Acids and Bases Lesson 1 Acid & Base Properties (Strong & Weak acids)
The Chemistry of Acids and Bases Chapter 17. 2 Strong and Weak Acids/Bases Acids and bases into...
-
Upload
jameson-thorndike -
Category
Documents
-
view
232 -
download
7
Transcript of The Chemistry of Acids and Bases Chapter 17. 2 Strong and Weak Acids/Bases Acids and bases into...
The Chemistry of The Chemistry of Acids and BasesAcids and Bases
Chapter 17Chapter 17
22Strong and Weak Acids/BasesStrong and Weak Acids/Bases• Acids and bases into STRONG or WEAK ones.Acids and bases into STRONG or WEAK ones.
STRONG ACIDSTRONG ACID
OrOr
HNOHNO33(aq) ---> H(aq) ---> H++(aq) + NO(aq) + NO33-- (aq) (aq)
HNOHNO33 is about 100% dissociated in water. is about 100% dissociated in water.
33
H2O
H—Cl
Cl-
H3O+
H2O
H—Cl
Cl-
H3O+
HNOHNO33, HCl, H, HCl, H22SOSO44 and HClO and HClO44 are are
among the strong acids.among the strong acids.
Strong and Weak Acids/BasesStrong and Weak Acids/Bases
44
• Weak acids are Weak acids are much lessmuch less than 100% than 100% ionized in water.ionized in water.
• One of the best known is acetic acid = One of the best known is acetic acid =
HCHC22HH33OO22 = CH = CH33COCO22H = HOAc = CHH = HOAc = CH33COOHCOOH
HCHC22HH33OO22 + H + H22OO <--> C<--> C22HH33OO22-- + H + H33OO
++
oror
HCHC22HH33OO22 <--> C<--> C22HH33OO22- - + H + H++
Strong and Weak Acids/BasesStrong and Weak Acids/Bases
55
Strong BasesStrong Bases
100% dissociated in water100% dissociated in water
NaOH(aq) ---> NaNaOH(aq) ---> Na++(aq) + OH(aq) + OH--(aq)(aq)
Strong and Weak Acids/BasesStrong and Weak Acids/Bases
Other common strong bases Other common strong bases
include KOH and include KOH and Ca(OH)Ca(OH)22..
CaOCaO
CaO (lime) + HCaO (lime) + H22O --> O -->
Ca(OH)Ca(OH)22 (slaked lime) (slaked lime)
66
Weak BasesWeak Bases
Less than 100% ionized in waterLess than 100% ionized in water
One of the best known weak bases isOne of the best known weak bases is
ammoniaammonia..
Strong and Weak Acids/BasesStrong and Weak Acids/Bases
NHNH3 3 (aq)(aq) + H + H22O O ((ll)) <--> <--> NHNH44++
(aq)(aq) + OH + OH-- (aq)(aq)
77
ACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIES
• The most general theory for common The most general theory for common
aqueous acids and bases is the aqueous acids and bases is the
BRØNSTED - LOWRYBRØNSTED - LOWRY theorytheory
• ACIDS DONATE HACIDS DONATE H++ IONS IONS
• BASES ACCEPT HBASES ACCEPT H++ IONS IONS
88
ACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIES
The Brønsted definition means NHThe Brønsted definition means NH33
is a is a BASEBASE in water — and water is in water — and water is
itself an itself an ACIDACID
BaseAcidAcidBaseNH 4
+ + OH -NH 3 + H 2 O
99
ACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIES
NHNH3 3 / NH/ NH44++ is a is a conjugate pairconjugate pair — related — related
by the gain or loss of Hby the gain or loss of H++
Every acid has a conjugate base - Every acid has a conjugate base - and vice-versa.and vice-versa.
NHNH33 is a is a BASEBASE in water — and water is in water — and water is itself an itself an ACIDACID..
1010
ACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIESA strong acid is 100% dissociated.A strong acid is 100% dissociated.
Therefore, a Therefore, a STRONG ACID STRONG ACID —— a good H a good H++ donor donor ——
must have a must have a WEAK CONJUGATE BASE WEAK CONJUGATE BASE —— a a
poor Hpoor H++ acceptor. acceptor.
HNOHNO33(aq)(aq) + H + H22OO((ll)) <--> <--> HH33OO++
(aq)(aq) + NO + NO33- -
(aq)(aq)
STRONG ASTRONG A basebase acid acid weak Bweak B
Notice that every A-B reaction has two Notice that every A-B reaction has two acids and two bases!acids and two bases!
1111ACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIES
We know that HNOWe know that HNO33 is a strong acid. is a strong acid.
1. It is a stronger acid than H1. It is a stronger acid than H33OO++
2. H2. H22O is a stronger base than NOO is a stronger base than NO33--
WEAK WEAK BASEBASE
STRONG STRONG ACIDACID
1212
ACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIES
Acetic acid is only 0.42% ionized when [HCAcetic acid is only 0.42% ionized when [HC22HH33OO22] = 1.0 ] = 1.0
M. It is a M. It is a WEAK ACIDWEAK ACID
Because [HBecause [H33OO++] is small, this must mean] is small, this must mean
1.1. H H33OO++ is a stronger acid than HC is a stronger acid than HC22HH33OO22
2.2. C C22HH33OO22-- is a stronger base than H is a stronger base than H22OO
1313
ACIDS AND BASESACIDS AND BASESACIDS AND BASESACIDS AND BASES
• MonoproticMonoprotic acidsacids can donate only one proton, can donate only one proton,
while while polyproticpolyprotic acidsacids can donate two or more can donate two or more
protons.protons.
• Monoprotic basesMonoprotic bases can accept only one proton, can accept only one proton,
while while polyprotic basespolyprotic bases can accept two or more can accept two or more
protons.protons.
• AmphiproticAmphiprotic substances can behave as either substances can behave as either
acids or bases.acids or bases.
1414
Acid-Base ReactionsAcid-Base ReactionsAcid-Base ReactionsAcid-Base Reactions
• Now we can describe Now we can describe reactions of acids with reactions of acids with bases and the bases and the direction of such direction of such reaction. reaction.
• Consider the acid HF Consider the acid HF reacting with the base reacting with the base NHNH33..
• HF + NHHF + NH33 <--> NH <--> NH44++ + F + F--
• Now we can describe Now we can describe reactions of acids with reactions of acids with bases and the bases and the direction of such direction of such reaction. reaction.
• Consider the acid HF Consider the acid HF reacting with the base reacting with the base NHNH33..
• HF + NHHF + NH33 <--> NH <--> NH44++ + F + F--
1515
Predicting the Direction of Predicting the Direction of Acid-Base Reactions Acid-Base Reactions
Predicting the Direction of Predicting the Direction of Acid-Base Reactions Acid-Base Reactions
Based on experiment, we can put acids and Based on experiment, we can put acids and bases on a chart.bases on a chart.
See Table 17.3 or Appendix H tables 16See Table 17.3 or Appendix H tables 16
ACIDSACIDS CONJUGATE BASESCONJUGATE BASES STRONGSTRONG weak weak
weak weak STRONG STRONG
This chart can be used to predict the This chart can be used to predict the direction of reactions between any A-B pair.direction of reactions between any A-B pair.
Reactions always go from the stronger A-B Reactions always go from the stronger A-B pair to the weaker A-B pair.pair to the weaker A-B pair.
1616
ACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIES
Predicting the direction of an acid-base reaction.Predicting the direction of an acid-base reaction.
ACID 1 BASE 2 ACID 2 BASE 1+ +
STRONG weak
Reactions always go from the Reactions always go from the
stronger A-B pairstronger A-B pair weaker A-B pair. weaker A-B pair.
1717
MORE ABOUT WATERMORE ABOUT WATERMORE 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 In pure water there can be AUTOIONIZATIONAUTOIONIZATION
1818
KKww = [H = [H33OO++] [OH] [OH--] = 1.00 x 10] = 1.00 x 10-14-14 at 25 at 25 ooCC
In a In a neutralneutral solution [Hsolution [H33OO++] = [OH] = [OH--]]
and so [Hand so [H33OO++] = [OH] = [OH--] = 1.00 x 10] = 1.00 x 10-7-7 M M
MORE ABOUT WATERMORE ABOUT WATERMORE ABOUT WATERMORE ABOUT WATERAutoionizationAutoionization
2H2H22OO HH33OO+ + OH OH--2H2H22OO HH33OO+ + OH OH--
1919
Calculating [HCalculating [H++] & [OH] & [OH--]]Calculating [HCalculating [H++] & [OH] & [OH--]]
You add 0.0010 mol of NaOH to 1.0 L of You add 0.0010 mol of NaOH to 1.0 L of pure water. Calculate [Hpure water. Calculate [H33OO++] and [OH] and [OH--].].
SolutionSolution
HOH <--> HOH <--> H H++ + OH + OH--
Le Chatelier predicts equilibrium shifts Le Chatelier predicts equilibrium shifts to the ____________. to the ____________.
[H[H++] < 10] < 10-7-7 at equilibrium. at equilibrium.
Set up a concentration table.Set up a concentration table.
leftleft
2020
You add 0.0010 mol of NaOH to 1.0 L of You add 0.0010 mol of NaOH to 1.0 L of pure water. Calculate [Hpure water. Calculate [H3OO++] and [OH] and [OH--].].
SolutionSolution HH22OO((ll)) <--> H <--> H++ + OH + OH--
Calculating [HCalculating [H++] & [OH] & [OH--]]Calculating [HCalculating [H++] & [OH] & [OH--]]
initialinitial 0 0 0.0010 0.0010
changechange +x +x +x +x
equilibequilib x x 0.0010 + x 0.0010 + x
KKww = [H = [H++][OH][OH--] = (x) (0.0010 + x)] = (x) (0.0010 + x)Because x << 0.0010 M, assume [OHBecause x << 0.0010 M, assume [OH--] = 0.0010 M] = 0.0010 M
[H[H33OO++] = K] = Kww / 0.0010 = 1.0 x 10 / 0.0010 = 1.0 x 10-11-11 M M
This solution is This solution is BASICBASIC because [H because [H33OO++] < [OH] < [OH--]]
2121
[H[H++], [OH], [OH--] and pH] and pH[H[H++], [OH], [OH--] and pH] and pH
A common way to express acidity and A common way to express acidity and basicity is with pH.basicity is with pH.
pH = log (1/ [HpH = log (1/ [H++]) = - log [H]) = - log [H++]]
In a neutral solution,In a neutral solution,
[H[H++] = [OH] = [OH--] = 1.00 x 10] = 1.00 x 10-7-7 at 25 at 25 ooCC
pH = -log (1.00 x 10pH = -log (1.00 x 10-7-7) = - (-7.000) ) = - (-7.000) = 7.000 = 7.000
2222
[H[H++], [OH], [OH--] and pH] and pH[H[H++], [OH], [OH--] and pH] and pH
What is the pH of the 0.0010 M NaOH What is the pH of the 0.0010 M NaOH solution? solution?
[H[H++] = 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
General conclusion —General conclusion —
Basic solution Basic solution pH > 7pH > 7
Neutral Neutral pH = 7pH = 7
Acidic solutionAcidic solution pH < 7pH < 7
2323
[H[H++], [OH], [OH--] and pH] and pH[H[H++], [OH], [OH--] and pH] and pH
If the pH of Coke is 3.12, it is ____________.If the pH of Coke is 3.12, it is ____________.
Because pH = - log [HBecause pH = - log [H++] then] then
log [Hlog [H++] = - pH] = - pH
Take antilog and getTake antilog and get
[H[H++] = 10] = 10-pH-pH
[H[H++] = 10] = 10-3.12-3.12 = =
7.6 x 107.6 x 10-4-4 M M
AcidicAcidic
2424
Other pX ScalesOther pX ScalesOther pX ScalesOther pX Scales
In generalIn general pX = - log XpX = - log X
and so and so pOH = - log [OHpOH = - log [OH--]]
KKww = [H = [H33OO++] [OH] [OH--] = 1.00 x 10] = 1.00 x 10-14-14 at 25 at 25 ooCC
Take the - log of both sidesTake the - log of both sides
- log (10- log (10-14-14) = - log [H) = - log [H33OO++] + (- log [OH] + (- log [OH--])])
pKpKww = 14.00 = pH + pOH = 14.00 = pH + pOH
2525
Equilibria Involving Equilibria Involving Weak Acids and BasesWeak Acids and Bases
Equilibria Involving Equilibria Involving Weak Acids and BasesWeak Acids and Bases
Aspirin is a good example Aspirin is a good example of a weak acid, of a weak acid, KKaa = 3.2 x 10 = 3.2 x 10-4-4
2626
Equilibria Involving Equilibria Involving Weak Acids and BasesWeak Acids and Bases
Equilibria Involving Equilibria Involving Weak Acids and BasesWeak Acids and Bases
AcidAcid Conjugate BaseConjugate Base
acetic, HCacetic, HC22HH33OO22 CC22HH33OO22--, acetate, acetate
ammonium, NHammonium, NH44++ NHNH33, ammonia, ammonia
bicarbonate, HCObicarbonate, HCO33-- COCO33
2-2-, carbonate, carbonate
A weak acid (or base) is one that A weak acid (or base) is one that ionizes to a VERY small extent (< 5%).ionizes to a VERY small extent (< 5%).
2727
• For a weak acid, HA, and its For a weak acid, HA, and its conjugate base, Aconjugate base, A--, ,
KKaa .. K Kbb = K = Kww
• Thus, one can calculate KThus, one can calculate Kaa from K from Kbb and Kand Kbb from K from Kaa..
• Notice as KNotice as Kaa increases, K increases, Kbb decreases decreases and visa versa.and visa versa.
• The stronger the acid the weaker The stronger the acid the weaker the conjugate base.the conjugate base.
Equilibria Involving Equilibria Involving Weak Acids and BasesWeak Acids and Bases
Equilibria Involving Equilibria Involving Weak Acids and BasesWeak Acids and Bases
2828
Consider acetic acidConsider acetic acid
HCHC22HH33OO22 + H + H22O <--> HO <--> H33OO++ + C + C22HH33OO22--
AcidAcid Conj. Conj. basebase
Equilibria Involving Equilibria Involving Weak Acids and BasesWeak Acids and Bases
Equilibria Involving Equilibria Involving Weak Acids and BasesWeak Acids and Bases
(K is designated K(K is designated Kaa for an for an ACIDACID))
Because [HBecause [H33OO++] and [C] and [C22HH33OO22--] are SMALL, ] are SMALL,
KKaa << 1. << 1.
K a [H 3O+ ][CC22HH33OO22
-- ]
[HCC22HH33OO22] 1.8 x 10-5
2929
Equilibria Involving Equilibria Involving Weak Acids and BasesWeak Acids and Bases
Equilibria Involving Equilibria Involving Weak Acids and BasesWeak Acids and Bases
Values of KValues of Kaa for acid and K for acid and Kbb for bases for bases
are found in are found in TABLE 17.4TABLE 17.4 — page 799 — page 799
Notice the relation of TABLE 17.4 to Notice the relation of TABLE 17.4 to
the table of relative acid/base the table of relative acid/base
strengths (Table 17.3).strengths (Table 17.3).
3030
Incr
easi
ng
Aci
d S
tren
gth
Increasin
g B
ase Stren
gth
Partial table 17.3
3131
Equilibria Involving A Weak AcidEquilibria Involving A Weak Acid
You have 1.00 M HOAc. You have 1.00 M HOAc.
Calculate the equilibrium concentrations of Calculate the equilibrium concentrations of HOAc, HHOAc, H33OO++, OAc, OAc--, and the pH., and the pH.
Step 1.Step 1. Define equilibrium concentrations. Define equilibrium concentrations.
[HOAc][HOAc] [H[H33OO++]] [OAc[OAc--]]
initialinitial 1.001.00 0 0 0 0
changechange -x -x +x +x +x +x
equilibequilib 1.00-x1.00-x x x x x
Note that we neglect [HNote that we neglect [H33OO++] from H] from H22O.O.
3232Equilibria Involving A Weak AcidEquilibria Involving A Weak Acid
This is a quadratic. Solve using quadratic This is a quadratic. Solve using quadratic formula or method of approximations formula or method of approximations
(see Appendix A.4).(see Appendix A.4).
You have 1.00 M HOAc. Calculate the equilibrium concentrations of HOAc, H3O+, OAc-, and the pH.
Step 2.Step 2. Write K Write Kaa expression expression
Ka 1.8 x 10 -5 = [H3O+][OAc- ]
[HOAc]
x2
1.00 - x
3333
And so x = [And so x = [HH33OO++] = [] = [OAcOAc--] = [K] = [Kaa • 1.00] • 1.00]1/21/2
Ka 1.8 x 10 -5 = x2
1.00
First assume x is very small because KFirst assume x is very small because Kaa is is
so small.so small.
Step 3.Step 3. Solve K Solve Kaa expression expression
You have 1.00 M HOAc. Calculate the equilibrium concentrations of HOAc, H3O+, OAc-, and the pH.
Equilibria Involving A Weak AcidEquilibria Involving A Weak Acid
Ka 1.8 x 10 -5 = [H3O+][OAc- ]
[HOAc]
x2
1.00 - x
3434
Ka 1.8 x 10 -5 = x2
1.00
Step 3.Step 3. Solve K Solve Kaa approximateapproximate expression expression
You have 1.00 M HOAc. Calculate the equilibrium concentrations of HOAc, H3O+, OAc-, and the pH.
Equilibria Involving A Weak AcidEquilibria Involving A Weak Acid
x = [Hx = [H33OO++] = [OAc] = [OAc--] = [K] = [Kaa • 1.00] • 1.00]1/21/2
x = x = [H[H33OO++] = [OAc] = [OAc--] = 4.2 x 10] = 4.2 x 10-3-3 M M
pH = - log [HpH = - log [H33OO++] = -log (4.2 x 10] = -log (4.2 x 10-3-3) = ) = 2.372.37
3535Equilibria Involving A Weak AcidEquilibria Involving A Weak Acid
For many weak acidsFor many weak acids
[H[H33OO++] = [conj. base] = [K] = [conj. base] = [Kaa • C • Coo]]1/21/2
where Cwhere C00 = initial concentration of acid = initial concentration of acid
Useful Rule of Thumb:Useful Rule of Thumb:
If 1000 • KIf 1000 • Kaa < C < Coo, then [H, then [H33OO++] = [K] = [Kaa • C • Coo]]1/21/2
Consider the approximate expressionConsider the approximate expression
x [H3O+
] = [K a • 1.00]1/2
Ka 1.8 x 10 -5 = x2
1.00
3636
Approximation RulesApproximation Rules
10 times K for .1M, so 1010 times K for .1M, so 10K < [X]K < [X]
100 times K for .10M, so 100100 times K for .10M, so 100K < [X]K < [X]
1000 times K for .100M , so 10001000 times K for .100M , so 1000K < [X]K < [X]
Look at the number of decimal places to Look at the number of decimal places to determine factor to multiply K bydetermine factor to multiply K by
3737Equilibria Involving A Weak AcidEquilibria Involving A Weak Acid
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++] ] = = [K[Kaa • C • Coo]]1/21/2 = 4.2 x 10 = 4.2 x 10-4-4 M, M, pH = 3.37pH = 3.37
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 M = 0.0007 M
pH = 3.47 pH = 3.47 NOT ValidNOT Valid
3838Equilibria Involving A Weak BaseEquilibria Involving A Weak Base
You have 0.010 M NHYou 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 concentrations. Define equilibrium concentrations.
[NH[NH33]] [NH[NH44++]] [OH[OH--]]
initialinitial 0.0100.010 0 0 0 0
changechange -x -x +x +x +x +x
equilibequilib 0.010 - x0.010 - x x x x x
3939
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 [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 !!
Kb 1.8 x 10-5 = [NH4
+][OH- ][NH3 ]
= x2
0.010 - x
You have 0.010 M NHYou 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 2.Step 2. Solve the equilibrium expression Solve the equilibrium expression
Equilibria Involving A Weak BaseEquilibria Involving A Weak Base
4040
You have 0.010 M NHYou 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
Equilibria Involving A Weak BaseEquilibria Involving A Weak Base
Step 3.Step 3. Calculate pH Calculate 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
4141
1. The pH of a 0.10 M nicotinic acid solution is 1. The pH of a 0.10 M nicotinic acid solution is 2.92. Calculate K2.92. Calculate Kaa and the % ionization. and the % ionization.
1.4x101.4x10-5 -5 1.2%1.2%
2. A solution of propionic acid, 2. A solution of propionic acid, CHCH33CHCH22COOH, is 0.20 M. KCOOH, is 0.20 M. Kaa = 1.3x10 = 1.3x10-5-5. . Calculate the pH and % ionization.Calculate the pH and % ionization.
2.79 0.80%2.79 0.80%
3. A solution of hydrazine, N3. A solution of hydrazine, N22HH44, is 0.025 M. , is 0.025 M. K Kbb = 8.5x10 = 8.5x10-5-5. . Calculate the pH and % Calculate the pH and % ionization.ionization.
11.15 5.6%11.15 5.6% (4.) (4.)
Equilibria Involving Equilibria Involving Weak Acids and BasesWeak Acids and Bases
Equilibria Involving Equilibria Involving Weak Acids and BasesWeak Acids and Bases
4242
4. Calculate the pH and 4. Calculate the pH and
concentrations of the arsenate concentrations of the arsenate
containing species in a 0.15 M containing species in a 0.15 M
solution of Hsolution of H33AsOAsO44. . KK11= 2.5x10= 2.5x10-4-4, ,
KK22 = 5.6x10 = 5.6x10-8-8, K, K33 = 3.0x10 = 3.0x10-13-13
2.21, 0.14, 6.1x102.21, 0.14, 6.1x10-3-3,, 5.8x105.8x10-8-8, 2.8x10, 2.8x10-18-18
4343
MX + HMX + H22O ----> acidic or basic solution?O ----> acidic or basic solution?
Consider NHConsider NH44ClCl
NHNH44Cl(aq) ----> NHCl(aq) ----> NH44++(aq) + Cl(aq) + Cl--(aq)(aq)
(a)(a) Reaction of ClReaction of Cl-- with H with H22OO
ClCl-- + + HH22O <---->O <----> HCl + HCl + OHOH--
basebase acidacid acidacid basebase
ClCl-- ion is a VERY weak base because its conjugate ion is a VERY weak base because its conjugate acid is strong. acid is strong.
Acid-Base Properties of SaltsAcid-Base Properties of Salts
Therefore, ClTherefore, Cl-- ----> neutral solution ----> neutral solution
XX
4444
NHNH44Cl(aq) ----> NHCl(aq) ----> NH44++(aq) + Cl(aq) + Cl--(aq)(aq)
(b)(b) Reaction of NHReaction of NH44++ with H with H22OO
NHNH44++ + H + H22O <---->O <----> NH NH33 + H + H33OO++
acidacidbasebase basebase acidacid
NHNH44++ ion is a moderate acid because its ion is a moderate acid because its
conjugate base is weak. conjugate base is weak.
Therefore, NHTherefore, NH44++ ----> acidic solution ----> acidic solution
See TABLE 17.5 for a summary of acid-See TABLE 17.5 for a summary of acid-base properties of ions.base properties of ions.
Acid-Base Properties of SaltsAcid-Base Properties of Salts
4545
Determine if the following solutions are acidic, basic, or neutral. KBr,CrCl3, NaNO2, KHCO3, Na2CO3, NH4C2H3O2
4646
Calculate the pH of a 0.10 M solution of NaCalculate the pH of a 0.10 M solution of Na22COCO33. .
NaNa++ + H + H22O ---> neutralO ---> neutral
COCO332-2- ++ HH22OO HCOHCO33
-- ++ OHOH--basebase
acidacid acidacid basebase
KKbb = 2.1 x 10 = 2.1 x 10-4-4
Step 1.Step 1. Set up concentration tableSet up concentration table
[CO[CO332-2-]] [HCO[HCO33
--]] [OH[OH--]]
initialinitial 0.10 0.10 0 0 0 0
changechange -x -x +x +x +x +x
equilibequilib 0.10 - x 0.10 - x x x x x
Acid-Base Properties of SaltsAcid-Base Properties of SaltsAcid-Base Properties of SaltsAcid-Base Properties of Salts
4747
Calculate the pH of a 0.10 M solution of NaCalculate the pH of a 0.10 M solution of Na22COCO33. .
NaNa++ + H + H22O ---> neutralO ---> neutral
COCO332-2- ++ HH22OO HCOHCO33
-- ++ OHOH--
basebase acidacid acidacid basebase
KKbb = 2.1 x 10 = 2.1 x 10-4-4
Step 2.Step 2. Solve the equilibrium expressionSolve the equilibrium expression
Acid-Base Properties of SaltsAcid-Base Properties of SaltsAcid-Base Properties of SaltsAcid-Base Properties of Salts
Assume 0.10 - x 0.10, because 100•KAssume 0.10 - x 0.10, because 100•Kbb < C < Coo
x = [HCOx = [HCO33--] = [OH] = [OH--] = 0.0046 M ] = 0.0046 M
Kb = 2.1 x 10-4 = [HCO3
-][OH- ]
[CO32 ]
x2
0.10 - x
4848
Calculate the pH of a 0.10 M solution of NaCalculate the pH of a 0.10 M solution of Na22COCO33. .
NaNa++ + H + H22O ---> neutralO ---> neutral
COCO332-2- ++ HH22OO HCOHCO33
-- ++ OHOH--
basebase acidacid acidacid basebase
KKbb = 2.1 x 10 = 2.1 x 10-4-4
Acid-Base Properties of SaltsAcid-Base Properties of SaltsAcid-Base Properties of SaltsAcid-Base Properties of Salts
Step 3.Step 3. Calculate the pHCalculate the pH[OH[OH--] = 0.0046 M] = 0.0046 MpOH = - log [OHpOH = - log [OH--] = 2.34] = 2.34pH + pOH = 14, pH + pOH = 14,
so so pH = 11.66pH = 11.66, and the solution is ________., and the solution is ________.Basic
4949
Review ProblemsReview ProblemsReview ProblemsReview Problems
• Calculate the pH and concentrations of Calculate the pH and concentrations of
the arsenate containing species in a 0.15 the arsenate containing species in a 0.15
M solution of HM solution of H33AsOAsO44. . KK11= 2.5x10= 2.5x10-4-4, K, K22 = =
5.6x105.6x10-8-8, K, K33 = 3.0x10 = 3.0x10-13-13
2.21, 0.14, 6.1x102.21, 0.14, 6.1x10-3-3,, 5.8x105.8x10-8-8, 2.8x10, 2.8x10-18-18
• Calculate the pH of a 0.050 M KCalculate the pH of a 0.050 M K22COCO33 solution. solution.
11.5011.50
5050
• Lewis base = Lewis base = electron pair donor electron pair donor (NH(NH33))
Lewis Acids & BasesLewis Acids & Bases
• Lewis acid = Lewis acid = electron pair electron pair acceptor (BFacceptor (BF33))
5151
A Lewis acid and base can interact A Lewis acid and base can interact by sharing an electron pair.by sharing an electron pair.
Lewis Acids & BasesLewis Acids & Bases
5252
HH
H
BASEACID
O—H••••
••O—H
H+
++
A Lewis acid and base can interact A Lewis acid and base can interact by sharing an electron pair.by sharing an electron pair.
Formation of Formation of hydronium ion hydronium ion is an is an excellent example.excellent example.
Lewis Acids & BasesLewis Acids & Bases
5353
Other good examples involve metal ions.Other good examples involve metal ions.
Lewis Acids & Bases
Colorless MColorless Mxx(NO(NO33))yy form colorful form colorful
hydrated [Mhydrated [Mxx(H(H22O)O)yy]]z+z+ compounds compounds
5454
Such bonds as the HSuch bonds as the H22O ---> Co bond are often O ---> Co bond are often called called COORDINATE COVALENT BONDSCOORDINATE COVALENT BONDS because both electrons are supplied by one of because both electrons are supplied by one of the atoms of the bond.the atoms of the bond.
Lewis Acids & BasesLewis Acids & Bases
Other good examples involve metal ions.Other good examples involve metal ions.
HH
Co2+ ••
BASEACID
O—H••••
••O—H
Co2+
5555
The combination of metal The combination of metal ions (Lewis acids) with ions (Lewis acids) with Lewis bases such as HLewis bases such as H22O O
and NHand NH33 ------> ------>
COMPLEX IONSCOMPLEX IONS
All metal ions form All metal ions form complex ions with water complex ions with water — and are of the type — and are of the type [M(H[M(H22O)O)xx]]n+n+ where x = 4 where x = 4
and 6.and 6.
Lewis Acids & BasesLewis Acids & Bases
[Cu(NH3)4]2+
5656
Add NHAdd NH33 to light blue [Cu(H to light blue [Cu(H22O)O)44]]2+2+ ------> ------>
light blue Cu(OH)light blue Cu(OH)22 and then deep blue and then deep blue
[Cu(NH[Cu(NH33))44]]2+2+
Lewis Acids & BasesLewis Acids & Bases
5757
[Ni(H[Ni(H22O)O)66]]2+2+ + 6 NH + 6 NH33 ---> [Ni(NH ---> [Ni(NH33))66]]2+2+
+ DMG+ DMG
Lewis Acids & BasesLewis Acids & Bases
5858
The FeThe Fe2+2+ in heme can interact with O in heme can interact with O22 or or
CO in a Lewis acid-base reaction.CO in a Lewis acid-base reaction.
Lewis Acids & BasesLewis Acids & Bases
5959
Many complex ions containing water undergo Many complex ions containing water undergo HYDROLYSISHYDROLYSIS to give acidic solutions. to give acidic solutions.
[Cu(H[Cu(H22O)O)44]]2+2+ + H + H22O ---> [Cu(HO ---> [Cu(H22O)O)33(OH)](OH)]++ + H + H33OO++
Lewis Acids & BasesLewis Acids & Bases
6060
Many complex ions containing water undergo Many complex ions containing water undergo HYDROLYSISHYDROLYSIS to give acidic solutions. to give acidic solutions.
This explains why water solutions of FeThis explains why water solutions of Fe3+3+, Al, Al3+3+, , CuCu2+2+, Pb, Pb2+2+, etc. are acidic., etc. are acidic.
Lewis Acids & BasesLewis Acids & Bases
6161
This explains This explains AMPHOTERICAMPHOTERIC nature of some nature of some
metal hydroxides.metal hydroxides.
Al(OH)Al(OH)33(s) + 3 H(s) + 3 H++ --> Al --> Al3+3+ + 3 H + 3 H22OO
Here Al(OH)Here Al(OH)33 is a Brønsted base. is a Brønsted base.
Al(OH)Al(OH)33(s) + OH(s) + OH-- --> Al(OH) --> Al(OH)44--
Here Al(OH)Here Al(OH)33 is a Lewis acid. is a Lewis acid. Al 3+ O—H -
••••
••
Lewis Acids & BasesLewis Acids & Bases
6262
Amphoterism of Al(OH)Amphoterism of Al(OH)33
Al(OH)Al(OH)33 on right on right Add NaOHAdd NaOH Add HClAdd HCl
See Kotz / Treichel 5See Kotz / Treichel 5thth, page 722, page 722
6363
1. The electronegativity of 1. The electronegativity of the O atoms causes the the O atoms causes the H attached to O to be H attached to O to be highly positive.highly positive.
2. The O—H bond is 2. The O—H bond is highly polar.highly polar.
3. The H atom of O—H is 3. The H atom of O—H is readily attracted to polar readily attracted to polar HH22O. O.
6464
Neutral Lewis AcidNeutral Lewis AcidNeutral Lewis AcidNeutral Lewis AcidCarbon dioxide is a neutral (molecular) Lewis acid.Carbon dioxide is a neutral (molecular) Lewis acid.
+1.5+1.5 -0.75-0.75-0.75-0.75
6565
Lewis Acids & BasesLewis Acids & Bases
Many complex ions are very stable.Many complex ions are very stable.
CuCu2+2+ + 4 NH + 4 NH33 <--><--> [Cu(NH[Cu(NH33))44]]2+2+
K for the reaction is calledK for the reaction is called
KKformationformation
or a “formation constant”.or a “formation constant”.
Here K = 6.8 x 10Here K = 6.8 x 101212. .
Rxn. is strongly product-favored.Rxn. is strongly product-favored.
6666
Formation of complex ions explains why you Formation of complex ions explains why you can dissolve a precipitate by forming a can dissolve a precipitate by forming a complex ion. complex ion.
AgCl(s) + 2 NHAgCl(s) + 2 NH33 Ag(NHAg(NH33))22+ + + Cl+ Cl--
Lewis Acids & BasesLewis Acids & Bases
AgCl(s)
6767
AgCl(s) AgCl(s) <--> <--> AgAg++ + Cl + Cl-- K Kspsp = 1.8 x 10 = 1.8 x 10-10-10
AgAg++ + 2 NH + 2 NH33 <--> Ag(NH <--> Ag(NH33))22++ K Kformform = 1.6 x 10 = 1.6 x 1077
--------------------------------------------------------------------------
AgCl(s) + 2 NHAgCl(s) + 2 NH33 <--><--> Ag(NHAg(NH33))22+ + + Cl+ Cl--
KKnetnet = K = Kspsp • K • Kformform = 2.9 x 10 = 2.9 x 10-3-3
Lewis Acids & BasesLewis Acids & Bases Formation of complex ions explains why Formation of complex ions explains why
you can dissolve a precipitate by forming a you can dissolve a precipitate by forming a complex ion.complex ion.
Why?Why?Why?Why?• Why are some Why are some
compounds acids?compounds acids?
• Why are some Why are some compounds bases?compounds bases?
• Why do acids and bases Why do acids and bases vary in strength?vary in strength?
• Can we predict Can we predict variations in acidity or variations in acidity or basicity?basicity?
• Why are some Why are some compounds acids?compounds acids?
• Why are some Why are some compounds bases?compounds bases?
• Why do acids and bases Why do acids and bases vary in strength?vary in strength?
• Can we predict Can we predict variations in acidity or variations in acidity or basicity?basicity?
6969
Trichloroacetic acid is much stronger acid Trichloroacetic acid is much stronger acid owing to the high electronegativity of Cl, owing to the high electronegativity of Cl, which withdraws electrons from the rest of which withdraws electrons from the rest of the molecule. This makes the O—H bond the molecule. This makes the O—H bond highly polar. The H of O—H is very positive.highly polar. The H of O—H is very positive.
Acetic acidAcetic acid Trichloroacetic acidTrichloroacetic acidKKaa = 1.8 x 10 = 1.8 x 10-5-5 KKaa = 0.3 = 0.3
7070
Oxyacid StrengthOxyacid StrengthOxyacid StrengthOxyacid Strength
• For oxyacids, XOFor oxyacids, XOnn(OH)(OH)mm, acid , acid
strength increases as n increases strength increases as n increases and is independent of m.and is independent of m.
• Rank in order of increasing Rank in order of increasing acidity: HBrOacidity: HBrO33, H, H33BOBO33, HIO, HIO44
BrO2(OH), B(OH)3, IO3(OH)
B(OH)B(OH)33 BrO BrO22(OH) IO(OH) IO33(OH)(OH)
7171
These ions are These ions are BASES.BASES.They become more and more basic as the They become more and more basic as the
negative charge increases.negative charge increases.
As the charge goes up, they interact more As the charge goes up, they interact more strongly with polar water molecules.strongly with polar water molecules.
NONO33--
COCO332-2-
Basicity of OxoanionsBasicity of OxoanionsBasicity of OxoanionsBasicity of Oxoanions
POPO443-3-
7272
Practice ProblemsPractice Problems1. Write the equation for the reaction of 1. Write the equation for the reaction of
hypochlorous acid and ammonia. Label the hypochlorous acid and ammonia. Label the acids and bases. Indicate the conjugate pairs.acids and bases. Indicate the conjugate pairs.
2. A 0.15 M weak acid solution is determined to 2. A 0.15 M weak acid solution is determined to be 2.0% dissociated. Calculate the Kbe 2.0% dissociated. Calculate the Kaa..
3. Calculate the pH of the following solutions:3. Calculate the pH of the following solutions:
a) 0.50 M nitric acida) 0.50 M nitric acid
b) 0.25 M potassium hydroxideb) 0.25 M potassium hydroxide
c) 0.15 M phosphoric acidc) 0.15 M phosphoric acid
d) 0.22 M sodium sulfited) 0.22 M sodium sulfite
e) 0.010 potassium cyanidee) 0.010 potassium cyanide
7373
Practice ProblemsPractice Problems
4. Calculate the carbonate ion concentration 4. Calculate the carbonate ion concentration in a 0.10 M carbonic acid solution.in a 0.10 M carbonic acid solution.
5. Determine whether a solution of Cs5. Determine whether a solution of Cs22SOSO33 is is acidic, neutral or basic.acidic, neutral or basic.
6. Write the Lewis acid/base reaction 6. Write the Lewis acid/base reaction between PHbetween PH33 and BF and BF33 using dot structures. using dot structures. Indicate the acid and the base.Indicate the acid and the base.
7474
Practice Problems AnswersPractice Problems Answers
1. HClO + NH1. HClO + NH33 <--> NH <--> NH44++ + ClO + ClO--
acidacid basebase acidacid basebase
2. 6.0 x 102. 6.0 x 10-5-5
3. a) .303. a) .30 b) 13.40b) 13.40 c) 1.52c) 1.52
d) 10.28d) 10.28 e) 10.70e) 10.70
4. 4.8 x 104. 4.8 x 10-11-11 5. basic 5. basic
7575
Practice Problems AnswersPractice Problems Answers
6. 6.
. H
H
P ... H . F
F
B ... F+ -->
. H
H P.
.. H
. F
F B.
. . F
The end.
7676
ammoniaammoniaNHNH33 + H + H22O <--> O <--> NHNH44
++ + OH + OH--
oror
NHNH44OH <--> OH <--> NHNH44+ + + OH + OH--
ReturnReturn
7777
Sample QuestionsSample QuestionsSample QuestionsSample Questions
Give the conjugate base of HCN. Give the conjugate base of HCN.
CNCN--
Give the conjugate acid of NOGive the conjugate acid of NO22--. .
HNOHNO22
Give the conjugate acid of NHGive the conjugate acid of NH33..
NHNH44++
The ion HCOThe ion HCO33-- has both a conjugate acid and has both a conjugate acid and
conjugate base. Give the formula of each.conjugate base. Give the formula of each.
HH22COCO33 CO CO332-2-
7878
Sample QuestionsSample QuestionsSample QuestionsSample Questions
Label the acid-base pairs: Label the acid-base pairs:
CNCN-- + HOH <--> HCN + OH + HOH <--> HCN + OH--
basebase acidacid acidacid basebase
HCOHCO331-1- + HOH <--> OH + HOH <--> OH-- + H + H22COCO33
basebase acidacid base base acidacid
HCOHCO331-1- + HOH <--> CO + HOH <--> CO33
2-2- + H + H33OO++
acidacid base base basebase acidacid
7979
Acid-Base ReactionsAcid-Base ReactionsAcid-Base ReactionsAcid-Base Reactions
Consider the acid HF reacting with Consider the acid HF reacting with the base NHthe base NH33..
HF + NHHF + NH33 <--> NH <--> NH44++ + F + F--
acid acid basebase acid acid basebase
Acids: HF > NHAcids: HF > NH44++
Bases: NHBases: NH33 > F > F--
Reaction favors productsReaction favors products
Consider the acid HF reacting with Consider the acid HF reacting with the base NHthe base NH33..
HF + NHHF + NH33 <--> NH <--> NH44++ + F + F--
acid acid basebase acid acid basebase
Acids: HF > NHAcids: HF > NH44++
Bases: NHBases: NH33 > F > F--
Reaction favors productsReaction favors products
8080
Incr
easi
ng
Aci
d S
tren
gth
Increasin
g B
ase Stren
gth
8181
Acid-Base ReactionsAcid-Base ReactionsAcid-Base ReactionsAcid-Base Reactions
Does the reaction between nitrous acid Does the reaction between nitrous acid and ammonia favor reactants or and ammonia favor reactants or products?products?
HNOHNO22 + NH + NH33 <--> NH <--> NH44++ + NO + NO22
--
acid acid basebase acid acid basebase
Acids: HNOAcids: HNO22 > NH > NH44++
Bases: NHBases: NH33 > NO > NO22--
Reaction favors productsReaction favors products
Does the reaction between nitrous acid Does the reaction between nitrous acid and ammonia favor reactants or and ammonia favor reactants or products?products?
HNOHNO22 + NH + NH33 <--> NH <--> NH44++ + NO + NO22
--
acid acid basebase acid acid basebase
Acids: HNOAcids: HNO22 > NH > NH44++
Bases: NHBases: NH33 > NO > NO22--
Reaction favors productsReaction favors products
8282
Acid-Base ReactionsAcid-Base ReactionsAcid-Base ReactionsAcid-Base Reactions
Does the reaction between hypochlorous Does the reaction between hypochlorous acid and fluoride ion favor reactants or acid and fluoride ion favor reactants or products? products?
HClO + FHClO + F-- <--> HF + ClO <--> HF + ClO--
acid acid basebase acid acid basebase
Acids: HF > HClOAcids: HF > HClO
Bases: ClOBases: ClO-- > F > F--
Reaction favors reactantsReaction favors reactants
Does the reaction between hypochlorous Does the reaction between hypochlorous acid and fluoride ion favor reactants or acid and fluoride ion favor reactants or products? products?
HClO + FHClO + F-- <--> HF + ClO <--> HF + ClO--
acid acid basebase acid acid basebase
Acids: HF > HClOAcids: HF > HClO
Bases: ClOBases: ClO-- > F > F--
Reaction favors reactantsReaction favors reactants
8383
Acid-Base ReactionsAcid-Base ReactionsAcid-Base ReactionsAcid-Base Reactions Does the reaction between ammonium Does the reaction between ammonium
chloride and sodium sulfate favor chloride and sodium sulfate favor reactants or products? reactants or products? NHNH44Cl + NaCl + Na22SOSO44 <--> ??? <--> ???
NHNH44++ + SO + SO44
2-2- <--> NH <--> NH33 + HSO + HSO441-1-
acid acid basebase basebase acid acid
Acids: HSOAcids: HSO441-1- > NH > NH44
++
Bases: NHBases: NH33 > SO > SO442-2-
Reaction favors reactantsReaction favors reactants
Does the reaction between ammonium Does the reaction between ammonium chloride and sodium sulfate favor chloride and sodium sulfate favor reactants or products? reactants or products? NHNH44Cl + NaCl + Na22SOSO44 <--> ??? <--> ???
NHNH44++ + SO + SO44
2-2- <--> NH <--> NH33 + HSO + HSO441-1-
acid acid basebase basebase acid acid
Acids: HSOAcids: HSO441-1- > NH > NH44
++
Bases: NHBases: NH33 > SO > SO442-2-
Reaction favors reactantsReaction favors reactants
8484
Sample QuestionsSample QuestionsSample QuestionsSample Questions
1. Determine the pH, pOH, and [OH1. Determine the pH, pOH, and [OH--] ]
[H[H++] = 2.5 x 10] = 2.5 x 10-6-6
pH = - log [HpH = - log [H++] = 5.60] = 5.60
pOH = 14 - 5.60 = 8.40pOH = 14 - 5.60 = 8.40
pOH = - log [OH-]pOH = - log [OH-] or 1.00 x 10or 1.00 x 10-14-14 = [H = [H++][OH][OH--]]
8.40 = - log [OH-] 8.40 = - log [OH-] 1.00 x 101.00 x 10-14-14 = (2.5 x 10 = (2.5 x 10-6-6)[OH)[OH--]]
[OH[OH--] = 10] = 10-8.40-8.40
[OH[OH--] = 4.0 x 10] = 4.0 x 10-9 -9 MM
8585
Sample QuestionsSample QuestionsSample QuestionsSample Questions
2. Determine the pOH, [H2. Determine the pOH, [H++], and [OH], and [OH--] ]
pH = 6.52pH = 6.52
pOH = 7.48pOH = 7.48
[H[H++] = 3.0 x 10] = 3.0 x 10-7 -7 MM
[OH[OH--] = 3.3 x 10] = 3.3 x 10-8 -8 MM
How can we determine the pH of a solution?Acid/Base indicators is one method.
8686Name of Indicator pH Interval Color ChangeMethyl violet 0.2 - 3.0 yellow to blue-violetThymol blue 1.2 - 2.8 red to yellowOrange IV 1.3 - 3.0 red to yellowMethyl orange 3.1 - 4.4 red to orange to yellowBromophenol blue 3.0 - 4.6 yellow to blue-violetCongo red 3.0 - 5.0 blue to redBromocresol green 3.8 - 5.4 yellow to blueMethyl red 4.4 - 6.2 red to yellowBromocresol purple 5.2 - 6.8 yellow to purpleBromothymol blue 6.0 - 7.6 yellow to bluePhenol red 6.8 - 8.2 yellow to redThymol blue 8.0 - 9.6 yellow to bluePhenolphthalein 8.3 - 10.0 colorless to redThymolphthalein 9.3 - 10.5 colorless to blueAlizarin yellow 10.0 - 12.0 yellow to redIndigo carmine 11.4 - 13.0 blue to yellow
8787
Sample QuestionsSample QuestionsSample QuestionsSample Questions
3. Determine the pH, pOH, [H3. Determine the pH, pOH, [H++], and [OH], and [OH--] ]
0.035 M HCl.0.035 M HCl.
Strong acid --> [HStrong acid --> [H++] = 0.035] = 0.035
pH = 1.46pH = 1.46
pOH = 12.54pOH = 12.54
[OH[OH--] = 2.9 x 10] = 2.9 x 10-13-13 MM
8888
Sample QuestionsSample QuestionsSample QuestionsSample Questions
4. Determine the pH, pOH, [H4. Determine the pH, pOH, [H++], and [OH], and [OH--] ] 0.15 M NaOH.0.15 M NaOH.
Strong base --> [OHStrong base --> [OH--] = 0.15 M] = 0.15 M
pOH = 0.82pOH = 0.82
pH = 13.18pH = 13.18
[H[H++] = 6.7 x 10] = 6.7 x 10-14-14 M M
8989
Equilibria Involving A Weak AcidEquilibria Involving A Weak Acid
You have 0.10 M HCYou have 0.10 M HC66HH44NONO22. .
Its pH is measured to be 2.92Its pH is measured to be 2.92
Calculate the equilibrium constant.Calculate the equilibrium constant.
Step 1.Step 1. Write the equation. Write the equation.
HA <--> HHA <--> H++ + A + A--
9090
Equilibria Involving A Weak AcidEquilibria Involving A Weak Acid
Step 2.Step 2. Make a chart. Make a chart.
HA <--> HHA <--> H++ + A + A--
0.100.10 0 0 0 0
- 0.0012
0.00120.0012
0.00120.0012
0.10
9191Equilibria Involving A Weak AcidEquilibria Involving A Weak Acid
Step 3.Step 3. Write K Write Kaa expression, without and expression, without and
with numbers.with numbers.
Ka = [H+][AA--]
[HAHA]
Ka =
(0.0012)(0.0012.0012)
(0.10 )
9292Equilibria Involving A Weak AcidEquilibria Involving A Weak Acid
Step 4.Step 4. Solve. Solve.
Ka = 1.4 x 10-5
Step 5.Step 5. Answer question(s). Answer question(s).
Ka = 1.4 x 10-5
Ka =
(0.0012)(0.0012.0012)
(0.10 )
9393
Sample QuestionsSample Questions
1. You have 0.050 M HCN. Its pH is 1. You have 0.050 M HCN. Its pH is measured to be 5.4. Calculate the measured to be 5.4. Calculate the equilibrium constant.equilibrium constant.
HCN <--> HHCN <--> H++ + CN + CN--
0.0500.050 0 0 0 0
- 4 x10- 4 x10-6-6 + 4 x10+ 4 x10-6-6 + 4 x10+ 4 x10-6-6
4 x104 x10-6-6 4 x10-60.0500.050
Ka = [H+][CN-]
[HCN]= = 3 x 10-10
(4 x 10-6)2
(0.050)
9494
Sample QuestionsSample Questions
2. You have 0.1000 M HA which is 1.30% 2. You have 0.1000 M HA which is 1.30% dissociated. dissociated.
Calculate the equilibrium constant.Calculate the equilibrium constant.
HA <--> HHA <--> H++ + A + A--
0.10000.1000 0 0 0 0
- 0.00130- 0.00130 + 0.00130+ 0.00130 + 0.00130+ 0.00130
0.001300.001300.001300.001300.09870.0987
Ka = [H+][A-]
[HA]= = 1.71 x 10-5
(0.00130)2
(0.0987)
9595
Equilibria Involving A Weak AcidEquilibria Involving A Weak Acid
You have 1.00 M HCYou have 1.00 M HC22HH33OO22. .
Calculate the equilibrium concentrations Calculate the equilibrium concentrations of HCof HC22HH33OO22, H, H++, C, C22HH33OO22 --, and the pH., and the pH.
Step 1.Step 1. Write the equation. Write the equation.
HCHC22HH33OO22 <--> H <--> H++ + C + C22HH33OO22--
9696
Equilibria Involving A Weak AcidEquilibria Involving A Weak Acid
Step 2.Step 2. Make a chart. Make a chart.
HCHC22HH33OO22 <--> H <--> H++ + C + C22HH33OO22--
1.001.00 0 0 0 0
-x
xx
+x+x
1.00-x
9797Equilibria Involving A Weak AcidEquilibria Involving A Weak Acid
Step 3.Step 3. Write K Write Kaa expression, without and expression, without and
with numbers.with numbers.
Ka = [H+][CC22HH33OO22
--]
[HCHC22HH33OO22]
1.8 x 10-5 = (x)(xx)
(1.00 - x)
9898Equilibria Involving A Weak AcidEquilibria Involving A Weak Acid
Step 4.Step 4. Solve. Solve.
1.8 x 10-5 = (x)(xx)
(1.00 - x)
1.8 x 10-5(1.00 - x) = x2
x2 + 1.8 x 10-5x - 1.8 x 10-5 = 0
x = 4.2 x 10-3 M
9999
Step 5.Step 5. Answer question(s). Answer question(s).
Equilibria Involving A Weak AcidEquilibria Involving A Weak Acid
x = [Hx = [H++] = [C] = [C22HH33OO22--] = 0.0042 M] = 0.0042 M
[HC[HC22HH33OO22] = 1.00 - 0.0042 = 1.00 M] = 1.00 - 0.0042 = 1.00 M
pH = 2.38pH = 2.38
100100Equilibria Involving A Weak AcidEquilibria Involving A Weak Acid
Note: Note: Assume x is very small because KAssume x is very small because Kaa
is so small.is so small.
1.8 x 10-5 = (x)(xx)
(1.00 - x)
1.8 x 10-5(1.00) = x2
x = 4.2 x 10-3 M
1.8 x 10-5 = (x)(xx)
(1.00)
General Rule: Simplfy if [HA] / Ka > 1000
101101
Sample QuestionsSample Questions
1. Calculate the pH of a 0.100 M HF solution.1. Calculate the pH of a 0.100 M HF solution.
HF <--> HHF <--> H++ + F + F--
0.1000.100 0 0 0 0 - x + x + x
x x0.100 - x
Ka = [H+][F-]
[HF]= = 7.2 x 10
- 4 x2
(0.100 - x)
X = 0.0081 M = {H+] pH = 2.09
102102
Sample QuestionsSample Questions
2. Calculate the pH of a 0.20 M HClO solution.2. Calculate the pH of a 0.20 M HClO solution.
HClO <--> HHClO <--> H++ + ClO + ClO--
0.200.20 0 0 0 0 - x + x + x
x x0.20
Ka = [H+][ClO-]
[HClO]= = 3.5 x 10
- 8 x2
(0.20)
X = 8.4 x 10 - 5 M = {H+] pH = 4.08
103103
Sample QuestionsSample Questions
3. Calculate the pH of a 0.20 M HCOOH solution.3. Calculate the pH of a 0.20 M HCOOH solution.
HA <--> HHA <--> H++ + A + A--
0.200.20 0 0 0 0 - x + x + x
x x0.20 - x
Ka = [H+][A-]
[HA]= = 1.8 x 10-4
x2
(0.20 - x)
X = 0.0059 M = {H+] pH = 2.23
104104
Sample QuestionsSample Questions
4. Calculate the pH of a 0.15 M H4. Calculate the pH of a 0.15 M H22COCO33 solution. solution.
HH22COCO33 <--> H <--> H++ + HCO + HCO33--
0.150.15 0 0 0 0- x + x + x
x x0.15
Ka = [H+][HCO3
-]
[H2CO3]= = 4.2 x 10-7
x2
0.15
X = 2.5 x 10-4 M
105105Sample QuestionsSample Questions Does the second equilibrium have a Does the second equilibrium have a
significant effect on the pH? significant effect on the pH?
HCOHCO331-1- <--> H <--> H++ + CO + CO33
2-2-
2.5 x 102.5 x 10-4 -4 2.5 x 102.5 x 10-4 -4 00
- y + y + y
y2.5 x 10-42.5 x 10-4
Ka = [H+][CO3
2-]
[HCO31-]
= = 4.8 x 10-11(2.5 x 10-4)y
2.5 x 10-4
y = 4.8 x 10-11 M pH = 3.60
NO!NO!
106106
Equilibria Involving A Weak BaseEquilibria Involving A Weak Base
You have 0.10 M BOH. Its pH is measured You have 0.10 M BOH. Its pH is measured to be 10.11 to be 10.11
Calculate the equilibrium constant.Calculate the equilibrium constant.
Step 1.Step 1. Write the equation. Write the equation.
BOH <--> BBOH <--> B++ + OH + OH--
[OH][OH]-- = 10= 10-pOH-pOH = 10= 10-3.89-3.89 = 1.3 x 10= 1.3 x 10-4 -4 MM
107107
Equilibria Involving A Weak BaseEquilibria Involving A Weak Base
Step 2.Step 2. Make a chart. Make a chart.
BOH <--> BBOH <--> B++ + OH + OH--
0.100.10 0 0 0 0
- 0.00013
0.000130.000130.00013
0.000130.00013
00.10
108108Equilibria Involving A Weak BaseEquilibria Involving A Weak Base
Step 3.Step 3. Write K Write Kbb expression, without and expression, without and
with numbers.with numbers.
Kb = [B+][OHOH--]
[BOHBOH]
Kb =
(0.00013)(0.00013.00013)
(0.10)
109109Equilibria Involving A Weak BaseEquilibria Involving A Weak Base
Step 4.Step 4. Solve. Solve.
Kb = 1.7 x 10-7
Kb =
(0.00013)(0.00013.00013)
(0.10)
Step 5.Step 5. Answer question(s). Answer question(s).
Kb = 1.7 x 10-7
110110
Sample QuestionSample Question You have 0.0200 M B. You have 0.0200 M B.
Its is 3.0% dissociated. Its is 3.0% dissociated.
Calculate the equilibrium constant.Calculate the equilibrium constant.
B + HOH <--> HBB + HOH <--> HB++ + OH + OH--
0.02000.0200 0 0 0 0- 6.0 x 10- 6.0 x 10-4-4 + 6.0 x 10+ 6.0 x 10-4-4 + 6.0 x 10+ 6.0 x 10-4-4
6.0 x 106.0 x 10-4-4 6.0 x 106.0 x 10-4-40.01940.0194
Kb = [HB+ ][OH-]
[B]= = 1.9x10-5
(6.0 x 10-4)2
(0.0194)
111111
Equilibria Involving A Weak BaseEquilibria Involving A Weak Base
You have 0.100 M NHYou have 0.100 M NH33
Calculate all equilibrium concentrations Calculate all equilibrium concentrations and the pH.and the pH.
Step 1.Step 1. Write the equation. Write the equation.
NHNH33 + HOH <--> NH + HOH <--> NH44++ + OH + OH--
112112
Equilibria Involving A Weak BaseEquilibria Involving A Weak Base
Step 2.Step 2. Make a chart. Make a chart.
NHNH33 + HOH <--> NH + HOH <--> NH44++ + OH + OH--
0.1000.100 0 0 0 0
-x
xx
+x+x
0.100
113113Equilibria Involving A Weak BaseEquilibria Involving A Weak Base
Step 3.Step 3. Write K Write Kbb expression, without and expression, without and
with numbers.with numbers.
Kb = [NH4
+][OHOH--]
[NHNH33]
1.8 x 10-5 = (x)(xx)
(0.100)
114114Equilibria Involving A Weak BaseEquilibria Involving A Weak Base
Step 4.Step 4. Solve. Solve.
1.8 x 10-5 = (x)(xx)
(0.100)
x = 1.3 x 10-3 M
115115
Step 5.Step 5. Answer question(s). Answer question(s).
Equilibria Involving A Weak BaseEquilibria Involving A Weak Base
[NH[NH44++] = [OH] = [OH--] = 0.0013 M] = 0.0013 M
[NH[NH33] = 0.099 M] = 0.099 M
pOH = 2.89pOH = 2.89
pH = 11.11pH = 11.11
116116
Sample QuestionsSample Questions Calculate the pH of a 0.015 M B solution.Calculate the pH of a 0.015 M B solution.
KKbb = 2.5 x 10 = 2.5 x 10-7-7
B + HOH <--> HBB + HOH <--> HB++ + OH + OH--
0.015 0 00.015 0 0- x + x + x
x x0.015
Kb = [HB+ ][OH-]
[B]= = 2.5x10-7
x2
(0.015)
X = 6.5 x 10-5 M = [OH-] pH = 9.79
117117
1. The pH of a 0.10 M nicotinic acid solution is 1. The pH of a 0.10 M nicotinic acid solution is 2.92. Calculate K2.92. Calculate Kaa and the % ionization. and the % ionization.
Sample Problems: Weak Acids and BasesSample Problems: Weak Acids and Bases
HA <--> HHA <--> H++ + A + A--
0.100.10 0 0 0 0
- 0.0012 + 0.0012 + 0.0012
0.0012 0.00120.10
Ka = [H+][A-]
[HA]= = 1.4 x 10-5
(0.0012)2
(0.10)
% ionization = 0.0012/0.10 x 100 = 1.2%
118118
2. A solution of propionic acid, 2. A solution of propionic acid, CHCH33CHCH22COOH, is 0.20 M. KCOOH, is 0.20 M. Kaa = 1.3x10 = 1.3x10-5-5. .
Calculate the pH and % ionization.Calculate the pH and % ionization.
Sample Problems: Weak Acids and BasesSample Problems: Weak Acids and Bases
HA <--> HHA <--> H++ + A + A--
0.200.20 0 0 0 0
- x + x + x
x x0.20
Ka = [H+][A-]
[HA]= = 1.3 x 10-5
x2
(0.20)
% ionization = 0.0016/0.20 x 100 = 0.80%
x = 1.6 x 10-3 = [H+], pH = 2.80
119119
3. A solution of hydrazine, N3. A solution of hydrazine, N22HH44, is 0.025 M. , is 0.025 M.
KKbb = 8.5x10 = 8.5x10-5-5. Calculate the pH and % ionization.. Calculate the pH and % ionization.
Sample Problems: Weak Acids and BasesSample Problems: Weak Acids and Bases
NN22HH44 + HOH <--> N + HOH <--> N22HH55++ + OH + OH--
0.0250.025 0 00 0- x + x + x
x x0.025
Kb = [N2H5
+][OH-]
[N2H4]= = 8.5 x 10-5
x2
(0.025)
% ionization = 0.0015/0.025 x 100 = 6.0%
x = 1.5 x 10-3 = [OH-], pOH = 2.82 pH = 11.18
1201204. Calculate the pH and concentrations of the 4. Calculate the pH and concentrations of the arsenate containing species in a 0.15 M solution of arsenate containing species in a 0.15 M solution of HH33AsOAsO44. . KK11= 2.5x10= 2.5x10-4-4, K, K22 = =
5.6x105.6x10-8-8, K, K33 = 3.0x10 = 3.0x10-13-13
HH33AsOAsO44 <--> H <--> H++ + + HH22AsOAsO44
--
0.15 0 00.15 0 0- x + x + x
x x0.15 - x
Ka = [H+][H2AsO4
-]
[H3AsO4]= = 2.5 x 10-4
x2
(0.15 - x)
[H3AsO4 ] = 0.14 M [H2AsO4-] = 6.0 x 10-3 M
x = 6.0 x 10-3 M = [H+], pH = 2.22
1211214. Calculate the pH and concentrations of 4. Calculate the pH and concentrations of
the arsenate containing species in a the arsenate containing species in a 0.15 M solution of H0.15 M solution of H33AsOAsO44. .
KK11= 2.5x10= 2.5x10-4-4, K, K22 = 5.6x10 = 5.6x10-8-8, K, K33 = 3.0x10 = 3.0x10-13-13
HH22AsOAsO44-- <--> H <--> H++ + +
HAsOHAsO442-2-
6.0 x 106.0 x 10-3-3 6.0 x 10 6.0 x 10-3-3 0 0- y + y + y
y 6.0 x 10-36.0 x 10-3
Ka = [H+][HAsO4
2-]
[H2AsO4-]
= = 5.6 x 10-86.0 x 10-3 y
6.0 x 10-3
y = 5.6 x 10-8 M = [HAsO42-]
1221224. Calculate the pH and concentrations 4. Calculate the pH and concentrations
of the arsenate containing species in a of the arsenate containing species in a 0.15 M solution of H0.15 M solution of H33AsOAsO44. .
KK11= 2.5x10= 2.5x10-4-4, K, K22 = 5.6x10 = 5.6x10-8-8, K, K33 = 3.0x10 = 3.0x10-13-13
HAsOHAsO442-2- <--> H <--> H++ + AsO + AsO44
3-3-
5.6 x 105.6 x 10-8-8 6.0 x 106.0 x 10-3-3 0 0- z + z + z
z 6.0 x 10-35.6 x 10-8
Ka = [H+][AsO4
3-]
[HAsO42-]
= = 3.0 x 10-136.0 x 10-3 z
5.6 x 10-8
z = 2.8 x 10-18 M = [AsO43-]
123123
Determine if the following solutions are Determine if the following solutions are acidic, basic, or neutral.acidic, basic, or neutral.
KBrKBr
KK++ + HOH <--> KOH + H + HOH <--> KOH + H++
BrBr-- + HOH <--> HBr + OH + HOH <--> HBr + OH--
Acid-Base Properties of SaltsAcid-Base Properties of Salts
CrCl3
Cr3+ + HOH <--> CrOH2+ + H+
Cl- + HOH <--> HCl + OH-
XXXX
neutral
XXacidic
124124
Determine if the following solutions are Determine if the following solutions are acidic, basic, or neutral.acidic, basic, or neutral.
NaNONaNO22
NaNa++ + HOH <--> NaOH + H + HOH <--> NaOH + H++
NONO22-- + HOH <--> HNO + HOH <--> HNO22 + OH + OH--
Acid-Base Properties of SaltsAcid-Base Properties of Salts
KHCO3
K+ + HOH <--> KOH + H+
HCO3- + HOH <--> H2CO3 + OH- Kb = 2.4x10-8
HCO3- + HOH <--> CO3
2- + H3O+ Ka = 4.8x10-11
XXbasic
XX
basic
125125
Determine if the following solutions are Determine if the following solutions are acidic, basic, or neutral.acidic, basic, or neutral.
NaNa22COCO33
NaNa++ + HOH <--> NaOH + H + HOH <--> NaOH + H++
COCO332-2- + HOH <--> HCO + HOH <--> HCO33
-- + OH + OH--
Acid-Base Properties of SaltsAcid-Base Properties of Salts
NH4C2H3O2
NH4+ + HOH <--> NH3 + H3O+ Ka = 5.6x10-10
C2H3O2- + HOH<-->HC2H3O2 +OH- Kb =5.6x10-10
XXbasic
neutral
126126
Acid-Base Properties of SaltsAcid-Base Properties of Salts
You have 0.10 M NaYou have 0.10 M Na22COCO33
Calculate the pH.Calculate the pH.
Step 1.Step 1. Write the equation. Write the equation.
COCO332-2- + HOH <--> HCO + HOH <--> HCO33
-- + OH + OH--
127127
Acid-Base Properties of SaltsAcid-Base Properties of Salts
Step 2.Step 2. Make a chart. Make a chart.
COCO332-2- + HOH <--> HCO + HOH <--> HCO33
-- + OH + OH--
0.100.10 0 0 0 0
-x
xx
+x+x
0.10 - x
128128Acid-Base Properties of SaltsAcid-Base Properties of Salts
Step 3.Step 3. Write K Write Kbb expression, without and expression, without and
with numbers.with numbers.
Kb = [HCO3
-][OHOH--]
[COCO332-2-]
2.1 x 10-4 = (x)(xx)
(0.10 - x)
129129Acid-Base Properties of SaltsAcid-Base Properties of Salts
Step 4.Step 4. Solve. Solve.
2.1 x 10-4 = (x)(xx)
(0.10 - x)
x = 4.5 x 10-3 M
130130
Step 5.Step 5. Answer question(s). Answer question(s).
Acid-Base Properties of SaltsAcid-Base Properties of Salts
[OH[OH--] = 0.0045 M] = 0.0045 M
pOH = 2.35pOH = 2.35
pH = 11.65pH = 11.65
131131
Sample QuestionsSample Questions
1. Calculate the pH of a 0.10 M K1. Calculate the pH of a 0.10 M K33POPO44 solution.solution.
POPO443-3- + HOH <--> HPO + HOH <--> HPO44
2-2- + OH + OH--
0.100.10 0 0 0 0- x + x + x
x x0.10 - x
Kb = [HPO4
2- ][OH-]
[PO43- ]
= = 0.028 x2
(0.10 - x)
X = 0.041 pH = 12.61
132132
Sample QuestionsSample Questions
2. Calculate the pH of a 0.10 M Na2. Calculate the pH of a 0.10 M Na22CC22OO44 solution.solution.
CC22OO442-2- + HOH <--> HC + HOH <--> HC22OO44
-- + OH + OH--
0.100.10 0 00 0- x + x + x
x x 0.10
Kb = [HCC22OO44
- ][OH-]
[CC22OO442- ]
= =1.6x10-10 x2
(0.10)
X = 4.0 x 10-6 pH = 8.60
133133
Sample QuestionsSample Questions3. Calculate the pH of a 0.045 M KCN 3. Calculate the pH of a 0.045 M KCN
solution. solution.
CNCN-- + HOH <--> HCN + OH + HOH <--> HCN + OH --
0.045 0 00.045 0 0
- x + x + x
x x 0.045
Kb = [HCNCN][OH-]
[CNCN- ]= = 2.5x10-5
x2
(0.045)
x = 1.1 x 10-3 = [OH-] pOH = 2.96, pH = 11.04