03_Ionic Equilibrium (2)

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Chapter

3

Ionic Equilibrium

Gilbert Newton Lewis

Born

23 October

1875

Died

23 March

1

946

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 ˆ ˆ †‡ ˆ ˆ H3O+ + Cl–

acid base conjugate acid conjugate base

H2O + NH3 ˆ ˆ †‡ ˆ ˆ NH4+ + 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.

(v) In carbonyl complexes, metal atoms act as Lew’s acids e.g. Ni in Ni(CO)4 etc.

And Lewis bases are

(i) Neutral molecules like NH3, RNH2 etc.

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(iii) Molecules with carbon-carbon multiple bonds can act as lewis base in some cases - e.g. C H in 2 4

(

2 4

)

Ag C H +

 

 

(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 + + − ˆ ˆ † ‡ ˆ ˆ

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

Q Ka cx2 ( 1 x 1) 1 = ∴ − ; or x Ka C = or x Ka V c n 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− = + = ×

Il l u s t r a t i o n s

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

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Solution CH3CH2COOH ˆ ˆ †‡ ˆ ˆ 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 Q ∴ 0.05 × α2 = 1.32 × 10–5; α =1.63 ×10-2 pH = – log [H+] = – log (cα) 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 ‡ ˆ ˆˆ ˆ † H+ + CNOc(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 × 10–2Ka = 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

1. 5.78 2. 2%

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 COOHˆ ˆ †‡ ˆ ˆ CH COO−+H+

If some amount of (CH3COONa) is added to the solution of acetic acid. Then CH COONa will provide 3

common acetate ions. Consequently equilibrium will shift towards reactant side and degree of dissociation becomes further smaller.

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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 ˆ ˆ † ‡ ˆ ˆ

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 + − + − + − → + ˆ ˆ † Q ‡ ˆ ˆ

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

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

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Il l u s t r a t i o n s

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 m illi m ole . M . Total Volum e − × = = = 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 Sˆ ˆ †‡ ˆ ˆ 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 NH4OH is suppressed and consequently lesser amount of OH– ions are

obtained which can bring about precipitation of hydroxides of group III radicals only.

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.

Q HA1 ˆ ˆ †‡ ˆ ˆ H+ + A1

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C1(1 – x1) C1x1 C1x1 (At equilibrium) HA2 ˆ ˆ †‡ ˆ ˆ 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 ) + + × = = − Q …(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.

Q H2S ˆ ˆ †‡ ˆ ˆ H+ + HS

C(1-x1)cx1 cx1(1 – y1) (At equilibrium) HS ˆ ˆ †‡ ˆ ˆ 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 ] + −− − + = = − (y1<<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)

From equation (i) and (ii) we can calculate the pH of aqueous solution of H2S.

Il l u s t r a t i o n s

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.

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2 H A‡ ˆ ˆˆ ˆ † H++HA− 3 1 2 4 10 [H ][HA ] K [H A] + − − = = × 2 1 [H ] cx, [HA ] cx; [H A] c(+ = − = = −x) Q 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 ; c1 = [HA] = 0.018 M

HA– ˆ ˆ †‡ ˆ ˆ H+ + A–2 1 0 0 (1 – x 1) x 1 x 1 ∴ 2 2 5 1 10 a [H ][A ] K [HA ] + − − − = × =

Q [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 ∴ [HA ] c (− = 11−x ) c1 = 1=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.

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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 (K )w

Water is weak electrolyte, hence

2 H O‡ ˆ ˆˆ ˆ † H++OH−

[

]

eq 2 H OH K H O + −         =

[

]

eq. 2 K H O =  H+ OH−

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

w

K depends on temperature. At 25° C value of K is 10w –14.

Hence H+  OH− = 10−14 at 25 C°

Thus H+ =2 10−14  

Or H+ = 10 at 25 C−7 °

Since [H+] = [OH]

At 25°C, PH of pure water

{

PH = −log H +

}

= −log 10

( )

−7 =7

 

Now at 25°C H+  OH− = 10−14 or −log H +−log OH −=14 or PH+POH =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 − − = −

A c i d i c

N e u t r a l

B a s i c

O

7

1 4

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

Thus on increasing temperature, value of K increases Pw H 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

+

+

ƒ

2 2 w K =H+  OH−=cx cx c x× = at 25°C 14 2 2 w K =10− =c x

(10)

or cx 10= −7 7 10 x C − =

For pure water c = 55.55 M

Thus x 10 7 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.

Il l u s t r a t i o n s

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−

( )

H 5 p = −log 10− =5 (ii) 10–6 M – HNO 3 HNO3 H10 6 NO3 + − − → +

( )

6 pH= −log 10− =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

[

NaOH

]

10 0.02 2 10 2 500 − = = = ×

(

)

OH 2 p = −log 2 10× − = −2 log 2

(11)

Hence pH 14 p= − OH = − −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.

Remember, that ionic strength is calculated as sum of concentrations, not activities of ions.

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 COOH molecules, CH3COO– ions and Na+ ions.3

Note: Dissociation of weak electrolyte is suppressed in the presence of common ion CH3COO– from CH3COONa.

When acid is added (say x

moles)-H+ + CH3COONa ˆ ˆ †

‡ ˆ ˆ CH3COOH + Na+

Hence pH pka log salt x

acid x

= +

+

and when base in added-

OH– + CH3COOH ˆ ˆ †

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Thus pH pka log salt x acid x + = + −

x is small [Salt] – x = [Salt] [Acid] + x = [Acid]

pH =pKa +log[[AcidSalt]] 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

4 4 2 H++NH OH→NH+ +H O K OH b salt x p p log base x + = + −

and when x moles of base are added

4 OH−+NH+ ˆ ˆ †‡ ˆ ˆ NH OH4 K OH b salt x p p log 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.

number of moles of acid or base added Buffer capacity

change in pH

=

Il l u s t r a t i o n s

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.

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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 (a–2) 0 2 2

(

14 9.14

)

pKb log

(

2

)

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

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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 (50×0.201)×0(20.1 ×0.1) = 4.699 + log

3 2

= 4.5229 (iv) When 25 ml of NaOH is added,

pH = 4.699 + log (50×0.251)×0(25.1 ×0.1) = 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 NH 9.26, +4

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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.

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 ‡ ˆ ˆˆ ˆ † H+ + In

∴ KHIn = [H ][In ]

[HIn]

+ −

[KHIn = Indicator constant or dissociation constant of indicator]

or [H+] = KHIn × [HIn]

[In ]−

Taking negative logarithm of both sides – log [H+] = – log KHIn – 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–.

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In order for the solution to show colour due to HIn, the minimum ratio of [In ] [HIn] − should be 1/10. ∴ 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.

( )

x y A B s ˆ ˆ †‡ ˆ ˆ xA( )yaq+ +yBx−

( )

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

In saturated solution [AxBy] = constant

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

e.g. for AgCl, AgCl(s) ˆ ˆ †‡ ˆ ˆ Ag aq+

( )

+Cl aq−

( )

sp

K = Ag+  Cl− = s2 where s is the solubility of AgCl in moles/l

For Ag CrO s 2 4

( )

‡ ˆ ˆˆ ˆ † 2Ag+(aq)+CrO24−(aq)

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 ‡ ˆ ˆˆ ˆ † xAxsy+ +yBysx−

[ ] [ ]

x y x y x y sp

K = xs ys =x y s +

e.g. for AgCl ˆ ˆ †‡ ˆ ˆ

s s Ag++Cl− Ksp = S2 Hence S= Ksp For Ag2CrO4 ˆ ˆ †‡ ˆ ˆ 24 2s s 2Ag CrO+ −

( ) ( )

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

for Ca PO3

(

4 2

)

ˆ ˆ †‡ ˆ ˆ 3Ca2++2PO34−

( ) ( )

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 Ksp and pure NaCl precipitates out from the solution. This process is called salting out.

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

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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 Ag2CrO4 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 ‡ ˆ ˆˆ ˆ † 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 ( Ksp −10 − = × = 1 × 10– 9 M

Ag2CrO4(s) ‡ ˆ ˆˆ ˆ † 2Ag+(aq) + CrO2 (aq) 4−

Ksp(Ag2CrO4) = [Ag+]2 × [ 2−

4

CrO ]

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

4

CrO as Ag2CrO4 would be

[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 CrO24− 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.

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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 ‡ ˆ ˆˆ ˆ † Mg2++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−   = =   after mixing Ag 105 100 200 − + ×   =   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: 3 Ag++2NH ˆ ˆ †‡ ˆ ˆ Ag NH

(

3 2

)

+

( )

AgCl s ˆ ˆ †‡ ˆ ˆ Ag++Cl− sp K = Ag+  Cl−

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(

)

[

]

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 equilibrium. Take

Ka(HCN) = 9 × 10– 10, Ksp(AgCN) = 4 × 10– 16.

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

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2

Cl−+H O×→HCl OH+ −

H2O ‡ ˆ ˆˆ ˆ † 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 ˆ ˆ †‡ ˆ ˆ 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.

Now CH3COOH ˆ ˆ ˆ †Ka ‡ ˆ ˆ ˆ CH3COO– + H+ ...(ii) Ka = ] COOH CH [ ] H [ ] COO CH [ 3 3 − +

and H2O ‡ ˆ ˆ ˆˆ Hˆ ˆ ˆ †ˆKw + + OH– ...(iii)

Kw = [H+] [OH]

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

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

or h w a K K K = Now 3 2 C(1 h) CH COO− H O − + ˆ ˆ † ‡ ˆ ˆ CH COOH OHCh3 Ch

+ 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 − = = = × ×

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w w a K K K H 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:

4 4 NH Cl→NH++Cl− 4 2 NH++H O Kh 4 NH OH H+ + ...(i) Kh = [NH4OH] [H+] / [ + 4 NH ] 4 NH +OH Kb 4 NH+ +OH− ...(ii) Kb = [NH+4 ] [OH–] / [NH4OH]

and H2O Kω H+ + OH...(iii)

Kw = [H+] [OH]

from all three equations h w

b

K K

K

=

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

4 2 C(1 h) NH+ H O − + ˆ ˆ †‡ ˆ ˆ Ch 4 Ch NH OH H+ +

(

)

2 h b K Ch Ch Ch K K C 1 h 1 h ω × = = = − − If h < < 1 Then 2 w 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 +logC 2 1 K

4. Salt of weak acid and weak base

e.g. CH3COONH4 3 4 3 4 CH COONH →CH COO−+NH+ 3 4 2 CH COO−+NH++H O Kh 3 4 CH COOH NH OH+ ...(i)

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Kh = ] NH [ ] COO CH [ ] OH NH [ ] COOH CH [ 4 3 4 3 + − 3 CH COOH Ka 3 CH COOH−+H+ ...(ii) Ka = ] COOH CH [ ] H [ ] COO CH [ 3 3 − + 4 NH OH Kb 4 NH++OH− ...(iii) Kb = ] OH NH [ ] OH [ ] NH [ 4 4 − + H2O Kω H++OH− ...(iv)

from all four equations. h w

a b K K K K = × How 3 4 2 C(1 h) C(1 h) CH COO− NH+ H O − − + + 3Ch Ch4 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 + − ×   = = =     a w a w a b b K K K K K K K × = = × or pH = 1 2 w a b K K K p p p  + −    = 7 + 21pKa −pKb

Il l u s t r a t i o n s

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,

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CH3COO– + H2O ˆ ˆ † ‡ ˆ ˆ CH3COOH + OH– C(1 – h) Ch Ch Kh =

(

)

2 2 Ch Ch 1 h− = (1 – h) → 1 Kh = 145 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 [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 ˆ ˆ †

‡ ˆ ˆ 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 × x2 or, x2 = 05 . 0 10 26 . 5 × −10 = 1.05 × 10– 8 or, x = 1.025 × 10– 4

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[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 ˆ ˆ †‡ ˆ ˆ Ca2 + + 2Lac– 0.13 × 2 M 2 × 2 × 0.13 M Lac– + H2O ˆ ˆ † ‡ ˆ ˆ 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 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)‡ ˆ ˆˆ ˆ † Ca2 + + 2 3 CO − 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+ ‡ ˆ ˆˆ ˆ † HA + H2O. 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 × 108 (d) 1.0 × 10– 6 Solution K = - + 8 a [HA] 1 1 K [A ][H ]= =1.0 10× − = 1.0 × 108 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 ˆ ˆ †‡ ˆ ˆ H+ + HCO2 − 0.001 – x x x Check for α α = Ka c α = 1.8 ×10 4 0.001 − = 0.42

α is > 0.1 which means x is not ignorable.

∴ Ka = 1.8 × 10– 4 =

2

x

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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] 3x 4 5 log [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 KNH3 =[NH ][NH ] 10+4 2− = −30. How many amide ions

are present per mm3 of pure liquid ammonia?

(a) 600 ions/mm3 (b) 6 × 106 ions/mm3 (c) 6 × 104 ions/mm3 (d) 60 ions/mm3

Solution 2NH3(l) ˆ ˆ †‡ ˆ ˆ + 4 NH + NH2− 1 – x x x

(Neglecting x in comparison to 1 since KNH3 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) Example 6

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(a) Kh = w b K K (b) Kh = 2 w b K K (c) Kh = 2 w 2 b K K (d) Kh = b 2 w K K

where Kb is effective dissociation constant of base Zn++

Solution Zn++ + 2H 2O ‡ ˆ ˆˆ ˆ † Zn(OH)2 + 2H+ ∴ Kh = + 2 2 ++ [Zn(OH) ][H ] [Zn ] Zn(OH)2 ‡ ˆ ˆˆ ˆ † Zn++ + 2OH– ∴ Kb = ++ 2 2 [Zn ][OH ] [Zn(OH) ] − Kw = [H+] [OH–] ∴ 2w 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, I.P. = 2 4 5 1 1 1 10 10 2 2 8 − −  ×   ×=  ÷  ÷     × 10 – 13 = 1.25 × 10– 14 As, I.P. < Ksp

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∴ 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, Ksp (Al(OH)3) = 1.3 × 10– 33,

Ksp (Cr(OH)3) = 6.3 × 10– 31)

(a) Ca2 + (b) Cr3 +

(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 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)

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(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–] 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?

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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)

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.

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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 = = [NH OH]4 0.02 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 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 ] + + = =

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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, α = 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

3O+, CH3COO– and CH3COOH at equilibrium. Solution

From the expression derived earlier we know that K = (2c ∴ α = K 1.8 10 5 9 10 5 0.9 10 4 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

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Exercise - I

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.

Ans. (a)

2. A : : AgBr is more soluble in water than KBr solution. R : : Common ion Br– decreases the solubility. Ans. (a)

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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)

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Exercise - 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)

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)

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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

(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

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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

Ans. (a)

Level

- II

26. Which of the following solutions will have pH close to 1.0?

(a) 100 ml of M/10 HCl + 100 ml of M/10 NaOH (b) 55 ml of M/10 HCl + 45 ml of M/10 NaOH (c) 10 ml of M/10 HCl + 90 ml of M/10 NaOH (d) 75 ml of M/5 HCl + 25 ml of M/5 NaOH

Ans. (d)

27. The correct order of increasing [H3O+] in the following aqueous solutions is

(a) 0.01 M H2S < 0.01 M H2SO4 < 0.01 M NaCl < 0.01 M NaNO2

(b) 0.01 M NaCl < 0.01 M NaNO2 < 0.01 M H2S < 0.01 M H2SO4

(c) 0.01 M NaNO2 < 0.01 M NaCl < 0.01 M H2S < 0.01 M H2SO4

(d) 0.01 M H2S < 0.01 M NaNO2 < 0.01 M NaCl < 0.01 M H2SO4 Ans. (c)

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(a) 0.1 M HCl + 0.2 M NaCl

(b) 100 ml of 0.2 M H2SO4 + 100 ml of 0.3 M NaOH

(c) 100 ml of 0.1 M CH3COOH + 100 ml of 0.1 M KOH

(d) 25 ml of 0.1 M HNO3 + 25 ml of 0.1 M NH3 Ans. (c)

29. The precipitate of CaF2 (Ksp = 1.7 × 10– 10) is obtained when equal volumes of the following are mixed

(a) 10– 4 M Ca2 + + 10– 4 M F(b) 10– 2 M Ca2 + + 10– 3 M F

(c) 10– 5 M Ca2 + + 10– 3 M F(d) 10– 3 M Ca2 + + 10– 5 M FAns. (b)

30. Solubility products of Al(OH)3 and Zn(OH)2 are 8.5 × 10– 23 and 1.8 × 10– 4 respectively. If equal moles of Al3 + and

Zn2 + ions are present in solution, which one will be precipitated first on addition of NH 4OH?

(a) Al(OH)3 (b) Zn(OH)2 (c) Both (a) & (b) (d) None of these Ans. (a)

31. What is the dissociation constant of NH4OH if at a given temperature its 0.1 N solution has pH = 11.27 and the ionic

product of water at this temperature is 7.1 × 10– 15.

(a) 1.96 × 10– 5 (b) 1.86 × 10– 5 (c) 1.76 × 10– 5 (d) 2.86 × 10– 5 Ans. (c)

32. Amongst the following hydroxides, the one which has the lowest value of Ksp at ordinary temperature (about 25°C) is

(a) Mg(OH)2 (b) Ca(OH)2 (c) Ba(OH)2 (d) Be(OH)2

Ans. (d)

33. Silicon tetrachloride hydrolyses almost completely. Its hydrolysis constant can be expressed in terms of ionic product of water, Kw and Kb of silicon hydroxide as

(a) w b K K (b) 2 w b K K (c) 4 w b K K (d) w 4 b K K Ans. (c)

34. An aqueous solution of a metal chloride, MCl2 (0.05 M) is saturated with H2S (0.1 M). The minimum pH at which

metal sulphide will precipitate is

Given : Ksp (MS) = 5 × 10– 21, K1 (H2S) = 1 × 10– 7, K2 (H2S) = 1.0 × 10– 14

(a) 1.25 (b) 1.50 (c) 2.50 (d) 3.25

Ans. (b)

35. A salt ML is formed by the neutralization of weak base MOH with strong acid LH. The dissociation constant of MOH is 2 × 10– 5. A 1 litre solution of ML having 0.1 M strength is diluted x times. So, as the pH of solution

becomes 6. Then the value of x will be

(a) 550 (b) 55.55 (c) 27.78 (d) 50

Ans. (d)

36. The volume of water needed to dissolve 1 mg of PbSO4 (Ksp = 1.44 × 10– 8 M2) at 25°C is approximately

(a) 10 ml (b) 27 ml (c) 43 ml (d) 80 ml

Ans. (b)

37. The pH at which Mg(OH)2 begins to precipitate from a solution of 0.10 M Mg2 + ions is same at which an indictor

MeOH present in solution changes its colour. The dissociation constant of indicator MeOH present in solution changes its colour. The dissociation constant of indicator will be (Given: Ksp of Mg(OH)2 = 1 × 10– 11)

(a) 1.0 × 10– 10 (b) 1.0 × 10– 3 (c) 1.0 × 10– 9 (d) 1.0 × 10– 5 Ans. (d)

38. What is the percent dissociation of 0.2 M HC2H3O2 in pure water and in 0.1 M HCl. (Ka of HC2H3O2 = 1.8 × 10– 5)

Figure

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