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11
HOMEWORK
Page 704
Day 1
#29, 31, 35, 37, 39, 43, 44, 45
Day 2
#47, 49, 51, 53, 57, 61, 63
Day 3
#65, 67, 69, 71, 75, 77, 79,83, 97,89
Day 4
#93,95,97.99.101,111,113,117
22The Chemistry The Chemistry of Acids and of Acids and BasesBases
The Chemistry The Chemistry of Acids and of Acids and BasesBases
33
Acid and BasesAcid and Bases
44
Acid and BasesAcid and Bases
55
Acid and BasesAcid and Bases
66Acids
Have a sour taste. Vinegar is a solution of acetic acid. CitrusHave a sour taste. Vinegar is a solution of acetic acid. Citrusfruits contain citric acid.fruits contain citric acid.
React with certain metals to produce hydrogen gasReact with certain metals to produce hydrogen gas..
React with carbonates and bicarbonates to produce carbon React with carbonates and bicarbonates to produce carbon dioxide gasdioxide gas
Have a bitter taste.Have a bitter taste.
Feel slippery. Many soaps contain bases.Feel slippery. Many soaps contain bases.
Bases
77
Some Properties of Acids
Produce H+ (as H3O+) ions in water (the hydronium ion is a
hydrogen ion attached to a water molecule)
Taste sour
Corrode metals
Electrolytes
React with bases to form a salt and water
pH is less than 7
Turns blue litmus paper to red “Blue to Red A-CID”
88
Some Properties of Bases
Produce OHProduce OH-- ions in water ions in water
Taste bitter, chalkyTaste bitter, chalky
Are electrolytesAre electrolytes
Feel soapy, slipperyFeel soapy, slippery
React with acids to form salts and waterReact with acids to form salts and water
pH greater than 7pH greater than 7
Turns red litmus paper to blue “Turns red litmus paper to blue “BBasic asic BBlue”lue”
99
Some Common Bases
NaOHNaOH sodium hydroxidesodium hydroxide lyelye
KOHKOH potassium hydroxidepotassium hydroxide liquid soapliquid soap
Ba(OH)Ba(OH)22 barium hydroxidebarium hydroxide stabilizer for plasticsstabilizer for plastics
Mg(OH)Mg(OH)22 magnesium hydroxidemagnesium hydroxide “MOM” Milk of magnesia“MOM” Milk of magnesia
Al(OH)Al(OH)33 aluminum hydroxidealuminum hydroxide Maalox (antacid)Maalox (antacid)
1010
Acid/Base definitions
• Definition #1: Arrhenius (traditional)
Acids – produce H+ ions (or hydronium ions H3O+)
Bases – produce OH- ions
(problem: some bases don’t have hydroxide ions!)
1111Arrhenius acid is a substance that produces H+ (H3O+) in water
Arrhenius base is a substance that produces OH- in water
1212
Acid/Base Definitions
• Definition #2: Brønsted – Lowry
Acids – proton donor
Bases – proton acceptor
A “proton” is really just a hydrogen atom that has lost it’s electron!
1313
A Brønsted-Lowry acid is a proton donorA Brønsted-Lowry base is a proton acceptor
acidconjugate
basebase conjugate
acid
1414
ACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIES
The Brønsted definition means NHThe Brønsted definition means NH33 is is aa BASEBASE in water — and water is in water — and water is itself anitself an ACIDACID
BaseAcidAcidBaseNH4
+ + OH-NH3 + H2OBaseAcidAcidBase
NH4+ + OH-NH3 + H2O
1515
Conjugate PairsConjugate Pairs
1616
Learning Check!
Label the acid, base, conjugate acid, and Label the acid, base, conjugate acid, and conjugate base in each reaction:conjugate base in each reaction:
HCl + OHHCl + OH-- Cl Cl-- + H + H22OO HCl + OHHCl + OH-- Cl Cl-- + H + H22OO
HH22O + HO + H22SOSO44 HSO HSO44-- + H + H33OO
++ HH22O + HO + H22SOSO44 HSO HSO44-- + H + H33OO
++
1717Acids & Base Acids & Base DefinitionsDefinitions
Lewis acid - a Lewis acid - a substance that substance that accepts an electron accepts an electron pairpair
Lewis base - a Lewis base - a substance that substance that donates an electron donates an electron pairpair
Definition #3 – Lewis Definition #3 – Lewis
1818
Formation ofFormation of hydronium ion hydronium ion is also an is also an excellent example.excellent example.
Lewis Acids & BasesLewis Acids & Bases
•Electron pair of the new O-H bond Electron pair of the new O-H bond originates on the Lewis base.originates on the Lewis base.
HH
H
BASE
••••••
O—HO—H
H+
ACID
1919
Lewis Acid/Base ReactionLewis Acid/Base Reaction
2020
Lewis Acid-Base Lewis Acid-Base Interactions in BiologyInteractions in Biology
• The heme group The heme group in hemoglobin in hemoglobin can interact with can interact with OO22 and CO. and CO.
• The Fe ion in The Fe ion in hemoglobin is a hemoglobin is a Lewis acidLewis acid
• OO22 and CO can act and CO can act as Lewis basesas Lewis bases
Heme group
2121The The pH scalepH scale is a way of is a way of expressing the strength expressing the strength of acids and bases. of acids and bases. Instead of using very Instead of using very small numbers, we just small numbers, we just use the NEGATIVE use the NEGATIVE power of 10 on the power of 10 on the Molarity of the HMolarity of the H++ (or (or OHOH--) ion.) ion.
Under 7 = acidUnder 7 = acid 7 = neutral 7 = neutral
Over 7 = baseOver 7 = base
2222
pH Scale
2323
pH of Common pH of Common SubstancesSubstances
2424
Strong Acids:Perchloric HClO4
Chloric, HClO3
Hydrobromic, HBrHydrochloric, HClHydroiodic, HINitric, HNO3
Sulfuric, H2SO4
Strong Acids:100% ionized (completely dissociated) in water.
HCl + H2O H3O+ + Cl-
2525What is a strong Base?
A base that is completely dissociated in water (highly soluble).
NaOH(s) Na+ + OH-
Strong Bases:
Group 1A metal hydroxides(LiOH, NaOH, KOH,RbOH, CsOH)
Heavy Group 2A metal hydroxides[Ca(OH)2, Sr(OH)2, andBa(OH)2]
2626
Weak Acids & Bases: “The Rest”
2727
Strong Acids:100% ionized (completely dissociated) in water.
HCl + H2O H3O+ + Cl-
Note the “one way arrow”.
Weak Acids:Only a small % (dissociated) in water.
HC2H3O2 + H2O H3O+ + C2H3O2
-
Note the “2-way” arrow.
Why are they different?
2828
Strong Acids:
HCl HCl HClHCl HCl
ADD WATER to MOLECULAR ACID
(H2O)
2929
Strong Acids:
(H2O)H3O+
H3O+
H3O+
H3O+
H3O+
Cl-
Cl-
Cl-
Cl-
Cl-
Note: No HCl molecules remain in solution, all have been ionized in water.
3030
HC2H3O2
HC2H3O2
HC2H3O2
HC
2H3O
2
HC2H3O2
H30
+ C2H3O2-
(H2O)
Weak Acid Ionization:
Note: At any given time only a small portion of the acid molecules are ionized and since reactions are running in BOTH directions the mixture composition stays the same.
This gives rise to an Equilbrium expression, Ka
H30+ C2H3O2
-HC2H3O2
3131
HNO3, HCl, H2SO4 and HClO4 are among the only known strong acids.
Strong and Weak Strong and Weak Acids/BasesAcids/Bases
The strength of an acid (or base) is determined by the amount of IONIZATION.
The strength of an acid (or base) is determined by the amount of IONIZATION.
3232
Strong and Weak Strong and Weak Acids/BasesAcids/Bases
• Generally divide acids and bases into STRONG or Generally divide acids and bases into STRONG or WEAK ones.WEAK ones.
STRONG ACID:STRONG ACID: HNOHNO3 3 (aq) + H(aq) + H22O (l) --->O (l) --->
HH33OO+ + (aq) + NO(aq) + NO33- - (aq)(aq)
HNOHNO33 is about 100% dissociated in water. is about 100% dissociated in water.
3333
• Weak acidsWeak acids are much less than 100% ionized in are much less than 100% ionized in
water.water.
One of the best known is acetic acid = CHOne of the best known is acetic acid = CH33COCO22HH
Strong and Weak Strong and Weak Acids/BasesAcids/Bases
Strong and Weak Strong and Weak Acids/BasesAcids/Bases
3434
• Strong Base:Strong Base: 100% dissociated in 100% dissociated in water.water.
NaOH (aq) ---> NaNaOH (aq) ---> Na+ + (aq) + OH(aq) + OH- - (aq)(aq)
Strong and Weak Strong and Weak Acids/BasesAcids/Bases
Other common strong Other common strong bases include KOH andbases include KOH and Ca(OH)Ca(OH)22..
CaO (lime) + HCaO (lime) + H22O -->O -->
Ca(OH)Ca(OH)22 (slaked lime) (slaked lime)CaOCaO
3535
More About WaterMore About Water
HH22O can function as both an ACID and a BASE.O can function as both an ACID and a BASE.
In pure water there can beIn pure water there can be AUTOIONIZATIONAUTOIONIZATION
Equilibrium constant for water = KEquilibrium constant for water = Kww
KKww = [H = [H33OO++] [OH] [OH--] =] = 1.00 x 101.00 x 10-14-14 at 25 at 25 ooCC
3636
More About WaterMore About Water
KKww = [H = [H33OO++] [OH] [OH--] = 1.00 x 10] = 1.00 x 10-14-14 at 25 at 25 ooCC
In a In a neutral neutral solution [Hsolution [H33OO++] = [OH] = [OH--]]
so Kso Kww = [H = [H33OO++]]22 = [OH = [OH--]]22
and so [Hand so [H33OO++] = [OH] = [OH--] = 1.00 x 10] = 1.00 x 10-7-7 M M
OH-
H3O+
OH-
H3O+
AutoionizationAutoionization
3737
Write equations for the following
• 1. sulfuric acid in water
• 2. ammonia in water
• 3. calcium hydroxide in water
• 4. hydroflouric acid in water
• 5. hydrogen sulfide in water
3838Calculating the pH
pH = - log [H+](Remember that the [ ] mean Molarity)
Example: If [H+] = 1 X 10-10
pH = - log 1 X 10-10
pH = - (- 10)
pH = 10
Example: If [H+] = 1.8 X 10-5
pH = - log 1.8 X 10-5
pH = - (- 4.74)
pH = 4.74
3939
Try These!Try These!
Find the pH of Find the pH of these:these:
1) A 0.15 M solution 1) A 0.15 M solution of Hydrochloric of Hydrochloric acidacid
2) A 3.00 X 102) A 3.00 X 10-7-7 M M solution of Nitric solution of Nitric acidacid
4040pH calculations – Solving for pH calculations – Solving for H+H+pH calculations – Solving for pH calculations – Solving for H+H+
If the pH of Coke is 3.12, [HIf the pH of Coke is 3.12, [H++] = ???] = ???
Because pH = - log [HBecause pH = - log [H++] then] then
- pH = log [H- pH = log [H++]]
Take antilog (10Take antilog (10xx) of both) of both sides and get sides and get
1010-pH -pH == [H[H++]][H[H++] = 10] = 10-3.12-3.12 = 7.6 x 10 = 7.6 x 10-4-4 M M *** to find antilog on your calculator, look for “Shift” or “2*** to find antilog on your calculator, look for “Shift” or “2nd nd
function” and then the log buttonfunction” and then the log button
4141pH calculations – Solving for pH calculations – Solving for H+H+
• A solution has a pH of 8.5. What is the A solution has a pH of 8.5. What is the Molarity of hydrogen ions in the Molarity of hydrogen ions in the solution?solution?
4242pOH
• Since acids and bases are Since acids and bases are opposites, pH and pOH are opposites, pH and pOH are opposites!opposites!
• pOH does not really exist, but it is pOH does not really exist, but it is useful for changing bases to pH.useful for changing bases to pH.
• pOH looks at the perspective of a pOH looks at the perspective of a basebase
pOH = - log [OHpOH = - log [OH--]]Since pH and pOH are on opposite Since pH and pOH are on opposite
ends,ends,pH + pOH = 14pH + pOH = 14
4343
pHpH [H+][H+] [OH-][OH-] pOHpOH
4444
[H[H33OO++], [OH], [OH--] and pH] and pHWhat is the pH of the What is the pH of the
0.0010 M NaOH solution? 0.0010 M NaOH solution?
[OH-] = 0.0010 (or 1.0 X 10[OH-] = 0.0010 (or 1.0 X 10-3-3 M) M)
pOH = - log 0.0010pOH = - log 0.0010
pOH = 3pOH = 3
pH = 14 – 3 = 11pH = 14 – 3 = 11
OR KOR Kww = [H = [H33OO++] [OH] [OH--]]
[H[H3OO++] = 1.0 x 10] = 1.0 x 10-11-11 M M
pH = - log (1.0 x 10pH = - log (1.0 x 10-11-11) = 11.00) = 11.00
4545The pH of rainwater collected in a certain region of the northeastern United States on a particular day was 4.82. What is the H+ ion concentration of the rainwater?
The OH- ion concentration of a blood sample is 2.5 x 10-7 M. What is the pH of the blood?
4646
[OH[OH--]]
[H[H++]] pOHpOH
pHpH
1010 -pOH
-pOH
1010 -pH-pH-Log[H
-Log[H++]]
-Log[OH
Log[OH
--]]
14 -
pOH
14 -
pOH
14 -
pH
14 -
pH
1.0
x 10
1.0
x 10-1
4-14
[OH[O
H-- ]]
1.0
x 10
1.0
x 10-1
4-14
[H[H
++ ]]
4747Calculating [H3O+], pH, [OH-], and pOH
A chemist dilutes concentrated hydrochloric acid to make two solutions: (a) 3.0 M and (b) 0.0024 M. Calculate the [H3O+], pH, [OH-], and pOH of the two solutions at 25°C.
4848Calculating [H3O+], pH, [OH-], and pOH
What is the [H3O+], [OH-], and pOH of a solution with pH = 3.67? Is this an acid, base, or neutral?
4949
• Weak base:Weak base: less than 100% ionized less than 100% ionized in waterin water
One of the best known weak bases is One of the best known weak bases is ammoniaammonia
NHNH3 3 (aq) + H(aq) + H22O (l) O (l) NH NH44+ + (aq) + OH(aq) + OH- - (aq)(aq)
Strong and Weak Strong and Weak Acids/BasesAcids/Bases
Strong and Weak Strong and Weak Acids/BasesAcids/Bases
5050
Weak BasesWeak Bases
5151
Equilibria Involving Equilibria Involving Weak Acids and BasesWeak Acids and Bases
Consider acetic acid, HCConsider acetic acid, HC22HH33OO22 (HOAc) (HOAc)
HCHC22HH33OO22 + H + H22O O H H33OO++ + C + C22HH33OO22 --
AcidAcid Conj. base Conj. 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 ACID) for ACID)
K gives the ratio of ions (split up) to molecules K gives the ratio of ions (split up) to molecules
(don’t split up) (don’t split up)
5252Ionization Constants for Ionization Constants for Acids/Bases Acids/Bases
AcidsAcids ConjugateConjugateBasesBases
Increase strength
Increase strength
5353
Equilibrium Constants Equilibrium Constants for Weak Acidsfor Weak Acids
Equilibrium Constants Equilibrium Constants for Weak Acidsfor Weak Acids
(Yes, all weak acids do this – DO NOT endeavor to (Yes, all weak acids do this – DO NOT endeavor to make this complicated!)make this complicated!)
Weak acid has KWeak acid has Kaa < 1 < 1
Leads to small [HLeads to small [H33OO++] and a pH of 2 - 7] and a pH of 2 - 7
Ka = XKa = X22 / or-x x = H / or-x x = H++ and conjugate base and conjugate base
5454
Calculate the H+ and pH for 0.250 M
HC2H3O3
5555Equilibria Involving A Weak Equilibria Involving A Weak AcidAcid
Equilibria Involving A Weak Equilibria Involving A Weak AcidAcid
You have 1.00 M HOAc. Calc. the You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, Hequilibrium concs. of HOAc, H33OO++, OAc, OAc--, ,
and the pH.and the pH.
5656Equilibria Involving A Weak Equilibria Involving A Weak AcidAcid
Equilibria Involving A Weak Equilibria Involving A Weak AcidAcid
You have 1.00 M HOAc. Calc. the You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, Hequilibrium concs. of HOAc, H33OO++, OAc, OAc--, ,
and the pH.and the pH.
Step 1.Step 1. Define equilibrium concs. in ICE Define equilibrium concs. in ICE
table.table.
[HOAc][HOAc] [H[H33OO++]] [OAc[OAc--]]
initialinitial
changechange
equilibequilib
1.001.00 00 001.001.00 00 00
-x-x +x+x +x+x-x-x +x+x +x+x
1.00-x1.00-x xx xx1.00-x1.00-x xx xx
5757Reaction of Weak Bases with Reaction of Weak Bases with WaterWater
The generic reaction for a base reacting with The generic reaction for a base reacting with water, producing its conjugate acid and water, producing its conjugate acid and hydroxide ion:hydroxide ion:
B + H2O BH+ + OH-
[ ][ ]
[ ]b
BH OHK
B
(Yes, all weak bases do this – DO NOT(Yes, all weak bases do this – DO NOTendeavor to make this complicated!)endeavor to make this complicated!)Kb = XKb = X22 /or-x X = OH /or-x X = OH- - and conjuate acid and conjuate acid
5858
Equilibrium Constants Equilibrium Constants for Weak Basesfor Weak Bases
Equilibrium Constants Equilibrium Constants for Weak Basesfor Weak Bases
Weak base has KWeak base has Kbb < 1 < 1
Leads to small [OHLeads to small [OH--] and a pH of 12 - 7] and a pH of 12 - 7
5959NOTE•Ka x Kb = Kw
•pH + pOH = 14
•H3O+ x OH- = 1 x 10-14
6060NOTE• On the AP test you might get the Ka
but need the Kb. If you are given an acid, you need Ka. If given a base, you need Kb. You must know if you are dealing with an acid or base.
6161
Relation Relation
of Kof Kaa, K, Kbb, ,
[H[H33OO++] ]
and pHand pH
6262Equilibria Involving A Weak Equilibria Involving A Weak AcidAcid
Equilibria Involving A Weak Equilibria Involving A Weak AcidAcid
Step 2.Step 2. Write KWrite Kaa expression expression
You have 1.00 M HOAc. Calc. the equilibrium concs. You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, Hof HOAc, H33OO++, OAc, OAc--, and the pH., and the pH.
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
This is a quadratic. Solve using quadratic This is a quadratic. Solve using quadratic formula.formula.
or you can make an approximation if x is very or you can make an approximation if x is very small! (Rule of thumb: 10small! (Rule of thumb: 10-5-5 or smaller is ok) or smaller is ok)or you can make an approximation if x is very or you can make an approximation if x is very small! (Rule of thumb: 10small! (Rule of thumb: 10-5-5 or smaller is ok) or smaller is ok)
6363Equilibria Involving A Weak Equilibria Involving A Weak AcidAcid
Equilibria Involving A Weak Equilibria Involving A Weak AcidAcid
Step 3.Step 3. Solve KSolve Kaa expression expression
You have 1.00 M HOAc. Calc. the equilibrium concs. You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, Hof HOAc, H33OO++, OAc, OAc--, and the pH., and the pH.
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
First assume x is very small because First 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.
6464
Equilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak Acid
Step 3.Step 3. Solve KSolve Kaa approximateapproximate expressionexpression
You have 1.00 M HOAc. Calc. the equilibrium concs. You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, Hof HOAc, H33OO++, OAc, OAc--, and the pH., and the pH.
Ka 1.8 x 10-5 = x2
1.00Ka 1.8 x 10-5 =
x2
1.00
x =x = [[HH33OO++] = [] = [OAcOAc--] = 4.2 x 10] = 4.2 x 10-3-3 M M
pH = - log [pH = - log [HH33OO++] = -log (4.2 x 10] = -log (4.2 x 10-3-3) =) = 2.372.37
6565Equilibria Involving A Weak Equilibria Involving A Weak AcidAcid
Equilibria Involving A Weak Equilibria Involving A Weak AcidAcid
Calculate the pH of a 0.0010 M solution of Calculate the pH of a 0.0010 M solution of formic acid, HCOformic acid, HCO22H.H.
6666Equilibria Involving A Weak Equilibria Involving A Weak AcidAcid
Equilibria Involving A Weak Equilibria Involving A Weak AcidAcid
Calculate the pH of a 0.0010 M solution of Calculate the pH of a 0.0010 M solution of formic acid, HCOformic acid, HCO22H.H.
HCOHCO22H + HH + H22O O HCO HCO22-- + H + H33OO++
KKaa = 1.8 x 10 = 1.8 x 10-4-4
Approximate solutionApproximate solution
[H[H33OO++] = 4.2 x 10] = 4.2 x 10-4-4 M, M, pH = 3.37pH = 3.37
Exact SolutionExact Solution
[H[H33OO++] = [HCO] = [HCO22--] = 3.4 x 10] = 3.4 x 10-4-4 M M
[HCO[HCO22H] = 0.0010 - 3.4 x 10H] = 0.0010 - 3.4 x 10-4-4 = 0.0007 M = 0.0007 M
pH = 3.47 pH = 3.47
6767Equilibria Involving A Weak Equilibria Involving A Weak BaseBase
You have 0.010 M NHYou have 0.010 M NH33. Calc. the pH.. Calc. the pH.
NHNH33 + H + H22O O NH NH44++ + OH + OH--
6868Equilibria Involving A Weak Equilibria Involving A Weak BaseBase
You have 0.010 M NHYou have 0.010 M NH33. Calc. the pH.. Calc. 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 concs. in ICE tableDefine equilibrium concs. in ICE table
[NH[NH33]] [NH[NH44++]] [OH[OH--]]
initialinitial
changechange
equilibequilib
0.0100.010 00 000.0100.010 00 00
-x-x +x+x +x+x-x-x +x+x +x+x
0.010 - x0.010 - x x x xx0.010 - x0.010 - x x x xx
6969Equilibria Involving A Weak Equilibria Involving A Weak BaseBase
You have 0.010 M NHYou have 0.010 M NH33. Calc. the pH.. Calc. 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 concs. in ICE tableDefine equilibrium concs. in ICE table
[NH[NH33]] [NH[NH44++]] [OH[OH--]]
initialinitial
changechange
equilibequilib
0.0100.010 00 000.0100.010 00 00
-x-x +x+x +x+x-x-x +x+x +x+x
0.010 - x0.010 - x x x xx0.010 - x0.010 - x x x xx
7070Equilibria Involving A Weak Equilibria Involving A Weak BaseBase
You have 0.010 M NHYou have 0.010 M NH33. Calc. the pH.. Calc. 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 concs. in ICE tableDefine equilibrium concs. in ICE table
[NH[NH33]] [NH[NH44++]] [OH[OH--]]
initialinitial
changechange
equilibequilib
0.0100.010 00 000.0100.010 00 00
-x-x +x+x +x+x-x-x +x+x +x+x
0.010 - x0.010 - x x x xx0.010 - x0.010 - x x x xx
7171Reaction of Weak Bases with Reaction of Weak Bases with WaterWater
The base reacts with water, producing its conjugate acid The base reacts with water, producing its conjugate acid and hydroxide ion:and hydroxide ion:
NHNH33 + H + H22O O NH NH44++ + OH + OH- - KKbb = 1.8 x 10 = 1.8 x 10-5-5
Kb = [NH4+][OH-
] [NH3]
7272Equilibria Involving A Weak Equilibria Involving A Weak BaseBase
You have 0.010 M NHYou have 0.010 M NH33. Calc. the pH.. Calc. the pH.
NHNH33 + H + H22O O NH NH44++ + OH + OH--
KKbb = 1.8 x 10 = 1.8 x 10-5-5
Step 2.Step 2. Solve the equilibrium expressionSolve the equilibrium expression
Assume x is small, soAssume x is small, so x = [OHx = [OH--] = [NH] = [NH44
++] = 4.2 x 10] = 4.2 x 10-4-4 M M
and [NHand [NH33] = 0.010 - 4.2 x 10] = 0.010 - 4.2 x 10-4-4 ≈ 0.010 M ≈ 0.010 M
The approximation is validThe approximation is valid !!
7373Equilibria Involving A Weak Equilibria Involving A Weak BaseBase
You have 0.010 M NHYou have 0.010 M NH33. Calc. the pH.. Calc. the pH.
NHNH33 + H + H22O O NH NH44++ + OH + OH--
KKbb = 1.8 x 10 = 1.8 x 10-5-5
Step 3.Step 3. Calculate pHCalculate pH
[OH[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
7474
Types of Acid/Base Reactions: Types of Acid/Base Reactions: SummarySummary
7575pH testing
• There are several ways to test pHThere are several ways to test pH
–Blue litmus paper (red = acid)Blue litmus paper (red = acid)
–Red litmus paper (blue = basic)Red litmus paper (blue = basic)
–pH paper (multi-colored)pH paper (multi-colored)
–pH meter (7 is neutral, <7 acid, >7 pH meter (7 is neutral, <7 acid, >7 base)base)
–Universal indicator (multi-colored)Universal indicator (multi-colored)
– Indicators like phenolphthaleinIndicators like phenolphthalein
–Natural indicators like red cabbage, Natural indicators like red cabbage, radishesradishes
7676Paper testing
• Paper tests like litmus paper and pH Paper tests like litmus paper and pH paperpaper
– Put a stirring rod into the solution Put a stirring rod into the solution and stir.and stir.
– Take the stirring rod out, and Take the stirring rod out, and place a drop of the solution from place a drop of the solution from the end of the stirring rod onto a the end of the stirring rod onto a piece of the paperpiece of the paper
– Read and record the color change. Read and record the color change. Note what the color indicates. Note what the color indicates.
– You should only use a small You should only use a small portion of the paper. You can use portion of the paper. You can use one piece of paper for several one piece of paper for several tests.tests.
7777pH paperpH paper
7878Behavior of Salts (Acidic or Basic)
7979Behavior of Salts Acidic, basic, or neutral?NaFNaOClKBrNH4FLi2SLi2CO3
Fe(NO3 )3
8080
NOTE•Ka > Kb acidic
•Kb < Ka basic
•Ka = Kb neutral
8181Behavior of Salts WRITE EQUATIONS to show WHY THE FOLLOWING ARE ACIDIC, BASIC, OR
NEUTRAL.NaFNaOClKBrNH4NO2
NH4FLi2SZnCl2Li2CO3
8282
Amphoteric Salts
8383
% DISSOCIATION OF ACIDS & Bases
• % Diss of an acid = H+ / original conc x 100
• % Diss of a base = OH- / original conc x 100
8484Determine the pH, pOH, hydrogen ion concentration, and hydroxideion concentration for 0.125 M Na2 CO3.
8585Determine the pH for 0.015 M sodium fluoride.
8686Determine the percent ionization for 0.250 M KNO2.
8787
What about polyprotic acids?
8888What about polyprotic acids? For example sulfuric acid:
11. Determine the pH for 1.2 M H2 SO4
8989What about polyprotic acids? For example sulfuric acid:
11. Determine the pH for 0.012 M H2 SO4
9090
pH meter
• Tests the voltage of the Tests the voltage of the electrolyteelectrolyte
• Converts the voltage to Converts the voltage to pHpH
• Very cheap, accurateVery cheap, accurate
• Must be calibrated with Must be calibrated with a buffer solutiona buffer solution
9191pH indicators
• Indicators are dyes that can be added that will change color in the presence of an acid or base.
• Some indicators only work in a specific range of pH
• Once the drops are added, the sample is ruined
• Some dyes are natural, like radish skin or red cabbage
9292
ACID-BASE REACTIONSACID-BASE REACTIONSTitrationsTitrations
ACID-BASE REACTIONSACID-BASE REACTIONSTitrationsTitrations
HH22CC22OO44(aq) + 2 NaOH(aq) --->(aq) + 2 NaOH(aq) --->
acidacid basebase
NaNa22CC22OO44(aq) + 2 H(aq) + 2 H22O(liq)O(liq)
Carry out this reaction using aCarry out this reaction using a TITRATIONTITRATION..
Oxalic acid,Oxalic acid,
HH22CC22OO44
9393Setup for titrating an acid with a baseSetup for titrating an acid with a base
9494
TitrationTitrationTitrationTitration
1. Add solution from the buret.1. Add solution from the buret.2. Reagent (base) reacts with 2. Reagent (base) reacts with
compound (acid) in solution compound (acid) in solution in the flask.in the flask.
3.3. Indicator shows when exact Indicator shows when exact stoichiometric reaction has stoichiometric reaction has occurred. (Acid = Base)occurred. (Acid = Base)
This is called This is called NEUTRALIZATION.NEUTRALIZATION.
9595
35.62 mL of NaOH is 35.62 mL of NaOH is
neutralized with 25.2 mL of neutralized with 25.2 mL of
0.0998 M HCl by titration to 0.0998 M HCl by titration to
an equivalence point. What an equivalence point. What
is the concentration of the is the concentration of the
NaOH?NaOH?
LAB PROBLEM #1: Standardize a LAB PROBLEM #1: Standardize a solution of NaOH — i.e., accurately solution of NaOH — i.e., accurately determine its concentration.determine its concentration.
LAB PROBLEM #1: Standardize a LAB PROBLEM #1: Standardize a solution of NaOH — i.e., accurately solution of NaOH — i.e., accurately determine its concentration.determine its concentration.
9696
PROBLEM: You have 50.0 mL of PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M 3.0 M NaOH and you want 0.50 M NaOH. What do you do?NaOH. What do you do?
PROBLEM: You have 50.0 mL of PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M 3.0 M NaOH and you want 0.50 M NaOH. What do you do?NaOH. What do you do?
Add water to the 3.0 M solution to lower Add water to the 3.0 M solution to lower its concentration to 0.50 M its concentration to 0.50 M
Dilute the solution!Dilute the solution!
9797
PROBLEM: You have 50.0 mL of 3.0 M PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. NaOH and you want 0.50 M NaOH. What do you do?What do you do?
PROBLEM: You have 50.0 mL of 3.0 M PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. NaOH and you want 0.50 M NaOH. What do you do?What do you do?
3.0 M NaOH 0.50 M NaOH
H2O
Concentrated Dilute
But how much water But how much water do we add?do we add?
9898
PROBLEM: You have 50.0 mL of 3.0 M PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. NaOH and you want 0.50 M NaOH. What do you doWhat do you do??
PROBLEM: You have 50.0 mL of 3.0 M PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. NaOH and you want 0.50 M NaOH. What do you doWhat do you do??
How much water is added?How much water is added?
The important point is that --->The important point is that --->
moles of NaOH in ORIGINAL solution = moles of NaOH in ORIGINAL solution = moles of NaOH in FINAL solutionmoles of NaOH in FINAL solution
9999
PROBLEM: You have 50.0 mL of 3.0 M NaOH PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do?and you want 0.50 M NaOH. What do you do?PROBLEM: You have 50.0 mL of 3.0 M NaOH PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do?and you want 0.50 M NaOH. What do you do?
Amount of NaOH in original solution = Amount of NaOH in original solution =
M • VM • V = =
(3.0 mol/L)(0.050 L) = 0.15 mol NaOH(3.0 mol/L)(0.050 L) = 0.15 mol NaOH
Amount of NaOH in final solution must also = Amount of NaOH in final solution must also = 0.15 mol NaOH0.15 mol NaOH
Volume of final solution =Volume of final solution =
(0.15 mol NaOH) / (0.50 M) = 0.30 L(0.15 mol NaOH) / (0.50 M) = 0.30 L
or or 300 mL300 mL
100100
PROBLEM: You have 50.0 mL of 3.0 M PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. NaOH and you want 0.50 M NaOH. What do you do?What do you do?
PROBLEM: You have 50.0 mL of 3.0 M PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. NaOH and you want 0.50 M NaOH. What do you do?What do you do?
Conclusion:Conclusion:
add 250 mL add 250 mL of waterof water to to 50.0 mL of 50.0 mL of 3.0 M NaOH 3.0 M NaOH to make 300 to make 300 mL of 0.50 M mL of 0.50 M NaOH.NaOH.
3.0 M NaOH 0.50 M NaOH
H2O
Concentrated Dilute
101101
A shortcutA shortcut
MM11 • V • V11 = M = M22 • V • V22
Preparing Solutions Preparing Solutions by Dilutionby Dilution
Preparing Solutions Preparing Solutions by Dilutionby Dilution
102102You try this dilution problem
• You have a stock bottle of hydrochloric acid, which is 12.1 M. You need 400 mL of 0.10 M HCl. How much of the acid and how much water will you need?