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General Chemistry IIGeneral Chemistry IICHEM 152 Unit 2CHEM 152 Unit 2
Week 8
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Week 8 Reading Assignment
Chapter 15 – Sections 15.6 (finding pH of acids), 15.7 (bases), 15.8 (salts)
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Methods for Calculating the pH of an Aqueous
SolutionWith a strong acid or base – we assume 100% dissociation and directly get the
concentration of H3O+ or OH¯ to get the pH.
But what happens when we are dealing with weak acids or bases?
To answer this question we need to apply the ideas we have learned
about chemical equilibrium.
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Equilibria Involving Equilibria Involving WeakWeak Acids Acids
Consider the equilibrium in water for the weak Consider the equilibrium in water for the weak acid, acetic acid acid, acetic acid CHCH33COOH:COOH:
CHCH33COOH(aq) + HCOOH(aq) + H22O(l) O(l) H H33OO++(aq) + CH(aq) + CH33COOCOO--(aq)(aq)
AcidAcid Conj. baseConj. base
Ka [H3O+][OAc- ]
[HOAc] 1.8 x 10-5Ka
[H3O+][OAc- ][HOAc]
1.8 x 10-5
(K is designated K(K is designated Kaa for for ACIDACID))
Because [HBecause [H33OO++] and [OAc] and [OAc--] are SMALL, ] are SMALL, KKaa << 1 << 1
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Equilibrium Constants Equilibrium Constants for for WeakWeak Acids Acids
Weak acid has KWeak acid has Kaa < 1 < 1
Leads to Leads to more [Hmore [H33OO++] than in pure water] than in pure water
and a and a pH of 2 - 7pH of 2 - 7
Acid Ionizatio
n Constant
In general:
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Ionization Constants for Ionization Constants for AcidsAcids
Increase strength
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Equilibrium Constants Equilibrium Constants for Weak Basesfor Weak Bases
Weak base has KWeak base has Kbb < 1 < 1
Leads to Leads to more [OHmore [OH¯̄] than in pure water ] than in pure water
and a and a pH of 7-12pH of 7-12
Base Ionizatio
n Constant
Write the ionization equation and the ionization constant for :
H2PO4- (acid) and C6H5NH2 (base).
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Ionization Constants for Ionization Constants for Bases Bases
Increase strength
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pH for a pH for a Weak AcidWeak Acid
Imagine that you have 1.00 M Imagine that you have 1.00 M CHCH33COOHCOOH. Calculate . Calculate the equilibrium concentrations of the equilibrium concentrations of CHCH33COOHCOOH, , HH33OO++,,
CHCH33COOCOO--, and the pH of the solution , and the pH of the solution (K(Kaa = 1.8 x 10 = 1.8 x 10-5-5). ). How do you do it?How do you do it?
Step 1.Step 1. Define equilibrium Define equilibrium
concentrations:concentrations:
[CH[CH33COOH] + HCOOH] + H22O O [H [H33OO++] + ] + [CH[CH33COOCOO--]]
I 1.00I 1.00 ~0 ~0 0 0
CC -x-x +x +x +x +x
E 1.00-x E 1.00-x x x x x
Note that we neglect [HNote that we neglect [H33OO++] from ] from HH22OO
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pH for a pH for a Weak AcidWeak Acid
Step 2.Step 2. Write K Write Kaa expression expression
Ka 1.8 x 10-5 = [H3O+][OAc- ]
[HOAc]
x2
1.00 - xKa 1.8 x 10-5 =
[H3O+][OAc- ][HOAc]
x2
1.00 - x
Step 3.Step 3. Solve K Solve Kaa expression to find x expression to find x
We can assume x is very small because We can assume x is very small because KKaa is so small. is so small.
Ka 1.8 x 10-5 = x2
1.00Ka 1.8 x 10-5 =
x2
1.00
Now we can more easily solve this Now we can more easily solve this approximate expression.approximate expression.
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pH for a pH for a Weak AcidWeak Acid
Step 4.Step 4. Solve K Solve Kaa approximate approximate expression expression
Ka 1.8 x 10-5 = x2
1.00Ka 1.8 x 10-5 =
x2
1.00
x = x = [[HH33OO++]] = = [[CHCH33COOCOO--]] = [K = [Kaa • 1.00] • 1.00]1/21/2
x = x = [[HH33OO++]] = = [[CHCH33COOCOO--]] = = 4.2 x 104.2 x 10-3-3 M M
pHpH= -log [= -log [HH33OO++]= -log (4.2 x 10]= -log (4.2 x 10-3-3)= )=
2.372.37
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pH for a pH for a Weak AcidWeak Acid
Consider the approximate Consider the approximate expression in the last problem: expression in the last problem:
x [H3O+ ] = [Ka • 1.00]1/2x [H3O+ ] = [Ka • 1.00]1/2
For many weak acidsFor many weak acids
[H[H33OO++]= [conjugate base]= [K]= [conjugate base]= [Kaa• •
CCoo]]1/21/2
where Cwhere C00 = initial concentration of = initial concentration of
acidacid
In general:In general:
If If 100•K100•Ka a < < CCoo, then , then [H[H33OO++]] = = [K[Kaa•C•Coo]]1/21/2
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Calculate the pH of a 0.0010 M Calculate the pH of a 0.0010 M solution of formic acid, solution of formic acid, HCOOHHCOOH. .
KKaa = 1.8 x 10 = 1.8 x 10-4-4
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Your TurnYour TurnCalculate the pH of a 0.0010 M solution Calculate the pH of a 0.0010 M solution of formic acid, of formic acid, HCOOHHCOOH. . KKaa = 1.8 x 10 = 1.8 x 10-4-4
Exact SolutionExact Solution
[H[H33OO++]= []= [HCOHCO22--] = 3.4 x 10] = 3.4 x 10-4-4 M M
[[HCOHCO22HH]= 0.0010-3.4 x 10]= 0.0010-3.4 x 10-4-4 = 0.0007 = 0.0007 M M pH = 3.46 pH = 3.46
Approximate solutionApproximate solution
[H[H33OO++] = [K] = [Kaa • C • Coo]]1/21/2 = 4.2 x 10 = 4.2 x 10-4-4 M, M, pH = 3.37pH = 3.37
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pH for a pH for a Weak BaseWeak BaseImagine that you have 0.010 M NHImagine that you have 0.010 M NH33. .
Calculate the pH.Calculate the pH.
NHNH33 + H + H22O O NH NH44++ + OH + OH--
KKbb = 1.8 x 10 = 1.8 x 10-5-5
Step 1.Step 1. Define equilibrium Define equilibrium
concentrations:concentrations:
[NH[NH33] + H] + H22O O [NH [NH44++] + ] +
[OH[OH--]]
InitialInitial 0.0100.010 0 0 0 0
ChangeChange -x -x +x +x +x +x
Equilib 0.010 - x Equilib 0.010 - x +x +x +x +x
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pH for a pH for a Weak BaseWeak Base
Step 2.Step 2. Solve the equilibrium Solve the equilibrium expressionexpression
Kb 1.8 x 10-5 = [NH4
+][OH- ][NH3 ]
= x2
0.010 - xKb 1.8 x 10-5 =
[NH4+][OH- ]
[NH3 ] =
x2
0.010 - x
Assume x is small (100•KAssume x is small (100•Kbb < C < Coo), so), so x = [OHx = [OH--] = [NH] = [NH44
++] = 4.2 x 10] = 4.2 x 10-4-4 M M
and and [NH[NH33] = 0.010 - 4.2 x 10] = 0.010 - 4.2 x 10-4-4 ≈ 0.010 ≈ 0.010
MMStep 3.Step 3. Calculate pH [OH Calculate pH [OH--] = 4.2 x 10] = 4.2 x 10-4-4 M M
so pOH = - log [OHso pOH = - log [OH--] = 3.37] = 3.37
Because pH + pOH = 14, Because pH + pOH = 14, pH = 10.63pH = 10.63
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Calculate the pH of a 0.15 M Calculate the pH of a 0.15 M solution of (CHsolution of (CH33))33N. N.
KKbb = 6.3 x 10 = 6.3 x 10-5-5
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MX + HMX + H22O O acidic or basic solution?acidic or basic solution?
Acid-Base Properties of Acid-Base Properties of SaltsSalts
ClCl-- ion is a VERY weak base because its ion is a VERY weak base because its conjugate acid is strong conjugate acid is strong
Therefore, Therefore, ClCl-- neutral solution neutral solution
Consider Consider NHNH44ClCl
NHNH44Cl(aq) Cl(aq) NH NH44++(aq) + Cl(aq) + Cl--(aq)(aq)
(a)(a) Reaction of ClReaction of Cl-- with H with H22OO
ClCl-- + H + H22O O HCl + OHHCl + OH--
basebase acidacid acidacid basebase
What happens with the pH if we dissolve a salt in water?
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NHNH44Cl(aq) Cl(aq) NH NH44++(aq) + Cl(aq) + Cl--(aq)(aq)
(b)(b) Reaction of NHReaction of NH44++ with H with H22OO
NHNH44++ + H + H22O O NHNH33 + H + H33OO++
acidacid basebase basebase acidacid
Acid-Base Properties of Acid-Base Properties of SaltsSalts
NHNH44++ ion is a moderate acid because ion is a moderate acid because its conjugate base is weak. its conjugate base is weak.
Therefore, Therefore, NHNH44++ acidic solution acidic solution
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A solution of NaCl would be:
1. acidic2. basic3. neutral
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A solution of K3PO4 would be:
1. acidic2. basic3. neutral
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A solution of NH4F would be:
1. acidic2. basic3. neutral
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Calculate the pH of a 0.10 M solution of NaCalculate the pH of a 0.10 M solution of Na22COCO33.. Step 1: Decide which ion will affect pH. Step 1: Decide which ion will affect pH.
NaNa++ + H + H22O O ? ?
COCO332-2- + H + H22O O ? ? KKbb = 2.1 x 10 = 2.1 x 10-4-4
Acid-Base Properties of SaltsAcid-Base Properties of Salts
Step 2Step 2.. Set up ICE tableSet up ICE table
[CO[CO332-2-] + H] + H22O O [HCO [HCO33
--]]+ [OH+ [OH--] ] initialinitial 0.100.10 00 00
changechange -x-x +x+x +x+x
equilibequilib 0.10-x0.10-x x x x x
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Acid-Base Properties of Acid-Base Properties of SaltsSalts
Kb = 2.1 x 10-4 = [HCO3
- ][OH- ]
[CO32 ]
x2
0.10 - xKb = 2.1 x 10-4 =
[HCO3- ][OH- ]
[CO32 ]
x2
0.10 - x
Assume 0.10 - x ≈ 0.10, because 100•Kb < Co
x =[HCO3-] =[OH-] = 0.0046 M
Step 2Step 2.. Solve the equilibrium expressionSolve the equilibrium expression
Step 3.Step 3. Calculate the pHCalculate the pH
[OH[OH--]= 0.0046 M, pOH= - log [OH]= 0.0046 M, pOH= - log [OH--] = ] =
2.342.34
pH + pOH = 14, so pH + pOH = 14, so pH = 11.66pH = 11.66
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Calculate the pH of a 0.15 M Calculate the pH of a 0.15 M solution of KNOsolution of KNO22. .
KKaa(HNO(HNO22) = 4.6 x 10) = 4.6 x 10-4-4
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Polyprotic AcidsSome acids donate more than one H+ per
molecule: H2SO4, H2CO3, H3PO4
These acids donate their protons in an stepwise manner:
H3PO4(aq) + H2O(l) H3O+(aq) + H2PO4-(aq)
Ka=7.5 x 10-3
H2PO4-(aq) + H2O(l) H3O+(aq) + HPO42-(aq)
Ka=6.2 x 10-8
HPO42-(aq) + H2O(l) H3O+(aq) + PO4
3- (aq) Ka=3.6 x 10-13
The resulting acids are weaker
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Molecular structure and acid/base strength
There are four main effects that influence the relative strength of an acid:
Size of the anionElectronegativity of the H-bearing atom
Inductive effectResonance stabilization of the anion
The first two factors influence binary acids
The latter two factors influence non-binary acids
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Binary Acid Strength
Factors affecting binary acid strength
The polarity of the H-A bond (most important factor when comparing atoms in the same
row in the periodic table) electronegativity
The bond energy of the H-A bond (dependent on the length
of the bond; important when comparing atoms in the same column in the periodic table)
size of the anion
HA type of acid
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Binary Acid Strength
Which of these is a stronger acid: HCl or HBr? Why?
Which of these is a stronger acid: H3P or H2S? Why?
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Strength of Non-binary Acids
For nonbinary acids –
Atoms with higher electronegativity can draw the electrons away from the bond with the
acidic hydrogen – making that bond more polar. This is known as the inductive effect.
Or if the anion of the acid has several equal resonance structures – the anion would be
rather stable – and the acid more likely to lose its acidic hydrogen. This is known as resonance
stablization.
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Strength of OxyacidsAcids in which the acidic hydrogen is bonded directly to oxygen in an H-O-Z bond are called
oxyacids.
For each of the following pairs, which acid is stronger – and why?
Strength?
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Why is Why is CHCH33COOH an COOH an
Acid?Acid?11.The electronegativity .The electronegativity
of the of the OO atoms atoms causes the H causes the H attached to O to be attached to O to be highly positive.highly positive.
2.The 2.The O—H O—H bond is bond is highly polar.highly polar.
3.The 3.The HH atom of atom of O—H O—H is readily attracted to is readily attracted to polar polar HH22OO. .
Example of Carboxylic Acid (-
COOH group) Weak Acid
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Trichloroacetic acid is much stronger acid owing to Trichloroacetic acid is much stronger acid owing to the high electronegativity of Cl, which withdraws the high electronegativity of Cl, which withdraws
electrons from the rest of the molecule. electrons from the rest of the molecule. This makes This makes the O—H bond highly polar.the O—H bond highly polar. The H of O—H is very The H of O—H is very
positive.positive.
Inductive effectInductive effect
Acetic acidAcetic acid Trichloroacetic acidTrichloroacetic acid
Which acid is stronger? Why?
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Weak Bases: AminesAmines are compounds that, like ammonia,
have a nitrogen atom with three of its valence electrons in covalent bonds and an
unshared electron pair on the nitrogen atom.
The lone pair of electrons can accept an H+.
H2O
H+
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Which of the following is a weak base?
CH3
NH
CH3 CH3
NH2CH3
(A) (B)
1. A2. B3. Both4. Neither
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1. What is the pH of a solution of 0.050 M (C2H5)2NH?
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2. What is the pH of a solution of 0.050 M HClO4?
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Summary Activity
3. Label the following as acidic, basic, or neutral solutions:
KCl NaNO2 (CH3)2NH2Cl
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Which of the following is the strongest acid?
HClO2, HBrO2, HIO2
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Which of the following is the strongest acid?
H3N, H2O, HF
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What kind of substance is
C2H5NH2?
1. Strong acid2. Strong base3. Weak Acid4. Weak base
