Important Ionic Equilibrium Formulas for JEE and NEET

 Page 1


  
IONIC EQUILIBRIUM
OSTWALD DILUTION LAW :
? Dissociation constant of weak acid (K
a
),
K
a 
= 
?
?
?
?
? ?
?
?
? 1
C
) ? 1 ( C
] C ][ C [
] HA [
] A ][ H [
2 ?
If ? << 1 , then 1 ? ? ?  1 or K
a
 = c?
2  
or
  
??= 
V K
C
K
a
a
? ?
? Similarly for a weak base ,
 
C
K
b
? ?
 
. Higher the value  of K
a
 / K
b
 , strong
is the acid / base.
Acidity and pH scale :
? pH = ? log ?
H
a
(where ?
H
a
 is the activity of H
+
 ions = molar concentration
for dilute solution).
[Note : pH can also be negative or > 14]
pH = ? log [H
+
] ; [H
+
] =
 
10
?pH
pOH = ? log [OH
?
] ; [OH
?
] = 10
?pOH
pKa = ? log Ka  ; Ka =  10
?pKa
pKb = ? log Kb  ; Kb = 10
?pKb
PROPERTIES OF WATER :
1. In pure water [H
+
] = [OH
?
] so it is Neutral.
2. Molar concentration / Molarity of water = 55.56 M.
3. Ionic product of water (K
W
) :
K
w 
= [H
+
][OH
?
] = 10
?14  
at 25° (experimentally)
pH
  
= 7 = pOH ? neutral
pH < 7  or pOH
 
 > 7 ? acidic
pH
 
 > 7 or pOH
 
< 7 ? Basic
4. Degree of dissociation of water :
taken initially moles of . No Total
d dissociate moles of . no
? ?
?= % 10 x 8 . 1 or 10 x 18
55 . 55
10
7 10
7
? ?
?
?
5. Absolute dissociation constant of water :
K
a
 = K
b
 = 
] O H [
] OH ][ H [
2
? ?
 = 
16
7 7
10 8 . 1
55 . 55
10 10
?
? ?
? ?
?
pK
a
 = pK
b
 = ? log (1.8 × 10
-16
) = 16 ? log 1.8 = 15.74
Page 2


  
IONIC EQUILIBRIUM
OSTWALD DILUTION LAW :
? Dissociation constant of weak acid (K
a
),
K
a 
= 
?
?
?
?
? ?
?
?
? 1
C
) ? 1 ( C
] C ][ C [
] HA [
] A ][ H [
2 ?
If ? << 1 , then 1 ? ? ?  1 or K
a
 = c?
2  
or
  
??= 
V K
C
K
a
a
? ?
? Similarly for a weak base ,
 
C
K
b
? ?
 
. Higher the value  of K
a
 / K
b
 , strong
is the acid / base.
Acidity and pH scale :
? pH = ? log ?
H
a
(where ?
H
a
 is the activity of H
+
 ions = molar concentration
for dilute solution).
[Note : pH can also be negative or > 14]
pH = ? log [H
+
] ; [H
+
] =
 
10
?pH
pOH = ? log [OH
?
] ; [OH
?
] = 10
?pOH
pKa = ? log Ka  ; Ka =  10
?pKa
pKb = ? log Kb  ; Kb = 10
?pKb
PROPERTIES OF WATER :
1. In pure water [H
+
] = [OH
?
] so it is Neutral.
2. Molar concentration / Molarity of water = 55.56 M.
3. Ionic product of water (K
W
) :
K
w 
= [H
+
][OH
?
] = 10
?14  
at 25° (experimentally)
pH
  
= 7 = pOH ? neutral
pH < 7  or pOH
 
 > 7 ? acidic
pH
 
 > 7 or pOH
 
< 7 ? Basic
4. Degree of dissociation of water :
taken initially moles of . No Total
d dissociate moles of . no
? ?
?= % 10 x 8 . 1 or 10 x 18
55 . 55
10
7 10
7
? ?
?
?
5. Absolute dissociation constant of water :
K
a
 = K
b
 = 
] O H [
] OH ][ H [
2
? ?
 = 
16
7 7
10 8 . 1
55 . 55
10 10
?
? ?
? ?
?
pK
a
 = pK
b
 = ? log (1.8 × 10
-16
) = 16 ? log 1.8 = 15.74
  
K
a 
× K
b
 = [H
+
] [OH
?
] = K
w
? Note: for a conjugate acid- base pairs
pK
a
 + pK
b
 = pK
w
 
 = 14 at 25ºC.
pK
a
 of H
3
O
+
 ions = ?1.74
pK
b
  of OH
?
 ions = ?1.74.
? pH Calculations of Different Types of Solutions:
(a) Strong acid solution :
(i) If concentration is greater than 10
?6
 M
In this case H
+
 ions coming from water can be neglected,
(ii) If concentration is less than 10
?6
 M
In this case H
+
 ions coming from water cannot be neglected
(b) Strong base solution :
Using similar method as in part (a) calculate first [OH
?
] and then use
[H
+
] × [OH
?
] = 10
?14
(c) pH of mixture of two strong acids :
Number of H
+
 ions from ?-solution = N
1
V
1
Number of H
+
 ions from ??-solution = N
2 
V
2
      
[H
+
] = N =
2 1
2 2 1 1
V V
V N V N
?
?
(d) pH of mixture of two strong bases :
 
[OH
?
] = N =
2 1
2 2 1 1
V V
V N V N
?
?
(e) pH of mixture of a strong acid and a strong base :
If N
1
V
1
 > N
2
V
2
, then solution will be acidic in nature and
[H
+
] = N =
2 1
2 2 1 1
V V
V N V N
?
?
If N
2
V
2
 > N
1
V
1
, then solution will be basic in nature and
[OH
?
] = N =
2 1
1 1 2 2
V V
V N V N
?
?
(f) pH of a weak acid(monoprotic) solution :
K
a
 = 
] HA [
] OH [ ] H [
? ?
 = 
? ?
?
1
C
2
if ??<<1  ???1 ?????? ? 1 ?????? K
a
 ? C?
2
Page 3


  
IONIC EQUILIBRIUM
OSTWALD DILUTION LAW :
? Dissociation constant of weak acid (K
a
),
K
a 
= 
?
?
?
?
? ?
?
?
? 1
C
) ? 1 ( C
] C ][ C [
] HA [
] A ][ H [
2 ?
If ? << 1 , then 1 ? ? ?  1 or K
a
 = c?
2  
or
  
??= 
V K
C
K
a
a
? ?
? Similarly for a weak base ,
 
C
K
b
? ?
 
. Higher the value  of K
a
 / K
b
 , strong
is the acid / base.
Acidity and pH scale :
? pH = ? log ?
H
a
(where ?
H
a
 is the activity of H
+
 ions = molar concentration
for dilute solution).
[Note : pH can also be negative or > 14]
pH = ? log [H
+
] ; [H
+
] =
 
10
?pH
pOH = ? log [OH
?
] ; [OH
?
] = 10
?pOH
pKa = ? log Ka  ; Ka =  10
?pKa
pKb = ? log Kb  ; Kb = 10
?pKb
PROPERTIES OF WATER :
1. In pure water [H
+
] = [OH
?
] so it is Neutral.
2. Molar concentration / Molarity of water = 55.56 M.
3. Ionic product of water (K
W
) :
K
w 
= [H
+
][OH
?
] = 10
?14  
at 25° (experimentally)
pH
  
= 7 = pOH ? neutral
pH < 7  or pOH
 
 > 7 ? acidic
pH
 
 > 7 or pOH
 
< 7 ? Basic
4. Degree of dissociation of water :
taken initially moles of . No Total
d dissociate moles of . no
? ?
?= % 10 x 8 . 1 or 10 x 18
55 . 55
10
7 10
7
? ?
?
?
5. Absolute dissociation constant of water :
K
a
 = K
b
 = 
] O H [
] OH ][ H [
2
? ?
 = 
16
7 7
10 8 . 1
55 . 55
10 10
?
? ?
? ?
?
pK
a
 = pK
b
 = ? log (1.8 × 10
-16
) = 16 ? log 1.8 = 15.74
  
K
a 
× K
b
 = [H
+
] [OH
?
] = K
w
? Note: for a conjugate acid- base pairs
pK
a
 + pK
b
 = pK
w
 
 = 14 at 25ºC.
pK
a
 of H
3
O
+
 ions = ?1.74
pK
b
  of OH
?
 ions = ?1.74.
? pH Calculations of Different Types of Solutions:
(a) Strong acid solution :
(i) If concentration is greater than 10
?6
 M
In this case H
+
 ions coming from water can be neglected,
(ii) If concentration is less than 10
?6
 M
In this case H
+
 ions coming from water cannot be neglected
(b) Strong base solution :
Using similar method as in part (a) calculate first [OH
?
] and then use
[H
+
] × [OH
?
] = 10
?14
(c) pH of mixture of two strong acids :
Number of H
+
 ions from ?-solution = N
1
V
1
Number of H
+
 ions from ??-solution = N
2 
V
2
      
[H
+
] = N =
2 1
2 2 1 1
V V
V N V N
?
?
(d) pH of mixture of two strong bases :
 
[OH
?
] = N =
2 1
2 2 1 1
V V
V N V N
?
?
(e) pH of mixture of a strong acid and a strong base :
If N
1
V
1
 > N
2
V
2
, then solution will be acidic in nature and
[H
+
] = N =
2 1
2 2 1 1
V V
V N V N
?
?
If N
2
V
2
 > N
1
V
1
, then solution will be basic in nature and
[OH
?
] = N =
2 1
1 1 2 2
V V
V N V N
?
?
(f) pH of a weak acid(monoprotic) solution :
K
a
 = 
] HA [
] OH [ ] H [
? ?
 = 
? ?
?
1
C
2
if ??<<1  ???1 ?????? ? 1 ?????? K
a
 ? C?
2
  
? ??=
C
K
a
 ( is valid if ?? < 0.1 or 10%)
On increasing the dilution
? C ?    ?????? ??????and [H
+
] ?? ??pH ?
RELATIVE STRENGTH OF TWO ACIDS :
2 a
1 a
2 2
1 1
c k
c k
c
c
acid by furnished ] H [
acid by furnished ] H [
2
1
?
?
?
?
??
?
?
?
? SALT HYDROLYSIS :
Salt of         Type of
         hydrolysis      k
h
        h     pH
(a)   weak acid  & strong base anionic      
a
w
k
k
   
c k
k
a
w
   7+
2
1
pk
a
+
2
1
log
 
c
(b)   strong acid  & weak base cationic      
b
w
k
k
    
c k
k
b
w
   7?
2
1
pk
b
?
2
1
log
 
c
(c)   weak acid  & weak base both     
b a
w
k k
k
b a
w
k k
k
 7+
2
1
pk
a
?
2
1
p
 
k
b
(d)   Strong acid & strong base    --------do not  hydrolysed-------      pH = 7
Hydrolysis of ployvalent anions or cations
For [Na
3
PO
4
] = C.
K
a1
 × K
h3
 = K
w
K
a1
 × K
h2
 = K
w
K
a3
 × K
h1
 = K
w
Generally pH is calculated only using the first step Hydrolysis
K
h1
 = 
h 1
Ch
2
?
? Ch
2
h = 
c
K
1 h
      ??[OH
?
] = ch = c K
1 h
? ??[H
+
] = 
C
K K
a3 W
?
So pH = C] log pK [pK
2
1
a3 w
? ?
Page 4


  
IONIC EQUILIBRIUM
OSTWALD DILUTION LAW :
? Dissociation constant of weak acid (K
a
),
K
a 
= 
?
?
?
?
? ?
?
?
? 1
C
) ? 1 ( C
] C ][ C [
] HA [
] A ][ H [
2 ?
If ? << 1 , then 1 ? ? ?  1 or K
a
 = c?
2  
or
  
??= 
V K
C
K
a
a
? ?
? Similarly for a weak base ,
 
C
K
b
? ?
 
. Higher the value  of K
a
 / K
b
 , strong
is the acid / base.
Acidity and pH scale :
? pH = ? log ?
H
a
(where ?
H
a
 is the activity of H
+
 ions = molar concentration
for dilute solution).
[Note : pH can also be negative or > 14]
pH = ? log [H
+
] ; [H
+
] =
 
10
?pH
pOH = ? log [OH
?
] ; [OH
?
] = 10
?pOH
pKa = ? log Ka  ; Ka =  10
?pKa
pKb = ? log Kb  ; Kb = 10
?pKb
PROPERTIES OF WATER :
1. In pure water [H
+
] = [OH
?
] so it is Neutral.
2. Molar concentration / Molarity of water = 55.56 M.
3. Ionic product of water (K
W
) :
K
w 
= [H
+
][OH
?
] = 10
?14  
at 25° (experimentally)
pH
  
= 7 = pOH ? neutral
pH < 7  or pOH
 
 > 7 ? acidic
pH
 
 > 7 or pOH
 
< 7 ? Basic
4. Degree of dissociation of water :
taken initially moles of . No Total
d dissociate moles of . no
? ?
?= % 10 x 8 . 1 or 10 x 18
55 . 55
10
7 10
7
? ?
?
?
5. Absolute dissociation constant of water :
K
a
 = K
b
 = 
] O H [
] OH ][ H [
2
? ?
 = 
16
7 7
10 8 . 1
55 . 55
10 10
?
? ?
? ?
?
pK
a
 = pK
b
 = ? log (1.8 × 10
-16
) = 16 ? log 1.8 = 15.74
  
K
a 
× K
b
 = [H
+
] [OH
?
] = K
w
? Note: for a conjugate acid- base pairs
pK
a
 + pK
b
 = pK
w
 
 = 14 at 25ºC.
pK
a
 of H
3
O
+
 ions = ?1.74
pK
b
  of OH
?
 ions = ?1.74.
? pH Calculations of Different Types of Solutions:
(a) Strong acid solution :
(i) If concentration is greater than 10
?6
 M
In this case H
+
 ions coming from water can be neglected,
(ii) If concentration is less than 10
?6
 M
In this case H
+
 ions coming from water cannot be neglected
(b) Strong base solution :
Using similar method as in part (a) calculate first [OH
?
] and then use
[H
+
] × [OH
?
] = 10
?14
(c) pH of mixture of two strong acids :
Number of H
+
 ions from ?-solution = N
1
V
1
Number of H
+
 ions from ??-solution = N
2 
V
2
      
[H
+
] = N =
2 1
2 2 1 1
V V
V N V N
?
?
(d) pH of mixture of two strong bases :
 
[OH
?
] = N =
2 1
2 2 1 1
V V
V N V N
?
?
(e) pH of mixture of a strong acid and a strong base :
If N
1
V
1
 > N
2
V
2
, then solution will be acidic in nature and
[H
+
] = N =
2 1
2 2 1 1
V V
V N V N
?
?
If N
2
V
2
 > N
1
V
1
, then solution will be basic in nature and
[OH
?
] = N =
2 1
1 1 2 2
V V
V N V N
?
?
(f) pH of a weak acid(monoprotic) solution :
K
a
 = 
] HA [
] OH [ ] H [
? ?
 = 
? ?
?
1
C
2
if ??<<1  ???1 ?????? ? 1 ?????? K
a
 ? C?
2
  
? ??=
C
K
a
 ( is valid if ?? < 0.1 or 10%)
On increasing the dilution
? C ?    ?????? ??????and [H
+
] ?? ??pH ?
RELATIVE STRENGTH OF TWO ACIDS :
2 a
1 a
2 2
1 1
c k
c k
c
c
acid by furnished ] H [
acid by furnished ] H [
2
1
?
?
?
?
??
?
?
?
? SALT HYDROLYSIS :
Salt of         Type of
         hydrolysis      k
h
        h     pH
(a)   weak acid  & strong base anionic      
a
w
k
k
   
c k
k
a
w
   7+
2
1
pk
a
+
2
1
log
 
c
(b)   strong acid  & weak base cationic      
b
w
k
k
    
c k
k
b
w
   7?
2
1
pk
b
?
2
1
log
 
c
(c)   weak acid  & weak base both     
b a
w
k k
k
b a
w
k k
k
 7+
2
1
pk
a
?
2
1
p
 
k
b
(d)   Strong acid & strong base    --------do not  hydrolysed-------      pH = 7
Hydrolysis of ployvalent anions or cations
For [Na
3
PO
4
] = C.
K
a1
 × K
h3
 = K
w
K
a1
 × K
h2
 = K
w
K
a3
 × K
h1
 = K
w
Generally pH is calculated only using the first step Hydrolysis
K
h1
 = 
h 1
Ch
2
?
? Ch
2
h = 
c
K
1 h
      ??[OH
?
] = ch = c K
1 h
? ??[H
+
] = 
C
K K
a3 W
?
So pH = C] log pK [pK
2
1
a3 w
? ?
  
BUFFER SOLUTION :
(a) Acidic Buffer : e.g. CH
3 
COOH and CH
3
COONa. (weak acid and salt
of its conjugate base).
pH= pK
a
 + log 
] Acid [
] Salt [
[Henderson's equation]
(b) Basic Buffer : e.g. NH
4
OH + NH
4
Cl. (weak base and salt of its conjugate
acid).
pOH = pK
b
  +  log 
] Base [
] Salt [
SOLUBILITY PRODUCT :
K
SP
 = (xs)
x
 (ys)
y 
= x
x
.y
y
.(s)
x+y
CONDITION FOR PRECIPITATION :
If ionic product K
I.P
  > K
SP
  precipitation occurs,
if K
I.P
 =  K
SP
 saturated solution (precipitation just begins or is just
prevented).