log(Kb), Kb = 10-pKb
• the stronger the acid, the smaller the pKa
larger Ka = smaller pKa
because it is the –log
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Finding the pH of a Strong Acid
• there are two sources of H3O+ in an aqueous solution of a strong acid – the acid and the water
• for the strong acid, the contribution of the water to the total [H3O+] is negligible
shifts the Kw equilibrium to the left so far that [H3O+]water is too small to be
significantexcept in very dilute solutions, generally < 1 x 10-4 M
• for a monoprotic strong acid [H3O+] = [HA] for polyprotic acids, the other ionizations can generally be ignored
• 0.10 M HCl has [H3O+] = 0.10 M and pH = 1.00
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Finding the pH of a Weak Acid
• there are also two sources of H3O+ in and aqueous solution of a weak acid – the acid and the water
• however, finding the [H3O+] is complicated by the fact that the acid only undergoes partial ionization
• calculating the [H3O+] requires solving an equilibrium problem for the reaction that defines the acidity of the acid
HA + H2O ⇌ A + H3O+
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[HNO2] [NO2-] [H3O+]
initial
change
equilibrium
Ex 15.6 Find the pH of 0.200 M HNO2(aq) solution @ 25°C
Write the reaction for the acid with water
Construct an ICE table for the reaction
Enter the initial concentrations – assuming the [H3O+] from water is ≈ 0
since no products initially, Qc = 0, and the reaction is proceeding forward
HNO2 + H2O NO⇌ 2 + H3O+
[HNO2] [NO2-] [H3O+]
initial 0.200 0 ≈ 0
change
equilibrium
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[HNO2] [NO2-] [H3O+]
initial 0.200 0 0
change
equilibrium
Ex 15.6 Find the pH of 0.200 M HNO2(aq) solution @ 25°C
represent the change in the concentrations in terms of x
sum the columns to find the equilibrium concentrations in terms of x
substitute into the equilibrium constant expression
+x+xx
0.200 x x x
x
xxK
12
3-2
a 1000.2HNO
]OH][[NO
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Ex 15.6 Find the pH of 0.200 M HNO2(aq) solution @ 25°C
x
xxK
12
3-2
a 1000.2HNO
OHNO
determine the value of Ka from Table 15.5
since Ka is very small, approximate the [HNO2]eq = [HNO2]init and solve for x
1
2
3-2
a 1000.2HNO
OHNO
xxK
1
24
1000.2106.4
x
3
14
106.9
1000.2106.4
x
x
Ka for HNO2 = 4.6 x 10-4
[HNO2] [NO2-] [H3O+]
initial 0.200 0 ≈ 0
change -x +x +x
equilibrium 0.200 x x0.200 x
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Ex 15.6 Find the pH of 0.200 M HNO2(aq) solution @ 25°C
Ka for HNO2 = 4.6 x 10-4
[HNO2] [NO2-] [H3O+]
initial 0.200 0 ≈ 0
change -x +x +x
equilibrium 0.200 x x
check if the approximation is valid by seeing if x < 5% of [HNO2]init
%5%8.4%1001000.2
106.91
3
the approximation is valid
x = 9.6 x 10-3
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Ex 15.6 Find the pH of 0.200 M HNO2(aq) solution @ 25°C
Ka for HNO2 = 4.6 x 10-4
[HNO2] [NO2-] [H3O+]
initial 0.200 0 ≈ 0
change -x +x +x
equilibrium 0.200-x x x
x = 9.6 x 10-3
substitute x into the equilibrium concentration definitions and solve
M 190.0106.9200.0200.0HNO 32 x
M 106.9OHNO 33
-2
x
[HNO2] [NO2-] [H3O+]
initial 0.200 0 ≈ 0
change -x +x +x
equilibrium 0.190 0.0096 0.0096
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Ex 15.6 Find the pH of 0.200 M HNO2(aq) solution @ 25°C
Ka for HNO2 = 4.6 x 10-4
substitute [H3O+] into the formula for pH and solve
[HNO2] [NO2-] [H3O+]
initial 0.200 0 ≈ 0
change -x +x +x
equilibrium 0.190 0.0096 0.0096
02.2106.9log
OH-logpH3
3
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Ex 15.6 Find the pH of 0.200 M HNO2(aq) solution @ 25°C
Ka for HNO2 = 4.6 x 10-4
[HNO2] [NO2-] [H3O+]
initial 0.200 0 ≈ 0
change -x +x +x
equilibrium 0.190 0.0096 0.0096
check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka
4
23
2
3-2
a
109.4190.0
106.9
HNO
OHNO
Kthough not exact, the answer is reasonably close
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Tro, Chemistry: A Molecular Approach
88
Percent Ionization• another way to measure the strength of an acid is to determine the
percentage of acid molecules that ionize when dissolved in water – this is called the percent ionization the higher the percent ionization, the stronger the acid
• since [ionized acid]equil = [H3O+]equil
%100acid ofmolarity initial
acid ionized ofmolarity IonizationPercent
%100[HA]
]O[HIonizationPercent
init
equil3
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Ex 15.9 - What is the percent ionization of a 2.5 M HNO2 solution?
Write the reaction for the acid with water
Construct an ICE table for the reaction
Enter the Initial Concentrations
Define the Change in Concentration in terms of x
Sum the columns to define the Equilibrium Concentrations
HNO2 + H2O NO⇌ 2 + H3O+
[HNO2] [NO2-] [H3O+]
initial
change
equilibrium
[HNO2] [NO2-] [H3O+]
initial 2.5 0 ≈ 0
change
equilibrium
+x+xx2.5 x x x
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determine the value of Ka from Table 15.5
since Ka is very small, approximate the [HNO2]eq = [HNO2]init and solve for x
5.2HNO
OHNO
2
3-2
a
xxK
5.2106.4
24 x
2
4
104.3
5.2106.4
x
x
Ka for HNO2 = 4.6 x 10-4
[HNO2] [NO2-] [H3O+]
initial 2.5 0 ≈ 0
change -x +x +x
equilibrium 2.5-x ≈2.5 x x
Ex 15.9 - What is the percent ionization of a 2.5 M HNO2 solution?
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Ex 15.9 - What is the percent ionization of a 2.5 M HNO2 solution?
HNO2 + H2O NO⇌ 2 + H3O+
[HNO2] [NO2-] [H3O+]
initial 2.5 0 ≈ 0
change -x +x +x
equilibrium 2.5 0.034 0.0342.5 x x x
substitute x into the Equilibrium Concentration definitions and solve
M 5.2034.05.25.2HNO2 x
M 034.0OHNO 3-2 x
x = 3.4 x 10-2
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Ex 15.9 - What is the percent ionization of a 2.5 M HNO2 solution?
HNO2 + H2O NO⇌ 2 + H3O+
[HNO2] [NO2-] [H3O+]
initial 2.5 0 ≈ 0
change -x +x +x
equilibrium 2.5 0.034 0.034
Apply the Definition and Compute the Percent Ionization
%4.1%1005.2
104.3
%100][HNO
]O[HIonizationPercent
2
init2
equil3
since the percent ionization is < 5%, the “x is small” approximation is valid
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Finding the pH of Mixtures of Acids
• generally, you can ignore the contribution of the weaker acid to the [H3O+]equil
• for a mixture of a strong acid with a weak acid, the complete ionization of the strong acid provides more than enough [H3O+] to shift the weak acid equilibrium to the left so far that the weak acid’s added [H3O+] is negligible
• for mixtures of weak acids, generally only need to consider the stronger for the same reasons as long as one is significantly stronger than the other, and their
concentrations are similar
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Ex 15.10 Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO2(aq)
Write the reactions for the acids with water and determine their Kas
If the Kas are sufficiently different, use the strongest acid to construct an ICE table for the reaction
Enter the initial concentrations – assuming the [H3O+] from water is ≈ 0
HF + H2O F⇌ + H3O+ Ka = 3.5 x 10-4
[HF] [F-] [H3O+]
initial 0.150 0 ≈ 0
change
equilibrium
HClO + H2O ClO⇌ + H3O+ Ka = 2.9 x 10-8
H2O + H2O OH⇌ + H3O+ Kw = 1.0 x 10-14
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[HF] [F-] [H3O+]
initial 0.150 0 0
change
equilibrium
Ex 15.10 Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO2(aq)
represent the change in the concentrations in terms of x
sum the columns to find the equilibrium concentrations in terms of x
substitute into the equilibrium constant expression
+x+xx
0.150 x x x
x
xxK
13
-
a 1050.1HF
]OH][[F
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Ex 15.10 Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO2(aq)
x
xxK
150.0HF
OHF 3-
a
determine the value of Ka for HF
since Ka is very small, approximate the [HF]eq = [HF]init and solve for x
150.0HF
OHF 3-
a
xxK
1
24
1050.1105.3
x
3
14
102.7
1050.1105.3
x
x
Ka for HF = 3.5 x 10-4
[HF] [F-] [H3O+]
initial 0.150 0 ≈ 0
change -x +x +x
equilibrium 0.150 x x0.150 x
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Tro, Chemistry: A Molecular Approach
99
Ex 15.10 Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO2(aq)
Ka for HF = 3.5 x 10-4
[HF] [F-] [H3O+]
initial 0.150 0 ≈ 0
change -x +x +x
equilibrium 0.150 x x
check if the approximation is valid by seeing if x < 5% of [HF]init
%5%8.4%1001050.1
102.71
3
the approximation is valid
x = 7.2 x 10-3
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Ex 15.10 Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO2(aq)
Ka for HF = 3.5 x 10-4
[HF] [F-] [H3O+]
initial 0.150 0 ≈ 0
change -x +x +x
equilibrium 0.150-x x x
x = 7.2 x 10-3
substitute x into the equilibrium concentration definitions and solve
M 143.0102.7150.0150.0HF 3 x
M 102.7OHF 33
- x
[HF] [F-] [H3O+]
initial 0.150 0 ≈ 0
change -x +x +x
equilibrium 0.143 0.0072 0.0072
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Ex 15.10 Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO2(aq)
Ka for HF = 3.5 x 10-4
substitute [H3O+] into the formula for pH and solve
14.2102.7log
OH-logpH3
3
[HF] [F-] [H3O+]
initial 0.150 0 ≈ 0
change -x +x +x
equilibrium 0.143 0.0072 0.0072
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Ex 15.10 Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO2(aq)
Ka for HF = 3.5 x 10-4
check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka
4
23
3-
a
106.3143.0
102.7
HF
OHF
Kthough not exact, the answer is reasonably close
[HF] [F-] [H3O+]
initial 0.150 0 ≈ 0
change -x +x +x
equilibrium 0.143 0.0072 0.0072
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NaOH → Na+ + OH-
Strong Bases
• the stronger the base, the more willing it is to accept H use water as the standard acid
• for strong bases, practically all molecules are dissociated into OH– or accept H’s strong electrolyte multi-OH strong bases completely
dissociated
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Example 15.11b – Calculate the pH at 25°C of a 0.0015 M Sr(OH)2 solution and determine if the solution is acidic, basic,
or neutral
pH is unitless. The fact that the pH > 7 means the solution is basic
[Sr(OH)2] = 1.5 x 10-3 M
pH
Check:
Solution:
Concept Plan:
Relationships:
Given:
Find:
]-][OHOH[ 3wK
3
14
3
3w
100.3
100.1]OH[
]-OH][OH[
K M 103.3]O[H 123
]OH[log pH 3-
11.48 pH
103.3log- pH 12
[H3O+][OH] pH[Sr(OH)2]
[OH]=2[Sr(OH)2]
[OH] = 2(0.0015)= 0.0030 M
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Practice - Calculate the pH of a 0.0010 M Ba(OH)2 solution and determine if it is acidic, basic, or neutral
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Practice - Calculate the pH of a 0.0010 M Ba(OH)2 solution and determine if it is acidic, basic, or neutral
[H3O+] = 1.00 x 10-14
2.0 x 10-3 = 5.0 x 10-12M
pH > 7 therefore basic
Ba(OH)2 = Ba2+ + 2 OH- therefore [OH-] = 2 x 0.0010 = 0.0020 = 2.0 x 10-3 M
pH = -log [H3O+] = -log (5.0 x 10-12)pH = 11.30
Kw = [H3O+][OH]
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Weak Bases
• in weak bases, only a small fraction of molecules accept H’s weak electrolyte most of the weak base molecules do not
take H from water much less than 1% ionization in water
• [HO–] << [weak base]
• finding the pH of a weak base solution is similar to finding the pH of a weak acid
NH3 + H2O ⇌ NH4+ + OH-
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[NH3] [NH4+] [OH]
initial
change
equilibrium
Ex 15.12 Find the pH of 0.100 M NH3(aq) solution
Write the reaction for the base with water
Construct an ICE table for the reaction
Enter the initial concentrations – assuming the [OH] from water is ≈ 0
since no products initially, Qc = 0, and the reaction is proceeding forward
NH3 + H2O NH⇌ 4+ + OH
[NH3] [NH4+] [OH]
initial 0.100 0 ≈ 0
change
equilibrium
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[NH3] [NH4+] [OH]
initial 0.100 0 0
change
equilibrium
Ex 15.12 Find the pH of 0.100 M NH3(aq) solution
represent the change in the concentrations in terms of x
sum the columns to find the equilibrium concentrations in terms of x
substitute into the equilibrium constant expression
+x+xx
0.100 x x x
x
xxK
13
4b 1000.1NH
]OH][[NH
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Ex 15.12 Find the pH of 0.100 M NH3(aq) solution
x
xxK
13
4b 1000.1NH
OHNH
determine the value of Kb from Table 15.8
since Kb is very small, approximate the [NH3]eq = [NH3]init and solve for x
1
3
4b 1000.1NH
OHNH
xxK
1
25
1000.11076.1
x
3
15
1033.1
1000.11076.1
x
x
Kb for NH3 = 1.76 x 10-5
[NH3] [NH4+] [OH]
initial 0.100 0 ≈ 0
change -x +x +x
equilibrium 0.100 x x0.100 x
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Ex 15.12 Find the pH of 0.100 M NH3(aq) solution
Kb for NH3 = 1.76 x 10-5
[NH3] [NH4+] [OH]
initial 0.100 0 ≈ 0
change -x +x +x
equilibrium 0.100 x x
check if the approximation is valid by seeing if x < 5% of [NH3]init
%5%33.1%1001000.1
1033.11
3
the approximation is valid
x = 1.33 x 10-3
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Ex 15.12 Find the pH of 0.100 M NH3(aq) solution
substitute x into the equilibrium concentration definitions and solve
M 099.01033.1100.0100.0NH 33 x
M 1033.1][OH]NH[ 3-4
x
Kb for NH3 = 1.76 x 10-5
[NH3] [NH4+] [OH]
initial 0.100 0 ≈ 0
change -x +x +x
equilibrium 0.100 x x x
x = 1.33 x 10-3
[NH3] [NH4+] [OH]
initial 0.100 0 ≈ 0
change -x +x +x
equilibrium 0.099 1.33E-3 1.33E-3
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Ex 15.12 Find the pH of 0.100 M NH3(aq) solution
use the [OH-] to find the [H3O+] using Kw
substitute [H3O+] into the formula for pH and solve
124.111052.7log
OH-logpH12
3
Kb for NH3 = 1.76 x 10-5
[NH3] [NH4+] [OH]
initial 0.100 0 ≈ 0
change -x +x +x
equilibrium 0.099 1.33E-3 1.33E-3
12-3
3-
14-
3
-3w
107.52]O[H
101.33
101.00]O[H
]][OHO[H
K
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Ex 15.12 Find the pH of 0.100 M NH3(aq) solution
Kb for NH3 = 1.76 x 10-5
[NH3] [NH4+] [OH]
initial 0.100 0 ≈ 0
change -x +x +x
equilibrium 0.099 1.33E-3 1.33E-3
check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Kb to the given Kb
5
23
3
4b
108.1099.0
1033.1
NH
OHNH
Kthough not exact, the answer is reasonably close
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Practice – Find the pH of a 0.0015 M morphine solution, Kb = 1.6 x 10-6
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Practice – Find the pH of a 0.0015 M morphine solution
Write the reaction for the base with water
Construct an ICE table for the reaction
Enter the initial concentrations – assuming the [OH] from water is ≈ 0
since no products initially, Qc = 0, and the reaction is proceeding forward
B + H2O BH⇌ + + OH
[B] [BH+] [OH]
initial 0.0015 0 ≈ 0
change
equilibrium
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[B] [BH+] [OH]
initial 0.0015 0 0
change
equilibrium
Practice – Find the pH of a 0.0015 M morphine solution
represent the change in the concentrations in terms of x
sum the columns to find the equilibrium concentrations in terms of x
substitute into the equilibrium constant expression
+x+xx
0.0015 x x x
x
xxK
3b 105.1B
]OH][[BH
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Practice – Find the pH of a 0.0015 M morphine solution
x
xxK
3b 105.1B
OHBH
determine the value of Kb
since Kb is very small, approximate the [B]eq = [B]init and solve for x
3b 105.1B
OHBH
xxK
1
26
105.1106.1
x
5
36
109.4
105.1106.1
x
x
Kb for Morphine = 1.6 x 10-6
[B] [BH+] [OH]
initial 0.0015 0 ≈ 0
change -x +x +x
equilibrium 0.0015 x x0.0015 x
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Practice – Find the pH of a 0.0015 M morphine solution
check if the approximation is valid by seeing if x < 5% of [B]init
%5%3.3%100105.1
109.43
5
the approximation is valid
x = 4.9 x 10-5
Kb for Morphine = 1.6 x 10-6
[B] [BH+] [OH]
initial 0.0015 0 ≈ 0
change -x +x +x
equilibrium 0.0015 x x
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Practice – Find the pH of a 0.0015 M morphine solution
substitute x into the equilibrium concentration definitions and solve
M 0015.0109.40015.00015.0Morphine 5 x
M 109.4][OH]BH[ 5- x
x = 4.9 x 10-5
Kb for Morphine = 1.6 x 10-6
[B] [BH+] [OH]
initial 0.0015 0 ≈ 0
change -x +x +x
equilibrium 0.0015 x x x
[B] [BH+] [OH]
initial 0.0015 0 ≈ 0
change -x +x +x
equilibrium 0.0015 4.9E-5 4.9E-5
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Practice – Find the pH of a 0.0015 M morphine solution
use the [OH-] to find the [H3O+] using Kw
substitute [H3O+] into the formula for pH and solve
69.9100.2log
OH-logpH10
3
10-3
5-
14-
3
-3w
100.2]O[H
109.4
101.00]O[H
]][OHO[H
K
Kb for Morphine = 1.6 x 10-6
[B] [BH+] [OH]
initial 0.0015 0 ≈ 0
change -x +x +x
equilibrium 0.0015 4.9E-5 4.9E-5
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Practice – Find the pH of a 0.0015 M morphine solution
check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Kb to the given Kb
6
25
b
106.10015.0
109.4
B
OHBH
Kthe answer matches the given Kb
Kb for Morphine = 1.6 x 10-6
[B] [BH+] [OH]
initial 0.0015 0 ≈ 0
change -x +x +x
equilibrium 0.0015 4.9E-5 4.9E-5
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Acid-Base Properties of Salts• salts are water soluble ionic compounds
• salts that contain the cation of a strong base and an anion that is the conjugate base of a weak acid are basic NaHCO3 solutions are basic
Na+ is the cation of the strong base NaOH HCO3
− is the conjugate base of the weak acid H2CO3
• salts that contain cations that are the conjugate acid of a weak base and an anion of a strong acid are acidic NH4Cl solutions are acidic
NH4+ is the conjugate acid of the weak base NH3
Cl− is the anion of the strong acid HCl
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Acid-Base Properties of SaltsAnions as Weak Bases
• every anion can be thought of as the conjugate base of an acid
• therefore, every anion can potentially be a base
A−(aq) + H2O(l) HA(⇌ aq) + OH−(aq)
• the stronger the acid is, the weaker the conjugate base is an anion that is the conjugate base of a strong acid is pH neutral
Cl−(aq) + H2O(l) HCl(aq) + OH−(aq) since HCl is a strong acid, this equilibrium lies practically completely to the
left an anion that is the conjugate base of a weak acid is basic
F−(aq) + H2O(l) HF(⇌ aq) + OH−(aq) since HF is a weak acid, the position of this equilibrium favors the right
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Ex 15.13 - Use the Table to Determine if the Given Anion Is Basic or Neutral
a) NO3−
the conjugate base of a strong acid, therefore neutral
b) NO2−
the conjugate base of a weak acid, therefore basic
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Relationship between Ka of an Acid and Kb of Its Conjugate Base
• many reference books only give tables of Ka values because Kb values can be found from them
][A
]OHA][H[ )(OH ) HA()O( H)(A
[HA]
]O][HA[ )(O H )(A )O( H)HA(
3b2
3a32
Kaqaqlaq
Kaqaqlaqwhen you add equations, you multiply the K’s )(OH )(OH )O(H 2 32 aqaql
w3ba
3ba
]OH][OH[
]A[
]OH][HA[
]HA[
]OH][A[
KKK
KK
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Ex 15.14 Find the pH of 0.100 M NaCHO2(aq) solutionNa+ is the cation of a
strong base – pH neutral. The CHO2
− is the anion of a weak acid – pH basic
Write the reaction for the anion with water
Construct an ICE table for the reaction
Enter the initial concentrations – assuming the [OH] from water is ≈ 0
CHO2− + H2O HCHO⇌ 2 + OH
[CHO2−] [HCHO2] [OH]
initial 0.100 0 ≈ 0
change
equilibrium
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0.100 x
Ex 15.14 Find the pH of 0.100 M NaCHO2(aq) solution
represent the change in the concentrations in terms of x
sum the columns to find the equilibrium concentrations in terms of x
Calculate the value of Kb from the value of Ka of the weak acid from Table 15.5
substitute into the equilibrium constant expression
+x+xxx x
x
xxK
12
2b 1000.1CHO
]OH][[HCHO
[CHO2−] [HCHO2] [OH]
initial 0.100 0 ≈ 0
change
equilibrium
114
14
b
wba
106.5108.1
100.1
K
KKK
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Ex 15.14 Find the pH of 0.100 M NaCHO2(aq) solution
since Kb is very small, approximate the [CHO2
−]eq = [CHO2−]init
and solve for x
1
211
1000.1106.5
x
6
111
104.2
1000.1106.5
x
x
Kb for CHO2− = 5.6 x 10-11
[CHO2−] [HCHO2] [OH]
initial 0.100 0 ≈ 0
change -x +x +x
equilibrium 0.100 x x0.100 x
x
xxK
12
2b 1000.1CHO
]OH][[HCHO
12
2b 1000.1CHO
]OH][[HCHO
xxK
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Ex 15.14 Find the pH of 0.100 M NaCHO2(aq) solution
check if the approximation is valid by seeing if x < 5% of [CHO2
−]init
%5%0024.0%1001000.1
104.21
6
the approximation is valid
x = 2.4 x 10-6
Kb for CHO2− = 5.6 x 10-11
[CHO2−] [HCHO2] [OH]
initial 0.100 0 ≈ 0
change -x +x +x
equilibrium 0.100 x x
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Ex 15.14 Find the pH of 0.100 M NaCHO2(aq) solution
substitute x into the equilibrium concentration definitions and solve
M 100.0104.2100.0100.0CHO 62 x
M 104.2][OH]HCHO[ 6-2
x
x = 2.4 x 10-6
Kb for CHO2− = 5.6 x 10-11
[CHO2−] [HCHO2] [OH]
initial 0.100 0 ≈ 0
change -x +x +x
equilibrium 0.100 −x x x
[CHO2−] [HCHO2] [OH]
initial 0.100 0 ≈ 0
change -x +x +x
equilibrium 0.100 2.4E-6 2.4E-6
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Ex 15.14 Find the pH of 0.100 M NaCHO2(aq) solution
use the [OH-] to find the [H3O+] using Kw
substitute [H3O+] into the formula for pH and solve
38.8102.4log
OH-logpH9
3
9-3
6-
14-
3
-3w
102.4]O[H
104.2
101.00]O[H
]][OHO[H
K
Kb for CHO2− = 5.6 x 10-11
[CHO2−] [HCHO2] [OH]
initial 0.100 0 ≈ 0
change -x +x +x
equilibrium 0.100 2.4E-6 2.4E-6
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Ex 15.14 Find the pH of 0.100 M NaCHO2(aq) solution
check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Kb to the given Kb
11
26
2
2b
108.5100.0
104.2
CHO
OHHCHO
Kthough not exact, the answer is reasonably close
Kb for CHO2− = 5.6 x 10-11
[CHO2−] [HCHO2] [OH]
initial 0.100 0 ≈ 0
change -x +x +x
equilibrium 0.100 2.4E-6 2.4E-6
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Polyatomic Cations as Weak Acids
• some cations can be thought of as the conjugate acid of a base others are the counterions of a strong base
• therefore, some cation can potentially be an acid MH+(aq) + H2O(l) MOH(⇌ aq) + H3O+(aq)
• the stronger the base is, the weaker the conjugate acid is a cation that is the counterion of a strong base is pH neutral a cation that is the conjugate acid of a weak base is acidic
NH4+(aq) + H2O(l) NH⇌ 3(aq) + H3O+(aq)
since NH3 is a weak base, the position of this equilibrium favors the right
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Metal Cations as Weak Acids
• cations of small, highly charged metals are weakly acidic alkali metal cations and alkali earth metal cations pH neutral cations are hydrated
Al(H2O)63+(aq) + H2O(l) Al(H⇌ 2O)5(OH)2+
(aq) + H3O+(aq)
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Ex 15.15 - Determine if the Given Cation Is Acidic or Neutral
a) C5N5NH2+
the conjugate acid of a weak base, therefore acidic
b) Ca2+
the counterion of a strong base, therefore neutral
c) Cr3+
a highly charged metal ion, therefore acidic
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Classifying Salt Solutions asAcidic, Basic, or Neutral
• if the salt cation is the counterion of a strong base and the anion is the conjugate base of a strong acid, it will form a neutral solution NaCl Ca(NO3)2 KBr
• if the salt cation is the counterion of a strong base and the anion is the conjugate base of a weak acid, it will form a basic solution NaF Ca(C2H3O2)2 KNO2
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Classifying Salt Solutions asAcidic, Basic, or Neutral
• if the salt cation is the conjugate acid of a weak base and the anion is the conjugate base of a strong acid, it will form an acidic solution NH4Cl
• if the salt cation is a highly charged metal ion and the anion is the conjugate base of a strong acid, it will form an acidic solution Al(NO3)3
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Classifying Salt Solutions asAcidic, Basic, or Neutral
• if the salt cation is the conjugate acid of a weak base and the anion is the conjugate base of a weak acid, the pH of the solution depends on the relative strengths of the acid and base
• Ion with higher K value dominates
NH4F = NH4+, F-
NH4+ is conjugate acid of weak base (NH3) = acidic
F- is conjugate base of weak acid = basic Ka of NH4
+ (5.68 x 10-10) is larger than Kb of the F− (2.9 x 10-11); therefore the solution will be acidic
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Ex 15.16 - Determine whether a solution of the following salts is acidic, basic, or neutral
a) SrCl2Sr2+ is the counterion of a strong base, pH neutralCl− is the conjugate base of a strong acid, pH neutralsolution will be pH neutral
b) AlBr3
Al3+ is a small, highly charged metal ion, weak acidCl− is the conjugate base of a strong acid, pH neutralsolution will be acidic
c) CH3NH3NO3
CH3NH3+ is the conjugate acid of a weak base, acidic
NO3− is the conjugate base of a strong acid, pH neutral
solution will be acidic
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Ex 15.16 - Determine whether a solution of the following salts is acidic, basic, or neutral
d) NaCHO2
Na+ is the counterion of a strong base, pH neutral
CHO2− is the conjugate base of a weak acid, basic
solution will be basic
e) NH4F
NH4+ is the conjugate acid of a weak base, acidic
F− is the conjugate base of a weak acid, basic
Ka(NH4+) > Kb(F−); solution will be acidic
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Practice - Lewis Acid-Base ReactionsLabel the Nucleophile and Electrophile
• BF3 + HF H⇌ +BF4-
• CaO + SO3 Ca⇌ +2SO4-2
• KI + I2 KI⇌ 3
F
B F
F
H F••
••
•• +
F
B F
F
F-1
H+1NucElec
O
S O
O
••
••Ca+2 O -2
•• •• +
O
S O
O
O
-2
Ca+2
ElecNuc
I I K+1 I -1••
••
•• •• +ElecNuc K+1 I I I -1