Ionic Equilibrium

 A . Very Short Answer Type Question.

1.

The degree of dissociation of a weak electrolyte is proportional to the square root of dilution i.e. on increasing dilution the degree of dissociation of weak electrolytes will be increased is called Ostwald’sdilution law,

For weak electrolytes,

Or, kc= α2v$\frac{{\alpha }^2}{\mathrm{v}}$. Where, kc = dissociation constant.

α = degree of dissociation

v = volume of solution.

Or, α = Kcv−−−−√$\sqrt{{\mathrm{K}}_{\mathrm{c}}\mathrm{v}}$

So, α ∝$\propto$√ v

 

2.

The fraction of the total number of molecules present as free ions in the solution is known as degree of ionization of electrolyte (α):

Or, α = No.ofmoleculesofionizedasionsTotalno.ofmoleuclesdissolved$\frac{\mathrm{N}\mathrm{o}.\mathrm{o}\mathrm{f}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{u}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{o}\mathrm{f}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{i}\mathrm{z}\mathrm{e}\mathrm{d}\mathrm{a}\mathrm{s}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{s}}{\mathrm{T}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}\mathrm{n}\mathrm{o}.\mathrm{o}\mathrm{f }\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{u}\mathrm{c}\mathrm{l}\mathrm{e}\mathrm{s }\mathrm{d}\mathrm{i}\mathrm{s}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{v}\mathrm{e}\mathrm{d}}$

 

3.

Solution
Here, pH = 5.5

So, pH = - logH+]

Or, 5.5 = -log[H+]

Or, [H+] = 10-5.5 = 3.162 * 10-6M.

So, Cu+ ion concentration of the solution is 3.162 * 10-6M.

 

4.

In a solution of a electrolyte, the stage in which the rate of dissociation of electrolyte is equal to the rate of recombination, forming a dynamic equilibrium is called ionic equilibrium.

 

5.

The degree of dissociation of a weak electrolyte is proportional to the square root of dilution i.e. on increasing dilution the degree of dissociation of weak electrolytes will be increased is called Ostwald’sdilution law.

Ostwald’s dilution law is not applicable to strong electrolyte because the entire strong electrolyte is dissociated into ions and hence equilibrium is not achieved. Moreover, in Ostwald's dilution law, for weak electrolyte, (1 – α) à 1

So, α ∝√v$\propto \surd \mathrm{v}$.

But, for strong electrolyte α = 1.

So, (1 – α) cannot be considered as 1.

 

 

6.

Experimentally it is formed that, the molar product of ions:

OH+, OH- in pure water at 25°C is:

[H+] = [OH]- for neutral water,

So, [H+][H+] = 10-14

Or, [H+] = 10-14/2 = 10-7M

So, the concentration of H+ and OH- ions in water is 10-7 mol/liter.

 

7.

pH is a scale to measure the acidity of a substance . It is defined as the negative logarithm of hydrogen ions present in a solution. Yes, the pH value of pure water varies with the temperature, since on increasing the temperature, the kinetic energy of water molecules increases , creating more collisions and thus more dissociation of molecules into ions take place. So, the concentration of hydrogen ion also increases. As, the concentration of H+ ion increases pH of water decreases, but it is still neutral.

 

8.

A saturated solution of sodium chloride contains an ionic equilibrium between dissociating sodium chloride molecules and recombining sodium chloride ions. But, when HCl is passed through the saturated solution due to common chloride ion, the common ion effect takes place, thus shifting the equilibrium to where the ion is being consumed. Thus, shifting the reaction towards the backward direction, i.e. precipitation of NaCl occurs.

 

9.

Soln:

Solubility product (K) = 8 * 10-45,

Let the solubility be ‘s’.

So, CuS

Initial

So, [Cu2+][s2-] = 8 * 10-45

Or, S2 = 8 * 10-45

So, S = 8∗10−45−−−−−−√$\sqrt{8\ast {10}^{-45}}$ = 8.944 * 10-22 Mol/liter.

So, the solubility is 8.994 * 10-22 mol/liter.

 

10.

From Ostwald’s dilution law,

Or, k = α2v$\frac{{\alpha }^2}{\mathrm{v}}$

Or, α = k.v−−−√$\sqrt{\mathrm{k}.\mathrm{v}}$

So, on increasing dilution, the degree of ionization increased by half the power of dilution.

 

11.

 Soln:

0.37g/liter of HCl contains 0.37g in a liter of solution.

So, Molarity of HCl = 0.3735.5∗11$\frac{0.37}{35.5}\ast \frac{1}{1}$ = 0.0104M.

So, [H+] = [HCl] = 0.0104.

Or, pH = -log[H+] = - log[0.0104] = 1.99

So, Required pH = 1.99

 

12.

 Soln:

So, pH = 2.82.

So, pH = -log[H+]

So, [H+] = 10-pH = 10-2.82M

So, Concentration of H+ ion = 1.5 * 10-3 Mol/liter.

 

13.

The product of concentrations of the ions furnished by sparingly soluble electrolyte raised to the power of their corresponding stiochiometricco – efficient is called solubility product.

Soln:

Solubility CaF2 = 2.05 * 10-4 mol/liter.

So,

At equilibrium, let the solubility be ‘s’. then,

s = 2.05 * 10-4 mol/ltr.

Here, [Ca2+] = 2.05 * 10-4M

[F2-] = 2 * 2.05 * 10-4

So, Ksp = [Ca2+][F-]2 = s.(2s)2

= 4s3 = 4 * (2.05 * 10-4)3

So, ksp = 3.446 * 10-11

 

14.

 Soln:

Here, pH = 4.70

So, pH = - 10g[H+]

So, [H+][OH-] = 1 * 10-14

So, [OH-] = 1∗10−141.995∗10−5$\frac{1\ast {10}^{-14}}{1.995\ast {10}^{-5}}$ = 5.012 * 10-10M

So, [OH-] = 5.012 * 10-10M and [H+] = 1.995 * 10-5M

 

15.

 Soln:

Here, concentration of acetic acid. i.e.

[CH3COOH] = 0.01M

Or, ka= 1.8 * 10-5

So, ka = [CH3COO−][H+][CH3COOH]$\frac{\left[\mathrm{C}{\mathrm{H}}_3\mathrm{C}\mathrm{O}{\mathrm{O}}^{-}\right]\left[{\mathrm{H}}^{+}\right]}{\left[\mathrm{C}{\mathrm{H}}_3\mathrm{C}\mathrm{O}\mathrm{O}\mathrm{H}\right]}$

Or, ka = [H+]2[CH3COOH]$\frac{{\left[{\mathrm{H}}^{+}\right]}^2}{\left[\mathrm{C}{\mathrm{H}}_3\mathrm{C}\mathrm{O}\mathrm{O}\mathrm{H}\right]}$

Or, [H+] = kax[CH3COOH]−−−−−−−−−−−−−√$\sqrt{\mathrm{k}{\mathrm{a}}^{\mathrm{x}}\left[\mathrm{C}{\mathrm{H}}_3\mathrm{C}\mathrm{O}\mathrm{O}\mathrm{H}\right]}$ = 0.01∗1.8∗10−5−−−−−−−−−−−√$\sqrt{0.01\ast 1.8\ast {10}^{-5}}$

= 4.24 * 10-4 mol.lit-1.

Now, pH = -log[H+] = -log(4.24 * 10-4) = 3.372.

So, required pH = 3.372.

 

16.

 Soln:

So, pH = 9.5, [H+] = 10-pH = 10-9.5 = 3.16 * 10-10M.

So, [H+] = 3.6 * 10-10 mol.ltr-1.

 

17.

Lewis acid is defined as the chemical species in a state, which can accept a pair of electrons. E.g: Lewis base is defined as the chemical species in any state, which can donate a pair of electron. E.g: NH3­.

 

18.

 When FeCl3 is dissolved in water, it gets hydrolyzed. Thus formed Fe3+ ion is again converted to Fe(OH)3, a weak base, thus, the concentration of OH- ion in a solution decreases. So, comparatively the no. of H+ ions in solution is increased, and hence the solution becomes acidic.

 

19.

 Soln:

Here, molarity of NaOH = 10-7M

Or, [OH-] = 10-7M

Molarity of OH- ions in water = 10-7M

Ie. [OH-] = 10-7M

So, Total [OH-] = 10-7 * 2M

Now, pOH = -loh[OH-] = -log[2 * 10-7]

= 6.699

Or, pH = 14 – pOH = 14 – 6.699 = 7.301.

 

20.

 When sodium carbonate (Na2CO3) is made aqueous is gets hydrolyzed furnishing Na+ and CO32-.

Na2CO3à 2Na+ + CO32-

H2O

The water resources are also in dynamic equilibrium with their respective ions. But, the CO32- ion combines with 2H+ ion to form H2CO3 – a weak base, thus comparatively increasing OH- ion concentration in solution. The formed acid doesn’t get completely ionized. Hence, the number of OH- ions in solution is high. So, the solution remains basic.

 

21.

 The aqueous solution of NH4Cl is acidic because, on hydrolysis on aq. medium, it forms corresponding acid HCl and corresponding base, NH4OH. Hence, HCl is strong acid and gets comparatively hydrolyzed and NH4OH is being weak acid, gets partially hydrolyzed, thus the concentration of H+ ion in solution is high, so the solution becomes acidic.

But in case of KCN, thus formed KOH and HCN, bases and acid, are strong and weak respectively. So, the bases and acids, are strong and weak respectively, so the base goes complete hydrolysis and the acid partial. So, due to more no. of OH- ion is solution the solution is basic.

 

22.

 Soln:

Kq of benzoic acid = 6.6 * 10-5

Molarity of benzoic acid = 0.1M

So, Kq = [H+]2[C6H5COOH]$\frac{{\left[{\mathrm{H}}^{+}\right]}^2}{\left[{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{C}\mathrm{O}\mathrm{O}\mathrm{H}\right]}$

Or, [H+] = Ka(C6H5COOH)−−−−−−−−−−−−−−√$\sqrt{{\mathrm{K}}_{\mathrm{a}}\left({\mathrm{C}}_6{\mathrm{H}}_5\mathrm{C}\mathrm{O}\mathrm{O}\mathrm{H}\right)}$

= 6.6∗10−5∗0.1−−−−−−−−−−√$\sqrt{6.6\ast {10}^{-5}\ast 0.1}$

= 2.57 * 10-3.

Or, pH = -10g[H+] = -log[2.57 * 10-3] = 2.59

So, pH is 2.59.

 

23.

Soln:

Here, ksp of BaSO4 = 1.1 * 10-10 mol2ltr-2.

Let, solubility of BaSO4 be S.mol.ltr, then,

BaSO4

Ksp = [Ba2+][SO42-]

Ksp = s2

Or, S = Ksp−−−√$\sqrt{{\mathrm{K}}_{\mathrm{s}\mathrm{p}}}$ = 1.1∗10−10−−−−−−−−√$\sqrt{1.1\ast {10}^{-10}}$

= 1.049 * 10-5 mol.ltr.

 

Or, gm/ltr = molarity * molecular wt.

= 1.049 * 10-5 * 233.33

= 2.447 * 10-3 gm/ltr.

So, required solubility is 2.447 * 10-3 gm/liter.

 

24.

 Soln:

Initial vol. (V1) = 25ml

Final vol. (V2) = 500ml

Initial strength (S1) = 110$\frac{1}{10}$N.

Final strength (S2) = ?

From normality equation,

Or, N1S1 = N2S2

Or, S2 = V1S1V2$\frac{{\mathrm{V}}_1{\mathrm{S}}_1}{{\mathrm{V}}_2}$

= 25∗110500$\frac{25\ast \frac{1}{10}}{500}$

= 0.005M.

For, complete ionization,

Or, [H+] = [HCl] = 0.005M

So, pH = - log[H+] = -log(0.005) = 2.301.

 

25.

 Soln:

Or, pH of NaOH = 10.6

Or, pOH = 14 – 10.6 = 3.4

[OH-] = 10-poH = 10 – 3.4

= 3.98 * 10-4M

So, [NaOH] = [OH-] = 3.98 * 10-4

So, gm/ltr of NaOH = Molarity * Mol. wt.

= 3.98 * 10-4 * 40

= 1.59 * 10-2 gm/liter.

So, 1.59 * 10-2 gm of NaOH should be dissolved per liter of solution.

 

26.

 Soln:

Here, 100cc of CaF2 soln: contains 1.7 * 10-3gm

So, 1000cc of CaF2 solution contains 1.7 * 10-3 * 1000100$\frac{1000}{100}$ gm

= 1.7 * 10-2gm

So, gm/liter of CaF2 = 1.7 * 10-2

Molarity of CaF2 = gmliterMol.wt$\frac{\frac{\mathrm{g}\mathrm{m}}{\mathrm{l}\mathrm{i}\mathrm{t}\mathrm{e}\mathrm{r}}}{\mathrm{M}\mathrm{o}\mathrm{l}.\mathrm{w}\mathrm{t}}$ = 1.7∗10−278$\frac{1.7\ast {10}^{-2}}{78}$ = 2.78 * 10-4

If ‘S’ be the solubility of CaF2then‘s’ = 2.18 * 10-4 Mol/liter.

CaF2

So, Ksp = [Ca2+][2F-]2

= S.(2s)2 = 4s3

= 4 * (2.18 * 10-4)3

= 4.14 * 10-11 mol2/liter2

So, the solubility product is 4.14 * 10-11 mol2/liter2.

 

27.

 Soln:

Here, Ksp of AgCl = 2.8 * 10-10

Let‘s’ be the solubility of AgCl at this temperature.

 

Ksp = [Ag+][Cl-]

= s.s = s2

Or, s = Ksp−−−√$\sqrt{{\mathrm{K}}_{\mathrm{s}\mathrm{p}}}$ = 2.8∗10−10−−−−−−−−√$\sqrt{2.8\ast {10}^{-10}}$

= 1.6 * 10-5 mol/liter.

So, the solubility is1.6 * 10-5 mol /liter

 

28.

 Soln:                                                                                            

Here, gm/liter of AgCl = 1.435 * 10-3 g/liter.

Molarity of AgCl = gmlitermol.wt$\frac{\frac{\mathrm{g}\mathrm{m}}{\mathrm{l}\mathrm{i}\mathrm{t}\mathrm{e}\mathrm{r}}}{\mathrm{m}\mathrm{o}\mathrm{l}.\mathrm{w}\mathrm{t}}$ = 1.435∗10−3143.5$\frac{1.435\ast {10}^{-3}}{143.5}$

= 10-5 mol/liter.

Let‘s’ be the solubility of AgCl, then s = 10-5M

Ksp = [Ag+][Cl-]

Or, s.s = s2

= (10-5)2 = 10-10 mol2lit-2.

So, solubility product is1 * 10-10 mol2/ltr2

 

29.

Soln:

Here, pH of HCN = 5.2

[HCN] = 0.1M

So, [H+] = 10-pH = 10-5.2 = 6.3 * 10-6M

So, Ka = [H+]2[HCN]$\frac{{\left[{\mathrm{H}}^{+}\right]}^2}{\left[\mathrm{H}\mathrm{C}\mathrm{N}\right]}$ = (6.3∗10−6)20.1$\frac{{\left(6.3\ast {10}^{-6}\right)}^2}{0.1}$ = 3.98 * 10-10

So, Ka of acid is 3.98 * 10-10

 

30.

 When an acid, dissociates to give a proton then, the conjugate part of that acid is called as its conjugate base. And when a base accepts a proton, then so formed species is known as its conjugate acid.

Thus, formed acids and bases form their corresponding bases and acids, taken in pair are known as conjugate acid – base pairs.

In the reaction,

CO32- + H2O

Here, CO32- accepts a proton to form HCO3-, so HCO3- is conjugate acid of CO32- base, and H2O dissociates to give OH-, giving a proton, so OH- is conjugate base of H2O acid. So, CO32- and HCO3-, and H2O and OH- are conjugate acid – base pairs.

 

31.

The product of concentrations of the ions furnished by sparingly soluble electrolyte raised to the power of their corresponding stiochiometric co - efficient is called solubility product.

For Ag2CrO=4

 

Ksp = [Ag+][CrO4-] is required expression.

 

32.

The effect by virtue of which, the degree of dissociation of weak electrolyte is decreased by a strong electrolyte, if both the electrolyte contains at least a common ion, in its solution.

 

33.

The phenomenon of interaction of anions and cations of the salt with H+ and OH- ions furnished by water yielding acidic or alkaline or even neutral solution is known salt hydrolysis.

 

34.

 Soln:

Here, Kb of BOH = 1.8 * 10-5

[BOH] = 0.1M

So, Kb = [OH−]2[BOH]$\frac{{\left[\mathrm{O}{\mathrm{H}}^{-}\right]}^2}{\left[\mathrm{B}\mathrm{O}\mathrm{H}\right]}$

Or, [OH-] = kb[BOH]−−−−−−−√$\sqrt{{\mathrm{k}}_{\mathrm{b}}\left[\mathrm{B}\mathrm{O}\mathrm{H}\right]}$

= 1.8∗10−5∗0.1−−−−−−−−−−√$\sqrt{1.8\ast {10}^{-5}\ast 0.1}$

= 1.34 * 10-3

Or, pH = -log[OH-]

= -log[1.34 * 10-3]

= 2.87

Or, pH = 14 – pOH = 14 – 2.87 = 11.13

So, Required pH = 11.13

 

35.

Soln:

The no. of moles of H+ ions furnished by 10-8M HCl is also 10-8 moles in a liter of solution. But, we also know, water self furnished 10-7 moles of H+ in a liter of solution.

So, total H+ ions in a liter of solution is 10-7 + 10-8

= 1.1 * 10-7 moles.

So, Total molarity = 1.1 * 10-7 M

Now, pH = -log[H+] = -log[1.1 * 10-7]

= 6.96

So, pH of 10-8 M HCl is 6.96 not 8.

 

36.

The effect by virtue of which, the degree of dissociation of weak electrolyte is decreased by a strong electrolyte, if both the electrolyte contains at least a common ion, in its solution.

 

 

37.

When an acid, dissociates to give a proton then, the conjugate part of that acid is called as its conjugate base. And when a base accepts a proton, then so formed species is known as its conjugate acid.

Thus, formed acids and bases form their corresponding bases and an acid, taken in pair is known as conjugate acid – base pairs.

 

 

38.

We know, the solubility product of water is the product of molar concentration of H+ and OH- ions furnished by water molecules in pure state. Then,

 

Kw = [H+][OH-]

Taking – ve logarithm on both sides,

Or, -logkw = -log[H+][OH-]

Or, pkw = -log[H+] – log[OH-]

So, pKw = pH + pOH.

 

39.

 Water in its molecular form, has a lone pairs of electrons around oxygen, which can be donated. So, being a electron pair donor, it is a lewis base.

 

Also, water in its aqueous form gets partially dissociated into H+ and OH- ions. So, H2O gives H+ ion in aqueous form, thus, it behaves as Bronsted acid.

 

40.

 Sodium carbonate on hydrolysis forms their respective base NaOH and H2CO3. Here, NaOH being strong base, dissociates completely to give OH- and Na+ ions, whereas, H2CO3 being weak and dissociates partially to ions. Hence, the concentration of OH- in the solution is high, so the solution becomes basic in nature.

But in case of NaCl, thus formed acids and bases are both strong and dissociates into complete ions. Hence, the no. of H+ and OH- ion in solution is equal to make solution neutral.

 

41.

When CuSO4 is hydrolyzed in water it gives corresponding acid H2SO4 and corresponding base Cu(OH)2. The acid H2SO4 is strong and hence completely dissociates to H+ and SO42- ions but CuOH)2 being weak base partially dissociates to the respective ions. Thus, the concentration of H+ ions in the solution is more than OH- ion and hence the solution is acidic.

 

42.

pH is a scale to measure the acidity of a substance . It is defined as the negative logarithm of hydrogen ions present in a solution. Yes, the pH value of pure water varies with the temperature, since on increasing the temperature, the kinetic energy of water molecules increases, creating more collisions and thus more dissociation of molecules into ions take place. So, the concentration of hydrogen ion also increases.As, the concentration of H+ ion increases, pH of water decreases, but it is still neutral

 

Soln:

Here, [H2SO4] = 0.1M

But, [H+] = 2 * [H2SO4] = 2 * 0.1 = 0.2M

pH = -10g[H+] = - log[0.2]

So, pH = 6.99 * 10-1

 

43.

 Soln:

pH of KOH = 10,

pOH = 14 – 10

= 4.

So, pH = -log[OH]

So, [OH-] = 10-pOH

= 10-4M

So, concentration of hydroxyl ion is 10-4M.

 

44.

 When CaCl2 is hydrolyzed in water it forms Ca(OH)2 a weak base and HCl, a strong acid. The base goes partial dissociation to give Ca2+ and OH- ion, but HCl goes complete ionization to give H+ and Cl- ions. Thus, the concentration of H+ ion is greater than OH- ion hence the solution is acidic.

 

 

 B Short question

1.

The main points of Arrehenius theory of ionization are:

a. When an electrolyte is dissolved in water it gives electrically charged particles called ions. The one with + ve charge is called cations and with negative charge is anion.

AB à A+ (cation) + B-(anion)

 

b. The thus formed ions try to recombine to form unionized molecules, forming a dynamic equilibrium, described by a constant known as ionization constant.

Or, K = [A+][B−][AB]$\frac{\left[{\mathrm{A}}^{+}\right]\left[{\mathrm{B}}^{-}\right]}{\left[\mathrm{A}\mathrm{B}\right]}$

c. In this solution, total no. of positive charge is equal to that of negative charge.

d. The movement of ions result in conduction of electricity.

e. The properties of an electrolyte depends on the ions furnished by it.

f. Each ion behaves osmotically as a molecule.

g. The degree of ionization is the fraction of total no. of molecules ionized.

 

2.

Ostwald’s dilution law;

Let AB a binary electrolyte gets dissolved in water with α degree of ionization.

Then,

[AB] = 1−αV$\frac{1-\alpha }{\mathrm{V}}$, [A+] = αv$\frac{\alpha }{\mathrm{v}}$, [B-‑] = αv$\frac{\alpha }{\mathrm{v}}$

Or, using law of mass action,

Or, Kc = ([A+][B−]/[AB]$(\left[{\mathrm{A}}^{+}\right]\left[{\mathrm{B}}^{-}\right]/\left[\mathrm{A}\mathrm{B}\right]$

= αv∗αv1−αv$\frac{\frac{\alpha }{\mathrm{v}}\ast \frac{\alpha }{\mathrm{v}}}{\frac{1-\alpha }{\mathrm{v}}}$

So, Kc = α2(1−α)v$\frac{{\alpha }^2}{\left(1-\alpha \right)\mathrm{v}}$

For, weak electrolyte, 1 – α ≈$\approx$ 1.

So, Kc = α2v$\frac{{\alpha }^2}{\mathrm{v}}$

Or, α = Kc∗v−−−−√$\sqrt{{\mathrm{K}}_{\mathrm{c}}\ast \mathrm{v}}$

So, α ∝$\propto$V√$\sqrt{\mathrm{V}}$.

Thus, the degree of dissociation of weak electrolyte is proportional to square root of dilution is the Ostwald’s dilution law.

 

3.

Bronsted Lowry concept of acids and bases:

 According to the Bronsted-Lowry concept, an acid is a proton-donor, and a base is a proton-acceptor.

The reaction of an acid with a base involves transfer of a proton from the acid to the base. So, an acid and a base should be present simultaneously in any system. The extent of an acid-base reaction is governed not only by the proton-donating ability of the acid, but also by the proton-accepting tendency of the base. Acids and bases classified on the basis of this concept are termed as Bronsted acids and bases.

 

HCl(aq) + H­2 O(l)⬌ H3 O (aq)++Cl- (aq)

 

 

 

 

 

In this reaction, HCl donates its one proton to become Cl-, and H2O accepts one proton to become H3O+. Thus, HCl is Bronsted acid and H2O is a Bronsted base. For the reverse reaction, H3O+is able to transfer its proton to Cl-. So, H3O+is a Bronsted acid and Cl- is a Bronsted base.

 

Lewis concept of acid and bases:

Lewis proposed broader and more general definitions of acids and bases which do not require presence of protons to explain the acid base behaviour. According to Lewis concept, an acid is a substance which can accept a pair of electrons.

A base is a substance which can donate a pair of electron.

Acid-base reactions according to this concept involve donation of electron pair by a base to an acid to form a co-ordinate bond..

Lewis acids are the species having vacant orbitals in the valence shell of one of its atoms. The following species can act as Lewis acids.

a) Molecules having an atom with incomplete octet e.g., BF3, AlCl3 etc.

b) Simple cations for e.g., H+, Ag+ etc.

 

 

4.

Common ion effect:

The effect by virtue of which, the degree of dissociation of weak electrolyte is decreased by a strong electrolyte, if both the electrolyte contains at least a common ion, in its solution.

Let, consider a weak acid, HA, and a strong salt SA. Then, in aqueous medium

 

Here,

Since A- ion being the common ion, the degree of dissociation of HA acid is decreased, i.e the equilibrium shifts in backward direction. This is common ion effect.

Applications:                                                                                                   

1. Precipitation of sulphides of group II and IIIB basic radicals

2. Precipitation of hydroxides of group IIIA basic radicals

3. Precipitation of group IV basic radicals

 

5.

Arrhenius concept of acids and bases: 

According to the Arrhenius concept, hydrogen chloride, acetic acid, and sulphuric acid, are

 acids because all these compounds give free H+ ions in aqueous solutions.

 

Compounds such as NaOH and NH4OH are bases, because these compounds give free OH- ions in aqueous solutions.

 

NaOH + H2 O (excess) à Na+ (aq) + OH-(aq)

NH4OH+ H2 O (excess) à NH4+(aq)+ OH-(aq)

 

 

 

Thus, According to Arrhenius concept of acids and bases, the neutralization of an acid with a base involves the reaction between H+(aq) and OH-(aq) i.e.,

 

H+(aq)+ OH-(aq)→neutralization$\stackrel{\mathrm{n}\mathrm{e}\mathrm{u}\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{l}\mathrm{i}\mathrm{z}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}}{\to }$ H2 O

From acid from base

 

 

 

However, the Arrhenius concept is applicable to the acid-base behavior only in the aqueous medium. It does not provide any explanation to the acid-base behavior in the absence of water.

This concept defines acids and bases as compounds-containing hydrogen and hydroxyl group respectively. There are however, many compounds, which act as acid even when there is no hydrogen in their molecule. Similarly, there are many bases, which do not contain hydroxyl group.

 

 

6.

Bronsted Lowry concept of acids and bases:

 According to the Bronsted-Lowry concept, an acid is a proton-donor, and a base is a proton-acceptor.

The reaction of an acid with a base involves transfer of a proton from the acid to the base. So, an acid and a base should be present simultaneously in any system. The extent of an acid-base reaction is governed not only by the proton-donating ability of the acid, but also by the proton-accepting tendency of the base. Acids and bases classified on the basis of this concept are termed as Bronsted acids and bases.

 

HCl(aq) + H­2 O⬌ H3 O (aq)++Cl- (aq)

 

 

 

In this reaction, HCl donates its one proton to become Cl-, and H2O accepts one proton to become H3O+. Thus, HCl is Bronsted acid and H2O is a Bronsted base. For the reverse reaction, H3O+is able to transfer its proton to Cl-. So, H3O+is a Bronsted acid and Cl- is a Bronsted base.

Every acid must form a base on donating its proton, and every base must form an acid on accepting a proton. The base that is produced when an acid donates its proton is called the conjugate base of the acid. The acid that is produced when a base accepts a proton is called the conjugate acid of the base. The above reaction can be written as,

 

 

HCl(aq) + H­2 O⬌ H3 O+ (aq)+Cl- (aq)

Acid1base2acid2 base1

 

 

 

In this Cl- is the conjugate base of the acid HCl and H2O is the conjugate base of the acid H3O+. The conjugate acid differs from conjugate base by one proton. A pair of an acid and a base which differ from one another by a proton constitute a conjugate acid base pair. Thus,

 

Figure 3_ B 6

 

 

 

 

Figure 3_ B 6

 

 

Although the Bronsted-Lowry concept of acids and bases is better than the Arrhenius concept, it cannot account for the acidic and basic character of compound not containing hydrogen. For example, acidic nature of oxides such as CO2, SO2 etc., and the basic nature of the compounds of the type CaO, Na2O etc

 

7.

 A buffer solution is the solution which can result the change in pH on addition of small amount of acid or base.

Mechanism of acidic buffer solution:

Let us consider an acidic buffer such as a solution createsthe change in pH.The solution contains large amount of of sodium ions, acetate ions and larger amount of un dissociates acetic acid along with a smallamount of hydrogen ion.

CH3COOH(aq ) ↔CH3COOH-

CH3COONa(aq) →CH3COO-(aq)+ Na+(aq)

Suppose, a few drops of HCL are added to this buffer solution. This would provide H+ ions. TheseH+ ions would combine with CH3COO-ions to form weaklyionized acetic acid molecules as given below.

 

CH3COO-(aq) +H+↔CH3COOH(aq)

Since the additional ions are neutralized by CH3COO_ions in the solution, there will be no change in itsPHvalue. On the other hand , if a few drops NAOH are added to the buffer solution ,it would provideOH_ions. These OH_ ions will combine with H+ Ions present in the buffer solution to form unionized water molecules. This would result in the greater ionization of acetic acid in order to restore the concentration ofH+to this original value.

H+(aq)+OH-↔H2O

Therefore,PHof the solution remains unchanged.

Similarly, in case of basic buffer solution containing NH4OHand NH4CL, there is a large concentration ofNH4+ions , CL-ions , undissociatedNH4OH and small amount ofOH- ions.

When a few drops of HCL are added, the additional H+ ions are neutralized by OH- ions present in the buffer.

As some ofOH-ions from NH4OH combine with H+ ions from the acid, it would result in thegreater ionization of NH4OH to restore the OH-ion concentration.

 

8.

The product of molar concentration of hydrogen and hydroxyl ion furnished by water and molecules is known as ionic product.

On increasing temperature the kinetic energy of water molecules increases, so due to more collision the no. of ions formed increases, and thus kw also increases.

Soln:

Vol. of HCl (V1) = 10cc

Strength of HCl(S1) = 12$\frac{1}{2}$N

Vol. of HNO3(V2) = 30cc

Strength of HNO3(S2) = 110$\frac{1}{10}$N

Vol. of H2SO4 (V3) = 60cc

Strength of HNO3 (S3) = 15N$\frac{1}{5}\mathrm{N}$

Let Vf and Sf be the final col. and strength. Then,

Or, VfSf = V1N1 + V2N2 + V3N3

Or, (10 + 30 + 60) * Sf = 10 * 12$\frac{1}{2}$ + 30 * 110$\frac{1}{10}$ + 60 * 15$\frac{1}{5}$

Or, 100 * Sf = 5 + 3 + 12 = 20

So, Sf = 10020$\frac{100}{20}$ = 5N

Molarity of H+ ion = 5N.

Or, pH = -log[H+]

= -log5 = - 0.698.

 

 

C. Long Answer Type Questions:

1.

 The postulates of Arrhenius theory of ionization are as follows:

(i) He explained the production of H+ from acid in H2O, but actually cannot exist independently.

It combines with H2O to form H3+O.

H+ + H2O à H3+O (Hydronium ion)

(ii) He explained the acid and base in their aqueous solution but couldn’t explain acidic and basic behavior in their original form.

(iii) He couldn’t explain satisfactorily the acidic nature of aluminum chloride (AlCl3) and basic nature of NH3.

(iv)He couldn’t explain the acidic nature of CO2.

Factors affecting the degree of ionization are as follows:

(i) Nature of electrolytes: strong electrolytes are almost completely ionized while the weaker ones are feasibly ionized.

(ii) Nature of solvent: The solvent plays a very important part in the process of ionization. It cuts the line of force binding the two ions and separates them in a solution. It is measured by dielectric constant.

(iii) Concentration: The ionization of an electrolyte is inversely proportional to the concentration of the solution. If the solution is less concentrated the solute will ionize to a greater extent. It is due to the fact that the number of molecules of solvent is very large as compared to the number of solute and more solvent molecules will ionize more solute molecules. So, ionization increased on dilution.

(iv)Temperature: The ionization directly depends upon the temperature. Higher the temperature, the greater will the ionization This is because the increase in temperature increases the velocity of molecule which in turn decreases the interionic forces of attraction between ions.

 

2..

 Ostwald’s dilution law;

Let AB a binary electrolyte gets dissolved in water with α degree of ionization.

Then,

[AB] = 1−αV$\frac{1-\alpha }{\mathrm{V}}$, [A+] = αv$\frac{\alpha }{\mathrm{v}}$, [B-‑] = αv$\frac{\alpha }{\mathrm{v}}$

Or, using law of mass action,

Or, Kc = ([A+][B−]/[AB]$(\left[{\mathrm{A}}^{+}\right]\left[{\mathrm{B}}^{-}\right]/\left[\mathrm{A}\mathrm{B}\right]$

= αv∗αv1−αv$\frac{\frac{\alpha }{\mathrm{v}}\ast \frac{\alpha }{\mathrm{v}}}{\frac{1-\alpha }{\mathrm{v}}}$

So, Kc = α2(1−α)v$\frac{{\alpha }^2}{\left(1-\alpha \right)\mathrm{v}}$

For, weak electrolyte, 1 – α ≈$\approx$ 1

So, Kc = α2v$\frac{{\alpha }^2}{\mathrm{v}}$

Or, α = Kc∗v−−−−√$\sqrt{{\mathrm{K}}_{\mathrm{c}}\ast \mathrm{v}}$

So, α ∝$\propto$V√$\sqrt{\mathrm{V}}$.

Thus, the degree of dissociation of weak electrolyte is proportional to square root of dilution is the Ostwald’s dilution law.

 

Soln.

 Numerical

HNIC(ka) =1.4∗10−5$=1.4\ast {10}^{-5}$

V = 2l

N = 0.10

 

3.

Bronsted Lowry concept of acids and bases:

 According to the Bronsted-Lowry concept, an acid is a proton-donor, and a base is a proton-acceptor.

The reaction of an acid with a base involves transfer of a proton from the acid to the base. So, an acid and a base should be present simultaneously in any system. The extent of an acid-base reaction is governed not only by the proton-donating ability of the acid, but also by the proton-accepting tendency of the base. Acids and bases classified on the basis of this concept are termed as Bronsted acids and bases.

 

HCl(aq) + H­2 O⬌ H3 O (aq)++Cl- (aq)

 

 

 

 

In this reaction, HCl donates its one proton to become Cl-, and H2O accepts one proton to become H3O+. Thus, HCl is Bronsted acid and H2O is a Bronsted base. For the reverse reaction, H3O+is able to transfer its proton to Cl-. So, H3O+is a Bronsted acid and Cl- is a Bronsted base.

Every acid must form a base on donating its proton, and every base must form an acid on accepting a proton. The base that is produced when an acid donates its proton is called the conjugate base of the acid. The acid that is produced when a base accepts a proton is called the conjugate acid of the base. The above reaction can be written as,

 

HCl(aq) + H­2 O⬌ H3 O+ (aq)+Cl- (aq)

Acid1base2acid2 base1

 

 

 

 

In this Cl- is the conjugate base of the acid HCl and H2O is the conjugate base of the acid H3O+. The conjugate acid differs from conjugate base by one proton. A pair of an acid and a base which differ from one another by a proton constitute a conjugate acid base pair. Thus,

 

 

 

 

Figure 3_C 3

 

 

Although the Bronsted-Lowry concept of acids and bases is better than the Arrhenius concept, it cannot account for the acidic and basic character of compound not containing hydrogen. For example, acidic nature of oxides such as CO2, SO2 etc., and the basic nature of the compounds of the type CaO, Na2O etc

 

 

Soln:

Given:

Molar concentration of HCN (C) = 0.01M

Dissociation constant of HCN (Ka) = 4.8 * 10-10.

We know that,

Degree of dissociation, α = KaC−−−√$\sqrt{\frac{{\mathrm{K}}_{\mathrm{a}}}{\mathrm{C}}}$

So, α = 4.8∗10−100.01−−−−−−√$\sqrt{\frac{4.8\ast {10}^{-10}}{0.01}}$

= 2.19 * 10-4

We also know,

[H+] = kaC−−−√$\sqrt{{\mathrm{k}}_{\mathrm{a}}\mathrm{C}}$

= 4.8∗10−10∗0.01−−−−−−−−−−−−√$\sqrt{4.8\ast {10}^{-10}\ast 0.01}$

= 2.19 * 10-6 mol/l.

pH value of a solution is given by the reaction,

or, pH = -log[H+]

So, pH = -log(2.19 * 10-6)

= 5.659.

 

4.

(i).

Arrhenius concept of acids and bases: 

According to the Arrhenius concept, hydrogen chloride, acetic acid, and sulphuric acid, are

 acids because all these compounds give free H+ ions in aqueous solutions.

 

Compounds such as NaOH and NH4OH are bases, because these compounds give free OH- ions in aqueous solutions.

 

NaOH + H2 O (excess) à Na+ (aq) + OH-(aq)

NH4OH+ H2 O (excess) à NH4+(aq)+ OH-(aq)

 

 

 

Thus, According to Arrhenius concept of acids and bases, the neutralization of an acid with a base involves the reaction between H+(aq) and OH-(aq) i.e.,

 

 

H+(aq)+ OH-(aq)→neutralization$\stackrel{\mathrm{n}\mathrm{e}\mathrm{u}\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{l}\mathrm{i}\mathrm{z}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}}{\to }$ H2 O(l)

From acid from base

 

 

 

However, the Arrhenius concept is applicable to the acid-base behaviour only in the aqueous medium. It does not provide any explanation to the acid-base behaviour in the absence of water.

This concept defines acids and bases as compounds-containing hydrogen and hydroxyl group respectively. There are however, many compounds, which act as acid even when there is no hydrogen in their molecule. Similarly, there are many bases, which do not contain hydroxyl group.

 

 

 

(ii).

Bronsted Lowry concept of acids and bases:

 According to the Bronsted-Lowry concept, an acid is a proton-donor, and a base is a proton-acceptor.

The reaction of an acid with a base involves transfer of a proton from the acid to the base. So, an acid and a base should be present simultaneously in any system. The extent of an acid-base reaction is governed not only by the proton-donating ability of the acid, but also by the proton-accepting tendency of the base. Acids and bases classified on the basis of this concept are termed as Bronsted acids and bases.

 

HCl(aq) + H­2 O⬌ H3 O (aq)++Cl- (aq)

 

 

In this reaction, HCl donates its one proton to become Cl-, and H2O accepts one proton to become H3O+. Thus, HCl is Bronsted acid and H2O is a Bronsted base. For the reverse reaction, H3O+is able to transfer its proton to Cl-. So, H3O+is a Bronsted acid and Cl- is a Bronsted base.

Every acid must form a base on donating its proton, and every base must form an acid on accepting a proton. The base that is produced when an acid donates its proton is called the conjugate base of the acid. The acid that is produced when a base accepts a proton is called the conjugate acid of the base. The above reaction can be written as,

 

HCl(aq) + H­2 O⬌ H3 O+ (aq)+Cl- (aq)

Acid1base2acid2 base1

 

 

 

In this Cl- is the conjugate base of the acid HCl and H2O is the conjugate base of the acid H3O+. The conjugate acid differs from conjugate base by one proton. A pair of an acid and a base which differ from one another by a proton constitute a conjugate acid base pair. Thus,

 

 

 

Figure 3_C 4 ii

 

Although the Bronsted-Lowry concept of acids and bases is better than the Arrhenius concept, it cannot account for the acidic and basic character of compound not containing hydrogen. For example, acidic nature of oxides such as CO2, SO2 etc., and the basic nature of the compounds of the type CaO, Na2O etc

 

 

(iii) .

Lewis concept of acid and bases:

Lewis proposed broader and more general definitions of acids and bases which do not require presence of protons to explain the acid base behaviour. According to Lewis concept ,An acid is a substance which can accept a pair of electrons.

A base is a substance which can donate a pair of electron.

Acid-base reactions according to this concept involve donation of electron pair by a base to an acid to form a co-ordinate bond..

Lewis acids are the species having vacant orbitals in the valence shell of one of its atoms. The following species can act as Lewis acids.

a) Molecules having an atom with incomplete octet e.g., BF3, AlCl3 etc.

b) Simple cations for e.g., H+, Ag+ etc.

 

(iv)

 Solubility principles:

When a sparingly soluble ionic solid is dissolved in water it undergoes ionization and the ions forming from solid phase pass into the solution till the solution becomes saturated and a dynamic equilibrium us established between the ions present in the saturated solution and the ions present in the solid phase at a constant temperature.

Consider a sparingly soluble substance AB which ionized as:

AB

At equilibrium,

K = [A+][B−][AB]$\frac{\left[{\mathrm{A}}^{+}\right]\left[{\mathrm{B}}^{-}\right]}{\left[\mathrm{A}\mathrm{B}\right]}$

Where, Ksp = solubility product. The solubility is the product of concentration of ions of sparingly soluble salts in its saturated solution to a particular temperature. It is represented as Ksp.

The value of Ksp changes with the change in temperature. The value of Ksp give the idea of precipitation of an electrolyte at a given temperature.

To precipitate in a chemical reaction, the ionic product must exceed the solubility product of sparingly soluble salt at particular temperature.

For precipitation.

IP < Kspà Unsaturated à No ppt.

IP = Kspà saturated à No. ppt.

IP > Kspà Super saturated à ppt.

Solubility product principle is very applicable in analytical chemistry. In salt analysis i.e. (qualitative analyisis) the solubility product principle is more applicable. The group separation of metal ions in slat analysis is done on the basis of solubility product principle.

 

(v)

Ostwald’s dilution law;

Let AB a binary electrolyte gets dissolved in water with α degree of ionization.

Then,

[AB] = 1−αV$\frac{1-\alpha }{\mathrm{V}}$, [A+] = αv$\frac{\alpha }{\mathrm{v}}$, [B-‑] = αv$\frac{\alpha }{\mathrm{v}}$

Or, using law of mass action,

Or, Kc = ([A+][B−]/[AB]$(\left[{\mathrm{A}}^{+}\right]\left[{\mathrm{B}}^{-}\right]/\left[\mathrm{A}\mathrm{B}\right]$

= αv∗αv1−αv$\frac{\frac{\alpha }{\mathrm{v}}\ast \frac{\alpha }{\mathrm{v}}}{\frac{1-\alpha }{\mathrm{v}}}$

So, Kc = α2(1−α)v$\frac{{\alpha }^2}{\left(1-\alpha \right)\mathrm{v}}$

For, weak electrolyte, 1 – α ≈$\approx$ 1.

So, Kc = α2v$\frac{{\alpha }^2}{\mathrm{v}}$

Or, α = Kc∗v−−−−√$\sqrt{{\mathrm{K}}_{\mathrm{c}}\ast \mathrm{v}}$

So, α ∝$\propto$V√$\sqrt{\mathrm{V}}$.

Thus, the degree of dissociation of weak electrolyte is proportional to square root of dilution is the Ostwald’s dilution law.

 

(vi).

Salt hydrolysis:

The process of dissolving salt in water to make its aqueous solution which may be acidic, basic or neutral is called hydrolysis of salt.

It is divided into four categories they are:

a. Salts of strong acid and strong base do not go hydrolysis. E.g: NaCl, NaNO3, etc.

Here, let us consider NaCl as an example. It is strong electrolyte, when dissolved in water, it goes complete dissociation to give Na+ and Cl- ions. But these ions do not have tendency to react with H+ or OH-. So, there is no change in concentration of H+ or OH- ions and hence the solution remains neutral.

b. Salts of weak acid and strong base:

Sodium acetate, CH3COONa, NaCN, KCN are examples of this type of salts. Here let us consider sodium cyanide (NaCN) it is the salt of weak acid, HCN and NaOH. It ionizes to form CN- anions. Being conjugate base of a weak acid. CN- is relatively strong base. Thus, the anion CN- accepts a H+ ion from water and undergoes hydrolysis.

So, the solution becomes basic due to the generation of OH- ions.

 

c. Salts of weak bases and strong acids:

Some salts of weak bases and strong acids undergo cationic hydrolysis and yield slightly acidic solution.

Ammonium chloride is a typical e.g. of this salts. It is the salt of a weak base NH4OH and strong acid HCl. It ionizes in aqueous solution to form the cation NH4+.

 

Here, NH4OH+ is a relatively strong acid. It accepts OH- ion from water (H2O) and forms the unionized NH4OH and H+ ion.

 

So, the accumulation of H+ ions in solution makes it acidic.

 

d. Salts of weak acids and weak bases:

The e.g. of this type of salts are ammonium acetate, ammonium cyanide etc. The resulting solution is neutral, basic or acidic depending on the relative hydrolysis of the anions and the cations.

 

(vii)

Arrehenius theory of ionization are:

a. When an electrolyte is dissolved in water it gives electrically charged particles called ions. The one with + ve charge is called cations and with negative charge is anion.

AB à A+ (cation) + B-(anion)

b. The thus formed ions try to recombine to form unionized molecules, forming a dynamic equilibrium, described by a constant known as ionization constant.

Or, K = [A+][B−][AB]$\frac{\left[{\mathrm{A}}^{+}\right]\left[{\mathrm{B}}^{-}\right]}{\left[\mathrm{A}\mathrm{B}\right]}$

c. In this solution, total no. of positive charge is equal to that of negative charge.

d. The movement of ions result in conduction of electricity.

e. The properties of an electrolyte depends on the ions furnished by it.

f. Each ion behaves osmotically as a molecule.

g. The degree of ionization is the fraction of total no. of molecules ionized.

 

(viii)

Common ion effect:

The effect by virtue of which, the degree of dissociation of weak electrolyte is decreased by a strong electrolyte, if both the electrolyte contains at least a common ion, in its solution.

Let, consider a weak acid, HA, and a strong salt SA. ,In aqueous medium , A- ion being the common ion, the degree of dissociation of HA acid is decreased, i.e the equilibrium shifts in backward direction. This is common ion effect.

 

Applications:                                                                                                   

1. Precipitation of sulphides of group II and IIIB basic radicals

2. Precipitation of hydroxides of group IIIA basic radicals

3. Precipitation of group IV basic radicals

 

 

5.

It is defined as the negative power of which 10 must be raised in order to express hydrogen ion concentration.[H+]=10−pH$\left[{\mathrm{H}}^{+}\right]={10}^{-\mathrm{p}\mathrm{H}}$

 

Soln

H2O dissociates as:

H2O ↔H2 + O2

So, Concentration of H+ is 10-3.

Now,

pH = -log[H+]

= -log[10-3]

= 2.

 

6.

Lewis concept of acid and bases:

Lewis proposed broader and more general definitions of acids and bases which do not require presence of protons to explain the acid base behaviour. According to Lewis concept ,An acid is a substance which can accept a pair of electrons.

A base is a substance which can donate a pair of electron.

Acid-base reactions according to this concept involve donation of electron pair by a base to an acid to form a co-ordinate bond..

Lewis acids are the species having vacant orbitals in the valence shell of one of its atoms. The following species can act as Lewis acids.

 

a) Molecules having an atom with incomplete octet e.g., BF3, AlCl3 etc.

b) Simple cations for e.g., H+, Ag+ etc.

 

 

7.

It is define as the ratio of velocity constant for forward reaction to the velocity constant for backward rexn. At equilibrium state

The ratio of total no. of molecules ionized to the total no of molecules dissolves is called ionization constant.

 Soln:

Dissociation constant of acetic acid (Ka) = 1.8 * 10-5

Molar concentration of acetic acid (C) = 0.1M

Now,

Degree of dissociation, α = KaC−−−√$\sqrt{\frac{{\mathrm{K}}_{\mathrm{a}}}{\mathrm{C}}}$

= 1.8∗10−50.1−−−−−−√$\sqrt{\frac{1.8\ast {10}^{-5}}{0.1}}$

= 1.34 * 10-2

We know that,

[H+] = Ka∗C−−−−−√$\sqrt{{\mathrm{K}}_{\mathrm{a}}\ast \mathrm{C}}$

= 1.8∗10−5∗0.1−−−−−−−−−−√$\sqrt{1.8\ast {10}^{-5}\ast 0.1}$

= 1.34 * 10-3

pH value of the solution is given by the reaction,

or, pH = -log[H+]

= -log[1.34 * 10-3]

= 2.87.

 

8.

(i) Bronsted Lowry concept of acids and bases:

 According to the Bronsted-Lowry concept, an acid is a proton-donor, and a base is a proton-acceptor.

The reaction of an acid with a base involves transfer of a proton from the acid to the base. So, an acid and a base should be present simultaneously in any system. The extent of an acid-base reaction is governed not only by the proton-donating ability of the acid, but also by the proton-accepting tendency of the base. Acids and bases classified on the basis of this concept are termed as Bronsted acids and bases.

 

HCl(aq) + H­2 O⬌ H3 O (aq)++Cl- (aq)

 

 

In this reaction, HCl donates its one proton to become Cl-, and H2O accepts one proton to become H3O+. Thus, HCl is Bronsted acid and H2O is a Bronsted base. For the reverse reaction, H3O+is able to transfer its proton to Cl-. So, H3O+is a Bronsted acid and Cl- is a Bronsted base.

Every acid must form a base on donating its proton, and every base must form an acid on accepting a proton. The base that is produced when an acid donates its proton is called the conjugate base of the acid. The acid that is produced when a base accepts a proton is called the conjugate acid of the base. The above reaction can be written as,

 

HCl(aq) + H­2 O⬌ H3 O+ (aq)+Cl- (aq)

Acid1base2acid2 base1

 

 

In this Cl- is the conjugate base of the acid HCl and H2O is the conjugate base of the acid H3O+. The conjugate acid differs from conjugate base by one proton. A pair of an acid and a base which differ from one another by a proton constitute a conjugate acid base pair. Thus,

 

 

 

 

 

 

 

Figure3_ C 8 i

 

Although the Bronsted-Lowry concept of acids and bases is better than the Arrhenius concept, it cannot account for the acidic and basic character of compound not containing hydrogen. For example, acidic nature of oxides such as CO2, SO2 etc., and the basic nature of the compounds of the type CaO, Na2O etc.

 

 

Lewis concept of acid and bases:

Lewis proposed broader and more general definitions of acids and bases which do not require presence of protons to explain the acid base behavior. According to Lewis concept, An acid is a substance which can accept a pair of electrons.

A base is a substance which can donate a pair of electron.

Acid-base reactions according to this concept involve donation of electron pair by a base to an acid to form a co-ordinate bond..

Lewis acids are the species having vacant orbitals in the valence shell of one of its atoms. The following species can act as Lewis acids.

a) Molecules having an atom with incomplete octet e.g., BF3, AlCl3 etc.

b) Simple cations for e.g., H+, Ag+ etc.

 

(ii)

Solubility principles:

When a sparingly soluble ionic solid is dissolved in water it undergoes ionization and the ions forming from solid phase pass into the solution till the solution becomes saturated and a dynamic equilibrium us established between the ions present in the saturated solution and the ions present in the solid phase at a constant temperature.

Consider a sparingly soluble substance AB which ionized as:

AB

At equilibrium,

K = [A+][B−][AB]$\frac{\left[{\mathrm{A}}^{+}\right]\left[{\mathrm{B}}^{-}\right]}{\left[\mathrm{A}\mathrm{B}\right]}$

Where, Ksp = solubility product. The solubility is the product of concentration of ions of sparingly soluble salts in its saturated solution to a particular temperature. It is represented as Ksp.

The value of Ksp changes with the change in temperature. The value of Ksp give the idea of precipitation of an electrolyte at a given temperature.

To precipitate in a chemical reaction, the ionic product must exceed the solubility product of sparingly soluble salt at particular temperature.

For precipitation.

IP < Kspà Unsaturated à No ppt.

IP = Kspà saturated à No. ppt.

IP > Kspà Super saturated à ppt.

Solubility product principle is very applicable in analytical chemistry. In salt analysis i.e. (qualitative analyisis) the solubility product principle is more applicable. The group separation of metal ions in slat analysis is done on the basis of solubility product principle.

 

 

9.

TheproductofmolarconcentrationofH+ionsandOHionsatparticulartemperatureisknownas$\mathrm{T}\mathrm{h}\mathrm{e}\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{d}\mathrm{u}\mathrm{c}\mathrm{t}\mathrm{o}\mathrm{f}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{o}\mathrm{f}{\mathrm{H}}^{+}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{O}\mathrm{H}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{a}\mathrm{t}\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{t}\mathrm{i}\mathrm{c}\mathrm{u}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{t}\mathrm{e}\mathrm{m}\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{u}\mathrm{r}\mathrm{e}\mathrm{i}\mathrm{s}\mathrm{k}\mathrm{n}\mathrm{o}\mathrm{w}\mathrm{n}\mathrm{a}\mathrm{s}$

                                                                   ionicproductofwater.$\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{i}\mathrm{c }\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{d}\mathrm{u}\mathrm{c}\mathrm{t }\mathrm{o}\mathrm{f }\mathrm{w}\mathrm{a}\mathrm{t}\mathrm{e}\mathrm{r}.$

 

 Soln:

pH of HCN = 5.2

Molar concentration of HCN (C) = 0.1M

Dissociation constant of HCN (Ka) = ?

Now,

 

10.

(i)

Common ion effect:

The effect by virtue of which, the degree of dissociation of weak electrolyte is decreased by a strong electrolyte, if both the electrolyte contains at least a common ion, in its solution.

 

(ii) pH of a solution:

It is defined as the negative power to which 10 must be raised in order to express the hydrogen ion concentration of solution.

(iii) Lewis base:

 According to lewis base concept, it is defined as a substance that can donate a pair of electrons.

The degree of dissociation of a weak electrolyte is proportional to the square root of dilution i.e. on increasing dilution the degree of dissociation of weak electrolytes will be increased is called Ostwald dilution law,

iv.

It is defined as the ratio of total no. of ions of molecules ionized to the total no. of molecules dissolved.

 

(v).

The degree of dissociation of a weak electrolyte is proportional to the square root of dilution i.e. on increasing dilution the degree of dissociation of weak electrolytes will be increased is called Ostwald dilution law,

 

Soln:

Concentration of Cl- ion = 0.003 g/liter.

= 0.00335.5$\frac{0.003}{35.5}$M

= 8.450 * 10-5M

Initial concentration of sodium carbonate [Na2CO3] = 2.0M

Solubility product of silver carbonate (Ksp) = 8.2 * 10-12

2AgCl(S) + Na2CO3(aq)

 

We know that,

Or, Ksp(Ag2CO3) = [Ag+]2[CO3-2]

So, [Ag+] = Ksp(Ag2CO3)[CO−23]−−−−−−−−−√$\sqrt{\frac{{\mathrm{K}}_{\mathrm{s}\mathrm{p}}\left(\mathrm{A}{\mathrm{g}}_2\mathrm{C}{\mathrm{O}}_3\right)}{\left[{\mathrm{C}\mathrm{O}}_3^{-2}\right]}}$

= 8.2∗10−122−−−−−−√$\sqrt{\frac{8.2\ast {10}^{-12}}{2}}$

= 2.0248 * 10-6M

Hence, Ksp of AgCl = [Ag+][Cl-]

= (2.0248 * 10-6)(8.450 * 10-5)

= 1.710 * 10-10.