Strength of Acids and Bases
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Transcript of Strength of Acids and Bases
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Strength of Strength of Acids and BasesAcids and Bases
Do they ionize 100%?Do they ionize 100%?
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Strong Acids :Strong Acids : Give up HGive up H++ easily easily
Dissociate completely (100%) in waterDissociate completely (100%) in water
HCl, HBr, HI, HNOHCl, HBr, HI, HNO33, H, H22SOSO44, HClO, HClO44, HClO, HClO33
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Weak acids: (all others)Weak acids: (all others)Hold onto HHold onto H++
Few molecules dissociateFew molecules dissociate
Ex: HCEx: HC2HH33OO2 2
Strong/Weak Acid Strong/Weak Acid AnimationAnimationhttp://educypedia.karadimov.info/library/acid13.swf
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HA
Let’s examine the behavior of an acid, HA, in aqueous solution.
What happens to the HA molecules in solution?
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HA
H+
A-
Strong Acid
100% dissociation of HA
Would the solution be conductive?
Oh yeah…
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HA
H+
A-
Weak Acid
Partial dissociation of HA
Would the solution be conductive?
Not really…
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HA
H+
A-
Weak Acid
HA H+ + A-
At any one time, only a fraction of
the molecules
are dissociated.
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Strong BasesStrong Bases: : Dissociate completely (100%) in waterDissociate completely (100%) in water
- Group I metal hydroxides (NaOH, LiOH, etc.)- Group I metal hydroxides (NaOH, LiOH, etc.)
- Some Group II metal hydroxides- Some Group II metal hydroxides
Ca(OH)Ca(OH)22, Ba(OH), Ba(OH)22, ,
Sr(OH)Sr(OH)22
Weak BasesWeak Bases
Only a few ions dissociateOnly a few ions dissociate
Ex: NHEx: NH 3 3 (ammonia) (ammonia)
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Strength and ReactivityStrength and Reactivity
Acids/bases of the same initial molar concentration Acids/bases of the same initial molar concentration can react differently and conduct electricity can react differently and conduct electricity differently if one is weak and the other strong.differently if one is weak and the other strong.
Ex: Ex: 2M HCl 2M HCl Strong Acid, Strong Acid,
very conductive very conductive very reactivevery reactive
2M 2M HCHC22HH33OO2 2 Weak Acid Weak Acid
Weak ConductionWeak Conduction
Salad Dressing!!!Salad Dressing!!!
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Conjugate Acid/Base PairsConjugate Acid/Base Pairs
Strong acid will have a weak conjugate baseStrong acid will have a weak conjugate baseStrong base will have a weak conjugate acidStrong base will have a weak conjugate acid
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HydrolysisHydrolysis
Opposite reaction to neutralizationOpposite reaction to neutralization
Salt + Water Salt + Water Acid + BaseAcid + Base
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Parent Acid/BaseParent Acid/Base
If you know the salt involved you should If you know the salt involved you should be able to determine which acid and base be able to determine which acid and base it would form if water is added.it would form if water is added.
Salt + Water Salt + Water Acid + BaseAcid + Base
Ex: Ex:
NaCl with water (HOH) would form HCl and NaOHNaCl with water (HOH) would form HCl and NaOH
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You Try ItYou Try It
Name the “parent” acid and base that Name the “parent” acid and base that would be produced from these salts.would be produced from these salts.
Ex:Ex: Potassium chloridePotassium chloride
Magnesium carbonateMagnesium carbonate
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pH and HydrolysispH and Hydrolysis
Salts can yield neutral, acidic or basic Salts can yield neutral, acidic or basic solutions depending on what type of acid solutions depending on what type of acid or base they produce.or base they produce.
SA/SB = NeutralSA/SB = Neutral
SA/WB = AcidicSA/WB = Acidic
WA/SB = BasicWA/SB = Basic
WA/WB = UndeterminedWA/WB = Undetermined
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The Acid and Base The Acid and Base Dissociation Constant, Dissociation Constant,
Ka & Kb Ka & Kb
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Setting up Ka and Kb ExpressionsSetting up Ka and Kb Expressions
weak acid:weak acid: CHCH33COOH + HCOOH + H22O ↔ HO ↔ H33OO++ + CH + CH33COOCOO--
Acid ionization constantAcid ionization constant: K: Kaa = [H = [H33OO++][CH][CH33COOCOO--]]
[CH[CH33COOH]COOH]
weak base:weak base: NHNH33 + H + H22O ↔ NHO ↔ NH44++ + OH + OH--
Base ionization constantBase ionization constant: K: Kbb = [NH = [NH44++][OH][OH--]]
[NH[NH33]]
Acid and base ionization constants are the measure of theAcid and base ionization constants are the measure of thestrengths of acids and bases.strengths of acids and bases.The larger the Ka/Kb value the stronger the acid or baseThe larger the Ka/Kb value the stronger the acid or base
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KaKa Weak acidsWeak acids::
only ionize to a small extent only ionize to a small extent come to a state of chemical equilibrium.come to a state of chemical equilibrium.
Determine how much it ionizes by calculating the equilibrium Determine how much it ionizes by calculating the equilibrium
constant (constant (KaKa) )
Larger Larger Ka:Ka:stronger acid stronger acid more ions found in solution more ions found in solution more easily donate a protonmore easily donate a proton
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HCOOH HCOOH (aq)(aq) ++ H H22O O (l)(l) < -- > H < -- > H33OO++ (aq)(aq) + HCOO + HCOO-- (aq)(aq)
Ka Ka = = [H[H33OO++][HCOO][HCOO--]]
[HCOOH][HCOOH]
Notice how the Ka ignores the water since Notice how the Ka ignores the water since we are dealing with dilute solutions of we are dealing with dilute solutions of acids, water is considered a constant and acids, water is considered a constant and doesn’t have a concentration valuedoesn’t have a concentration value
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Practice:Practice:
1.1. A solution of a weak acid, “HA”, is made A solution of a weak acid, “HA”, is made up to be 0.15 M. Its pH was found to be up to be 0.15 M. Its pH was found to be 2.96. Calculate the value of 2.96. Calculate the value of KaKa..
Steps to follow:Steps to follow:1.1. Write balanced equationWrite balanced equation
2.2. Calculate [HCalculate [H++] using 10] using 10-pH-pH
3.3. Set up chart for equilibrium (ICESet up chart for equilibrium (ICE))
4.4. Solve using Ka expressionSolve using Ka expression
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AnswerAnswer
1.1. HA HA (aq)(aq) + H + H22O O (l)(l) < -- >H < -- >H33OO++ (aq)(aq) + A + A--(aq)(aq)
2.2. [H[H33O+]= 10 O+]= 10 -2.96-2.96 = 0.0011 M = 0.0011 M
HA H2O < -- > H3O+ A-
0.15
-.0011
0.139 0.0011 0.0011
I
C
E
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4.4. Ka = [Ka = [HH33OO++][ A][ A--]]
[HA][HA]
= = [0.0011][0.0011][0.0011][0.0011]
[0.139][0.139]
= 8.7 x 10 = 8.7 x 10 -6-6
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Percent IonizationPercent IonizationThe fraction of acid molecules that dissociateThe fraction of acid molecules that dissociate
compared with the initial concentration of the acid.compared with the initial concentration of the acid.
Percent Ionization = Percent Ionization = [H[H33OO++]] x 100% x 100%
[HA [HA ii]]
For the previous question:For the previous question:
Percent Ionization = Percent Ionization = [0.0011][0.0011] x 100% =0.73 % x 100% =0.73 %
[0.15][0.15]
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Practice:Practice:KaKa, for a hypothetical weak acid, HA, at 25°C , for a hypothetical weak acid, HA, at 25°C
is 2.2 x 10is 2.2 x 10-4-4..
a)a) Calculate [HCalculate [H33OO++] of a 0.20 M solution of HA.] of a 0.20 M solution of HA.
b) Calculate the percent ionization of HA.b) Calculate the percent ionization of HA.
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AnswerAnswer
a) HA(aq) + H2O(l) < -- >H3O+(aq) + A-(aq)a) HA(aq) + H2O(l) < -- >H3O+(aq) + A-(aq)
Ka Ka = = [H[H33OO++] [A-] ] [A-] = 2.2 x 10= 2.2 x 10-4-4
[HA][HA]
2.2 x 102.2 x 10-4-4 = = ((xx)()(xx)) (0.20M - x)(0.20M - x)
xx22 = (2.2 x 10 = (2.2 x 10-4-4) (0.20 M)) (0.20 M)x x = 0.0066 M= 0.0066 M
The [HThe [H33OO++] is 0.0066 M.] is 0.0066 M.
Because it is a 1:1 ratio they are both the same concentration (x)
Since Ka is rather small this number can be disregarded
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b) % ionization = b) % ionization = 0.0066 M0.0066 M x 100% = 3.3% x 100% = 3.3%
0.20 M0.20 M
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The pH of 0.100 M H2C2O4 is 1.28.
Calculate the Ka for the weak acid.
H2C2O4 ⇄ H+ + HC2O4-
I 0.100 CE
The pH of 0.100 M H2C2O4 is 1.28.
Calculate the Ka for the weak acid.
H2C2O4 ⇄ H+ + HC2O4-
I 0.100 CE
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The pH of 0.100 M H2C2O4 is 1.28.
Calculate the Ka for the weak acid.
H2C2O4 ⇄ H+ + HC2O4-
I 0.100 CE
pH = 1.28
The pH of 0.100 M H2C2O4 is 1.28.
Calculate the Ka for the weak acid.
H2C2O4 ⇄ H+ + HC2O4-
I 0.100 CE
pH = 1.28
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The pH of 0.100 M H2C2O4 is 1.28.
Calculate the Ka for the weak acid.
H2C2O4 ⇄ H+ + HC2O4-
I 0.100 CE
pH = 1.28
[H+] = 10-1.28
The pH of 0.100 M H2C2O4 is 1.28.
Calculate the Ka for the weak acid.
H2C2O4 ⇄ H+ + HC2O4-
I 0.100 CE
pH = 1.28
[H+] = 10-1.28
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The pH of 0.100 M H2C2O4 is 1.28.
Calculate the Ka for the weak acid.
H2C2O4 ⇄ H+ + HC2O4-
I 0.100 CE
pH = 1.28
[H+] = 10-1.28
[H+] = 0.05248 M
The pH of 0.100 M H2C2O4 is 1.28.
Calculate the Ka for the weak acid.
H2C2O4 ⇄ H+ + HC2O4-
I 0.100 CE
pH = 1.28
[H+] = 10-1.28
[H+] = 0.05248 M
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The pH of 0.100 M H2C2O4 is 1.28.
Calculate the Ka for the weak acid.
H2C2O4 ⇄ H+ + HC2O4-
I 0.100 CE 0.05248 0.05248
pH = 1.28
[H+] = 10-1.28
[H+] = 0.05248 M
The pH of 0.100 M H2C2O4 is 1.28.
Calculate the Ka for the weak acid.
H2C2O4 ⇄ H+ + HC2O4-
I 0.100 CE 0.05248 0.05248
pH = 1.28
[H+] = 10-1.28
[H+] = 0.05248 M
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The pH of 0.100 M H2C2O4 is 1.28.
Calculate the Ka for the weak acid.
H2C2O4 ⇄ H+ + HC2O4-
I 0.100 C - 0.05248E 0.04752 0.05248 0.05248
pH = 1.28
[H+] = 10-1.28
[H+] = 0.05248 M
The pH of 0.100 M H2C2O4 is 1.28.
Calculate the Ka for the weak acid.
H2C2O4 ⇄ H+ + HC2O4-
I 0.100 C - 0.05248E 0.04752 0.05248 0.05248
pH = 1.28
[H+] = 10-1.28
[H+] = 0.05248 M
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1. The pH of 0.100 M H2C2O4 is 1.28.
Calculate the Ka for the weak acid.
H2C2O4 ⇄ H+ + HC2O4-
I 0.100 C - 0.05248E 0.04752 0.05248 0.05248
Ka = [H+][HC2O4-] =(0.05248)2
[H2C2O4] 0.04752
Ka = 5.8 x 10-2
1. The pH of 0.100 M H2C2O4 is 1.28.
Calculate the Ka for the weak acid.
H2C2O4 ⇄ H+ + HC2O4-
I 0.100 C - 0.05248E 0.04752 0.05248 0.05248
Ka = [H+][HC2O4-] =(0.05248)2
[H2C2O4] 0.04752
Ka = 5.8 x 10-2
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KbKb
When using weak bases, the same rules When using weak bases, the same rules apply as with weak acids, except you are apply as with weak acids, except you are solving for pOH and using [OHsolving for pOH and using [OH--]]
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If the pH of 0.40 M NH3 @ 25 oC is 11.427,
calculate the Kb.
NH3 + H2O ⇄ NH4+ + OH-
I 0.40 CE
If the pH of 0.40 M NH3 @ 25 oC is 11.427,
calculate the Kb.
NH3 + H2O ⇄ NH4+ + OH-
I 0.40 CE
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If the pH of 0.40 M NH3 @ 25 oC is 11.427,
calculate the Kb.
NH3 + H2O ⇄ NH4+ + OH-
I 0.40 CE
pH = 11.427pOH = 2.573
If the pH of 0.40 M NH3 @ 25 oC is 11.427,
calculate the Kb.
NH3 + H2O ⇄ NH4+ + OH-
I 0.40 CE
pH = 11.427pOH = 2.573
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If the pH of 0.40 M NH3 @ 25 oC is 11.427,
calculate the Kb.
NH3 + H2O ⇄ NH4+ + OH-
I 0.40 CE
pH = 11.427pOH = 2.573[OH-] =10-2.573
[OH-] =0.002673 M
If the pH of 0.40 M NH3 @ 25 oC is 11.427,
calculate the Kb.
NH3 + H2O ⇄ NH4+ + OH-
I 0.40 CE
pH = 11.427pOH = 2.573[OH-] =10-2.573
[OH-] =0.002673 M
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If the pH of 0.40 M NH3 @ 25 oC is 11.427,
calculate the Kb. NH3 + H2O ⇄ NH4
+ + OH-
I 0.40 C - 0.002673E 0.3973 0.002673 0.002673
[OH-] = 0.002673 M
If the pH of 0.40 M NH3 @ 25 oC is 11.427,
calculate the Kb. NH3 + H2O ⇄ NH4
+ + OH-
I 0.40 C - 0.002673E 0.3973 0.002673 0.002673
[OH-] = 0.002673 M
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If the pH of 0.40 M NH3 @ 25 oC is 11.427,
calculate the Kb. NH3 + H2O ⇄ NH4
+ + OH-
I 0.40 C - 0.002673E 0.3973 0.002673 0.002673
[OH-] = 0.002673 M
Kb = [NH4+][OH-] = (0.002673)2
[NH3] 0.3973
Kb = 1.8 x 10-5
If the pH of 0.40 M NH3 @ 25 oC is 11.427,
calculate the Kb. NH3 + H2O ⇄ NH4
+ + OH-
I 0.40 C - 0.002673E 0.3973 0.002673 0.002673
[OH-] = 0.002673 M
Kb = [NH4+][OH-] = (0.002673)2
[NH3] 0.3973
Kb = 1.8 x 10-5
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The pH of 0.20 M NaCN is 11.456, calculate the Kb for CN-.
The pH of 0.20 M NaCN is 11.456, calculate the Kb for CN-.
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The pH of 0.20 M NaCN is 11.456, calculate the Kb for CN-.
CN- + H2O ⇄ HCN+ OH-
pH = 11.456pOH = 2.55[OH-] = 10-2.55
[OH-] = .002858M
The pH of 0.20 M NaCN is 11.456, calculate the Kb for CN-.
CN- + H2O ⇄ HCN+ OH-
pH = 11.456pOH = 2.55[OH-] = 10-2.55
[OH-] = .002858M
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The pH of 0.20 M NaCN is 11.456, calculate the Kb for CN-.
CN- + H2O ⇄ HCN + OH-
I 0.20C - 0.002858E 0.1971 0.002858 0.002858
The pH of 0.20 M NaCN is 11.456, calculate the Kb for CN-.
CN- + H2O ⇄ HCN + OH-
I 0.20C - 0.002858E 0.1971 0.002858 0.002858
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The pH of 0.20 M NaCN is 11.456, calculate the Kb for CN-.
CN- + H2O ⇄ HCN + OH-
I 0.20C - 0.002858E 0.1971 0.002858 0.002858
Kb = [HCN][OH-] = (0.002858)2= 4.1 x 10-5
[CN-] 0.1971
The pH of 0.20 M NaCN is 11.456, calculate the Kb for CN-.
CN- + H2O ⇄ HCN + OH-
I 0.20C - 0.002858E 0.1971 0.002858 0.002858
Kb = [HCN][OH-] = (0.002858)2= 4.1 x 10-5
[CN-] 0.1971
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Calculate the pH of 0.020 M H3BO3 (Ka = 3.8 x 10-10)
(weak acid dissociates one H+ at a time)
H3BO3 ⇄ H+ + H2BO3-
I 0.020 MC - xE 0.020 - x x x
Calculate the pH of 0.020 M H3BO3 (Ka = 3.8 x 10-10)
(weak acid dissociates one H+ at a time)
H3BO3 ⇄ H+ + H2BO3-
I 0.020 MC - xE 0.020 - x x x
disregard small Ka
x2 = 3.8 x 10-10
0.020
x = [H+] = 2.76 x 10-6 M
pH = -Log[2.76 x 10-6]
pH = 5.42
2 sig figs due to molarity and Ka