Ionic Equilibrium
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Ionic Equilibrium
Strong electrolyte
Those electrolytes which are almost completely ionized and produce large no. of ions in the soln is called strong electrolyte.
eg: NaCl, HCl, HNO3, H2SO4, KCl, KOH etc.
Weak electrolyte
Those electrolyte which are almost partially ionized and produce less no. of ions in the soln is called weak electrolyte.
eg: NH4OH, HCOOH, HCN, NH4OH, H2CO3 etc.
Postulates of Arrhenius theory of Ionization
- When electrolytes are dissolving in water they produce electrically charge particles called ions, cation and anion.
- An electrolyte soln is electrically neutral due to the equal no. of +ve and -ve charge in the soln.
- The property of soln of an electrolyte is the property of ions in the soln.
- The process of ionization is not a complete process; only the fraction of total molecule undergoes ionization which is denoted by α and called degree of ionization.
- The ions of an electrolytes can move freely.
Degree of ionization
Degree of ionization is defined as the no. of mole of ionized electrolyte to the total no. of mole of electrolyte. It is denoted by α.
For strong electrolyte α ≅ 1.
For weak electrolyte α <<< 1.
Dissociation / Ionization constant (Kd / Ki / K)
Ionization constant define as the ratio of product concn of ions produce in the soln to the concn of unionized electrolyte. It is denoted by Kd / Ki / K.
Kd = [A+] × [B−][AB]
Ostwald’s dilution law
It states that at a constant temp degree of ionization of weak electrolyte is directly proportional to the square root of its dilution (volume).
Let us consider an electrolyte AB which undergoes ionization as:
Let’s initial concn of an electrolyte AB is C mol/l.
After sometime α% of C mol/l molecule ionized then:
| AB | A+ | B− | |
|---|---|---|---|
| t = 0 | C mol/l | 0 | 0 |
| t = equilibrium | (C − Cα) | Cα | Cα |
Kd = [A+] × [B−][AB]
Kd = Cα × CαC(1 − α)
Kd = Cα21 − α …(2)
For strong electrolyte α = 1 and 1 − α = 0, the eqn (2) becomes Kd = ∞ which is undefined.
For weak electrolyte α <<< 1 and 1 − α = 1, the eqn (2) becomes:
Kd = Cα2
α = √(Kd/C) …(3)
α ∝ 1/√C …(4)
Eqn (4) shows that degree of ionization weak electrolyte is inversely proportional to the square root of its concn.
Concn (C) = n / V
For 1-mole C = 1 / V
putting value of C in eqn (4)
α ∝ √V
Hence, degree of ionization of weak electrolyte is directly proportional to the square root of dilution.
Limitations of Ostwald’s dilution law
- Ostwald’s dilution law is only applicable for weak electrolyte not for strong electrolyte.
We have:
For strong electrolyte α = 1 and 1 − α = 0 then eqn is too be written as Kd = ∞ which is undefined.
For weak electrolyte α <<< 1 and 1 − α = 1 the eqn is too be written as:
c) Calculate the degree of ionization of HCN having concn 0.01 M [Ka of HCN = 4.8 × 10−10]. Also calculate H+ concn and pH of the soln.
Given,
for HCN
Ka of HCN = 4.8 × 10−10
C = 0.01
HCN undergoes ionization as:
| HCN | H+ | CN− | |
|---|---|---|---|
| t = 0 | 0.01 M | 0 | 0 |
| t = equilibrium | (0.01 − 0.01α) | 0.01α | 0.01α |
[H+] = [CN−] = 0.01α
[HCN] = (0.01 − 0.01α)
Ka = [H+][CN−][HCN]
= 0.01α × 0.01α0.01 − 0.01α
= 0.01α21 − α
For weak electrolyte α <<< 1 and 1 − α ≅ 1
Ka = 0.01α2
4.8 × 10−10 = 0.01α2
α = √[(4.8 × 10−10)/0.01]
α = 2.19 × 10−4
Now,
[H+] = 0.01α
= 0.01 × 2.19 × 10−4
= 2.19 × 10−6 M
[CN−] = 0.01α = 2.19 × 10−6 M
[HCN] = 0.01 − 0.01α
= 0.01 − 2.19 × 10−6
= 9.99 × 10−3
pH = −log[H+]
= −log(2.19 × 10−6)
= 5.65
Concept of Acids and Bases
1) Arrhenius concept of Acid and base
The chemical substance which can produce H+ ions in its aq soln is called acid eg HCl, HNO3, H2SO4, HCOOH etc.
The chemical substance which can produce OH− ions in its aq soln is called base eg NaOH, KOH, Ca(OH)2, Mg(OH)2 etc.
2) Bronsted and Lowry concept of Acid and Base
The chemical substance which can donate proton to the other substance is called acid and the chemical substance which can accept proton from other substance is called base.
C.A : B C.B : C.A
acid / accept conjugate / conjugate acid
C.A:B pair by donating H+
Conjugate acid base pair
The pair which are differ only by proton and formed by donating proton and by accepting proton is called conjugate acid base pair.
| Compound | C.A | C.B |
|---|---|---|
| H2O | H3O+ | OH− |
| NH3 | NH4+ | NH2− |
| HS− | H2S | S2− |
| HCO3− | H2CO3 | CO32− |
| HSO4− | H2SO4 | SO42− |
Amphoteric substance
The chemical substance which acts as an acid as well as a base during chemical rxn is called amphoteric substance eg water (H2O), NH3, HCO3− etc.
acid base C.B C.A
Here H2O act as an acid and a base during chemical rxn.
3) Lewis concept of acid and Base
Lewis acid are species (charged or uncharged) that can accept pair of electron or [Unreadable in source]. In other words, pair of electron acceptor is Lewis acid. They are also called electrophile eg BF3, AlCl3, FeCl3, H+ etc.
Lewis base are species (charged or uncharged) that can donate pair of electron to other substance. In other words, pair of electron donor is Lewis base. They are also called nucleophile eg NH3, H2O, NH2−, OH−, Cl− etc.
Limitation
1) Arrhenius concept of Acid and base
- This concept is only limited in aq soln.
- According to this concept when an acid is dissolve in water H+ ion is produce only in aq form but H+ ion is highly reactive and immediately produce hydronium ion (H3O+).
H+ + H2O → H3O+
Hydronium ion
- It is failed to explain the acidic and basic character of some compounds which doesn’t contain hydrogen such as compounds SO2 etc. are acidic nature but some compounds which doesn’t contain OH− ion such as calcium oxide, magnesium oxide, BaO etc.
- It is failed to explain the acidic and basic character of electron deficient compounds such as BF3, AlCl3, FeCl3 etc.
2) Bronsted and Lowry concept of Acid and base
- This concept is limited to proton transfer acid base rxn and ignore the non-proton transfer acid base reaction and acidic oxide.
base acid salt
- It is failed to explain the acidic character of some compounds which doesn’t contain hydrogen such as CO2, SO2, SO3 etc and basic character of some compound which doesn’t contain OH− ion such as calcium oxide, magnesium oxide, BaO etc.
- It is failed to explain the acidic character of electron deficient compound such as BF3, AlCl3, FeCl3 etc.
3. Lewis concept of acid and base
- According to this concept there is formation of co-ordinate covalent bond b/w acid and base. However acids such as HCl, HNO3, H2SO4, HNO3 etc. [Unreadable in source].
- Lewis concept cannot explain acid base behaviour of such type of compound.
Auto Ionization of water
Water itself get ionized by producing H+ ion or OH− ion (hydroxyl ion) called auto ionization of water.
By applying law of mass action:
Keq = [H+][OH−]H2O
Keq × H2O = [H+][OH−]
Kw = [H+][OH−]
where Kw is ionic product of water.
Experimentally the ionic product of water is found to be 1.0 × 10−14.
Hence ionic product of H2O is define as the product of the molar conc of its ion.
pH and pOH scale
pH of a soln is defined as the negative logarithm of molar concn of H+ ion.
[H+] = 10−pH
Similarly pH of a soln is defined as the negative logarithm of a molar concn of OH− ion.
[OH−] = 10−pOH
pH scale is that scale in which the value of pH is computed with corresponding molar concn of hydrogen H+ ion.
Relation b/w pH and pOH
let us consider water itself get ionized as:
We have,
taking log on both side:
log Kw = log[H+] + log[OH−]
Taking −ve common on both side
−log Kw = −log[H+] − log[OH−]
−log(1.0 × 10−14) = −log[H+] + −log[OH−]
pH + pOH = 14
Hydrolysis of Salt
The interaction of radical of salt with water is called hydrolysis of salt. It is of four type.
Salt of strong acid and strong base
When the salt of strong acid and strong base hydrolyzed then the soln is neutral. Due to the formation of equal amount of H+ ion and OH− ion in the soln.
eg: NaCl, KCl, NaNO3, KNO3, Na2SO4, K2SO4 etc.
Salt of strong acid and weak base
When the salt of strong acid and weak base hydrolyzed then the soln is acidic. Due to the formation of greater amount of H+ ion then OH− ion.
eg: NH4Cl, NH4NO3, (NH4)2SO4, FeCl3 etc.
Salt of weak acid and strong base
When the salt of weak acid and weak acid hydrolyzed then the soln is basic. Due to the formation of greater amount of OH− ion then H+ ion.
eg: CH3COONa, KCN, CH3COONa, HCOONa etc.
Salt of weak acid and weak base
When the salt of weak acid and weak base hydrolyzed then the soln is neutral. Due to the formation of equal amount of H+ ion and OH− ion in the soln.
eg: CH3COONH4, HCOONH4, (NH4)2CO3 etc.
Calculate pH, pOH, H+ and OH− ion of 0.02 N H2SO4
N = M × Z
0.02 = M × 2
M = 0.01 M
∴ molarity of H2SO4 = 0.01 M
H2SO4 ionized as:
0.01 M 2 × 0.01 M 0.01 M
∴ Total conc of [H+] = 2 × 0.01 M = 0.02 M
Now pH = −log[H+] = −log(0.02) = 1.69
pH + pOH = 14
pOH = 14 − pH = 14 − 1.69
pOH = 12.31
[OH−] = 10−pOH
= 10−12.31
= 4.89 × 10−13 M
Solubility product and Solubility product Principle
Dynamic eq
Some electrolyte such as silver chloride (AgCl), barium sulphate, barium carbonate, CaF2, [Unreadable in source] etc dissolved in limited amount in water & produce ions are called sparingly soluble electrolyte.
Let us consider an electrolyte AgCl dissolve in water then:
soln
Applying law of mass action:
Keq = [Ag+] × [Cl−][AgCl]
Keq × [AgCl] = [Ag+] × [Cl−]
Ksp = [Ag+][Cl−]
Where Ksp is solubility product constant or simply solubility product.
Hence, solubility product of sparingly soluble electrolyte such as AgCl, BaSO4, CaF2, Mg(OH)2 etc is define as the product of molar concn of its ion at saturated state.
ppt only takes when ionic product is greater then that of solubility product otherwise not. This principle is called theory of ppt or solubility product principle.
Depending upon the ionic product soln is divided into three types:
- IP = Ksp then the soln is saturated.
- IP < Ksp then the soln is unsaturated.
- IP > Ksp the soln is supersaturated and ppt out or salting out.
Application of solubility product principle
1. Calculate the solubility of sparingly soluble electrolyte at a particular temperature.
a) For AB electrolyte [eg: AgCl, BaSO4, CaCO3 etc]
Let S mol/l be the solubility of AgCl and it ionized as:
S mol/l S mol/l S mol/l
∴ [Ag+] = S mol/l
[Cl−] = S mol/l
Solubility product of AgCl = [Ag+] × [Cl−]
Ksp = S × S
Ksp = S2
S = √Ksp
B. For AB2 / A2B type electrolyte [Eg: Mg(OH)2, CaF2, Na2S]
let S mol/l be the solubility of Mg(OH)2 and it ionized as:
S mol/l S mol/l 2S mol/l
[Mg2+] = S mol/l
[OH−] = 2S mol/l
Solubility product of Mg(OH)2 = [Mg2+] × [OH−]2
Ksp = (S) × (2S)2
Ksp = 4S3
S = ∛(Ksp/4)
C. For AB3 / A3B type salt [Eg: Fe(OH)3, Na3P, AlCl3, K3[Fe(CN)6] etc.]
let S mol/l is a solubility of Fe(OH)3 and it ionizes as:
S mol/l S mol/l 3S mol/l
[Fe3+] = S mol/l
[OH−] = 3S mol/l
Solubility product of Fe(OH)3 = [Fe3+] × [OH−]3
Ksp = (S) × (3S)3
Ksp = 27S4
S = ⁴√(Ksp/27)
4) For A4B / AB4 type salt [Eg K4[Fe(CN)6]]
let S mol/l is a solubility of K4[Fe(CN)6] and it ionized as:
S mol/l 4S mol/l S mol/l
[K+] = 4S mol/l
[Fe(CN)64−] = S mol/l
Solubility product of K4[Fe(CN)6] = [K+]4 × [Fe(CN)64−]
Ksp = (4S)4 × S
Ksp = 256S5
S = ⁵√(Ksp/256)
5. For A3B2 / A2B3 type of salt [Eg Ca3(PO4)2]
let S mol/l be the solubility of Ca3(PO4)2 and it ionized as:
S mol/l 3S mol/l 2S mol/l
[Ca2+] = 3S mol/l
[PO43−] = 2S mol/l
Solubility product of Ca3(PO4)2 = [Ca2+]3 × [PO43−]2
Ksp = (3S)3 × (2S)2
Ksp = 27S3 × 4S2
Ksp = 108S5
S = ⁵√(Ksp/108)
5) Soln
W = 0.00143 g
V = 1 litre
Mol wt = 108 + 35.5 = 143.5
g/l = 0.00143 / 1 = 0.00143 g/l
we know,
M × mol wt = g/l
M = g/l / mol wt
M = 0.00143 / 143.5
M = 9.96 × 10−6 M
Now AgCl ionized as:
9.96 × 10−6 M 9.96 × 10−6 M 9.96 × 10−6 M
∴ [Ag+] = [Cl−] = 9.96 × 10−6 M
Solubility product of AgCl = [Ag+] × [Cl−]
Ksp = 9.96 × 10−6 × 9.96 × 10−6
= 9.92 × 10−11
2. To predict the precipitation of mixing solution
It helps to predict whether the sparingly soluble salt ppt or not when two soln of known concn are mixed. Since ppt is only taken place when I.P is greater the Ksp otherwise no.
3. Precipitation of soluble salt
what happens when HCl gas is passed through a saturated soln of NaCl and why?
When HCl gas is passed through a saturated soln of NaCl then concn of Cl− ion is increases and hence I.P exceed the Ksp and pure NaCl ppt out.
HCl → H+ + Cl−
common-ion
Common-ion effect
The suppression (decrease) of degree of ionization of weak electrolyte on addition of strong electrolyte having a common ion is called common ion effect.
eg
HCl → H+ + Cl−
Now, H2S is a weak electrolyte when small amount of HCl (which is strong electrolyte) is added degree of ionization of weak electrolyte (H2S) is decreases due to formation of H+ ion as common ion.
NH4Cl (strong) → NH4+ + Cl−
common ion
Here NH4OH is a weak electrolyte when small amount of NH4Cl (which is strong electrolyte) is added degree of dissociation of weak electrolyte (NH4OH) is decrease due to formation of NH4+ ion as common ion.
Application of common-ion effect
1. To qualitative analysis of inorganic salts
During group separation of inorganic salt in qualitative analysis, metal cations are precipitate as their sulphide in different condition by using solubility product principle and common ion effect in which pH is carried out altering the ionic product and solubility product.
a) Role of HCl in IInd group metal
All 2nd group metal (Cu2+, Pb2+, Hg2+ etc) are precipitate as their sulphide. Dilute HCl is added to the original soln of salt before passing H2S gas. The main role of HCl is to suppress the sulphide S2− ion concn in soln by producing H+ ion as common-ion.
HCl → H+ + Cl−
common-ion
b) Role of NH4Cl in 3rd group metal
All 3rd group metal (Fe3+, Al3+, Cr3+ etc) are ppt in the hydroxide form. Dilute NH4Cl is added to the original soln of salt before passing NH4OH gas. The main role of NH4Cl is to suppress the hydroxyl (OH−) ion concn in soln by producing NH4+ ion as common-ion.
NH4Cl → NH4+ + Cl−
common-ion
Buffer soln
The soln that can resists the change in pH on addition either acid or base is called buffer soln. It is of:
- Acidic buffer soln
- Basic buffer soln
1) Acidic buffer soln
If the pH of buffer soln is less than 7 is known as acidic buffer soln.
2) Basic buffer soln
If the pH of buffer soln is greater than 7 is known as a basic buffer soln.
Mechanism of buffer soln
1. Mechanism of acidic buffer soln
Mixture of weak acid and its salt with strong base.
Consider an acidic buffer soln prepared by mixing CH3COOH and CH3COONa.
CH3COONa → CH3COO− + Na+
Common ion
a) Effect of add of an acid
When small amount of acid is added to it then excess H+ ion combines with CH3COO− ion to form CH3COOH. Thus there is no change in pH.
acid buffer soln
b) Effect of add of base
When small amount of base is added to it then excess OH− ion combines with H+ ion to form H2O. Thus there is no change in pH.
base buffer
2) Mechanism of basic buffer soln
Mixture of weak base and its salt with strong acid.
Consider a basic buffer soln prepared by mixing NH4OH and NH4Cl.
NH4Cl → NH4+ + Cl−
common ion
a) Effect of add of an acid
When small amount of acid is added to it then excess H+ ion combines with OH− ion to form H2O. Thus there is no change in pH.
acid buffer soln
b) Effect of add of base
When small amount of a base is added to it then excess OH− ions combines with NH4+ ion to form NH4OH. Thus there is no change in pH.
base buffer soln
Numericals
58) What will be the resultant pH when 200 ml of aqueous solution of HCl (pH = 2) is mixed with 300 ml of an aqueous solution of NaOH (pH = 12)?
HCl
V1 = 200 ml
pH = 2
[H+] = 10−pH
= 10−2 M = 0.01 M
M1 = N1 = 10−2 M
NaOH
V2 = 300 ml
pH = 12
pH + pOH = 14
pOH = 14 − pH
= 14 − 12 = 2
[OH−] = 10−pOH = 10−2 M = 0.01 M
M2 = N2 = 10−2 M
Vmix = V1 + V2 = 200 + 300 = 500 ml
Nmix Vmix = M1V1 − M2V2
Nmix × 500 = 0.01 × 200 − 0.01 × 300
Nmix = −1 / 500
Nmix = −2 × 10−3
Since value of Nmix value is −ve so the resultant soln is basic in nature.
2 × 10−3 M 2 × 10−3 M 2 × 10−3 M
[OH−] = 2 × 10−3 M
pOH = −log(2 × 10−3)
= 2.69
pH + pOH = 14
pH = 14 − 2.69
pH = 11.30
ii)
HCl
V1 = n
pH = 1
[H+] = 10−pH
= 10−1
M1 = N1 = 10−1 M
NaOH
V2 = n
pH = 12
pH + pOH = 14
pOH = 14 − pH = 2
[OH−] = 10−2 M
M2 = N2 = 10−2 M
Vmix = V1 + V2 = n + n = 2n ml
NmixVmix = M1V1 − M2V2
Nmix × 2n = 10−1 × n − 10−2 × n
Nmix = (10−1 − 10−2) / 2
= 0.045
since the value of mix is +ve so the resultant soln is acidic or basic.
0.045 0.045 0.045
[OH−] = 0.045
pOH = −log[OH−]
= −log 0.045
= 1.34
pH + pOH = 14
pH = 14 − 1.34
pH = 12.66
4) Soln
HCl
V1 = n
pH = 3
[H+] = 10−pH
= 10−3 M
M1 = N1 = 10−3 M
NaOH
V2 = n
pOH = 5
[OH−] = 10−pOH
= 10−5 M
M2 = N2 = 10−5 M
Vmix = V1 + V2 = n + n = 2n ml
Nmix = Mmix = ?
Nmix × Vmix = M1V1 − M2V2
Nmix × 2n = 10−3 × n − 10−5 × n
Nmix = (10−3 − 10−5) / 2
Nmix = 4.95 × 10−4 M
Since the value of Nmix is +ve so, the resultant soln is Acidic in nature.
4.95 × 10−4 4.95 × 10−4 4.95 × 10−4
[H+] = 4.95 × 10−4
pH = −log(4.95 × 10−4)
pH = 3.30
42) Calculate the pH of an aqueous soln containing 10−7 moles of NaOH per litre.
Soln
Molarity of NaOH = 10−7 M
NaOH ionized as:
10−7 M 10−7 M 10−7 M
10−7 M 10−7 M
Total conc of [OH−] = [OH−]NaOH + [OH−]H2O
= 10−7 + 10−7
= 2 × 10−7 M
pOH = −log(2 × 10−7)
pOH = 6.69
pH + pOH = 14
pH = 14 − pOH
= 14 − 6.69
= 7.31
Discussion
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