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Chapter 3
Ionic Equilibrium Gilbert Newton Lewis
Born
23 October 1875
Died
23 March 1946
Nationality
American
Field
Chemistry
Known for
Acid, Base Theory
There is no scientist in American history who has contributed more extensively to all fields in chemistry than Gilbert Newton Lewis. His thinking was far ahead of his time and his theories have had profound influence on modern chemistry Lewis researched standard electrode potentials, conductivity, free energy and other thermodynamic constants for the elements. These tables are still being used. His ability to organize and apply the scattered laws of thermodynamics brought about the evolution of physical chemistry into the science as it is known today. Lewis once defined physical chemistry as encompassing "everything that is interesting. Lewis work on acid base theory is most generalized theory. Lewis did not believe only that an electron completely transfers from one atom to another, as in the positive-negative theory. He describes the partial transfer of two electrons, one from each of the two bonding atoms, so that there is a shared pair of electrons between them. This eliminates the need for the formation of oppositely charged atoms when there was no indication of individually charged atoms (ions) in a compound. This was the first description of covalent bonding. Lewis' research on isotopes is an example of his wide-ranging and prolific interests. He published twentysix papers on heavy hydrogen and heavy water, isotopes of lithium, and neutron physics. He predicted the existence of naturally occurring heavy water before he isolated it.
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Chapter-3 / Ionic Equilibrium
3.1 Concepts of Acid and Base Arrhenius concept Acid is a substance which is capable of furnishing H+ ions in aqueous solution. e.g. - HCl, H2SO4 etc. While base is a substance which furnishes OH– ions e.g. - NaOH, NH4OH etc. Limitations. Arrhenius theory failed to explain. (i)
behaviour of acids/bases in non aqueous solutions.
(ii)
neutralization reaction giving salt in absence of a solvent.
(iii)
acid character of certain salts like AlCl3,BF3 etc.
(iv)
existence of H+ in water.
Bronsted Lowery Concept Acid is a substance which is capable of donating a proton while base is a substance which is capable of accepting a proton. This is also called protonic theory of acids-bases. HCl
H2O
acid
base
conjugate acid
H2O +
NH3
NH +4
acid
base
↽⇀
↽⇀
H3O+
+
Cl– conjugate base
+
conjugate acid
OH– conjugate base
conjugate base of an acid is a species formed by the loss of a proton from acid. Acid → H+ + conjugated base Similarly conjugate acid is formed from a base by gain of H+. Base + H+ → conjugated acid Weak acid has a strong conjugate base and vice-versa. A Bronsted - Lowery acid base reaction always proceeds in the direction from the stronger to the weaker acid base combination. e.g. HI
+
OH–
→
H2O
+
I–
Strong
Strong
Weak
Weak
acid
base
acid
base
Lewis concept Acid is a substance which can accept a pair of electrons while base is a substance which can donate a pair of electrons. Hence Lewis acids are (i)
Molecules in which central atom has incomplete octet e.g. BF3, AlCl3 and FeCl3 etc.
(ii)
Simple cations like Ag+, H+ etc.
(iii) Molecules in which central atom has vacant d-orbitals e.g.- SiF4, SnCl4 etc. (iv)
Molecules in which atoms of different electronegativities are joined by multiple bond. e.g. CO2, SO3 etc.
(v)
In carbonyl complexes, metal atoms act as Lew’s acids e.g. Ni in Ni(CO)4 etc.
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And Lewis bases are (i)
Neutral molecules like NH3, RNH2 etc.
(ii)
Negatively charged anions. e.g. CN–, Cl– etc.
(iii) Molecules with carbon-carbon multiple bonds can act as lewis base in some cases - e.g. C 2 H 4 in Ag ( C2 H 4 )
(iv)
+
In complex compounds, the ligands act as Lewis bases e.g. CO in Ni(CO)4 etc.
3.2 Strong and weak electrolytes Electrolytes which dissociate completely into ions in aqueous solutions are called strong electrolytes e.g. HCl, H2SO4, NaCl etc. Whereas electrolytes which dissociate to a lesser extent are called weak electrolytes eg. CH3COOH, NH4OH etc.
3.3 Ionisation of Weak Acids Let us consider the ionization of weak acid HA, having initial concentration ‘c’ in mol/litre H+ + A−
↽⇀
HA c c(1–x )
0 cx
0 cx
At equilibrium
It Ka is ionization constant for acid, then Ka =
[H + ][A − ] [HA] cx.cx cx 2 = c(1 − x) (1 − x)
=
for very weak acids. x < < 1 ∵
Ka =
cx 2 (∴ 1 − x ≃ 1) 1
or
x=
Ka C
or
x=
Ka × V n c = n V
or
x α V (Ostwald’s dilution law)
From above calculation, we can calculate Ka = Ka × c c
(i)
[H+ ] = cx = c
(ii)
[A − ] = cx = Ka × c
(iii)
[HA] = c(1 − x) ≅ c (x 7
(c) < 7
(d) 0
Ans. (c) 3.
If temperature is raised degree of dissociation of water will (a)
increase
(b) decrease
(c) remain same
(d) none of these
(c) expand
(d) none of these
(c) lesser than 14
(d) depends on temp.
Ans. (d) 4.
On increasing temperature, PH range will (a)
remain same
(b) shrink
Ans. (b) 5.
Value of PH + POH at any temp. is (a)
always 14
(b) greater than 14
Ans. (d) Passage – 2 PH value of solution is – log [H+]. If solution is very dilute than, concentration of H+ from water should also be considered. While at very higher concentration (above 1M). It should be taken O. 1.
PH value of 10–5 N – H2SO4 is (a)
5
(b) 5/2
(c) 5 × 2
(d) 4.7
(c) Just above 7
(d) Just below 7
Ans. (d) 2. PH value of 10–8M – HCl should be (a)
8
(b) 6
Ans. (d) 3.
If 100 ml of 10–4M – HCl and 200 ml of 10–4M HNO3 are mixed together. Resultant PH will be (a)
4
(b) above 4
(c) below 4
(d) None of these
Ans. (a)
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4.
PH value of 10–2M – Ca(OH)2 solution is (a)
2
(b) 4
(c) 11.7
(d) 12.3
Ans. (d) 5.
When 1 or mol of
M N M HCl, 200 ml of H 2SO 4 and 200 ml of NaOH are mixed together. 10 10 10
Resultant PH will be (a)
0
(b) 7
(c) 1
(d) 2
Ans. (b) Passage – 3 Consider an ionic solid that dissolves in water according to the equation MnXm(s) ↽ ⇀ nMm +(aq) + mXn –(aq) The equilibrium constant for this reaction, Ksp = [Mm +]n [Xn –]m is known as the solubility product of MnXm. The form of this equilibrium is important in understanding effects such as the influence of pH, complex formation and common ion effect. Equilibrium constants in solution should be written correctly using activities and not concentrations. The difference between these quantities is large in concentrated ionic solutions and Ksp is quantitatively reliable as a guide of solubilities only for very dilute solutions, if solubility product of AB type salt is 4 × 10– 10 at 18°C. 1.
If mol. wt. of AB is 143.5 g/mol., then solubility in g/lit of AB is (a)
14.35 gm/lit
(b) 2.87 × 10– 3 gm/lit
(c) 1.43 gm/lit
(d) 28.7 gm/lit
Ans. (b) 2.
If ppt. of AB is washed with 5 lit water, loss in wt. of ppt. of AB is (a)
10– 4 mol/lit
(b) 10– 4 gm/mol
(c) 10– 4 mol
(d) 10– 4 gm
Ans. (c) 3.
The % error in washing of 2 mol of ppt. of AB with 100 cc of N/100 solution of BC2 is (a)
4 × 10–7%
(b) 4 × 10– 2%
(c) 14 × 10– 3%
(d) 2 × 10– 5%
Ans. (a) 4.
The solubility of Calomel in water at 25°C is s moles/lit. Its solubility product is (a)
4s3
(b) 12s3
(c) 10s5
(d) s2
Ans. (a) Passage – 4 A solution which remains in equilibrium with undissolved solute, in contact, is said to be saturated. The concentration of a saturated solution at a given temperature is called solubility. The product of concentration of ions in a saturated solution of an electrolyte at a given temperature, is called solubility product (Ksp). For the electrolyte AxBy with solubility S, AxBy(s) ↽ ⇀ xAy + + yBx –
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Chapter-3 / Ionic Equilibrium
The solubility product (Ksp) is given as, Ksp = xx × yy × Sx + y. While calculating the solubility of a sparingly soluble salt in the presence of some strong electrolyte containing a common ion, the common ion concentration is practically equal to that of strong electrolyte. If in a solution, the ionic product of an electrolyte exceeds its Ksp value at a particular temperature, then precipitation occurs. 1.
The solubility of PbSO4 in water is 0.0303 g/l at 25°C, its solubility product at that temperature is (a)
10– 4 M2
(b) 9.18 × 10– 4 M2
(c) 10– 8 M2
(d) 9.18 × 10– 8 M2
Ans. (c) 2.
The solubility of BaSO4 in 0.1 M BaCl2 solution is (Ksp of BaSO4 = 1.5 × 10– 9) (a)
1.5 × 10– 9 M
(b) 1.5 × 10– 8 M
(c) 2.25 × 10– 16 M
(d) 2.25 × 10– 18 M
Ans. (b) 3.
If S0, S1, S2 and S3 are the solubilities of AgCl in water, 0.01 M CaCl2, 0.01 M NaCl and 0.05 M AgNO3 solutions respectively, then (a)
S0 > S1 > S2 > S3
(b) S0 > S2 > S1 > S3
(c) S0 > S1 = S2 > S3
(d) S0 > S2 > S3 > S1
Ans. (b) 4. pH of 0.1 M (NH4)2HPO4 solution is approximately equal to (Kb of NH4OH = 10– 5, K a , K a and K a 1 2 3 of H3PO4 are 10– 4, 10– 8 and 10– 10 respectively) (a)
8.52
(b) 4.52
(c) 9.52
(d) 7.52
Ans. (a)
Passage – 5 In case of a weak acid, a + is calculated using Ostwald’s dilution law, H i.e., a + = H
Ka × C
where Ka is the dissociation constant of weak acid with concentration C. In case of a mixture of two or more strong acids/bases, the pH (or pOH) is calculated by using the following relation,
∑ NV (for mixture of acids) ∑V ∑ NV (for mixture of bases) or a − = OH ∑V where, ∑ NV is the total equivalents of strong acids or bases and ∑ V is a
H+
=
the total volume of the
solution. In a mixture of a strong acid (concentration C1) and a weak acid (concentration C2 and dissociation constant
K a 2 ), HA → C1
C1
HB ↽ ⇀
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H+ + A –
(strong acid)
C1 H+ + B– (weak acid)
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C2 – x
C1 + x
x
Total a + for this mixture can be calculated from the following expression: H
Ka 2 =
( C1 + x ) x ( C2 − x )
where x is the number of moles of HB dissociated at equilibrium. In a mixture of two or more weak acids with concentrations C1 and C2 and dissociation constants
K a1 and K a 2 respectively, total a a
H+
H+
can be calculated by using the following expression:
K a1 C1 +K a 2 C2
pH of a buffer is calculated by using Henderson’s equation. In the titration of a weak acid with a strong base, one uses the concept of buffer before the point of neutralization and at the point of neutralization, concept of salt hydrolysis is used. 1.
At any temperature, a solution will be neutral if its pH is equal to (a)
7
(b)
1 pKa 2
(c)
1 pKw 2
(d)
1 pKb 2
Ans. (c) 2.
pH of 10– 9 M CH3COOH at 25°C is (Ka for CH3COOH = 1.8 × 10– 5, log 1.67 = 0.2041) (a)
2.8
(b) 6.8
(c) 7.8
(d) 4.8
Ans. (b) 3.
pH of a mixture obtained by mixing 100 ml of 0.1 M CH3COOH (Ka = 1 × 10– 5) with 100 ml of 0.2 M HCOOH (Ka = 1 × 10– 4) is approximately (log 1.05 = 0.0212 and log 3.25 = 0.5119) (a)
2.9
(b) 3.9
(c) 4.9
(d) 5.9
Ans. (a) 4.
100 ml of a 0.1 M solution of acetic acid is titrated with 0.1 M Ca(OH)2 solution pH of the solution in the titration flask at the titre values of 25 and 50 ml, respectively, are (pKa of CH3COOH = 4.74) (a)
2.37, 4.74
(b) 4.74, 8.98
(c) 2.37, 8.98
(d) 4.74, 2.37
Ans. (b)
Matching Type Questions List I
1. A.
List II
100 ml 1 M HCl + 100 ml 1 M NH4OH
p.
pH = 1/2 pKw
B.
100 ml 1 M NH4OH + 100 ml 1 M CH3COOH
q.
pH = 7 – 1/2 (pKb + log C)
C.
100 ml 1 N Ca(OH)2 + 100 ml 1 M HCl
r.
pH = 7 + 1/2 (pKa – pKb)
D.
100 ml 1 N CH3COOH + 100 ml ½ N NaOH
s.
pH = pKa + log
(a)
A – q, B – r, C – p, D – s
(b)
(c)
A – r, B – s, C – q, D – p
(d) A – s, B – p, C – q, D – r
[salt ] [Acid ]
A – p, B – s, C – r, D – q
Ans (a)
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Chapter-3 / Ionic Equilibrium
Compound
2. A.
MnO2
Nature p.
Acidic
B.
NaHCO3
q.
Basic
C.
NaCl
r.
Neutral
D.
NH4Cl
s.
Salt
(a)
A – p, q, s; B – p, q; C – p, r; D – r, s
(b)
(c)
A – p, q; B – p, q, s; C – r, s; D – p, r
(d) A – p, q, r; B – p, q, s; C – p, r; D – q, r, s
A – q, s; B – q, r; C – r, s; D – p, s
Ans. (c) Solution
3.
Nature
A.
Pure H2O at 35°C
p.
pH > 7
B.
Solution with [CH3COO–] = [CH3COOH]
q.
pH < 7
C.
10– 7 N HCl
r.
Acidic
D.
Pure H2O at 0°C
s.
Neutral
(a)
A – q, t; B – p, q; C – q, r; D – r, s
(b)
(c)
A – q, t; B – q, r; C – q, r; D – p, s
(d) A – r, s; B – p, s; C – q, s; D – p, q
A – p, s; B – q, r; C – p, r; D – p, s
Ans. (c)
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Exercise - V Subjective Type 1.
For an organic monoprotic acid solution of concentration c0 moles per litre, if Ka has a value comparable to Kw show that the hydronium concentration is given by
K K a c0 [H + ] = w + + + [H ] K a + [H ] where H+ stands for the hydronium ion. If [H+] = 10– 3 M & c0 = 10– 1 M in a solution of some organic monoprotic acid, what according to the above equation must be the order of magnitude of Ka? ~ 10−5 Ans. K a −
2.
The Kw of water at two different temperatures is T
25°C 50°C
Kw
1.08 × 10– 14
5.474 × 10– 14
Assuming that ∆H of any reaction is independent of temperature, calculate the enthalpy of neutralization of a strong acid and strong base. Ans. – 51.96 kJ/eq. 3.
The equilibrium constant of the reaction 2Ag(s) + 2I– + 2H2O ↽ ⇀ 2AgI(s) + H2(g) + 2OH– is 1.2 × 10– 23 at 25°C. Calculate the pH of a solution at equilibrium with the iodine ion concentration = 0.10 and the pressure of H2 gas = 0.60 atm.
Ans. 1.65 4.
Ethylenediamine, H2N – C2H4 – NH2, can interact with water in two steps, giving OH– in each step. Calculate the concentration of OH– and [H3N – C2H4 – NH3]2 + in a 0.15 M aqueous solution of the amine. K1 = 8.5 × 10– 5, K2 = 2.7 × 10– 8 for the base.
Ans. [OH–] = 3.57 × 10– 3 M, [H2en]2 + = 2.7 × 10– 8 M 5.
What is the solubility of AgCl in 0.20 M NH3? Given : Ksp(AgCl) = 1.7 × 10–
10
M2, K1 = [Ag(NH3)+]/[Ag+] [NH3] = 2.33 × 103 M–
1
and K2 =
[Ag(NH3 ) +2 ] /[Ag(NH3] = 7.14 × 103 M– 1. Ans. 9.66 × 10– 3 6.
Show that solubility of a sparingly soluble salt M2 + A2 – in which A2 – ions undergoes hydrolysis is given by
+ + 2 H H S= K sp 1 + + K2 K1K 2
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Chapter-3 / Ionic Equilibrium
where K1 and K2 are the dissociation constant of acid H2A. Ksp is solubility product of MA. 7.
Potash alum, KAl(SO4)2 . 12 H2O is a strong electrolyte and it is 100% dissociated into K+, Al3 + and 3 + SO 24 ions. The solution is acidic because of the hydrolysis of Al , but not so acidic as might be
− + expected, because the SO 24 can sponge up some of the H2O by forming HSO 4 . A solution of alum
was made by dissolving 11.4 gm of KAI(SO4)2 . 12H2O in enough water to make 0.1 litre of solution. Calculate the pH of this solution. Given : Al3 + + 2H2O ↽ ⇀ Al(OH)2 + + H3O+;
HSO −4 + H2O ↽ ⇀ H3O+ + SO 24− ;
Kh = 1.4 × 10– 5 M K2 = 1.26 × 10– 2 M
Ans. 3.53 8.
A solution of volume V contains n1 moles of QCl and n2 moles of RCl where QOH and ROH are two weak bases of dissociation constants K1 and K2 respectively. Show that the pH of the solution is given by pH =
K K 1 V log 1 2 2 K w ( n1K 2 + K1n 2 )
State assumptions, if any. 9.
The ‘fixing’ of photographic film involves dissolving unexposed silver bromide in a thiosulphate
(S2O32- ) solution:
3− AgBr(s) + 2S2 O32- (aq) ↽ ⇀ Ag ( S2 O3 ) (aq) + Br– (aq) 2 3− , calculate the equilibrium constant 2
Using Ksp = 5.4 × 10– 13 for AgBr and Kf = 2.9 × 1013 for Ag ( S2 O3 )
K for the above reaction and calculate the molar solubility of AgBr in 0.10 M Na2S2O3. Ans. K = 15.7, x = 0.044 10. An acid base indicator has a Ka of 3 × 10– 5. The acid form of the indicator is red & the basic form is blue. By how much must the pH change in order to change the indicator form 75% red to 75% blue? Ans. – 0.95 11.
Ka for butyric acid is 2.0 × 10–5. Calculate pH and hydroxyl ion concentration in 0.2M aqueous solution of sodium butyrate.
Ans. 9 12. Calculate the pH of an aqueous solution of 1.0M ammonium formate assuming complete dissociation. (pKa of formic acid = 3.8 and pKb of ammonia = 4.8). Ans. 6.5 13.
Calculate the concentrations of [Ag+] ion, [Br–] ion, [Cl–] ion, [Ag(NH3)2]+ ion, NH +4 ion and hydroxide ion in a solution which results from shaking excess AgCl and AgBr with 0.02 M ammonia solution. Assume that no monoamine complex is formed. Given, KSP of AgCl = 1 × 10– 10, KSP of AgBr = 5 × 10– 13, Kb of NH3 = 1.8 × 10– 5 and Kf of [Ag(NH3)2]+ = 1 × 108.
Ans. [Ag+] = 6 × 10– 8 M, [Br–] = 8 × 10– 6 M, [Cl–] = 1.7 × 10– 3 M, [Ag(NH3)2]+ = 1.7 × 10– 3 M, [ NH +4 ] = 6 × 10– 4 M, [OH–] = 6 × 10– 4 M
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14. The vapour pressure of 0.01 molal solution of weak base BOH in water at 20°C is 17.356mm. Calculate Kb for base. Aqueous tension at 20°C is 17.540mm. Assume molality and molarity same. Ans. 9.74 × 10–4 15. Predict whether or not AgCl will be precipitated from a solution which is 0.02 M in NaCl and 0.05 M in
(
)
KAg(CN)2. Given K inst Ag (CN )−2 = 4.0 × 10– 19 M2 and KSP(AgCl) = 2.8 × 10– 10 M2. Ans. Precipitate will occur.
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Chapter-3 / Ionic Equilibrium
Exercise - VI IIT-JEE Problems 1.
500 mL of 0.2 M aqueous solution of acetic acid is mixed with 500 mL of 0.2 M HCl at 25°C. (i)
Calculate the degree of dissociation of acetic acid in the resulting solution and pH of the solution.
(ii)
If 6 g of NaOH is added to the above solution, determine the final pH (assuming there is no
Formatted: Space Before: 3 pt, After: 0 pt, Line spacing: Multiple 1.2 li
change in volume on mixing, Ka of acetic acid is 1.75 × 10– 5 mol/L. Ans. α = 1.75 × 10– 2%; pH = 1 2.
Formatted: Space Before: 3 pt, Line spacing: Multiple 1.2 li
The average concentration of SO2 in the atmosphere over a city on a certain day is 10 ppm, when the average temperature is 298 K. Given that the solubility of SO2 in water at 298 K is 1.3653 mol/L and pKa of H2SO3 is 1.92, estimate the pH of rain on that day.
Ans. 0.5 3.
The solubility of Pb(OH)2 in water is 6.7 × 10– 6 M. Calculate the solubility of Pb(OH)2 in a buffer solution of pH = 8.
Ans. 1.2 × 10– 3 M 4.
What will be the resultant pH when 200 mL of an aqueous solution of HCl (pH = 2.0) is mixed with 300 mL of an aqueous solution of NaOH (pH = 12.0)?
Ans. pH = 11.3 5.
Given, Ag(NH 3 ) +2 → Ag + + 2NH3 , Kc = 6.2 × 10– 8 and Ksp of AgCl = 1.8 × 10– 10 at 298 K. If ammonia is added to a water solution containing excess of AgCl(s) only. Calculate the concentration of the complex in 1.0 M aqueous ammonia.
Ans. [Ag(NH3)2]+ = 0.054 M 6.
An acid type indicator, H/n differs in colour from its conjugate base (In–). The human eye is sensitive to colour differences only when the ratio [In–] / [H/n] is greater than 10 or smaller than 0.1. What should be the minimum change in the pH of the solution to observe a complete colour change? (Ka = 1.0 × 10– 5)
Ans. 4 to 6 7.
A sample of AgCl was treated with 5.00 mL of 1.5 M Na2CO3 solution to give Ag2CO3. The remaining solution contained 0.0026 g of Cl– ions per litre. Calculate the solubility product of AgCl. [Ksp (Ag2CO3) = 8.2 × 10– 12]
Ans. 96 gm
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8.
What is the pH of a 0.50 M aqueous NaCN solution? pKb of CN– = 4.70
Ans. 11.5 9.
Calculate the pH of an aqueous solution of 1.0, M ammonium formate assuming complete dissociation (pKa of formic acid = 3.8 and pKb of ammonia = 4.8)
Ans. 6.5 10.
An aqueous solution of a metal bromide MBr2 (0.05 M) is saturated with H2S. What is the minimum pH which MS will precipitate. Ksp for MS = 6.0 × 10– 21, conc. of saturated H2S = 0.1 M and K1 = 10– 7 and K2 = 1.3 × 10– 13, for H2S.
Ans. 0.983 11.
The pH of blood stream is maintained by a proper balance of H2CO3 and NaHCO3 concentrations. What volume of 5 M NaHCO3 solution should be mixed with a 10 mL maintain a pH of 7.4? Ka for H2CO3 in blood is 7.8 × 10– 7
Ans. 78.36 12.
The solubility product (Ksp) of Ca(OH)2 at 25°C is 4.52 × 10– 5. A 500 mL of saturated solution of Ca(OH)2 is mixed with equal volume of 0.4 M NaOH. How much Ca(OH)2 in milligrams is precipitated.
Ans. 7.58 13.
A 40.0 mL solution of a weak base, BOH is titrated with 0.1 N HCl solution. The pH of the solution is found to be 10.04 and 9.14 after the addition of 5.0 mL and 20.0 mL of the acid respectively. Find out the dissociation constnat of the base.
Ans. 1.81 × 10– 5 14.
The solubility product of Ag2C2O4 at 25°C is 1.29 × 10– 11 mol3 L– 3. A solution of K2C2O4 containing 0.1520 mole in 500 mL water is shaken at 25°C with excess of Ag2CO3 till the following equilibrium is reached: Ag2CO3 + K2C2O4 → Ag2C2O4 + K2CO3 At equilibrium, the solution contains 0.0358 mole of K2CO3. Assuming the degree of dissociation of K2C2O4 and K2CO3 to be equal, calculate the solubility product of Ag2CO3.
Ans. 3.9 × 10– 12 15.
What is the pH of a 1.0 M solution of acetic acid? To what volume must one litre of this solution be diluted so that the pH of the resulting solution will be twice the original value? Given : Ka = 1.8 × 10–
5
Ans. 2.37; V = 2.77 × 104 L
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16.
Chapter-3 / Ionic Equilibrium
Freshly precipitated aluminium and magnesium hydroxides are stirred vigorously in a buffer solution containing 0.25 mol/L of NH4Cl and 0.05 M of ammonium hydroxide. Calculate the concentration of aluminium and magnesium ions in solution. Kb [NH4OH] = 1.8 × 10– 5 Ksp [Mg(OH)2] = 8.9 × 10– 12 Ksp [Al(OH)3] = 6 × 10– 32
Ans. [Al+ 3] = 1.28 × 10– 15; [Mg2 +] = 0.686 17.
How many gram-mole of HCl will be required to prepare one litre of buffer solution (containing NaCN and HCl) of pH 8.5 using 0.01 gram formula weight of NaCN? KHCN = 4.1 × 10– 10
Ans. 8.85 × 10– 3 18.
What is the pH of the solution when 0.20 moles of HCl is added to one litre of a solution containing (i)
1 M each of acetic acid and acetate ion
(ii)
0.1 M each of acetic acid and acetate ion
Assume the total volume is one litre. Ka for acetic acid = 1.8 × 10– 5. Ans. (i) 4.57; (ii) 1 19.
The solubility of Mg(OH)2 in pure water is 9.57 × 10– 3 g/L. Calculate its solubility (in g/L) in 0.02 M Mg(NO3)2 solution.
Ans. 8.7 × 10– 4 gm/L 20.
The concentration of hydrogen ions is a 0.2 molar solution of formic acid is 6.4 × 10– 3 mol/L. To this solution, sodium format is added so as to adjust the concentration of sodium formate to one mole per litre. What will be the pH of this solution? The dissociation constant of formic acid is 2.4 × 10– 4 and the degree of dissociation of sodium formate is 0.75.
Ans. pH = 4.2 21.
A solution contains a mixture of Ag(0.10 M) and Hg2 + (0.10 M) which are to be separated by selective precipitation. Calculate the maximum concentration of iodine ion at which one of them gets precipitated almost completely. What percentage of that metal ion is precipitated? Ksp : AgI = 8.5 × 10– 17, HgI2 = 2.5 × 10– 26
Ans. [I–] = 5 × 10– 13 M 22.
The dissociation constant of a weak acid HA is 4.9 × 10– 8. After making the necessary approximations, calculate (i)
percentage ionization
(ii)
pH
-60-
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(iii) OH– concentration in a decimolar solution of the acid. Water has a pH of 7.
Ans. (i) 0.07%; (ii) 4.15; (iii) 1.4 × 10– 10
*****
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Chapter-3 / Ionic Equilibrium
Answers
Formatted
Exercise - I True/False 1. 1. True
2.
True
3.
Fill in the Blanks 1. 1. pKw 4. neutralization
False
4.
False
2. anionic
5.
decrease
5.
3.
False
Formatted
strong
Formatted: Justified, Indent: Left: 0", First line: 0", Space Before: 0 pt, Tab stops: 2.63", Left + 3", Left + 4.75", Left + 5.19", Left + Not at 1.75" + 2.06" + 3.5" + 3.81" + 5.25" + 5.56" Formatted
Assertion and Reason 1. 1. (a) 2.
(a)
3.
(d)
4.
(b)
5.
(c)
Formatted
(d)
3.
(a)
4.
(c)
5.
8. 13. 18. 23.
(d) (a) (d) (c)
9. 14. 19. 24.
(d) (c) (a) (a)
10. 15. 20. 25.
(b) (c) (b) (a)
28. 33. 38. 43. 48.
(c) (c) (c) (d) (d)
29. 34. 39. 44. 49.
(b) (b) (d) (d) (a)
30. 35. 40. 45. 50.
(a) (d) (b) (b) (a)
2.
(a, b)
3.
(c, d)
4.
(c)
5.
... [4]
Formatted
... [5]
Formatted
... [6]
Formatted
... [7]
Formatted
... [8]
Formatted
... [9]
Formatted
... [10]
Formatted
... [11]
Formatted
... [12]
Formatted
... [13]
Formatted
... [14]
Formatted
... [15]
Formatted: Font: Not Bold
8. 13. 18.
(b, c) (a, b, c) (a, b, c)
9. 14. 19.
(b, d) (a, b, c) (a, b, c)
10. 15. 20.
(a, b) (a, b) (b, c, d)
Exercise - IV Passage – 1 1. -62-
... [3]
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Exercise - III More Than One Choice Correct 1. 1. (a, c, d) (b) 6. (a, b) 7. (b, c) 11. (c, d) 12. (b, c, d) 16. (a, b, c) 17. (a, c, d) 21. (a, b, d)
... [2]
Formatted: Justified, Indent: Left: 0", First line: 0", Space Before: 0 pt, Tab stops: 1.38", Left + 1.78", Left + 2.78", Left + 3.22", Left + 4.22", Left + 4.63", Left + 5.56", Left + 6", Left + Not at 1.75" + 2.06" + 3.5" + 3.81" + 5.25"
Exercise - II Only One Option is correct Level - I 1. 1. (a) 2. (c) 6. (a) 7. (c) 11. (a) 12. (d) 16. (d) 17. (d) 21. (d) 22. (a) Level - II 26. (d) 27. (c) 31. (c) 32. (d) 36. (b) 37. (d) 41. (c) 42. (c) 46. (c) 47. (a) Level - II 26.
... [1]
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Formatted: Indent: Left: 0", First line: 0", Tab stops: 0.31", Left + Not at 0.41" Formatted
... [16]
Formatted
... [17]
Formatted
... [18]
Formatted
... [19]
Formatted
... [20]
Formatted: Font: Not Bold Sri Chaitanya IIT-JEE Head Office: 11, Central Market, West Punjabi Bagh, New Delhi-110 026. Phones: 25226309/10. Web: srichaitanya.edu.in, E-mail:
[email protected]
Sri Chaitanya: The Final Word in IIT-JEE
IIT/TYP/U2/Chemistry/07
Passage – 1 1.
(c)
2.
(c)
3.
(d)
4.
(b)
5.
(d)
Formatted: Font: Formatted
Passage – 2 1.
(d)
2.
(d)
3.
(a)
4.
(d)
(b)
2.
(c)
3.
(a)
4.
(a)
Passage – 4 1. (c)
2.
(b)
3.
(b)
4.
(a)
5.
(b)
Passage – 3 1.
Passage – 5 1. (c)
2.
(b)
3.
(a)
3.
(c)
4.
Formatted: Font: (Default) Palatino Linotype, Not Bold Formatted: Font: Formatted: Left, Indent: Left: 0", Hanging: 0.32", Line spacing: single, Adjust space between Latin and Asian text, Adjust space between Asian text and numbers, Tab stops: 0.41", Left + 1.38", Left + 1.78", Left + 2.78", Left + 3.22", Left + 4.22", Left + 4.63", Left + 5.56", Left + 6", Left + Not at 0.38" + 0.75" + 1.97" + 2.25" + 3.63" + 3.88" + 5.25" + 5.5"
(b)
Matching Type Questions 1.
(a)
2.
(c)
... [21]
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Passage – 2 1.
Formatted
... [22]
Formatted: Font:
Passage – 3 1.
Formatted: Left, Indent: Left: 0", Hanging: 0.32", Line spacing: single, Adjust space between Latin and Asian text, Adjust space between Asian text and numbers, Tab stops: 0.41", Left + 1.38", Left + 1.78", Left + 2.78", Left + 3.22", Left + 4.22", Left + 4.63", Left + 5.56", Left + 6", Left + Not at 0.38" + 0.75" + 1.97" + 2.25" + 3.63" + 3.88" + 5.25" + 5.5"
Passage – 4 1. Matching Type Questions 1.
Formatted
Exercise - V
... [23]
Formatted: Font: (Default) Palatino Linotype, Not Bold
1.
~ 10 −5 Ka −
2.
– 51.96 kJ/eq.
Formatted
... [24]
3.
1.65
Formatted
... [25]
4.
[OH–] = 3.57 × 10– 3 M, [H2en]2 + = 2.7 × 10– 8 M
Formatted: Font:
5.
9.66 × 10– 3
Formatted
... [26]
7.
3.53
Formatted
... [27]
9.
K = 15.7, x = 0.044
Formatted
... [28]
10. – 0.95
Formatted
... [29]
11. 9 12. 6.5
Formatted: Font: (Default) Arial, 15 pt
13.
[Ag+] = 6 × 10–
Formatted: Font:
Formatted 8
M,
[Br–] = 8 × 10–
6
M,
[Cl–] = 1.7 × 10–
3
M,
[Ag(NH3)2]+ = 1.7 × 10–
3
M,
[ N H 4+ ] = 6 × 10– 4 M, [OH–] = 6 × 10– 4 M 14. 9.74 × 10–4 Sri Chaitanya IIT-JEE Head Office: 11, Central Market, West Punjabi Bagh , New Delhi-110 026. Phones: 25226309/10. Web: srichaitanya.edu.in, E-mail:
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-63-
Formatted: Space Before: 0 pt
... [30]
Sri Chaitanya: The Final Word in IIT-JEE
Chapter-3 / Ionic Equilibrium
15. Precipitate will occur.
Exercise - VI
Formatted: Font: (Default) Arial, 15 pt
1.
α = 1.75 × 10– 2%; pH = 1
2.
0.5
3.
1.2 × 10– 3 M
4.
pH = 11.3
5.
[Ag(NH3)2]+ = 0.054 M
6.
4 to 6
7.
96 gm
8.
11.5
9.
6.5
10.
0.983
11.
78.36
12.
7.58
13.
1.81 × 10– 5
14.
3.9 × 10– 12
15.
2.37; V = 2.77 × 104 L
16.
[Al+ 3] = 1.28 × 10– 15; [Mg2 +] = 0.686
17.
8.85 × 10– 3
18.
(i) 4.57; (ii) 1
19.
8.7 × 10– 4 gm/L
20.
pH = 4.2
21.
[I–] = 5 × 10– 13 M
22.
(i) 0.07%; (ii) 4.15; (iii) 1.4 × 10– 10
Formatted: Space Before: 0 pt, Tab stops: 0.32", Left + 0.63", Left + 1.38", Left + 1.78", Left + 2.78", Left + 3.22", Left + 4.22", Left + 4.63", Left + 5.56", Left + 6", Left + Not at 0.28" + 0.55" + 0.83"
Exercise - V Subjective Type 1.
Exercise - VI IIT-JEE Level Problem
1.
***** Formatted
*****
-64-
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-96-
Chapter-1 3 / Ionic Equilibrium
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Sri Chaitanya: The Final Word in IIT-JEE
IIT/OYPTYP/U1U2/Chemistry/07
Exercise - I General Type (Fill in the blanks/ True or False/ Assertion & Reason ) True/False 1.
True
2.
True
3.
False
4.
False
5.
False
Fill in the Blanks 1.
pKw
2.
decrease
3.
strong
4.
neutralization
5.
anionic
Assertion and Reason 1. 2. 3. 4. 5.
(a) (a) (d) (b) (c)
Exercise - II Level – I 1.
(a)
2.
(d)
3.
(a)
4.
(c)
5.
(c)
6.
(a)
7.
(c)
8.
(d)
9.
(d)
10.
(b)
11.
(a)
12.
(d)
13.
(a) Sri Chaitanya IIT-JEE Head Office: 11, Central Market, West Punjabi Bagh , New Delhi-110 026. Phones: 25226309/10. Web: srichaitanya.edu.in, E-mail:
[email protected]
-97-
Formatted: Tab stops: 6.88", Right + Not at 7"
Sri Chaitanya: The Final Word in IIT-JEE
14.
(c)
15.
(c)
16.
(d)
17.
(d)
18.
(d)
19.
(a)
20.
(b)
21.
(d)
22.
(a)
23.
(c)
24.
(a)
25.
(a)
Chapter-1 3 / Ionic Equilibrium
Level - II 26.
(d)
27.
(c)
28.
(c)
29.
(b)
30.
(a)
31.
(c)
32.
(d)
33.
(c)
34.
(b)
35.
(d)
36.
(b)
37.
(d)
38.
(c)
39.
(d)
40.
(b)
41.
(c)
42.
(c)
43.
(d)
44.
(d)
45.
(b)
46.
(c)
47.
(a)
48.
(d)
49.
(a)
50.
(a)
-98-
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IIT/OYPTYP/U1U2/Chemistry/07
Sri Chaitanya: The Final Word in IIT-JEE
Exercise - III 1.
(a, c, d)
2.
(a, b)
3.
(c, d)
4.
(c)
5.
(b)
6.
(a, b)
7.
(b, c)
8.
(b, c)
9.
(b, d)
10.
(a, b)
11.
(c, d)
12.
(b, c, d)
13.
(a, b, c)
14.
(a, b, c)
15.
(a, b)
16.
(a, b, c)
17.
(a, c, d)
18.
(a, b, c)
19.
(a, b, c)
20.
(b, c, d)
21.
(a, b, d)
Exercise - IV Passage – 1 1.
(c)
2.
(c)
3.
(d)
4.
(b)
5.
(d)
Passage – 2 1.
(d)
2.
(d)
3.
(a)
4.
(d) Sri Chaitanya IIT-JEE Head Office: 11, Central Market, West Punjabi Bagh , New Delhi-110 026. Phones: 25226309/10. Web: srichaitanya.edu.in, E-mail:
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-99-
Formatted: Tab stops: 6.88", Right + Not at 7"
Sri Chaitanya: The Final Word in IIT-JEE
5.
Chapter-1 3 / Ionic Equilibrium
(b)
Passage – 3 1.
(b)
2.
(c)
3.
(a)
4.
(a)
Passage – 4 1.
(c)
2.
(b)
3.
(b)
4.
(a)
Passage – 5 1.
(c)
2.
(b)
3.
(a)
4.
(b)
Matching Type Questions 1.
(a)
2.
(c)
3.
(c)
*****
Formatted: Centered
-100-
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Sri Chaitanya: The Final Word in IIT-JEE
IIT/OYPTYP/U1U2/Chemistry/07
Exercise - V
~ 10−5 Ka −
1. 2.
– 51.96 kJ/eq. 3.
4.
1.65
[OH–] = 3.57 × 10– 3 M, [H2en]2 + = 2.7 × 10– 8 M 5.
9.66 × 10– 3 7.
9.
3.53
K = 15.7, x = 0.044 10. – 0.95 11. 9 12. 6.5
13. [Ag+] = 6 × 10– 8 M, [Br–] = 8 × 10– 6 M, [Cl–] = 1.7 × 10– 3 M, [Ag(NH3)2]+ = 1.7 × 10– 3 M, [ N H 4+ ] = 6 × 10– 4 M, [OH–] = 6 × 10– 4 M 14. 9.74 × 10–4 15. Precipitate will occur. Exercise - VI
1. 3. 5.
α = 1.75 × 10– 2%; pH = 1 1.2 × 10– 3 M
9.
96 gm
6.5
6. 0.983
12.
7.58
13. 1.81 × 10– 5 14. 15. 2.37; V = 2.77 × 104 L
16.
17. 8.85 × 10
–3
21. [I–] = 5 × 10– 13 M
22.
3.9 × 10– 12 [Al+ 3] = 1.28 × 10– 15; [Mg2 +] = 0.686
18.
19. 8.7 × 10– 4 gm/L
4 to 6
8. 11.5
10.
11. 78.36
0.5
pH = 11.3
[Ag(NH3)2]+ = 0.054 M 7.
2.
4.
(i) 4.57; (ii) 1 20.
pH = 4.2
(i) 0.07%; (ii) 4.15; (iii) 1.4 × 10– 10
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