03_Ionic Equilibrium.pdf

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Sri Chaitanya IIT-JEE Head Office: 11, Central Market, West Punjabi Bagh , New Delhi-110 026. Phones: 25226309/10.

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 twenty-six 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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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, H} 2SO4

etc. While base is a substance which furnishes OH– _{ions e.g. - NaOH, NH} 4OH 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 ↽⇀ H3O

+ ₊ _Cl–

acid base conjugate acid conjugate base

H₂O + NH₃ _↽⇀

4

NH+ + OH–

acid base conjugate acid 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– → _H

2O + 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.

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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. 2 4

C H in Ag C H

(

2 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

x cx cx HA H A c 0 0 c(1– ) At equilibrium + ₊ − ⇀ ↽

It Ka is ionization constant for acid, then

a [H ][A ] K [HA] + − = 2 cx.cx cx c(1 x) (1 x) = = − −

for very weak acids. x < < 1

∵ 2 a cx K ( 1 x 1) 1 = ∴ − ≃ or Ka x C = or Ka V n x c n V ×   =  =   

or xα V (Ostwald’s dilution law) From above calculation, we can calculate

(i) [ ] cx c a a c c K H+ = = = K × (ii) [A−]=cx= Ka c× (iii) [HA]=c(1 x)− ≅c (x<<1) (iv) [ ] Kw /[n ] Kw Ka c OH− = + = ×

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Illustr ations

Illustration 1

The ionization constant of propionic acid is 1.32 × 10–5_{. Calculate the degree of ionization of the}

acid in its 0.05 M solution an also its pH. Solution CH3CH2COOH ↽⇀ CH3CH2COO– + H+ c(1 – α) cα cα 5 3 2 2 a 3 2 [CH CH COO ][H ] c .c K 1.32 10 c ( 1 a 1) [CH CH COOH] c(1 ) − + − α α = × = = = α − ≈ − α ∵ ∴ 0.05 × α2_{= 1.32 × 10}–5_; α = -2 1.63 × 10 pH = – log [H+_{] = – log (c}α₎ 2 0 05 1 63 10 log ( . . − ) = − × × =3.09 Illustration 2

The pH of 0.1 M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.

Solution: HCNO _↽⇀ _H+ ₊ _CNO− c(1 – x) c x c x pH = 2.34 – log [H+_{] = 2.34} or [H+_{] = 4.57 × 10}–3 or c x = 4.57 × 10–3 ∴ 0.1 × x = 4.57 × 10–3 _or _{x = 4.57 × 10}–2 ∴ Ka = c x 2 = (4.57 × 10–2)2 × 0.1 = 2.09 × 10–2

Practice Exercise

1. The pH of 0.005 M codeine (C18H21NO2) solution is 9.95. Calculate its (pKb) ionization

constant.

2. Determine degree of dissociation of 0.05 M NH3 at 25°C in a solution of pH = 11.

Answers

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3.4 Common Ion Effect

In presence of common ion dissociation of weak electrolyte is further suppressed. e.g. CH3COOH is weak electrolyte. It dissociates as

3 3

CH COOH_↽⇀CH COO−+H+

If some amount of (CH3COONa) is added to the solution of acetic acid. Then CH COONa3 will provide common acetate ions. Consequently equilibrium will shift towards reactant side and degree of dissociation becomes further smaller.

Similarly addition of NH4Cl will led to the suppression of dissociation of NH4OH. Let us consider the ionization of weak acid HA in presence of strong acid HX.

(1 x) cx cx At + – HA H + A c 0 0 c − equilibrium ⇀ ↽

At equilibrium if we add some moles of strong acid HX, this will create common ion effect due to ‘H+_{’ furnished by it. Due to common ion effect, degree of dissociation decreases.}

(1 x) cx cx n – – n n HA H A c HX H X 0 + − + − + − → + ⇀ ∵ ↽

In above ionization of HA, x’ is degree of ionization in presence of strong acid HX.

∴ +T -a [H ] [A ] K = [HA] cx n cx c x ( + ) = (1 – ) Ka=nx (cx<<n and x<1) x Ka n ′ = [H ]+_T=cx′+n ∵ ≃n (cx′ <<n) [A−]=cx′ [HA]=c(1−x′)≅c [OH−]=Kw H/[ +].

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Illustr ations

Illustration 3

Calculte the pH of resulting solution when 50 mL of 0.20 M HCl is mixed with 50 mL of 0.20 M CH3COOH.

Solution

After mixing, volume becomes 100 mL.

∴ 50 0 2 0 1 100 HCl milli mole . M . Total Volume − × = = = 3 50 0 2 0 1 100 CH COOH . M = × = .

The pH will be decided by strong acid, since dissociation of CH3COOH is insignificant in

presence of HCl due to common ion effect. Thus pH of solution of 0.1 M HCl = 1.

Practice Exercise

3. The ionization constant of chloroacetic acid is 1.35 × 10–3_{. What will be pH of 0.1 M acid and}

its 0.1 M sodium salt solution?

Answers

3. 7.94

Use of common ion effect in qualitative analysis

H2S is a weak electrolyte.

(

2

)

2

H S_↽⇀2H++S− _{. In presence of HCl acid its dissociation is further}

suppressed. Ksp values of sulphides of groups II basic radicals are quite lesser that Ksp values of sulphides of group IV radicals. That’s why in II group H2S gas is passed in presence of HCl, otherwise sulphides of groups IV will also precipitate out along with sulphides of group II radicals.

Similarly NH4OH is added in presence of NH4Cl in qualitative tests of group III radicals, in presence of common NH4

+_{ions dissociation of NH}

4OH is suppressed and consequently lesser

amount of OH– ions are obtained which can bring about precipitation of hydroxides of group III radicals only.

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3.5 Mixture of two weak acids

Let us consider the mixture of aqueous solutions of two weak acids HA1 and HA2 whose

concentrations and are C1 and C2 and ionization constants are Ka1 and Ka2.

∵ _HA1 ↽⇀ H+ + A1— C1 — — C1(1 – x1) C1x1 C1x1 (At equilibrium) HA2 ↽⇀ H+ + A2— C2 — — C2(1 – x2) C2x2 C2x2 (At equilibrium) – T 1 1 1 2 2 1 1 1 1 1 1 [H ] [A ] (C x C x ) C x Ka [HA ] C (1 x ) + ₊ _× = = − ∵ …(i) and – T 2 1 1 2 2 2 2 2 2 2 2 [H ] [A ] (C x C x ) C x Ka [HA ] C (1 x ) + ₊ _× = = − …(ii)

Dividing Eq. 1 by Eq. 2 1 1 1

2 2 2

Ka C x ,

Ka =C x on putting the value of x1 in terms of x2 in equation (i), we can calculate x2 from

above discussion we can calculate the pH of solution mixture of two acids.

3.6 Ionisation of polyprotic acids

Let us consider the ionization of H2S in its aqueous solution. It C is concentration in mole/litre C

and Ka1 and Ka2 are ionization constants for first and second step ionization of H2S.

∵ H2S ↽⇀ H+ + HS—

C(1-x1) cx1 cx1(1 – y1) (At equilibrium)

HS _↽⇀ _H+ ₊ _S– –

cx1 (1–y1) cx1 y1 x1 y1 (At equilibrium)

T 1 1 1 1 1 1 2 1 [H ] [HS ] (cx cx y ).{cx (1 y )} Ka [H S] c(1 x ) + − ₊ ₋ = = − (x1<<1) T 1 1 1 1 1 2 1 1 [H ] [S ] (cx cx y ).cx y Ka cx (1 y ) [HS ] + −− − + = = − 1 (y <<1)

because generally Ka1>>Ka ,2 hence we can assume that cx y1 1<<cx1

∴ 1 1 1 2 1 1 c x .cx Ka cx c = = …(i) 2 2 1 1 1 1 1 1 Cx y Ka Cx y Cx = = …(ii)

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Illustr ations

Illustration 4

K1 and K2 for dissociation of H2A are 4 × 10–3 and 1 × 10–5. Calculate concentration of A2–ion in 0.1

M H2A solution. Also report [H+] and pH.

Solution: 2 H A_↽⇀H++HA− 3 1 2 4 10 [H ][HA ] K [H A] + − − = = × 2 1 [H ] cx, [HA ] cx; [H A] c(+ = − = = −x) ∵ or 2 3 4 10 1 1 cx.cx cx c( x) ( x) − × = = − − (c = 0.1 M) or 2 3 0 1 4 10 1 1 . x ( x ( x) − × × = −

− should not be neglected)

∴ x = 0.18

∴ [H+_{] = c x = 0.1 × 0.18 = 0.018 M}

∴ pH = 1.7447

∴ [HA–_{] = c x = 0.1 × 0.18 = 0.018 M}

[H2A] = c (1 – x) = 0.1 (1 – 0.18) = 0.082 M

Now HA–_{further dissociates to H}+_{and A}–2_{; c}₁_{= [HA}–_{] = 0.018 M}

HA– _↽⇀ H+ ₊ _A–2 1 0 0 (1 – x 1) x 1 x 1 ∴ 2 2 5 1 10 a [H ][A ] K [HA ] + − − − = × =

∵ [H+_{] already in solution = 0.018 and thus, dissociation of HA}–_{further suppresses due to}

common ion effect and 1− ≈x 1

∴ 5 1 1 1 1 1 0 018 1 10 0 018 1 . c x . x c ( x ) − × × = = × − ∴ 5 4 1 1 10 5 55 10 0 018 x . . − − × = = × ∴ 2 4 5 1 1 0 018 5 55 10 10 [A ] c x− = = . × . × − = − M

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∴ [HA ] c (− = 11−x )1 =c1=0.018 M

Practice Exercise

4. The first ionization constant of H2S is 9.1 × 10–8. Calculate the concentration of HS– ion in its

0.1 M solution and how will this concentration be effected if the solution is 0.1 M in HCl also. If the second dissociation constant of H2S is 1.2 × 10–13, calculate the concentration of S2–

under both conditions.

Answers

4. 1.54 × 10–5_{M, 9.1 × 10}–8_{M, 1.2 × 10}–13_{, 1.092 × 10}–19_M

3.7 Ionisation of Water

Ionic product of water (Kw) Water is weak electrolyte, hence

H O2 H OH +₊ − ⇀ ↽

[

]

eq 2 H OH K H O + −         = Keq.

[

H O2

]

H OH + −     =_ _{ } _

Since [H2O] = constant = Kw=K [H O] [H ][OH ]eq 2 = + − w

K depends on temperature. At 25° C value of Kw is 10–14. Hence __H+_{ }_OH−_₌₁₀−14_{at 25 C}_°     Thus __H+_2₌₁₀−14   Or 7 H+ 10 at 25 C−  ₌ _°   Since [H+_{] = [OH}–_] At 25°C, PH_{of pure water}

{

H

}

( )

7 P = −log H_ +_ = −log 10− =7 Now at 25°C 14 H+ OH− 10−    ₌     or −log H_ +_−log OH_ −_=14 or H OH P +P =14 at 25°C.

Therefore PH_{range at 25° C will be 0 to 14}

H H H P below 7 acidic P above 7 basic While P 7 neutral − − = −

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Acidic

Neutral

Basic

O

7

14

As temperature increases, degree of dissociation of water also increases, therefore value of Kw increases.

Thus on increasing temperature, value of Kw increases PH range will shrink and PH value of water would be less than 7 but water will be chemically neural.

(

)

2

Again

H O

H

OH

Initially

c

0

0 at equilib. c 1 x cx

cx

+

₊

−

⇌

2 2 w K =_H+_{ } OH−_=cx cx× =c x at 25°C 14 2 2 w K =10− =c x or _cx₌₁₀−7 7 10 x C − =

For pure water c = 55.55 M Thus 7 10 x 55.55 − =

3.8 pH and pOH

PH_value

It is negative logarithm of [H+_{] ion concentration.}

Thus PH_{value of solution = –log[H}+_]

(i) PH_{has a great importance in agriculture.}

(ii) PH_{value plays vital role in biological reactions.}

(iii) Human blood has PH_{value 7.36 – 7.42.}

(iv) PH_{value plays an important role in qualitative analysis.}

(v) Food preservation also requires a definite PH_value.

Illustr ations

Illustrations 5 Find the PH_{values of}

(i) 10–5_{M – HCl} (ii) 10–6_{M – HNO} 3 Solution: (i) 10–5_{M – HCl} 5 10 HCl→H+− +Cl−

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Sri Chaitanya IIT-JEE Head Office: 11, Central Market, West Punjabi Bagh , New Delhi-110 026. Phones: 25226309/10. H

( )

5 p = −log 10− =5 (ii) 10–6_{M – HNO} 3 HNO3 H₁₀6 NO3 + − − → + _pH_{= −}_{log 10}

( )

−6 ₌₆ Illustration 6 Find resultant pH_{if 200 ml of}M 10 HCl and 300ml of M

10 NaOH are mixed together.

Solution

Millimoles of HCl = 20 Millimoles of NaOH = 30

Millimoles of NaOH left = 10 and volume = 500 ml

Thus

[

]

10 2 NaOH 0.02 2 10 500 − = = = × OH

(

2

)

p = −log 2 10× − = −2 log 2 Hence _pH₌₁₄₋_pOH_{= − −}₁₄

(

_{2 log 2}

)

= 12 + log 2 = 12.3

Warning: In real solutions not concentrations, but ion activities should be used for calculations.

Especially pH definition uses not minus logarithm of concentration, but minus logarithm of activity. In diluted solutions activity is for all practical purposes identical to concentration, but when the concentration goes higher activity starts first to be lower then the concentration, than - once the concentration rises - higher the concentration. As a rule of thumb if the concentration of charged ions present in the solution is below 0.001 M you don't have to be concerned about activities and you can use classic pH definition. All calculations done using concentrations are wrong. Ions are charged so they interact in the solution attracting and repelling each other with coulomb forces. These interactions influence ions behavior and doesn't allow to treat every ion in the solution independently. Whole phenomenon - although investigated for over 100 years - is still not fully understood and described. To be precise in our equilibrium calculations instead of using concentrations we should use ions activities. Activities are not a theoretical construct - they can be measured for every solution. In fact whenever we put pH electrode into solution we are not measuring [H+_{] but activity of H}+_ions.

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3.9 Buffer solutions

The solutions which resists the change in its pH value on addition of small amount of acid or base are called buffer solutions. On adding small amount of acid or base there is no significant Change in pH of the buffer solution.

(a) Acidic Buffer:

It contains mixture of weak acid and its salt with strong base. e.g. mixture of CH3 COOH and CH3COONa.

For acidic buffer of CH COOH and CH COONa3 3

We have in solution - CH COOH3 molecules, CH3COO– ions and Na+ ions.

Note: Dissociation of weak electrolyte is suppressed in the presence of common ion CH3COO–

from CH3COONa.

When acid is added (say x moles)-

H+_{+ CH}₃_COONa_↽⇀_CH3COOH + Na+ Hence a salt x pH pk log acid x − = + +

and when base in added-

OH–_{+ CH}₃_COOH_↽⇀_CH3COO– + H2O Thus a salt x pH pk log acid x + = + − ∵_{x is small} [Salt] – x = [Salt] [Acid] + x = [Acid] ∴ ] Acid [ ] Salt [ log a p

pH==== K ++++ or _pH _pKa _log[Conjugate Base] [Acid]

= +

(b) Basic Buffer:

It is the mixture of weak base and its salt with strong acid. e.g. NH4OH and NH4Cl.

For basic buffer solution of NH4OH and NH4Cl. We have in solution- NH4OH, NH₄+ ions and Cl–

ions.

Note: Dissociation of weak electrolyte NH4OH is suppressed in the presence of common ion NH₄+

from NH4Cl.

When x moles of acid are added H NH OH4 NH4 H O2

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Sri Chaitanya IIT-JEE Head Office: 11, Central Market, West Punjabi Bagh , New Delhi-110 026. Phones: 25226309/10. _pOH _pKb _logsalt x base x + = + −

and when x moles of base are added

OH NH4 −₊ + _⇀ ↽ NH OH4 _pOH _pKb _logsalt x base x − = + + pOH ₌ ] Base [ ] Salt [ log b

pK ++++ ∵ [Salt−x]~_[Salt]&[Base+x]~_[Base]

(c) Aqueous solution of salt of weak acid and weak base can also behave like buffer solution.

Buffer capacity:

Defined as the number of moles of acid or base required to be added to one litre of buffer solution in order to change its pH by unity.

Buffer capacity number of moles of acid or base added change in pH

=

Illustr ations

Illustration 7

A 40 ml solution of weak base BOH is titrated with 0.1 N–HCl solution. The pH of solution is found to be 10.04 and 9.14 after addition of 5 ml and 20 ml if acid respectively. Find Kb for weak base. Solution: Let m eq of BOH is a I. BOH + HCl → BCl + _H2O a 0.5 0 0 (a–0.5) 0 0.5 0.5 ] Base [ ] Salt [ log b p pOH= K + (14 – 10.04) = ] 05 . 0 a [ ] 5 . 0 [ log b pK − + ...(i)

II. BOH + HCl → BCl + _H2O

a 2 0 0

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(

)

(

)

Kb 2 14 9.14 p log a 2 − = + − ...(ii)

On solving equation (i) and (ii) Kb = 1.81×10–5

Illustration 8

Calculate pH of buffer prepared by dissolving 30 gm of Na2CO3 in 500 ml of aqueous solution

containing 150 ml of 1M, HCl (Ka for 11

3

HCO−=5.63 10× − ). Solution:

_{Na2CO3 +} HCl → NaCl + _NaHCO3

30 1000

106× 150×1

Initial = 283 m moles = 150 m moles 0 0

after reaction 133 0 150 150 H

(

11

)

133 p log 5.63 10 log 150 − = − × + = 10.249 – 0.052 = 10.197 Illustration 9

CH3COOH (50 ml, 0.1 M) is titrated against 0.1 M NaOH solution. Calculate the pH after the

addition of 0 ml, 10 ml, 20 ml, 25 ml, 40 ml, 50 ml and 60 ml of NaOH. Ka of CH3COOH is

2 × 10– 5_.

Solution:

(i) When 0 ml of NaOH is added, the pH calculation should be done due to acetic acid only.

∴ [H+_{] =} 5 6 a c 2 10 0.1 2 10 K × = × − × = × − pH = – log 2 1 10 2× −6=− [log 2 – 6] = 3 – 0.15 = 2.85

(ii) When 10 ml of NaOH is added, it reacts with CH3COOH to produce salt, CH3CO2Na and

water. Some CH3CO2H would be left behind and CH3CO2Na is produced. So, the solution is that

of an acidic buffer. ∴ pH = pKa + log ] Acid [ ] Salt [ = 4.699 + log ) 1 . 0 10 ( ) 1 . 0 50 ( 1 . 0 10 × − × × = 4.699 + log 4 1 = 4.0969 (iii) When 20 ml of NaOH is added,

pH = pKa + log ) 1 . 0 20 ( ) 1 . 0 50 ( 1 . 0 20 × − × × = 4.699 + log 3 2 = 4.5229

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Sri Chaitanya IIT-JEE Head Office: 11, Central Market, West Punjabi Bagh , New Delhi-110 026. Phones: 25226309/10. (iv) When 25 ml of NaOH is added,

pH = 4.699 + log ) 1 . 0 25 ( ) 1 . 0 50 ( 1 . 0 25 × − × × = 4.699 (v) When 40 ml of NaOH is added,

pH = 4.699 + log ) 1 . 0 40 ( ) 1 . 0 50 ( 1 . 0 40 × − × × = 4.699 + log 4 = 5.3011 (vi) When 50 ml of NaOH is added

Here, if we use the buffer equation, pH would be = ∞

The buffer equation cannot be used, as there is no acid. Therefore we will use the hydrolysis equation. ∴ [H+_{] =} c K Kw a c = 2 1 . 0

[∵ Total volume is 100 ml and millimoles of salt is 50 × 0.1]

[H+_{] =} 1 . 0 2 10 2 10−14× × −5× pH = 8.699

(vii) When 60 ml of NaOH is added, the excess of OH–_{ions from NaOH would suppress the}

hydrolysis of CH3COO– ion. So we can ignore the contribution of OH– ion from the hydrolysis of

CH3COO– ion and pH calculation should be done with OH– from NaOH only.

∴ [OH–_{] =} 110 1 110 10 1 . 0 × ₌ pOH = 2.0414 ∴ pH = 14 – 2.0414 = 11.9586

Practice Exercise

5. 50 mL of 0.1 M NaOH is added to 75 mL of 0.1 M NH4Cl to make a basic buffer. If pKa of

+ 4

NH 9.26, calculate pH.

6. (a) Determine the pH of a 0.2 M solution of pyridine C5H5N . Kb = 1.5 × 10– 9.

(b) Predict the effect of addition of pyridinium ion C5H5NH+ on the position of the

equilibrium. Will the pH be raised or lowered?

(c) Calculate the pH of 1.0 L of 0.10 M pyridine solution to which 0.3 mol of pyridinium chloride C5H5NH+Cl, has been added, assuming no change in volume.

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Answers

5. 9.56 6. (a) pH = 9.239, (b) lowered (c) pH = 4.699

3.10 Indicators

Indicators are the substances, which indicates the end-point of a titration by changing their colour. They are in general, either weak organic acids or weak organic bases having characteristically different colours in the ionized and unionized forms. For example, methyl orange is a weak base (having red colour in ionized form and yellow colour in the unionized form) and phenolphthalein is a weak acid.

Let us consider the equilibrium between the ionized and unionized form of an acid indictor (HIn).

HIn _↽⇀_H+_{+ In}–

∴ KHIn = [H ] [In ]

[HIn]

+ −

[KHIn = Indicator constant or dissociation constant of indicator]

or [H+_{] = K}_HIn_×[HIn]

[In ]−

Taking negative logarithm of both sides – log [H+_{] = – log K}_HIn_{– log}[HIn]

[In ]−

∴ pH = pKHIn + log [In ]

[HIn]

−

pH = pKHIn + log [Ionised form]

[Unionised form]

Case I.

In order for the solution to show colour of In–_{, the minimum ratio of}[In ]

[HIn]

−

should be 10.

∴ pH = pKHIn + log(10) = pKHIn + 1

In fact pH = pKHIn + 1 is the minimum pH up to which the solution has a distinct colour

characteristic of In–_{. At pH greater than this value, some more indicator will be present in the}

ionized form. Thus at pH ≥ pKHIn + 1, the solution has a colour characteristic of In–.

Case II.

In order for the solution to show colour due to HIn, the minimum ratio of [In ]

[HIn]

−

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∴ pH = pKHIn + log 1

10 = pKHIn – 1

In fact pH = pKHIn – 1 is the maximum pH up to which the solution has a distinct colour

characteristic of HIn. At pH smaller than this value, some more indicator will be present in the unionized form. Thus at pH ≤ pKHIn – 1, the solution has a colour characteristic of HIn.

Therefore, in between the pH range pKHIn – 1 to pKHIn + 1, transition of colour takes place for any

acid-base indicator. pKHIn± 1 is called the range of indicator.

Practice Exercise

4. A certain solution has a hydrogen ion concentration 4 × 10– 3_{M. For the indicator thymol}

blue, pH is 2.0 when half the indicator is in unionized form. Find the % of indicator in unionized form in the solution with [H+_{] = 4 × 10}– 3_M.

Answers

7. [HIn] = 28.57%

3.11 Solubility Product

It is the product of the molar concentrations of the ions in a saturated solution of an sparingly soluble salt with each concentration term raised to the power equal to the number of times that ion appears in balanced equation that represents equilibrium. It is denoted by Ksp.

A Bx y

( )

s ↽⇀ ( )

( )

y x aq xA + +yB− aq x y y x x y A B K A B + −         =    

In saturated solution [AxBy] = constant

y x x y x y K A B_ _=_A + _{ }B −_ y x x y sp K =_A+ _{ }B−_

e.g. for AgCl, AgCl(s) _↽⇀ Ag+

( )

aq +Cl−

( )

aq

Ksp Ag Cl

+ −

   

=_ _{ } _{ = s}2 _{where s is the solubility of AgCl in moles/l}

For Ag CrO2 4

( )

s ↽⇀ 2Ag (aq) CrO (aq)

2

4−

+ ₊

2 ₂

sp 4

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Ksp depends on temperature and it is the highest limit of ionic product of the electrolyte in solutions.

Applications of Solubility Product

(i) It helps in predicting the formation of a precipitate.

If ionic product > Ksp, precipitation occurs and if ionic product < Ksp no precipitate is formed.

(ii) Calculation of solubility of sparingly soluble salt - let solubility of salt AxBy in water at a

particular temperature is S mole per litre. Then at equilibrium.

AxBy↽⇀ y x xs ys xA ++yB −

[ ] [ ]

x y x y x y sp K = xs ys =x y s +

e.g. for AgCl _↽⇀

s s Ag++Cl− _{Ksp = S2} Hence S= Ksp For Ag2CrO4↽⇀ 2₄ 2s s 2Ag CrO+ −

( ) ( )

2 3 sp K = 2S S =4S 1/ 3 sp K S 4   =    for Ca3

(

PO4 2

)

↽⇀ 2 3 4 3Ca++2PO−

( ) ( )

3 2 5 sp K = 3S 2S =108S 1/ 5 sp K S 108   =   

(iii) In qualitative analysis

The separation and identification of various basic radicals into different groups is based upon (a) solubility product principle and (b) common ion effect.

(iv) Purification of common salt.

Saturated solution of impure common salt is prepared and insoluble impurities are filtered off. HCl gas is passed through this solution. Thus, ionic product of Na_ + _{ }Cl−_{ exceeds the}

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Selective Precipitation

Let us have a solution containing more than one ion capable of forming a precipitate with another ion, which is added to the system. The added ion will selectively form precipitate with any one of the ions present in the solution. This process of selectively or preferentially precipitating an ion from a solution of more than one ion is called selective precipitation. The selective precipitation of ions from a solution in the form of a salt, (which is partially soluble) can be done by adding precipitating agent drop by drop.

If the stoichiometry of the precipitated salts is same, then the salt with minimum solubility product (and hence also the minimum solubility) will precipitate first, followed by the salt of next higher solubility and so on. For example, in a solution containing Cl–_{, Br}–_{and I}–_{ions, when Ag}+

ions are added, then out of the three, the least soluble silver salt is precipitated first. If the addition of Ag+_{ions is continued further, eventually a stage will be reached when the next lesser}

soluble salt starts precipitating along with the least soluble salt and so on.

If the stoichiometry of the precipitated salts is not the same, then from the solubility product data alone, we will not be straight away able to predict, which ion will precipitate first. Let us take a solution containing 0.1 M each of Cl–_and 2−

4

CrO ion and the precipitating ion used is Ag=_{. We are given the solubility products of AgCl and Ag}₂_CrO₄_{as 1 × 10}– 10_M2_{and 1 × 10}– 13_M3

respectively.

We know that the precipitation takes place only if the ionic product exceeds solubility product. Though the solubility product of Ag2CrO4 is less than that of AgCl, yet it is AgCl (lesser

soluble) that precipitates first when Ag+_{ions are added to the solution. Thus order to predict}

which ion precipitates first, we will first calculate the concentration of Ag+_{required to make each}

of their ionic products equal to solubility products.

( )

AgCl s _↽⇀_Ag+_{(aq) + Cl}– _(aq)

Ksp(AgCl) = [Ag+] × [Cl–]

Minimum concentration of Ag+_{required to precipitate 0.1 M Cl}–_{as AgCl would be}

[Ag+_{] =} 1 . 0 10 1 ] Cl [ ) AgCl ( K_sp −10 − × = = 1 × 10– 9_M

Ag2CrO4(s) ↽⇀ 2Ag+(aq) + CrO2₄−(aq)

Ksp(Ag2CrO4) = [Ag+]2 × [CrO2₄−]

Minimum concentration of Ag+_{required to precipitate 0.1 M} 2−

4

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[Ag+_{] =}

(

)

1 . 0 10 1 ] CrO [ CrO Ag K 13 2 4 4 2 sp − − = × = 1 × 10– 6 M

When AgNO3 is added to the solution, the minimum of the two concentrations of Ag+

needed to start the precipitation will be reached first and thus the corresponding ion (Cl–_{in this}

case) will be precipitated in preference to the other.

When the concentration of Ag+_{becomes equal to 1 × 10}– 9_{M, AgCl starts precipitating.}

During the course of precipitation, concentration of Cl–_{decreases and the corresponding}

concentration of Ag+_{to start the precipitation increase. A stage will reach when the concentration}

of Ag+_{becomes equal to 1 × 10}– 6_{m, which is required to precipitate} 2−

4

CrO ion. The addition of more of AgNO3 causes the precipitation of both the ions together but at this stage, practically

whole of Cl–_{ions have been precipitated.}

Illustr ations

Illustration 10

Calculate pH at which Mg(OH)2 begins to precipitate from a solution containing 0.1 M Mg2+ ions.

Ksp of Mg(OH)2 = 10–11. Solution:

( )

2 Mg OH _↽⇀ 2 Mg ++2OH− 2 2 11 sp K =_Mg+ _{ }OH−_ =10− 2 ₁₁ 0.1×_OH−_ =10− 5 OH− 10−  ₌   OH p =5 H P = − =14 5 9 Illustration 11

A mixture of water and AgCl is shaken until a saturated solution is obtained. Now 100 ml of this solution is mixed with 100 ml of 0.03 M.NaBr. should a precipitate form?

_{Ksp of AgCl and AgBr are 10–10 and 5×10–13 respectively.} Solution: 5 sp Ag+ K of AgCl 10 M−  ₌ ₌  

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Sri Chaitanya IIT-JEE Head Office: 11, Central Market, West Punjabi Bagh , New Delhi-110 026. Phones: 25226309/10. after mixing 5 10 100 Ag 200 − + ×  ₌   and Br 100 0.03 200 − ×  ₌   Thus __Ag+_{ }_Br−_₌_{7.5 10}_× −8    

Which is greater than 5×10–13 hence precipitate of AgBr will be obtained. Illustration 12

Find the concentration of NH3 solution whose 1 litre can dissolve 0.1 mole of AgCl. Ksp of AgCl =

10–10 and Kf of

(

)

7 3 2 Ag NH + 1.6 10   ₌ _×   Solution: Ag 2NH3 +₊ _⇀ ↽ Ag NH

(

_{3 2}

)

+ AgCl s

( )

_↽⇀_Ag++Cl− Ksp Ag Cl + −     =_ _{ } _

(

)

[

]

4 2 3 f 2 3 Ag NH K Ag+ NH     =     Thus

(

)

[

]

3 2 sp f 2 3 Ag NH Cl K K NH + ₋         × =

[

]

10 7 2 3 0.1 0.1 10 1.6 10 NH − _× _× ₌ ×

[

3

]

3 0.1 0.1 NH 6.25 2.5 1.6 10− × = = = ×

Since 0.2 moles NH3 is needed to dissolve 0.1 moles Ag+. Thus required [NH3] = 2.7 M

Practice Exercise

8. The solubility of PbSO4 water is 0.038 g/L. Calculate the solubility product constant PbSO4.

9. Equal volumes of 0.02 M AgNO3 and 0.02 M HCN were mixed. Calculate [Ag+] at

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Answers

8. 1.6 × 10– 8 _9._[Ag+_{] = 6.667 × 10}– 5_M

3.12 Salt Hydrolysis

Salts are the non-water product of an acid base neutralization. There are four possible acid base reactions that produce salts. They are the reaction of a:

1. Strong acid with a strong base, e.g., HCl + NaOH → NaCl + H2O

2. Weak acid with a strong base, e.g., CH3COOH + NaOH → CH3COONa + H2O

3. Strong acid with a weak base, e.g., HCl + NH4OH → NH4Cl + H2O

4. Weak acid with a weak base, e.g., CH3COOH + NH4OH → CH3COONH4 + H2O

The combination of salt with water to form acidic or basic solution is called hydrolysis of salt. Actually hydrolysis of cations and anions takes place not that of salt.

One should remember that solvation and hydrolysis are different phenomenon’s.

1. Salt of strong acid and strong base: e.g. NaCl

NaCl→Na++Cl− Na++H O₂ ×→NaOH+H+

Cl H O2 HCl OH

−₊ __×_→ ₊ −

H2O ↽⇀ H+ + OH–

Since NaOH and HCl both are strong electrolytes hence no hydrolysis occurs in case of salt of strong acid and strong base. Thus pH value of solution would be 7 at 25°C.

2. Salt of weak acid and strong base:

e.g. CH3COONa, salt will dissociate completely in the solution as:

_CH3COONa→_{CH3COO– +Na+}

_CH3COO–_{+ H2O}_↽⇀_{CH3COOH + OH}– _...(i)

Kh = ] COO CH [ ] OH [ ] COOH CH [ 3 3 − −

Where Kh is called hydrolysis constant.

Note: Due to the presence of OH– ions. Solution becomes basic.

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Sri Chaitanya IIT-JEE Head Office: 11, Central Market, West Punjabi Bagh , New Delhi-110 026. Phones: 25226309/10. Ka = ] COOH CH [ ] H [ ] COO CH [ 3 3 − +

and H2O _↽Kw⇀_{H+ + OH–} _...(iii)

Kw = [H+] [OH–]

From equations (i), (ii) and (iii)

equation (i) = equation (iii) ÷ equation (ii)

or w h a K K K = Now 3 2 C(1 h) CH COO− H O − + ⇀ ↽ 3 Ch Ch

CH COOH OH+ − where C = initial conc. of CH3COO–

Where h be the degree of hydrolysis.

(

)

2 h Ch Ch Ch K C 1 h 1 h × = = − − If h < < 1 Then Kh = Ch2 & h w a K K h C K C = = × Now w w a a K K C [OH ] Ch C K C K − ₌ ₌ ₌ × × _H Kw Kw Ka C OH + − ×  ₌ ₌   _ _   Or 1 7 1 2 2 w a a K K K H p = _p +p +log C_= + _p +log C_

3. Salt of strong acid and weak base:

NH Cl4 NH4 Cl + − → + NH4 H O2 +₊ Kh 4 NH OH+H+ ...(i) Kh = [NH4OH] [H+] / [NH+₄] NH4+OH Kb 4 NH++OH− ...(ii) Kb = [NH+₄] [OH–] / [NH4OH] and H2O Kω H+_{+ OH}– _...(iii) Kw = [H+] [OH–]

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from all three equations w h b K K K =

Note : Due to presence of H+, solution becomes acidic.

4 2 C(1 h) NH+ H O − + ⇀ ↽ 4 Ch Ch NH OH H+ +

(

)

2 h b K Ch Ch Ch K K C 1 h 1 h ω × = = ₋ = ₋ If h < < 1 Then w 2 b K Ch K = or h = w b K K ×C [H+_{] = Ch =} w b K C K × or _pH 1 _pKw _pKb _{log C} 2   = _ − − _{ = 7 –}      _p b₊_log_C 2 1 K

4. Salt of weak acid and weak base

e.g. CH3COONH4 CH COONH3 4 CH COO3 NH4 − + → + CH COO3 NH4 H O2 −₊ +₊ Kh 3 4 CH COOH+NH OH ...(i) Kh = ] NH [ ] COO CH [ ] OH NH [ ] COOH CH [ 4 3 4 3 + − CH COOH3 Ka 3 CH COOH−+H+ ...(ii) Ka = ] COOH CH [ ] H [ ] COO CH [ 3 3 − + NH OH4 Kb 4 NH++OH− ...(iii) Kb = ] OH NH [ ] OH [ ] NH [ 4 4 − + H2O Kω H++OH− ...(iv)

from all four equations. w

h a b K K K K = ×

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Sri Chaitanya IIT-JEE Head Office: 11, Central Market, West Punjabi Bagh , New Delhi-110 026. Phones: 25226309/10. How 3 4 2 C(1 h) C(1 h) CH COO− NH+ H O − − + + ₃ ₄ Ch Ch CH COOH+NH OH Kh =

( ) ( ) ( )

2 2 h 1 h h 1 C h 1 C Ch Ch − = − × − × or 1 w h a b K h K h= = K .K − How

[

3

]

a 3 CH COO H K CH COOH − +         =

[

]

(

)

a 3 a a 3 K CH COOH K Ch K h H C 1 h 1 h CH COOH + − ×  ₌ ₌ ₌   _ _ ₋ ₋   w a w a a b b K K K K K K K × = = × or pH₌1 2 w a b K K K p p p  ₊ ₋    = 7 + _p a−p b_ 2 1 K K

Illustr ations

Illustration 13

When 0.2 M acetic acid is neutralized with 0.2 M NaOH in 0.5 litre of water the resulting solution is slightly alkaline. Calculate the pH of the resulting solution. Ka for CH3COOH = 1.8 × 10– 5.

Solution

0.2 M acetic acid will form 0.2 M CH3COONa in 0.5 litre of water. Hence, concentration of sodium

acetate, [CH3COONa] = 0.1 mol L– 1.

CH3COO– + H2O ↽⇀ CH3COOH + OH– C(1 – h) Ch Ch Kh =

(

)

2 2 Ch Ch 1 h− = (1 – h) → 1 Kh = 5 14 a w 10 8 . 1 10 1 K K − − × × = = 5.5 × 10– 10 So, Kh = Ch2 = 5.5 × 10– 10 or, h2₌ 1 . 0 10 5 . 5 × −10 = 55 × 10– 10 or, h = 7.42 × 10– 5

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[OH–_{] = Ch = 7.42 × 10}– 5_{× 0.1 = 7.42 × 10}– 6_M [H+_{] =} 6 14 w 10 42 . 7 10 1 ] OH [ K − − − = ×_× = 1.3477 × 10– 9 M pH = – log [H+_{] = – log (1.3477 × 10}– 9_{) = 8.87} Illustration 14

Calculate the pH at the equivalence point when a solution of 0.1 M acetic acid is titrated with a solution of 0.1 M sodium hydroxide.

Ka for acetic acid = 1.9 × 10– 5

Solution

Concentration of sodium acetate =

2 1 . 0

= 0.05 M as equal volumes of the acid and the base will be used.

The equilibrium is

CH3COO– + H2O ↽⇀ CH3COOH + OH–

C(1 – x) Cx Cx

where x is the degree of hydrolysis, and Kh =

(

1 x

)

Cx2 − We know that, Kh = 5 14 a w 10 9 . 1 10 1 K K − − × × = = 5.26 × 10– 10 So, Kh = Cx2 as (1 – x) → 1 5.26 × 10– 10_{= 0.05 × x}2 or, x2₌ 05 . 0 10 26 . 5 × −10 = 1.05 × 10– 8 or, x = 1.025 × 10– 4 [OH–_{] = Cx = 1.025 × 10}– 4_{× 0.05 = 5.125 × 10}– 6_M [H+_{] =} 6 14 10 125 . 5 10 1 − − × × = 1.95 × 10– 9_M pH = – log [H+_{] = – log (1.95 × 10}– 9_{) = 8.71}

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Illustration 15

Calcium lactate is a salt of a weak organic acid and represented as Ca(Lac)2. A saturated solution

of Ca(Lac)2 contains 0.13 mole of this salt in 0.50 litre solution. The pOH of this solution is 5.60.

Assuming a complete dissociation of salt, calculate Ka of the lactic acid.

Solution Ca(Lac)2 ↽⇀ Ca2 +_{+ 2Lac}– 0.13 × 2 M 2 × 2 × 0.13 M Lac– _{+ H}₂_O_↽⇀_{HLac + OH}– At equilibrium (0.52 – x) x x Kh =

(

)

0.52 x x 52 . 0 x2 2 = − as x is small [OH–_{] = 10}– 5.6_{= 2.51 × 10}– 6_{= x} Kh = 52 . 0 10 51 . 2 10 51 . 2 × −6× × −6 = 12.12 × 10– 12 Ka = 12 14 h w 10 12 . 12 10 K K − − × = = 8.26 × 10– 4

Practice Exercise

9. Calculate the percent hydrolysis in a 0.06 M solution of KCN [Ka(HCN) = 6 × 10– 10]

10. 0.25 M solution of pyridinium chloride C5H6N+Cl– was found to have a pH of 2.699. What is

Kb for pyridine, C5H5N?

Answers

9. 1.667% 10. Kb = 6.25 × 10– 10

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Miscellaneous

Miscellaneous Problems

Objective Type

Example 1

Two litre of a saturated solution of CaCO3 is evaporated to dryness due to which 14.0 mg of residue is left.

The solubility product for CaCO3 is

(a) 4.9 × 10– 6 _{(b) 4.9 × 10}– 4 _{(c) 4.9 × 10}– 9 _{(d) 4.9 × 10}– 10

Solution

Moles of CaCO3 in residue =

3 14 10 100 − × = 14 × 10– 5

Moles of CaCO3 in 1 litre solution = 7 × 10– 5

CaCO3(s)↽⇀ Ca2 + + CO₃2− 7 × 10– 5_{7 × 10}– 5 Ksp = [Ca2 +] × [CO2₃−] = 7 × 10– 5_{× 7 × 10}– 5_{= 4.9 × 10}– 9 Ans. (c) Example 2

Consider the reaction A–_{+ H}3O+_↽ ⇀_{HA + H}2O. The Ka value for acid HA is 1.0 × 10– 8_{. What is the value}

of K for this reaction.

(a) 1.0 × 106 _{(b) 1.0 × 10}– 8 _{(c) 1.0 × 10}8 _{(d) 1.0 × 10}– 6 Solution K = _- ₊ ₈ a [HA] 1 1 K [A ] [H ]= =1.0 10× − = 1.0 × 10 8 Ans. (c) Example 3

The ionization constant of HCO2H is 1.8 × 10– 4. What is the percent ionization of a 0.001 M solution?

(a) 66% (b) 42% (c) 34% (d) 58% Solution HCO2H ↽⇀ H+ + HCO₂− 0.001 – x x x Check for α α = Ka c α = 4 1.8 ×10 0.001 − = 0.42

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Sri Chaitanya IIT-JEE Head Office: 11, Central Market, West Punjabi Bagh , New Delhi-110 026. Phones: 25226309/10. ∴ Ka = 1.8 × 10– 4 = 2 x 0.01 x− 5.55 x2_{+ x – 0.01 = 0} 2 1 1 4 5.55 0.01 x 2 5.55 − ± − × × = × x = 3.4 × 10– 4 ∴ % ionization = 4 2 2 ionized HCO H 3.4 10 100 100 34% total HCO H 0.001 − × × = × = Ans. (c) Example 4

A weak base (BOH) with Kb = 10– 5 is titrated with a strong acid HCl. At 3/4th of the equivalence point, pH of

the solution is

(a) 5 + log 3 (b) 5 – log 3 (c) 14 – 5 – log 3 (d) 8.523 Solution

Let the initial equivalent of BOH are x. BOH + HCl → BCl + H2O x 3/4x 0 0 3 x x x 4 4 − = 0 3x 4 3 x 4

pOH = pKb + log [Salt] 5 log3x 4

[Base] 4 x × = + × pH = 14 – 5 – log 3 = 8.523 Ans. (c, d) Example 5

At – 50°C, the self-ionization constant (ion product) of NH3 is 3

+ 30

NH 4 2

K =[NH ] [NH ] 10− = − . How many amide ions are present per mm3_{of pure liquid ammonia?}

(a) 600 ions/mm3 _{(b) 6 × 10}6_ions/mm3 _{(c) 6 × 10}4_ions/mm3 _{(d) 60 ions/mm}3

Solution

2NH3(l) ↽⇀ NH+₄ + NH₂− 1 – x x x (Neglecting x in comparison to 1 since

3 NH K is very small) 3 2 30 NH K =x =10− thus, x = 10– 15_{M =} 2 [NH ]− [NH ]2− = 10– 15 moles/lit = 15 6 10 10 − moles/mm3 [NH ]2− = 10– 21 × 6 × 1023 ions/mm3 = 600 ions/mm3 Ans. (a)

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Example 6

The hydrolysis constant for ZnCl2 will be

(a) Kh = w b K K (b) Kh = 2 w b K K (c) Kh = 2 w 2 b K K (d) K h = b 2 w K K where Kb is effective dissociation constant of base Zn++

Solution Zn++_{+ 2H}₂_O_↽ ⇀_Zn(OH)2 + 2H+ ∴ Kh = + 2 2 ++ [Zn(OH) ] [H ] [Zn ] Zn(OH)2↽⇀ Zn++ + 2OH– ∴ Kb = ++ 2 2 [Zn ] [OH ] [Zn(OH) ] − Kw = [H+] [OH–] ∴ 2 w b K K = Kh Ans. (b) Example 7

The precipitate of Ag2CrO4 (Ksp = 1.9 × 10– 12) is obtained when equal volumes of the following are mixed

(a) 10– 4_{M Ag}+_{+ 10}– 4_M 2 4 CrO − (b) 10– 2_{M Ag}+_{+ 10}– 3_M 2 4 CrO− (c) 10– 5_{M Ag}+_{+ 10}– 3_M 2 4 CrO − (d) 10– 4_{M Ag}+_{+ 10}– 5_M 2 4 CrO− Solution

Precipitation occurs when the ionic product exceeds the Ksp value. When equal volumes of two solutions

are mixed the concentration of each is reduced to half. Therefore, In first case,

Ionic product, I.P. =

2 4 4 1 1 1 10 10 2 2 8 − −     × × =         × 10 – 12_{= 1.25 × 10}– 13 As, I.P. < Ksp ∴ no precipitation occurs In second case, I.P. = 2 2 3 1 1 1 10 10 2 2 8 − −     × × =         × 10 – 7_{= 1.25 × 10}– 8 As, I.P. > Ksp ∴ precipitation occurs In third case, I.P. = 2 5 3 1 1 1 10 10 2 2 8 − −     × × =         × 10 – 13_{= 1.25 × 10}– 14 As, I.P. < Ksp ∴ no precipitation occurs In fourth case,

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Sri Chaitanya IIT-JEE Head Office: 11, Central Market, West Punjabi Bagh , New Delhi-110 026. Phones: 25226309/10. I.P. = 2 4 5 1 1 1 10 10 2 2 8 − −     × × =         × 10 – 13_{= 1.25 × 10}– 14 As, I.P. < Ksp ∴ no precipitation occurs Ans. (b) Example 8

A buffer solution is prepared which is 0.50 M CH3COOH and 0.25 M CH3COONa. Which of the following

ions can be maintained at a concentration of 0.10 M or greater without precipitating as the hydroxide from this solution?

(Given : Ksp (Ca(OH)2) = 5.5 × 10– 6_{, K}sp (Al(OH)3) = 1.3 × 10– 33_,

Ksp (Cr(OH)3) = 6.3 × 10– 31)

(a) Ca2 + _{(b) Cr}3 +

(c) Al3 + _{(d) All ions will be precipitated}

Solution

pH = pKa + log [salt] [acid]

= 4.74 + log 0.25 0.5 = 4.438 pOH = 9.562

[OH–_{] = 2.74 × 10}– 10

Ionic product of Ca(OH)2 = [Ca2 +_{] [OH}–_]2_{= (0.1) × (2.74 × 10}– 10₎2_{= 7.5 × 10}– 21

Ionic product of Al(OH)3 = (0.1) × (2.74 × 10– 10)3 = 2.05 × 10– 30

Ionic product of Cr(OH)3 = (0.1) × (2.74 × 10– 10)3 = 2.05 × 10– 30

In case of only Ca(OH)2 ionic product < Ksp

Ans. (a) Example 9

Aqueous tension at 20°C is 16 mm of Hg. A one molar solution of weak base BOH show the vapour pressure of 15.6 mm of Hg at 20°C. If a salt of BOH with acetic acid (Ka = 1.27 × 10– 5 at 20°C) is dissolved in

water then the solution will be

(a) Neutral (b) Basic (c) Acidic (d) Buffer

Solution

Mole fraction of solute, XBOH = 2 H O P 16 15.6 0.4 0.025 16 16 P° ∆ ₌ − ₌ ₌

Mole fraction of solvent, XH O₂ =0.975

Molality ≈ Molarity = XH O2 1000 _1.425

0.975 18

× = ×

As for a very dilute solution molality and molarity can be assumed to be same. BOH B+_{+ OH}–

1 0 0 1 – α α α 1 + α = 1.425; α = 0.425

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Now, Kb = 2 1 0.425 0.425 3.14 10 1 1 0.425 − α × = = × − α − Since, Kb > Ka

Hence salt solution will be basic due to hydrolysis Ans. (b)

Example 10

0.15 mole of pyridinium chloride has been added to 500 cm3_{of 0.2 M pyridine solution. What is the pH of}

the resulting assuming no change in volume? (Kb for pyridine = 1.5 × 10– 9 M)

(a) 9 (b) 5 (c) 6 (d) 8

Solution

pOH = pKb + log [salt]

[base] = – log (1.5 × 10– 9_{) + log} 3 0.15 500 0.2 10× × − = 9 – log 1.5 + log 1.5 = 9 pH = 14 – 9 = 5 Ans. (b)

Subjective Type

Example 1

Find the pH of 0.001M acetic acid solution, if it is 1% ionised at this dilution Solution 1 0.01, c 0.001 100 α = = = [H+] = c α = 0.001 × 0.01 = 0.00001 = 10–5 pH = 5 Example 2

How many moles of Ca (OH)2 must be dissolved to produce 250 ml of an aq. solution of pH = 10.65? Solution

pH + pOH = 14 pOH = 14 – 10.65 = 3.35 pOH = – log [OH–] 3.35 = – log [OH–]

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Sri Chaitanya IIT-JEE Head Office: 11, Central Market, West Punjabi Bagh , New Delhi-110 026. Phones: 25226309/10. or [OH–] = 4.47 × 10–4

No. of moles of OH– in 250 ml. =

4 4 4.47 10 1.12 10 4 − − × ₌ _×

No. of moles of Ca (OH)2 dissolved = 1 _{1.12 10}4 2

−

× × = 0.56 × 10–4 Example 3

What is the pH of a 500 ml aqueous solution containing 0.05 mol of NaOH? Solution

Wt. of NaOH = No of moles × Mol. wt. = 0.05 × 40 = 2g Volume = 500 ml, Eq. wt. of NaOH = 40

∴ N Wt. of NaOH 1000 2 1000 1 0.1N Eq. wt volume 40 500 10

× ×

= = = =

× ×

∴ pOH = – log10 [OH–]

∴ pOH = 1

∴ pH = 14 – pOH = 14 – 1 = 13 Example 4

Calculate the pH of a 0.001 M solution of Ba (OH)2 assuming it to be complete ionised. Solution

2 2

Ba(OH) →Ba++2OH−

Thus 1 mole of Ba(OH)2 gives 2 moles of OH– ions 3 2

[OH ]− =2[Ba(OH) ]= ×2 0.001= ×2 10 g ions / litre− 14 12 3 1 10 [H ] 5 10 2 10 − + − − × = = × × pH = – log (5 × 10–12) = – (0.699 – 12) = – (– 11.301) = 11.301 Example 5

Two acids A1 and A2 have their dissociation constant as 0.00018 and 0.0037 respectively at 25°C. What would be the relative dilution of the acids so that the solution become isohydric?

Solution

For isohydric solutions

1 1 2 2 V V α = α ... (i)

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According to Ostwald’s law when degree of ionisation is very low, 2 1 K V1 1 0.00018V1 α = = 2 2 K V2 2 0.0037V2 α = =

Substituting the above values in equation (i) 2 1 1 2 2 2 V 0.00018V 0.0037V V =

Thus the relative dilution of acids A1 and A2 is 1 : 205.5. Example 6

Calculate the pH of a solution obtained by mixing equal volumes of 0.10N ammonium nitrate and 0.02N ammonium hydroxide, Kb for NH4OH is 1.8 × 10–5.

Solution Before mixing

[NH4NO3] = 0.10N = 0.10M [NH4OH] = 0.02N = 0.02M

After mixing. Since the volume after mixing becomes double to that of either of the individual component, its concentration in the resulting solution becomes half, i.e.,

4 3 0.10 [NH NO ] 0.05M 2 = = 4 0.02 [NH OH] 0.01M 2 = =

Now substituting the values in the following equation pOH = pKb + log [Salt]

[Base] = – log 1.8 × 10–5 + log = 4.7447 + 0.6990 = 5.4437

Now since pH = 14 – pOH = 14 – 5.4437 = 8.5563 ≈8.56 Example 7

What is [H+] in a solution obtained when 0.1M ammonia solution is just neutralised by a strong HCl? The dissociation constant of ammonia is 1.8 × 10–5.

Solution Conc. of NH3 solution = 0.1 M, 4 2 4 NH H O NH OH H c(1 h) ch ch +₊ __→ ₊ + − 14 4 w 15 b K 10 h 0.745 10 K c 1.8 10 0.1 − − = = = × × × × [H+] = ch = 0.1 × 0.745 × 10–4 = 7.45 × 10–6

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Sri Chaitanya IIT-JEE Head Office: 11, Central Market, West Punjabi Bagh , New Delhi-110 026. Phones: 25226309/10. Example 8

In 0.3M solution of NH4Cl, H+ ion concentration is 1.3 × 10–5M. What is the dissociation constant of NH3? Solution 4 4 NH Cl→NH++Cl− 4 2 3 3 NH (aq) H O NH (aq) H O (0.3 x)M xM xM + ₊ __→ ₊ + − 5 x=[H ] 1.3 10+ = × − w 3 3 h b 4 K [NH ] [H O ] K K [NH ] + + = = 14 5 2 b 10 x x (1.3 10 ) K 0.3 0.3 − _× _× − = = 14 b 5 2 0.3 10 K (1.3 10 ) − − × = = × 1.8 × 10–5 Example 9

The dissociation constant of CH3COOH is 1.6 × 10–5. The degree of dissociation d, of 0.01 molar CH3COOH in presence of 0.1 M HCl is equal to

(1) 0.016 (2) 0.16 (3) 0.04 (4) 0.4 Solution 3 3 CH COOH CH COO H at start 1 0 0 at equilibrium c(1 ) c c 0.01 (1 ) 0.01 0.01 [In presence of 0.1M HCl] 0.01 (1 ) 0.01 (0.01 0.1) − + → + − α α α − α α α − α α α + 3 3 [CH COO ] [H ] K [CH COOH] − + = 5 0.01 (0.01 0.1) 1.6 10 0.1 (1 ) − α × α + × = − α On solving, αααα = 0.016% Example 10

Find the percentage ionisation of 0.2M acetic acid solution whose dissociation constant is 1.8 × 10–5_{. Also determine concentration of H}₃_{O+, CH}₃_{COO– and CH}₃_{COOH at equilibrium.}

Solution

From the expression derived earlier we know that K = (2c

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∴ α = 5 5 4 K 1.8 10 9 10 0.9 10 c 0.2 − − − × = = × = × = 0.910−2=0.95 10× −2 (% Ionisation = 0.95 × 10–2 × 100 = 0.95%

We know that concentration of [H+] or [H3O+] c( = 0.2 × 0.95 × 10–2 = 0.190 × 10–2

= 1.9 × 10–3

Also concentration of [CH3COO–] = c( = 1.9 × 10–3

Concentration of unionised CH3COOH = c (1 – .0095) = 0.2 (1 – 0.0095) = 0.2 × 0.9905

= 0.198 M

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Exercise

Exercise ---- IIII

General Type

(Fill in the blanks/ True or False/ Assertion & Reason

)

True/False

1. All ligands are Lewis base. Ans. (True)

2. If the solubility of Li3Na3(AlF6)2 is ‘s’ then its solubility product will be 2916 s8.

Ans. (True)

3. In the titration of weak acid HA with NaOH, at half neutralization point pH = 1/2 pKa. Ans. (False)

4. An equimolar solution of NH4OH & HCl can act as buffer.

Ans. (False)

5. Dissociation constant of weak acid increases on dilution. Ans. (False)

Fill in the Blanks

1. pH + pOH = _____________ at all temperature. Ans. pKw

2. pH of pure water _______________ with increase in temperature but water remains neutral. Ans. decrease

3. 0.1 M solution of two salts of weak acid & strong base NaX & NaY have pH 9 & 11 respectively, hence HX is _______________ acid than HY.

Ans. strong

4. Hydrolysis is reverse of _____________ . Ans. neutralization

5. Na2CO3 solution is alkaline due to _____________ hydrolysis.

Ans. anionic

Assertion and Reason

(a) If both A and R are true and R is the correct explanation of A. (b) If both A and R are true but R is not the correct explanation of A (c) If A is true but R is false.

(d) If A is false but R is true. 1. A : : CuCO4 solution is acidic.

R : : Only cationic hydrolysis takes place.

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Ans. (a)

2. A : : AgBr is more soluble in water than KBr solution. R : : Common ion Br–_{decreases the solubility.}

Ans. (a)

3. A : : Equal volume of 0.002 M solution of NaIO3 & Cu(ClO3)2 are mixed, then Cu(IO3)2 (Ksp = 7.4 × 10– 8_{) will be precipitated.}

R : : Prepicipitation will take place when ionic product exceeds solubility product. Ans. (d)

4. A : : Phenolphthalein is suitable indicator for weak acid & strong base titration. R : : Indicator changes colour at pH = pKa.

Ans. (b)

5. A : : 0.1 N HCl is more acidic than 0.1 N H2SO4.

R : : HCl is stronger acid than H2SO4

Ans. (c)

*****

Formatted: Indent: Left: 0", Hanging: 0.63", Line spacing: Multiple 1.2 li

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Exercise

Exercise ---- II

II

Objective Type

(Only One Choice Correct)

Level – I

1. _{pH of Ca(OH)2 is 12. Milliequivalents of Ca(OH)2 present in 100 mL solution will be:}

(a) 1 (b) 0.5 (c) 0.05 (d) 5

Ans. (a)

2. Which of the following solutions will have pH of 4.74:

(a) 100 mL of 1M CH3COOH (pKa = 4.74) at the equivalent point using NaOH (b) 50 ml of 1M CH3COONa + 25 mL of 1M HCl

(c) 50mL of 1M CH3COOH + 25 mL of 1M NaOH (d) Both (b) and (c) true

Ans. (d)

3. 10mL of 10–6M HCl solution is mixed with 10ml 10– 6_{M NaOH. pH will change approximately:}

(a) by one unit (b) by 0.3 unit (c) by 0.5 unit (d) by 0.1 unit Ans. (a)

4. Which one of the following statements is not true ? (a) The conjugtate base of H2PO4– is HPO42– (b) pH + pOH = 14 for all aqueous solutions at 25°C (c) The pH of 1 × 10–8 M HCl is 8

(d) [H+_{] = [OH}–_{] in pure water at any temperature}

Ans. (c)

5. Which one of the following substances has highest proton affinity ?

(a) H2O (b) H2S (c) NH3 (d) PH3

Ans. (c)

6. Which one of the following is an amphoteric oxide ?

(a) ZnO (b) Na2O (c) SO2 (d) B2O3

Ans. (a)

7. _{pKa (CH3COOH) is 4.74, x mol of lead acetate and 0.1 mol of acetic acid in one L solution make a} solution of pH = 5.04. Hence x is

(a) 0.2 (b) 0.05 (c) 0.1 (d) 0.02

Ans. (c)

8. _{pH of a mixture containing 0.10 M X– (base) and 0.20 M HX with pKb (X}–) = 4 is

(a) 4 + log 2 (b) 4 – log 2 (c) 10 + log 2 (d) 10 – log 2 Ans. (d)

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9. Equimolar solution of which one of the following is NOT a buffer solution ?

(a) 0.8 M H2S + 0.8 M KHS (b) 2 M C6H5NH2 + 2 M C6H5 NH+3Br– (c) 3 M H2CO3 + 3 M KHCO3 (d) 0.05 M KClO4 + 0.05 M HClO4

Ans. (d)

10. The solubility product of a salt AB is 1 × 10–8 in solution in which concentration of A is 10–3 M. The salt will be precipitated when concentration of B becomes more than

(a) 10–6 M (b) 10–5 M (c) 10–4 M (d) 10–3 M

Ans. (b)

11. The solubility in water of a sparingly soluble salt AB2 is 1.0 × 10–5 mol L–1. Its solubility product number will be

(a) 4 × 10–15 (b) 4 × 10–10 (c) 1 × 10–15 (d) 1 × 10– 10

Ans. (a)

12. The solubility of A2X3 is S mol L–1. Its solubility product is:

(a) 6S4 (b) 64S4 (c) 36S5 (d) 108 S5

Ans. (d)

13. The solubility product of MA, MB, MC and MD are 1.8 × 10–10, 4 × 103, 4 × 10–8 and 6 × 10–5 respectively. If a 0.01 M solution of MX is added drop wise to a mixture containing A, B, C and D ions then the one to be precipitated first will be:

(a) MA (b) MB (c) MC (d) MD

Ans. (a)

14. Some chemist at ISRO wished to prepare a saturated solution of a silver compound and they wanted it to have the highest concentration of silver ion possible. Which of the following compounds would they used:

AgCl; Ksp = 1.8 × 10–10 AgBr; Ksp = 5.0 × 10–13 Ag2CrO4 ; Ksp = 2.4 × 10–12

(a) AgCl (b) AgBr _{(c) Ag2CrO4} (d) Any of them

Ans. (c)

15. The solubility product of AgI at 25ºC is 1.0 × 10–16 mol2L–2. The solubility of AgI in 10–4 N solution of KI at 25ºC is approximately (in mol L–1)

(a) 1.0 × 10–8 (b) 1.0 × 10–16 (c) 1.0 × 10–12 (d) 1.0 × 10– 10

Ans. (c)

16. A mixture of weak acid (say acetic acid) and its salt with a strong base (say sodium acetate) is a buffer solution. Which other pair of substances from the following may have a similar property?

(a) HCl and NaCl (b) NaOH and NaNO3 (c) KOH and KCl (d) NH4OH and NH4Cl

Ans. (d)

17. A precipitate is formed when (a) The solution becomes saturated

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Sri Chaitanya IIT-JEE Head Office: 11, Central Market, West Punjabi Bagh , New Delhi-110 026. Phones: 25226309/10. (b) The ionic product is less than the solubility product

(c) The ionic product is nearly equal to the solubility product (d) The ionic product exceeds the solubility product Ans. (d)

18. For pure water

(a) pH increases and pOH decreases with increase in temperature (b) pH decreases and pOH increases with increase in temperature (c) Both pH and pOH increases with increase in temperature (d) Both pH and pOH decreases with increase in temperature Ans. (d)

19. Which of the following is the electron deficient molecule?

(a) _B2H6 (b) _C2H6 _{(c) PH3} _{(d) SiH4}

Ans. (a)

20. If pH of a solution decreases from 5 to 2 then it is

(a) Diluted 1000 times (b) Concentrated 1000 times (c) Diluted 100 times (d) Concentrated 100 times Ans. (b)

21. The pH of 10-8 N-HCl is approximately

(a) 8 (b) 7.02 (c) 7 (d) 6.96

Ans. (d)

22. pH at which an acid indicator with Ka = 1 × 10–5 changes colour when the indicator is 1 × 10–3 M is

(a) 5 (b) 3 (c) 8 (d) 4

Ans. (a)

23. A 50.00mL sample of acetic acid was titrated with 0.1200M KOH, and 38.62 mL of base were required to reach the equivalence point. What was the pH of the titration mixture when 19.31mL of base had been added? [pKa (acetic acid) = 4.74]

(a) 2.94 (b) 3.54 (c) 4.74 (d) 5.74

Ans. (c)

24. pH of which of the following solution is affected by dilution: (a) 0.01M CH3COONa

(b) 0.01M NH4HCO3

(c) buffer of 0.01M CH3COONa and 0.01M CH3COOH (d) 0.01M CH3COONH4

Ans. (a)

25. An acid-base indicator has a Ka of 3.0 × 10–5. The acid form of the indicator is red and the basic form is blue. Then:

(a) pH is 4.05 when indicator is 75% red (b) pH is 6.00 when indicator is 75% blue (c) pH is 5.00 when indicator is 75% red (d) pH is 4.05 when indicator is 75% blue