ionic equlibrium type 1

24 Pages·2011·0.61 MB·English

ffffoooo////uuuu ffffooooppppkkkkjjjjrrrr HHHHkkkkhhhh#### ttttuuuu]]]] uuuugggghhhh aa aa vvvvkkkkjjjjEEEEHHHHkkkk ss ss ddddkkkkeeee]]]] ffffooooiiiiffffrrrr nnnn[[ss[[ss kkkk NNNNkkkkMMssMMss +s+s +s+s rrrrjjqqjjqq rraarraa eeee////;;;;eeee eeeeuuuu ddddjjjj '''';;;;kkkkeeeeAAAA
iiii##qq##qq """"kkkk ffffllllggaaggaa llllddaaddaa YYYYiiii ddddjjjj]]]] llllggggrrrr ss ss ffffooooiiiiffffrrrr vvvvuuuuddssddss ]]]] ^^^^ccccuuuukkkk^^^^ uuuu NNNNkkkkMMssMMss +s+s +s+s ////;;;;;;ss;;ss ddddkkkk]]ss]]ss jjjj????kkkkccqqccqq jjjj jjjjkkkk[[[[kkkk ss ss VVVVddssddss AAAAAAAA
jjjjffffpppprrrr%%%% eeeekkkkuuuuoooo ////kkkkeeeeZZZZ iiiizzzz....kkkksrsrsrsrkkkk
llllnnnn~~~~xxxxqqqq#### JJJJhhhh jjjj....kkkkNNNNkkkkssssMMMM++++nnnnkkkklllltttthhhh eeeeggggkkkkjjjjkkkktttt
STUDY PACKAGEThis is TYPE 1 Package
please wait for Type 2
Subject : CHEMISTRY
Topic : IONIC EQULIBRIUM
R
Index
1. Key Concepts
2. Exercise I
3. Exercise II
4. Exercise III
5. Exercise IV
6. Answer Key
7. 34 Yrs. Que. from IIT-JEE
8. 10 Yrs. Que. from AIEEE
Student’s Name :______________________
Class :______________________
Roll No. :______________________
ADDRESS: R-1, Opp. Raiway Track,
New Corner Glass Building, Zone-2, M.P. NAGAR, Bhopal
(cid:1)
: (0755) 32 00 000, 98930 58881, www.tekoclasses.com
THE KEY M
U
Fundamentals of Acids, Bases & Ionic Equilibrium
I
R
Acids & Bases B
I
L
When dissolved in water, acids release H+ ions, base release OH– ions. U
Q
Arrhenius Theory E
When dissolved in water, the substances which release C
I
(i) H+ ions are called acids (ii) OH− ions are called bases N
O
Bronsted & Lowry Concept 4 I
2
Acids are proton donors, bases are proton acceptors of
2
Note that as per this definition, water is not necessarily the solvent. e
g
When a substance is dissolved in water, it is said to react with water e.g. a
P
HCl + H O → H O+ + Cl− ; HCl donates H+ to water, hence acid.
2 3
NH + H O → NH + + OH− ; NH takes H+ from water, hence base. AL
3 2 4 3 P
For the backward reaction, NH + donates H+, hence it is an acid; OH− accepts H+, hence it is base. HO
m 4 B
s.coConjugNatHe 3a c(ibda saen)d & b aNsHes4+ (acid) from conjugate acid base pair. 81 ,
e 8
ss To get conjugate acid of a given species add H+ to it. e.g. conjugate acid of N H is N H +. 8
a 2 4 2 5 5
cl To get conjugate base of any species subtract H+ from it. e.g. Conjugate base of NH is NH −. 0
o 3 2 3
k Note: Although Cl− is conjugate base of HCl, it is not a base as an independent species. In fact, 9
w.te anions of all strong acid like Cl−, NO3−, ClO4− etc. are neutral anions. Same is true for cations 0 98
ww of strong bases like K+, Na+, Ba++ etc. When they are dissolved in water, they do not react 0,
with water (i.e. they do not undergo hydrolysis) and these ions do not cause any change in 0
0
: pH of water (others like CN− do). 0
e 0
sit Some examples of : 2
web Basic Anions : CH3COO−, OH−, CN− (Conjugate bases of weak acids) 5)- 3
m Acid Anions: HSO3−, HS− etc. Note that these ions are amphoteric, i.e. they can 75
fro HbeSh−a v +e HbotOh as a nS a2−c i+d HandO a+s a( bfuansect. ieo.ngi.n gfo ars H an2 PaOcid4−) : H: (0
e 2 3 P
g HS− + H O H S + OH− (functioning as a base) )
ka Acid Catio2ns : NH2+, H O+ etc.(Conjugate acids of weak bases) Sir
ac 4 3 K.
P Note : Acid anions are rare. R.
dyLewis Concept : Acids are substances which accept a pair of electrons to form a coordinate S.
tu bond and bases are the substances which donate a pair of electrons to A (
d S form a coordinate bond. RIY
a H F H F A
o K
wnl | | | | R.
oe.g. H − N: + B − F → H − N→ B − F G
D A
| | | | H
E U
E H F H F S
FR (Lewis base) (Lewis acid) or :
Important : Ca + S → Ca2+ + S2− is not a Lewis acid−base reaction since dative bond is not t
c
e
formed. r
Di
Lewis Acids : As per Lewis concept, following species can acts as Lewis Acids : S,
E
(i) Molecules in which central atom has incomplete octet. (e.g. BF , AlCl etc.) S
3 3 S
(ii) Molecules which have a central atom with empty d− orbitals (e.g. SiX , GeX , A
4 4 L
PX , TiCl etc.) C
3 4 O
(iii) Simple Cations: Though all cations can be expected to be Lewis acids, Na+, Ca++, K+ etc. K
E
show no tendency to accept electrons. However H+, Ag+ etc. act as Lewis acids. T
(iv) Molecules having multiple bond between atoms of dissimilar electronegativity. M
U
e.g. CO2, SO2, SO3 → (O=C=O+OH— → —O−C=O or HCO3—) RI
Lewis acid Lewis base | B
I
OH L
U
Q
Lewis bases are typically :
E
C
(i) Neutral species having at least one lone pair of electrons. NI
•• •• •• O
e.g. NH2−NH2, R−•O•−H 24 I
(ii) Negatively charged species (anions). e.g. CN−, OH−, Cl− etc. f
o
3
e
pH and pOH pH = −log [H O+], pOH = −log [OH−] ag
10 3 10 P
Note : * pH of very dilute (~ 10−8M or Lower) acids (or bases) is nearly 7 (not simply
L
−log[acid] etc. due to ionization of water. A
P
* pH of strong acids with concentration > 1M is never negative. It is zero only. O
H
m * At 25°C, if pH = 7, then solution is neutral, pH > 7 than solution is basic. B
s.coAutoprotolysis of water (or any solvent) 81 ,
e 8
ss Autoprotolysis (or self−ionization) constant (K ) = [H O+] [OH−] 8
a w 3 5
cl Hence, pH + pOH = pK at all temperatures 0
o w 3
k 9
w.teConditiAont 2 o5f° nCe,u Ktra =li t1y0 − 1[4H. 3KO+ ]i n=c [rOeaHse−s] w(foitrh w inactreera asse sino ltveemnpt)erature. Accordingly, the neutral 0 98
ww point of watWer (pH = 7 aWt 25°C) also shifts to a value lower than 7 with increase in 0,
0
temperature. 0
e: Important: K = 10−14 is a value at (i) 25°C (ii) for water only. If the temperature 00
bsit changes or if soWme other solvent is used, autoprotolysis constant will not be same. 32
weIonisation Constant 5)-
m * For dissociation of weak acids (eg. HCN), HCN + H O H O+ + CN− the equilibrium 075
o 2 3 (
fr [H O+][CN−] H:
e constant expression is written as K = 3 P
kag a [HCN] Sir)
ac * For the Polyprotic acids (e.g. H3PO4), sucessive ionisation constants are denoted K.
P by K1, K2, K3 etc. For H3PO4, R.
d Study K1 = [H3O[H+3]P[HO24P]O4−] K2 = [H3[OH+3]P[OH4P−O]24−] K3 =[H3[OHP+]O[P24−O]34−] RIYA (S.
a Similarly, K denotes basic dissociation constant for a base. A
o b K
wnl Also, pKa = −log10Ka, pKb = −log10Kb R.
o Some Important Results: [H+] concentration of G
D A
Case (i) A weak acid in water H
E U
E S
FR (a) if α= KCa is < 0.1, then [H+] ≈ Kac0 . tor :
c
e
r
(b) General Expression : [H+]=0.5(−K + K2+4K c ) Di
Similarly for a weak base, substitute [OH−] aand K ainsteada ofo [H+] and K respectively in ES,
b a S
these expressions. S
A
L
C
Case (ii)(a) A weak acid and a strong acid [H+] is entirely due to dissociation of strong acid O
K
E
T
(b) A weak base and a strong base [H+] is entirely due to dissociation of strong base M
Neglect the contribution of weak acid/base usually. U
I
R
Condition for neglecting : If c = concentration of strong acid, c = concentration of B
0 1 I
weak acid then neglect the contribution of weak acid if K ≤ 0.01 c 2/ c L
a 0 1 U
Q
Case (iii) Two (or more) weak acids E
Proceed by the general method of applying two conditions C
I
N
(i) of electroneutrality (ii) of equilibria. O
The accurate treatement yields a cubic equation. Assuming that acids dissociate to 4 I
2
a negligible extent [ i.e. c −x ≈ c ] [H+] = (K c + K c + ...+ K )1/2 f
0 0 1 1 2 2 w o
4
Case (iv) When dissociation of water becomes significant: e
g
Dissociation of water contributes significantly to [H+] or [OH−] only when for Pa
(i) strong acids (or bases) : 10−8M < c < 10−6M. Neglecting ionisation of water at
0 L
10−6M causes 1% error (approvable). Below 10−8M, contribution of acid (or base) can be neglected A
P
and pH can be taken to be practically 7. O
H
m Weak acids (or bases) : When K c < 10−12, then consider dissociation of water as well. B
s.coHYDROLYSIS a 0 81 ,
sse * Salts of strong acids and strong bases do not undergo hydrolysis. 88
a 5
kocl * dSiassltosl voef da, sittr doinssgo acciaitdess a tno dg wiveea NkH ba+s eios ngsi vaen da nN aHcid+i c+ sHoluOti o n. e N.gH. N +H H4ClO w+h.en 930
w.te Kh = [NH3][H3O+] / [NH4+] = Kw/K4b of conjugate4 base o2f NH4+ 3 3 0 98
ww Important! In general : Ka(of an acid)xKb(of its conjugate base) = Kw 0,
If the degree of hydrolysis(h) is small (<<1), h = K c . 0
h 0 0
: 0
website Otherwise h = −Kh+ K2c2h0+4Khc0 , [H+] = c0h 5)- 32 0
m * Salts of strong base and weak acid give a basic solution (pH>7) when dissolved in water, e.g. 75
0
o NaCN,CN− + H O HCN + OH− [OH−] = c h, h = K c (
fr 2 0 h 0 H:
e P
g * Salts of weak base and weak acid )
ka Assuming degree of hydrolysis to be same for the both the ions, Sir
ac K = K / (K .K ), [H+] = [K K /K ]1/2 K.
dy P Nohte: Ewxact tareatbment of this caasew is dbifficult to solve. So use this assumption in general cases. S. R.
u Also, degree of anion or cation will be much higher in the case of a salt of weak acid and weak base. This (
t A
S is because each of them gets hydrolysed, producing H+ and OH− ions. These ions combine to form Y
d RI
a water and the hydrolysis equilibrium is shifted in the forward direaction. A
o K
wnl Buffer Solutions are the solutions whose pH does not change significantly on adding a small quantity of R.
o strong base or on little dilution. G
D A
These are typically made by mixing a weak acid (or base) with its conjugate base (or acid). e.g. H
E U
E CH COOH with S
FR CH33COONa, NH3(aq) with NH4Cl etc. or :
t
c
If K for acid (or K for base) is not too high, we may write : e
a b r
Henderson's Equation Di
pH = pK + log {[salt] / [acid]} for weak acid with its conjugate base. S,
a E
or pOH = pK + log {[salt] / [base]} for weak base with its conjugate acid. S
b S
A
Important : For good buffer capacity, [salt] : [acid ratios should be as close to one as possible. In such a case, L
C
pH = pK . (This also is the case at midpoint of titration) O
a K
Buffer capacity = (no. of moles of acid (or base) added to 1L) / (change in pH) E
T
Indicators. Indicator is a substance which indicates the point of equivalence in a titration by undergoing a M
change in its colour. They are weak acids or weak bases. U
I
R
B
I
Theory of Indicators. The ionized and unionized forms of indicators have different colours. If 90 % or more ofL
U
a particular form (ionised or unionised) is present, then its colour can be distinclty seen.In general, for an indicator Q
E
which is weak acid, HIn H+ + In–, the ratio of ionized to unionized form can be determined from C
I
N
[In−] O
pH = pKa + log [HIn] 24 I
f
o
So, for detectable colour change, pH = pKa ± 1 e 5
This roughly gives the range of indicators. Ranges for some popular indicators are g
a
Table 1 : Indicators P
Indicators pH range Colour
L
A
acid medium basic medium P
O
Methyl Orange 3.1-4.4 pink yellow H
m B
s.co MLitemthuysl red 54..52--76..53 rreedd byelulleow 81 ,
sse Phenol red 6.8-8.4 yellow red 88
a 5
cl Phenolphathlene 8.3-10 colourless pink 0
ko Thymol blue 1.2-2.8 red yello 93
w.te 0 98
wwEquivalence point. The point at which exactly equivalent amounts of acid and base have been mixed. 0,
0
: Acid Base Titration. For choosing a suitable indicator titration curves are of great help. In a titration curve, 0 0
e 0
sit change in pH is plotted against the volume of alkali to a given acid. Four cases arise. 2
b 3
we(a) Strong acid vs strong base. The curve is almost vertical over the pH range 3.5-10. This abrupt change 5)-
m corresponds to equivalence point. Any indicator suitable. 75
0
fro(b) Weak acid vs strong base. Final solution is basic 9 at equivalence point. Vertical region (not so sharp) H: (
e lies in pH range 6.5-10. So, phenolphathlene is suitable. P
g )
ka(c) Strong acid vs weak base. Final solution acidic. Vertical point in pH range 3.8-7.2. Methyl red or Sir
ac methyl orange suitable. K.
P(d) Weak acid vs weak base. No sharp change in pH. No suitable indicator. R.
udyNote : at midpoint of titration, pH = pK , thus by pH measurements, K for weak acids (or K for weak bases) (S.
t a a b A
S can be determined. Y
oad Polyprotic acids and bases. Usually K2, K3 etc. can be safely neglected and only K1 plays a significant role. KARI
l
wnSolubility product (Ksp). For sparingly soluble salts (eg. Ag2C2O4) an equilibrium which exists is R.
o Ag C O 2Ag+ (aq.) C O 2– (aq.) G
D 2 2 4 2 4 A
Then K = [Ag+]2[C O 2–] H
E sp 2 4 U
EPrecipitation. Whenever the product of concentrations (raised to appropriate power) exceeds the solubility S
FR product, precipitation occurs. or :
t
c
Common ion effects. Suppression of dissociation by adding an ion common with dissociation products. e.g. e
r
Ag+ or C O 2– in the above example. Di
2 4 S,
Simultaneous solubility. While solving these problems, go as per general method i.e. E
S
S
(i) First apply condition of electroneutrality and A
L
(ii) Apply the equilibria conditions. C
O
K
E
T
THE ATLAS M
U
I
R
B
I
L
U
Q
E
C
I
N
O
4 I
2
f
o
6
e
g
a
P
L
A
P
O
H
m B
s.co 81 ,
e 8
ss 8
a 5
cl 0
o 3
k 9
w.te 0 98
ww 0,
0
0
: 0
e 0
sit 2
b 3
we 5)-
m 75
0
o (
fr H:
e P
g )
ka Sir
ac K.
P R.
dy S.
u (
t A
S Y
d RI
a A
o K
l
wn R.
o G
D A
H
E U
E S
FR or :
t
c
e
r
Di
S,
E
S
S
A
L
C
O
K
E
T
GLOSSARY M
U
Amphoteric substance. A molecule which can act both as an acid and as a base.
I
R
B
Autoprotolysis constant. The equilibrium constant for the reaction in which one solvent molecule loses a I
L
proton to another, as 2H O H O+ + OH–. U
2 3 Q
E
Amphiprotic solvent. A solvent which possesses both acidic and basic properties. C
I
N
Aprotic solvent. A solvent which is neither appreciably acidic or basic.
O
Bronsted acid. A substance which furnishes a proton. 4 I
2
f
Bronsted base. A substance which accepts a proton. 7 o
e
Buffer capacity. A measure of the effectiveness of a buffer in resisting changes in pH; the capacity is greater the ag
P
concentrations of the conjugate acid-base pair.
L
Buffer solution. A solution which contains a conjugated acid-base pair. Such a solution resists large changes in A
P
pH when H O+ or OH– ions are added and when the solution is diluted. O
3 H
m B
s.coCofh paorsgitei-vbea clhaanrcgee emquusatt eioqnua. lT thhee etoqtuaal tcioonn ceexnptrreastsioinng o tfh nee egleactitvroe ncehuatrrgaeli.ty principle; i.e., the total concentration 81 ,
e 8
ssCommon-ion effect. The effect produced by an ion, say from a salt, which is the same ion produced by the 8
a 5
cldissociation of a weak electrolyte. The "common" ion shifts the dissociation equilibrium in accordance with 0
o 3
w.tekLCeeCnhtraatel lmiere'st aplr aintcoimple. .A cation which accepts electrons from a ligand to form a complex ion. 0 989
wwConjugate acid-base pair. An acid-base pair which differ only by a proton, as HCl and Cl–. 0,
0
0
: Diprotic acid. An acid which furnishes two protons. 0
e 0
bsitElectrolyte. A compound which produces positive and negative ions in solution. Strong electrolytes are completely 32
wedissociated, whereas weak electrolytes are only partially dissociated. 5)-
m 75
Hydrolysis. An acid-base reaction of a cation or anion with water. 0
o (
e frIsoelectric point. The pH at which there is an exact balance of positive and negative charge on an amino acid. PH:
g )
kaIndicator. A visual acid-base indicator is a weak organic acid or base which shows different colors in the Sir
acmolecular and ionic forms. K.
dy PLigand. An anion or neutral molecule which forms a complex ion with a cation by donating one or more pairs of S. R.
uelectrons. (
t A
S Y
d Nonelectrolyte. A substance which does not dissociate into ions in solution. RI
a A
lopH. The negative logarithm of the hydrogen ion concentration. K
wn R.
opK. The negative logarithm of an equilibrium constant. G
D A
Polyprotic acid. An acid which furnishes two or more protons. H
E U
E S
FRRthae ningdei coaft aonr ±in 1d uicnaitt.or. That portion of the pH scale over which an indicator changes color, roughly the pK of or :
t
c
e
Salt. The product other than water which is formed when an acid reacts with a base; usually an ion solid. r
Di
Simultaneous equilibria. Equilibria established in the same solution in which one molecule or ions is a participant S,
E
in more than one of the equilibria. S
S
A
Solubility product constant, K . The constant for the equilibrium established between a slightly soluble salt L
sp C
and its ions in solution. O
K
Stability constant. The equilibrium constant for a reaction in which a complex is formed. Also called a formation E
T
constant.
EXERCISE I M
U
I
IONIZATION CONSTANTS AND pH R
B
Q.1.1 Calculate I
L
U
(i) K for H O (K = 10–14) Q
a 2 w
(ii) Kb for B(OH)4– , Ka (B(OH)3) = 6 × 10–10 C E
(iii) K for HCN , K (CN–) = 2.5 × 10–5 I
N
a b
O
Q.1.2 Calculate the ratio of degree of dissociation (α /α ) when 1 M acetic acid solution is diluted to 1 4 I
2 1 100 f 2
o
times. [Given K =1.8 × 10–5] 8
a e
g
a
Q.1.3 Calculate the ratio of degree of dissociation of acetic acid and hydrocyanic acid (HCN) in 1 M their P
respective solution of acids.[Given K =1.8×10−5; K =6.2×10−10]
L
a(CH3COOH) a(HCN) A
P
O
Q.1.4 Calculate : H
m B
(a) K for a monobasic acid whose 0.10 M solution has pH of 4.50.
s.co(b) Kab for a monoacidic base whose 0.10 M solution has a pH of 10.50. 81 ,
e 8
ssQ.1.5 Calculate pH of following solutions : 8
a 5
cl (a) 0.1 M HCl (b) 0.1 M H SO (50 ml) + 0.4 M HCl 50 (ml) 0
o 2 4 3
w.tek ((ce)) 100.1– 8M M C HHC3lCOOH (Ka= 1.8 × 10–5) ((fd)) 100.1–1 M0 M N NHa4OOHH (Kb= 1.8 × 10–5) 0 989
ww (g) 10–6 M CH3COOH (h) 10–8 M CH3COOH 0,
(i) 0.1 M HA + 0.1 M HB [ Ka (HA) = 2 × 10–5 ; Ka (HB) = 4 × 10–5 ] 00
: (j) Decimolar solution of Baryta (Ba(OH) ), diluted 100 times. 0
site (k) 10–3 mole of KOH dissolved in 100 L2 of water. 2 0
b 3
we (l) 0.5 M HCl (25 ml) + 0.5 M NaOH (10 ml) + 40 ml H2O 5)-
m (m) equal volume of HCl solution (PH = 4) + 0.0019 N HCl solution 75
0
o (
e frQ.1.6 pTohien tv aolfu we aotfe Kr awt atht itsh ete pmhpyesrioaltougreic, awl hteemrep tehrearteu raer e( 3e7q°uCal) nisu 2m.5b6er × o 1f 0H–+1 4a.n Wd hOaHt i−s? the pH at the neutral PH:
g )
ka Sir
acQ.1.7 Calculate the number of H+ present in one ml of solution whose pH is 13. K.
PQ.1.8 Calculate change in concentration of H+ ion in one litre of water, when temperature changes from 298 K R.
dy to 310 K. Given K (298) = 10–14 K (310) = 2.56 × 10–14. S.
tu w w A (
S Y
d Q.1.9 RI
loa(i) Kw for H2O is 9.62 × 10–14 at 60°C. What is pH of water at 60°C. KA
wn(ii) What is the nature of solution at 60°C whose R.
o (a) pH = 6.7 (b) pH = 6.35 G
D A
H
EQ.1.10pH of a dilute solution of HCl is 6.95. Calculate molarity of HCl solution. U
E S
FRQ.1.11 The pH of aqueous solution of ammonia is 11.5. Find molarity of solution. K (NH OH) = 1.8 × 10–5. or :
b 4 t
c
e
Q.1.12The solution of weak monoprotic acid which is 0.01 M has pH = 3. Calculate K of weak acid. r
a Di
S,
Q.1.13Boric acid is a weak monobasic acid. It ionizes in water as E
S
S
B(OH) + H O B(OH)− + H+ : K = 5.9 × 10–10 A
3 2 4 a L
C
Calculate pH of 0.3 M boric acid. O
Q.1.14Calculate [H+] and [CHCl COO−] in a solution that is 0.01 M in HCl and 0.01 M in CHCl COOH. K
E
2 2
Take (K = 2.55 × 10–2). T
a
Q.1.15Calculate the percent error in the [H O+] concentration made by neglecting the ionization of water in a M
3
10–6M NaOH solution. U
I
R
Q.1.16Calculate [H+], [CH COO–] and [ C H O –] in a solution that is 0.02 M in acetic acid and 0.01M B
3 7 5 2 I
in benzoic acid. K (acetic) = 1.8 × 10–5 , K (benzoic) = 6.4 × 10–5. L
a a U
Q
Q.1.17At 25°C , the dissociation constant of HCN and HF are 4 × 10–10 and 6.7 × 10–4. Calculate the pH ofE
C
a mixture of 0.1 M HF and 0.1 M HCN. I
N
O
POLYPROTIC ACIDS & BASES 4 I
2
Q.2.1 Determine the [S2−] in a saturated (0.1M) H S solution to which enough HCl has been added to produce f
a [H+] of 2 × 10−4 . K1 = 10−7 , K2 = 10−124. ge 9 o
a
Q.2.2 Calculate [H+], [H PO − ], [HPO 2−] and [PO 3−] in a 0.01M solution of H PO . P
2 4 4 4 3 4
Take K = 7.225 × 10−3, K = 6.8 × 10−8 , K = 4.5 × 10−13.
1 2 3 AL
P
Q.2.3 Calculate the pH of a 0.1M solution of H NCH CH NH ; ethylenediamine (en). Determine the O
2 2 2 2 H
s.com 7en.1 H ×2 12+0.– C8 roenscpeencttriavteiloyn. in the solution. Kb1 and Kb2 values of ethylenediamine are 8.5 × 10–5 and 81 , B
e 8
assQ.2.4 What are the concentrations of H+, HSO−,SO2−and H SO in a 0.20 M solution of sulphuric acid ? 58
cl 4 4 2 4 0
o 3
www.tek Given : HH2SSOO−44 → H H+ ++ +SHOS42O− −4; K; s2t =ro 1n.g3 × 10–2 M 0, 0 989
Q.2.5 What are the concentration of H+, H C O , HC O− and C O2− in a 0.1 M solution of oxalic acid ? 00
: 2 2 4 2 4 2 4 0
site [K1 = 5.9 ×10–2 M and K2 = 6.4 × 10–5 M ] 2 0
b 3
weQ.2.6 Nicotine, C H N , has two basic nitrogen atoms and both can react with water to give a basic solution 5)-
m 10 14 2 75
Nic (aq) + H O (l) NicH+ (aq) + OH– (aq) 0
o 2 (
fr NicH+ (aq) + H O (l) NicH 2+ (aq) + OH– (aq) H:
kage Kb1 is 7 × 10–7 and Kb2 i2s 1.1 × 10–10. C2alculate the approximate pH of a 0.020 M solution. Sir) P
acQ.2.7 Ethylenediamine, H N–C H –NH , can interact with water in two steps, giving OH– in each step. K.
P Calculate the concen2tratio2n o4f OH2– and [H N–C H –NH ]2+ in a 0.15 M aqueous solution of the R.
udy amine. K = 8.5 × 10–5, K = 2.7 × 10–8 for3 the b2ase4. 3 (S.
t 1 2 A
S Y
ad BUFFER SOLUTION ARI
o K
l
wnQ.3.1 Determine [OH–] of a 0.050 M solution of ammonia to which has been added sufficient NH Cl to make R.
o 4 G
D the total [NH+] equal to 0.100.[K =1.8 × 10–5] A
E 4 b(NH3) UH
E S
FRQ.3.2 Calculate the pH of a solution prepared by mixing 50.0 mL of 0.200 M HC2H3O2 and 50.0 mL of or :
0.100 M NaOH.[K =1.8 × 10–5] t
a(CH COOH) c
3 e
r
Di
Q.3.3 A buffer of pH 9.26 is made by dissolving x moles of ammonium sulphate and 0.1 mole of ammonia into
S,
100 mL solution. If pK of ammonia is 4.74, calculate value of x. E
b S
+ S
Q.3.4 50 mL of 0.1 M NaOH is added to 75 mL of 0.1 M NH Cl to make a basic buffer. If pK of NH is A
4 a 4 L
C
9.26, calculate pH. O
K
E
T
Q.3.5
M
(a) Determine the pH of a 0.2 M solution of pyridine C H N . K = 1.5 × 10−9 U
5 5 b RI
(b) Predict the effect of addition of pyridinium ion C H NH+ on the position of the equilibrium. Will the pH B
5 5 I
be raised or lowered ? L
U
(c) Calculate the pH of 1.0 L of 0.10 M pyridine solution to which 0.3 mol of pyridinium chloride C H NH+Cl, Q
5 5 E
has been added, assuming no change in volume. C
I
N
Q.3.6 Calculate the pH of a solution which results from the mixing of 50.0 ml of 0.3 M HCl with 50.0 ml ofO
0.4 M NH . [K (NH ) = 1.8 × 10−5] 4 I
3 b 3 2
f
o
Q.3.7 Calculate the pH of a solution made by mixing 50.0 ml of 0.2M NH4Cl & 75.0 ml of 0.1 M NaOH. 10
[ K (NH ) = 1.8 × 10 −5 ] e
b 3 g
a
P
Q.3.8 A buffer solution was prepared by dissolving 0.02 mol propionic acid & 0.015 mol sodium propionate
in enough water to make 1.00 L of solution .(K for propionic acid is 1.34 × 10−5) AL
a P
(a) What is the pH of the buffer? O
H
m(b) What would be the pH if 1.0 × 10−5 mol HCl were added to 10 ml of the buffer ? B
es.co((cd)) WAlhsoa tr ewpoourltd t hbee ptehrec penHt cifh 1an.0g e× i1n0 p−H5 mofo ol rNigaiOnaHl b wueffreer aind dcaesde tso ( b1)0 a mndl o(cf )t.he buffer. 881 ,
ss 8
a 5
koclQ.3.9 cAh sloorluotaicoent awtaes C mlCadHe uCpO tOo Nbea 0 .. W01h Mat iins [cHhl+o]r oina ctheeti cso alcuitdio, Cn l?C HK2 C=O 1O.5H × a1n0d− 3a.lso 0.002 M in sodium 930
w.te 2 a 0 98
ww INDICATORS 0,
0
: Q.4.1 A certain solution has a hydrogen ion concentration 4 × 10−3 M. For the indicator thymol blue, pH is 2.0 0 0
e 0
sit when half the indicator is in unionised form. Find the % of indicator in unionised form in the solution with 2
b [H+] = 4 × 10−3 M. 3
we 5)-
m Q.4.2 At what pH does an indicator change colour if the indicator is a weak acid with K = 4 × 10−4. For 75
ind 0
o which one(s) of the following neutralizations would the indicator be useful ? Explain. (
fr H:
ge (a) NaOH + CH3COOH (b) HCl + NH3 (c) HCl + NaOH ) P
kaQ.4.3 What indicator should be used for the titration of 0.10 M KH BO with 0.10 M HCl ? Sir
ac K (H BO ) = 7.2 × 10−10 . 2 3 K.
P a 3 3 R.
udyQ.4.4 Bromophenol blue is an indicator with a Ka value of 6 × 10−5 . What % of this indicator is in its basic (S.
t form at a pH of 5 ? A
S Y
load Q.4.5 Ais nb laucei.d B bya hseo win dmicuactho rm huasst ath Ke ap oHf c3h ×an 1g0e− i5n. Torhdee ar ctoid c fhoarnmg eo ft hteh ein idnidciactaotro fro irsm re 7d5 &% t rheed btaos 7ic5 f%or bmlue? KARI
wn R.
o G
D HYDROLYSIS A
H
E U
EQ.5.1 What is the OH− concentration of a 0.08 M solution of CH COONa. [K (CH COOH)=1.8 × 10−5] S
FR 3 a 3 or :
t
Q.5.2 Calculate the pH of a 2.0 M solution of NH Cl. [K (NH ) = 1.8 × 10−5] c
4 b 3 e
r
Di
Q.5.3 0.25 M solution of pyridinium chloride C H N+Cl− was found to have a pH of 2.699. What is K for S,
5 6 b E
pyridine, C H N ? S
5 5 S
Q.5.4 Calculate the extent of hydrolysis & the pH of 0.02 M CH COONH . A
3 4 L
[K (NH )= 1.8 × 10−5, K (CH COOH)=1.8 × 10−5] C
b 3 a 3 O
K
Q.5.5 Calculate the percent hydrolysis in a 0.06 M solution of KCN. [K (HCN) = 6 × 10−10] E
a T