CHEM*131 (W 02)
PART A
QUESTION 3
REVIEW QUESTIONS FOR MIDTERM EXAM
Given:
hypochlorous acid (HClO, a weak acid)
Ka (HClO) = 3.0 × 10!8 M (at 25oC)
NIE:
HClO(aq) + H2O(R) º ClO !(aq) + H3O+(aq)
[ClO !] [H3 O %]
Kc '
Kc '
ˆ
' K a (HClO) ' 3.0 × 10!8 M
[HClO]
º HClO(aq) + H2O(RR)
(a) H3O+(aq) + ClO ! (aq)
Kc '
PAGE - 1
[HClO]
[H3 O %] [ClO !]
1
K a (HClO)
1
K a (HClO)
'
'
1
!8
3.0 × 10
H3O+(aq) + ClO !(aq)
!
' 3.3 × 107 M ! 1
M
(LARGE)
HClO(aq) + H2O(R)
(H3O+(aq), being a strong acid, drives the reaction to completion.)
º HClO(aq) + OH! (aq)
(b) ClO ! (aq) + H2O(RR)
Kc '
[HClO] [OH !]
[ClO !]
1.0 × 10!14 M 2
Kc ' Kb (ClO !) '
(c) HClO(aq) + OH! (aq)
Kc '
Kc '
ˆ
!8
3.0 × 10
[HClO] [OH !]
!
Kb (ClO )
K a (HClO)
' 3.3 × 10!7 M
M
º H2O(RR) + ClO ! (aq)
[ClO !]
1
Kw
' Kb (ClO !) '
'
'
1
K b (ClO !)
3.0 × 10!8 M
!14
1.0 × 10
HClO(aq) + OH!(aq)
!
M
2
'
K a (HClO)
Kw
' 3.0 × 106 M !1
H2O(R) + ClO !(aq)
(OH!!(aq), being a strong base, drives the reaction to completion.)
(LARGE)
CHEM*131 (W 02)
REVIEW QUESTIONS FOR MIDTERM EXAM
To write:
To calculate:
PART A
QUESTION 4
PAGE - 2
a balanced net ionic equation
equilibrium constant (Kc)
(a) HBr(aq) + Na2CO3! NaHCO3 buffer [Ka(HCO3!): 5.6 × 10!11]
H3O+ (aq)
HBr(aq):
2!
Na 2CO3(aq):
CO3 (aq)
NaHCO3(aq): HCO3!(aq)
[HCO3!]
Kc '
%
[H3 O ]
'
[CO32!]
(b) C6H5NH2(aq) + HCl(aq)
C6H5NH2(aq):
HCl(aq):
º
HCO3!(aq) + H2O(R)
1
'
!11
' 1.8 × 1010
5.6 × 10
[K b(C 6H5NH2): 3.8 × 10!10]
Cl!(aq)
(spectator ion)
C6H5NH2(aq) + H3O+(aq) º C6H5NH3+(aq) + H2O(R)
[C6 H5 NH3%]
'
[C6 H5 NH2 ] [H3 O %]
Kc '
K a (HCO3!)
Na (aq)
(spectator ion)
Na + (aq)
(spectator ion)
(organic weak base)
(strong acid)
H3O+ (aq)
WB + SA:
Kc '
1
(spectator ion)
+
(conjugate weak base)
(conjugate weak acid)
H3O+(aq) + CO32!(aq)
SA + WB:
Br!(aq)
(strong acid)
K b (C6 H5 NH2 )
'
Kw
1
Ka (C 6 H5 NH3%)
3.8 × 10!10
!14
1.0 × 10
'
Kb (C6 H5 NH2 )
Kw
' 3.8 × 104
! CH3NH2 buffer [Kb(CH3NH2): 4.2 × 10!4]
(c) Ca(OH)2(aq) + CH3NH3Cl!
Ca(OH)2(aq): OH!(aq)
CH3NH3Cl(aq):
+
CH3NH3 (aq)
CH3NH2(aq):
!
!
!
(organic weak acid)
Cl (aq)
(spectator ion)
(spectator ion)
(conjugate weak base, organic amine molecule)
OH!(aq) + CH3NH3+(aq) º CH3NH2(aq) + H2O(R)
SB + WA:
Kc '
Ca2+(aq)
(strong base)
[CH3 NH2 ]
!
[OH ]
[CH3 NH3%]
'
1
Kb (CH3 NH2 )
'
1
!4
' 2.4 × 103
4.2 × 10
In all cases, the large Kc values indicate the completion of the reaction due to the presence of either
strong base, (Ca(OH)2(aq)) or strong acids (HBr(aq) or HCl(aq)).
ACID must react with BASE, neither SA reacts with WA nor SB reacts with WB.
CHEM*131 (W 02)
REVIEW QUESTIONS FOR MIDTERM EXAM
PART A
QUESTION 5
PAGE - 3
Î [HC7H5O3] = 2.00 × 10!2 M
(salicylic acid, a weak acid)
!3
Ï Ka [HC7H5O3] = 1.49 × 10 M (at 25oC)
Ð [NaOH] = 2.00 × 10!2 M
(a strong base)
Given:
To find: pH at the equivalence point
HC7H5O3(aq)
º
OH!(aq)
+
!2
!2
I: 2.00 × 10 M
C: s 2.00 × 10!2 M
!
E:
2.00 × 10 M
s 2.00 × 10!2 M
!
C7H5O3!(aq) +
H2O(R)
!
r 2.00 × 10!2 M
2.00 × 10!2 M ÷ 2
!
= 1.00 × 10!2
M (‡ volume double)
At equivalence point,
All HC7H5O3(aq) is consumed and converted into C7H5O3!(aq).
(i.e. Theoretically no more HC 7H5O3(aq) LEFT, but only C7H5O3!(aq) is present.)
Since C7H5O3!(aq) is a weak base, which only partially ionizes in water until the equilibrium is reached.
(1)
(2)
C7H5O3!(aq) +
I:
C:
E:
º
H2O(R)
HC7H5O3(aq)
!
1.00 × 10!2
M
sx
!2
(1.00 × 10 ) ! x
where
Kc '
+
OH!(aq)
!
!
rx
x
rx
x
x = [OH!] produced = [HC7H5O3] produced = [C7H5O3!] ionized
[HC7 H5 O3 ] [OH !]
!
[C7 H5 O3 ]
Kw
!
' Kb (C 7 H5 O3 ) '
(x) (x)
(1.00 × 10!2 ) ! x
'
K a (HC7 H5 O3 )
1.00 × 10!14 M 2
1.49 × 10!3 M
x = 2.591 × 10!7 M = [OH!]
pOH = ! log [OH ]
!
ˆ
pOH = ! log (2.591 × 10!7) = 6.5866
pH at the equivalence point = 14.000 ! pOH = 14.000 ! 6.5866 = 7.413
CHEM*131 (W 02)
REVIEW QUESTIONS FOR MIDTERM EXAM
PART A
QUESTION 5
Salicylic acid (WA) + NaOH (SB)
PAGE - 4
Best INDICATOR chosen for the titration should have: pKIn – pH at equivalence point
Choose the indicator whose pKIn is close to 7.413 (i.e. pH at the equivalence point)
1.
methyl red
(MR, pKIn 4.9)
ÿ not suitable (color change far before equivalence point)
2.
bromothymol blue
(BB, pKIn 7.1)
ÿ best (very close to pH at equivalence point)
3.
phenolphthalein
(Ph, pKIn 9.3)
ÿ O.K. (colour change after equivalence point)
pH
Ph (pKIn 9.3)
BB (pK In 7.1)
n(HC 7 H 5 O3 ) = n(C7 H 5 O3 -)
WA = WB
pH = pK a
1:1 Buffer
equivalence point
(pH 7.413)
(or endpoint of HIn)
n(HC 7 H 5 O3 ) = n(OH- )
WA = SB
C 7 H 5O 3 - only
MR (pK In 4.9)
half-equivalence point
(pH 2.827)
1.00 x 10-2
-2
2.00 x 10
NaOH (M)
In general,
Î WA + SB:
Ï SA + SB:
Ð WB + SA:
bromothymol blue (pKIn 7.1); phenolphthalein (pKIn 9.3)
methyl red (pKIn 4.9);
bromothymol blue (pKIn 7.1); phenolphthalein (pKIn 9.3)
methyl red (pKIn 4.9); methyl orange (pKIn 3.4)
CHEM*131 (W 02)
REVIEW QUESTIONS FOR MIDTERM EXAM
PART A
QUESTION 6
Given:
Î (CH3)3N(aq) = 0.100 M
Ï Kb (CH3)3N(aq) = 6.1 × 10!5 M
Ð [HClO 4(aq)] = 0.100 M = [H 3O+]
PAGE - 5
(a weak base)
(a strong acid)
(CH3)3N(aq) + H3O+(aq) º (CH3)NH+(aq) + H2O(R)
WB + SA
Kc '
[(CH3 )3 NH %]
[(CH3 )3 N] [H3 O %]
1
'
Ka [(CH3 )3 NH %]
(a) pH at half-equivalence point (or half-neutralization point)
At half-equivalence point, pH = pKa [(CH3)3NH+]
Kw
K a [(CH3 )3 NH %] '
Kb [(CH3 )3 N]
pKa = ! log Ka
ˆ
'
1.0 × 10!14 M 2
6.1 × 10!5 M
' 1.64 × 10!10 M
pKa = ! log Ka = ! log (1.64 × 10!10) = 9.79
pH at half-equivalence point = pKa [(CH3)3NH+] = 9.79
pH
half-equivalence point
complete neutralization
9.79
WB = SA
half-neutralization
5.54
WB = WA
[(CH3)3N] = [(CH3)3NH+]
pH = pKa
1:1 Buffer
0.100 M HClO4
[(CH3)3N] = [H3O+]
~100% (CH ) NH+
33
equivalence point
CHEM*131 (W 02)
REVIEW QUESTIONS FOR MIDTERM EXAM
PAGE - 6
PART A
QUESTION 6
(b) pH at equivalence point
(1)
All Weak Base (CH3)3N(aq) are theoretically converted into Weak Acid (CH3)3NH+ (aq).
(2)
(3)
Since (CH)3NH+ (aq) is a weak acid, which only partially dissociates in water until the equilibrium is reached.
Since the volume of the solution is doubled, the concentration of (CH3)3NH+ (aq) is reduced by half to 0.0500 M.
(CH3)3NH+(aq)
+
H2O(R)
I: 0.100 M / 2 = 0.0500 M
C:
sx
E: (0.0500 ! x) M
º
H3O+(aq)
(CH3)3N(aq) +
!
!
rx
x
rx
x
x = [H3O+] produced = [(CH3)3N] produced = [(CH3)3NH+] dissociated
where
Kc '
[(CH3 )3 N] [H3 O %]
[(CH3 )3 NH %]
' K a [(CH3 )3 NH %] '
(x) (x)
(0.0500 ! x) M
'
Kw
Kb [(CH3 )3 N]
1.0 × 10!14 M 2
6.1× 10!5 M
x = 2.86 × 10!6 M = [H3O+]
pH = ! log [H3O+]
ˆ
pH at equivalence point = ! log (2.86 × 10!6) = 5.543 = 5.54
CHEM*131 (W 02)
REVIEW QUESTIONS FOR MIDTERM EXAM
PART A
QUESTION 7
Given:
There are only
(a) WA (e.g.
(b) WB (e.g.
(c) WA (e.g.
three different methods to prepare buffer:
HA)
+
SB (e.g. NaOH)
A !)
+
SA (e.g. HCl)
HA)
+
WB (e.g. A !)
0.10 M aqueous solutions available
formic acid
(HCOOH)
sodium formate
(HCOONa)
perchloric acid
(HClO 4)
sodium hydroxide
(NaOH)
(a) WA + SB:
NIE:
Kc '
!
PAGE - 7
HCOOH(aq)
HCOO!(aq)
H3O+(aq)
OH!(aq)
weak acid
weak base
strong acid
strong base
(WA)
(WB)
(SA)
(SB)
formic acid + sodium hydroxide
º
HCOOH(aq) + OH!(aq)
[HCOO !]
'
[HCOOH] [OH !]
HCOO!(aq) + H2O(R)
1
'
K b (HCOO !)
K a (HCOOH)
Kw
A strong base, OH!(aq) (being a limiting reagent), converts some weak acid, HCOOH(aq), into its
conjugate base buffer component, HCOO!(aq).
(b) WB + SA:
NIE:
sodium formate + perchloric acid
Kc '
!
º
HCOO!(aq) + H3O+(aq)
[HCOOH]
!
%
[HCOO ] [H3 O ]
'
HCOOH(aq) + H2O(R)
1
K a (HCOOH)
A strong acid, H3O+ (aq) (being a limiting reagent), converts some weak base, HCOO!(aq), into its
conjugate acid buffer component, HCOOH(aq).
(c) WA + WB: formic acid + sodium formate
NIE:
HCOOH(aq) + H2O(R) º HCOO!(aq) + H3O+(aq)
OR
HCOO!(aq) +
H2O(R) º
Kc = Ka (HCOOH)
OR
HCOOH(aq) + OH!(aq)
Kc = Kb (HCOO!)
CHEM*131 (W 02)
REVIEW QUESTIONS FOR MIDTERM EXAM
PAGE - 8
PART A
QUESTION 8
Given:
[HNO2] = 0.10 M
[NaNO2] = 0.10 M
(HNO2 is a weak acid)
(NO2! is its conjugate base)
Ka (HNO2) = 4.6 × 10!4
ˆ pKa (HNO2) = ! log Ka = ! log (4.6 × 10!4) = 3.337
To find: ratio of NO2! : HNO2 at pH = 3.0
pH ' pKa % log
[WB]
[WA]
pH ' pKa (HNO2 ) % log
3.0 ' 3.337 % log
!0.337 ' log
ˆ
[NO2! ]
[HNO2 ]
[HNO2 ]
[NO2!]
[HNO2 ]
[NO2!]
[HNO2 ]
' 0.46 '
0.46
1
'
46
100
'
46
46 % 100
× 100% ' 31.5%
% HNO2 '
100
46 % 100
× 100% ' 68.5%
!
% NO2
ˆ
[NO2!]
ratio of NO 2! : HNO2 = 0.46 : 1.0