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Some Properties of Acids
• Produce 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
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 “Basic Blue”Turns red litmus paper to blue “Basic Blue”
Arrhenius acid is a substance that produces (H3O+) in water
Arrhenius base is a substance that produces OH- in water
Acid/Base definitionsDefinition 1: Arrhenius
Acid/Base Definitions
Definition #2: Brønsted – LowryDefinition #2: Brønsted – Lowry
Acids – proton donorAcids – proton donor
Bases – proton acceptorBases – proton acceptor
A “proton” is really just a hydrogen A “proton” is really just a hydrogen atom that has lost it’s electron!atom that has lost it’s electron!
A Brønsted-Lowry acid is a proton donorA Brønsted-Lowry base is a proton acceptor
acidconjugate
basebase conjugate
acid
ACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIES
The Brønsted definition means NHThe Brønsted definition means NH33 is a BASE is a BASE
in water — and water is itself an ACIDin water — and water is itself an ACID
BaseAcidAcidBaseNH4
+ + OH-NH3 + H2OBaseAcidAcidBase
NH4+ + OH-NH3 + H2O
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
++
AcidAcidAcidAcid
AcidAcidAcidAcid
BaseBaseBaseBase
BaseBaseBaseBase
Conj.Conj.BaseBaseConj.Conj.BaseBase
Conj.Conj.BaseBaseConj.Conj.BaseBase
Conj.Conj.AcidAcidConj.Conj.AcidAcid
Conj.Conj.AcidAcidConj.Conj.AcidAcid
Acids & Base DefinitionsAcids & Base Definitions
Lewis acid - a substance Lewis acid - a substance that accepts an electron that accepts an electron pairpair
Lewis base - a substance Lewis base - a substance that donates an electron that donates an electron pairpair
Definition #3 – Lewis Definition #3 – Lewis
Lewis Acids & BasesLewis Acids & BasesFormation of hydronium ion is also an Formation of hydronium ion is also an
excellent example.excellent example.
•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
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 neutral solution [HIn a neutral solution [H33OO++] = [OH] = [OH--]]
and so [Hand so [H33OO++] = [OH] = [OH--] = 1.00 x 10] = 1.00 x 10-7-7 M M
OH-
H3O+
OH-
H3O+
AutoionizationAutoionization
More About WaterMore About WaterHH22O can function as both an ACID and a BASE.O can function as both an ACID and a BASE.
In pure water there can be AUTOIONIZATIONIn pure water there can be AUTOIONIZATION
Equilibrium constant for water = KEquilibrium constant for water = Kww
KKww = 1.00 x 10 = 1.00 x 10-14-14 = [H = [H33OO++] [OH] [OH--] at 25 ] at 25 ooCC
Take -logs of both sidesTake -logs of both sides
14 = pH + pOH14 = pH + pOH
The pH scale is a way of The pH scale is a way of expressing the strength of expressing the strength of acids and bases. Instead of acids and bases. Instead of using very small numbers, we using very small numbers, we just use the NEGATIVE power just use the NEGATIVE power of 10 on the Molarity of the of 10 on the Molarity of the HH33OO++ (or OH (or OH--) ion.) ion.
Under 7 = acidUnder 7 = acid7 = neutral7 = neutral
Over 7 = base Over 7 = base
Calculating the pHpH = - log [H3O+]
((Remember that the [ ] mean Molarity or concentration)
Example: If [H3O+] = 1 X 10-10
pH = - log 1 X 10-10
pH = - (- 10)
pH = 10
Example: If [H3O+] = 1.8 X 10-5, what is the pH?
pH = - log 1.8 X 10-5
pH = - (- 4.74)
pH = 4.74
pH calculations – Solving for HpH calculations – Solving for H33OO++pH calculations – Solving for HpH calculations – Solving for H33OO++
If the pH of Coke is 3.12, what is the [If the pH of Coke is 3.12, what is the [H3O+] ?] ?
Because pH = Because pH = -- log [ log [H3O+] then] then
-- pH = log [ pH = log [H3O+]]
Take antilog (10Take antilog (10xx) of both) of both sides and get sides and get
1010--pH pH == [[H3O+]]
[[H3O+] = 10] = 10--3.123.12 = 7.58 x 10 = 7.58 x 10-4-4 M M
*** to find antilog on your calculator, look for “Shift” *** to find antilog on your calculator, look for “Shift” or “2or “2nd nd function” and then the log buttonfunction” and then the log button
Strong and Weak Acids/BasesStrong and Weak Acids/Bases
• Generally acids and bases are divided into Generally acids and bases are divided into STRONG or WEAK ones.STRONG or WEAK ones.
• Strong acids – 6 (HNOHNO33, HCl, HBr, HI, HClO, HCl, HBr, HI, HClO44, , HH22SOSO44) all others are weak) all others are weak
• Strong Bases – group 1 and 2 hydroxides (all others are weak) except Be(OH)2
Strong acids and basesStrong acids and bases
Find the pH of these:Find the pH of these:
1)1) A 0.15 M solution of A 0.15 M solution of Hydrochloric acidHydrochloric acid
2) 2) A 3.00 X 10A 3.00 X 10-7-7 M M solution of Nitric acidsolution of Nitric acid
3) What is the pH of 3) What is the pH of 0.0034M H0.0034M H22SOSO44??
pH = - log [pH = - log [H3O+]]
pH = - log 0.15pH = - log 0.15
pH = - (- 0.82)pH = - (- 0.82)
pH = 0.82pH = 0.82
pH = - log 3 X 10pH = - log 3 X 10-7-7
pH = - (- 6.52)pH = - (- 6.52)
pH = 6.52pH = 6.52
• Strong acids and bases - are 100 % ionized.Strong acids and bases - are 100 % ionized.• No equilibrium is set upNo equilibrium is set up• [Acid] = [H30+] ( 1:1 ratio)[Acid] = [H30+] ( 1:1 ratio)• ACID ACID H A + HH A + H22O O H3O+ + A-• BASE B + HH22O O BH BH++ + OH + OH--
Finding pH Finding pH
pH of strong basespH of strong bases
What 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
What is the pH of a 2 x 10-3 M HNO3 solution?
HNO3 is a strong acid – 100% dissociation.
HNO3 (aq) + H2O (l) H3O+ (aq) + NO3- (aq)
pH = -log [H+] = -log [H3O+] = -log(0.002) = 2.7
Start
End
0.002 M
0.002 M 0.002 M0.0 M
0.0 M 0.0 M
What is the pH of a 1.8 x 10-2 M Ba(OH)2 solution?
Ba(OH)2 is a strong base – 100% dissociation.
Ba(OH)2 (s) Ba2+ (aq) + 2OH- (aq)
Start
End
0.018 M
0.018 M 0.036 M0.0 M
0.0 M 0.0 M
pH = 14.00 – pOH = 14.00 - (-log(0.036) )= 12.5615.4
Weak acidsWeak acids are much less than 100% ionized in water.are much less than 100% ionized in water. Equilibrium is set up. Common weak acids are acetic acid Equilibrium is set up. Common weak acids are acetic acid
and weak base is ammoniaand weak base is ammonia
Weak Acids/BasesWeak Acids/Bases
Equilibria Involving Equilibria Involving Weak Acids and BasesWeak Acids and Bases
Consider acetic acid, HCConsider acetic acid, HC22HH33OO22 (HOAc) (HOAc)
HCHC22HH33OO2(aq)2(aq) + H + H22OO(l)(l) H H33OO++ (aq)(aq) + C + C22HH33OO22 ––(aq)(aq)
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 KK is designated Kaa for ACID) for ACID)
K gives the ratio of ions (split up) to molecules (don’t split up)K gives the ratio of ions (split up) to molecules (don’t split up)
Ionization Constants for Acids/BasesIonization Constants for Acids/Bases
AcidsAcids ConjugateConjugateBasesBases
Increase strength
Increase strength
Equilibrium Constants Equilibrium Constants for Weak Acidsfor Weak Acids
Equilibrium Constants Equilibrium Constants for Weak Acidsfor Weak Acids
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
Equilibrium Constants Equilibrium Constants for Weak Basesfor Weak Bases
Equilibrium Constants Equilibrium Constants for Weak Basesfor Weak Bases
Kb – base dissociation constant and Kb – base dissociation constant and
is temp dependentis temp dependent
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
Equilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak AcidYou 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. Define equilibrium concs. in ICE Step 1. 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
Equilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak Acid
Step 2. Write KStep 2. Write 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)
Equilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak Acid
Step 3. Solve KStep 3. Solve 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.
Equilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak Acid
Step 3. Solve KStep 3. Solve Kaa approximate approximate 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 = 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) = 2.37) = 2.37
Equilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak AcidCalculate the pH of a 0.0010 M solution of formic Calculate the pH of a 0.0010 M solution of formic
acid, HCOacid, 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, pH = 3.37 M, pH = 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
Equilibria Involving A Weak BaseEquilibria Involving A Weak BaseYou 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. Define equilibrium concs. in ICE tableStep 1. Define 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
Equilibria Involving A Weak BaseEquilibria Involving A Weak BaseYou 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. Solve the equilibrium expressionStep 2. Solve the equilibrium expression
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, 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 !!
Equilibria 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. Calculate pHStep 3. Calculate pH[OH[OH--] = 4.2 x 10] = 4.2 x 10-4-4 M Mso pOH = - log [OHso pOH = - log [OH--] = 3.37] = 3.37Because pH + pOH = 14,Because pH + pOH = 14,
pH = 10.63pH = 10.63
Percent Ionization for Weak Acids Most weak acids ionize < 50% Percent ionization (p)
General Weak Acid: HA(aq) + H2O(l) H3O+ (aq) + A- (aq)
• p varies depending on concentration: increase [HA ] decreases p
• This is caused by Le Chatelier’s Principle
• Remember, for strong acids we assume complete ionization (100%)
100xacidofionconcentrat
onHydroniumiofionconcentrat=p
100)(3 x][HA
]O[H=p
(aq)
aq+
ExamplesThe pH of a 0.10mol/L methanoic acid (HCOOH)
solution is 2.38. Calculate the percent ionization of methanoic acid Ans: 4.2%
Calculate the acid ionization constant (Ka )of acetic acid if a 0.1000mol/L solution at equilibrium at SATP has a percent ionization of 1.3% (Hint: ICE table Ans: 1.7x10-5
Relationship Between Ka and Kb for Conjugate Base Pairs
• Recall: Conjugate Pairs – an acid and base that differ by one hydrogen
• Lets consider the hypothetical weak acid, HA, and its conjugate base, A -
HA (aq) + H2O (l) H3O+ (aq) + A- (aq)
HA
]
aq
3
][
(aq)(aq)][AO[H=K
)(
+
a
• Now consider the hypothetical weak base, A- in water
A- (aq) + H2O (l) HA (aq) + OH- (aq)
• Now let’s put that together
HA (aq) + H2O (l) H3O+ (aq) + A- (aq) Ka
A- (aq) + H2O (l) HA (aq) + OH- (aq) Kb
Add both equations
2H2O (l) H3O+ (aq) + OH- (aq) Kw
Relationship Between Ka and Kb for Conjugate Base Pairs
)(
]
]aq[A
(aq)(aq)][HA[OH=Kb
Relationship Between Ka and Kb for Conjugate Base Pairs
HA aq
aqaq3
][
]][AO[H=K
)(
)()(+
a
OHHA
aq
aqaq
][A
]][[=K
)(
)()(
b
ba KxK OHHA
HA aq
aqaq
aq
aqaq3
][A
]][[x
][
]][AO[H=
)(
)()(
)(
)()(+
OH aqaq3 ]][O[H= )()(+ Kw
Recall: Autoionization of water H2O(l) + H2O(l) H3O+(aq) + OH-(aq)
Kw=1.00x10-14 **must remember this value**
Relationship Between Ionization Constants for Conjugate Base Pairs
For acids and bases whose chemical formulas differ by only one hydrogen (conjugate pairs) the following apply:
Kw = Ka x Kb Kb =Kw/Ka Ka = Kw/Kb
• Therefore if only the Ka value is available in the table, we can determine the conjugate pairs Kb by using the
above equations Note: these equations show the larger the Ka the
smaller the KbStronger acid weaker conjugate baseWeaker acid stronger conjugate base
Learning check!
1. What is the value of the base ionization constant (Kb) for the acetate
ion, C2H3O2- (aq)
Ans: 5.6x10-10
2. Calculate the percent ionization of propanoic acid, HC3H5O2(aq), if a 0.050
mol/L solution has a pH of 2.78Ans. 3.3%
Rapid changes in pH can kill fish and other
organisms in lakes and streams.
Soil pH is affected and can kill plants and create
sinkholes