Equilibrium

A. When volume is suddenly increased, the vapour pressure decreases.

Explanation: When volume is increased suddenly there is no sudden increase in number of vapour substance. Thus, same number of particle occupies larger volume and thus have lower pressure. (After some time, the number of vapour molecules increases and vapour pressure slowly rises from the decreased value)

B. At equilibrium state,

Rate of evaporation = Rate of condensation

When volume is increased, the vapour pressure decreases and so, the rate of evaporation will be greater than rate of condensation.

This can be understood by the Le Chatelier’s principle, which states that, when an equilibrium state experiences a disturbance (here pressure decrease) the system will try to cancel the effect of change in the system.

C. When equilibrium is finally restored, the system will have initial vapour pressure. Vapour pressure does not depend on volume of container, it only depends on nature of substance and temperature. Thus, the sudden decrease in partial pressure is slowly raised so that a new equilibrium is obtained. Partial pressure of the vapour will be same as the initial.

The equilibrium constant K_c is the ratio of product of concentration of Product raise to their coefficients and product of concentration of reactants raised to their coefficients in balanced chemical equation.

i.e., For a chemical change,

aA + bB ⇌ cC + dD

Equilibrium constant K_c for the reaction,

For given reaction,

2SO₂(g) + O₂(g) ⇌2SO₃(g)

Equilibrium constant,

Given that, Equilibrium concentrations of each substance are,

[SO₂] = 0.60M = 0.60 mol litre^-1

[O₂] = 0.82M =0.82 mol litre^-1

[SO₃] = 1.90M = 1.90 mol litre^-1

Thus,

K_c =

⇒ K_c = 12.299 M^-1

Or K_c = 12.299 L mol^-1

K_p value at equilibrium for reaction,

I₂ (g) ⇌2I (g)

is given as,

Given, 40% volume is occupied by I atoms.

∴ 60 % volume is occupied by I₂.

Let V be total volume, then volume occupied by I and I₂ are 0.4V and 0.6V respectively.

Partial pressure of I₂ = (60/100) × 10⁵ = 60 kPa

Partial pressure of I = (40/100) × 10⁵ = 40 kPa

Substituting in equation for K_p,

K_p = = 2.67 × 10⁴ KPa

For a chemical change,

aA + bB ⇌ cC + dD

Equilibrium constant K_c for the reaction,

K_c =

The concentration of solid substance and pure substance are 1.

(i) K_c =

(ii) K_c =

= [NO₂]⁴ [O₂]

(iii) K_c =

=

(iv) K_c =

=

(v) K_c =

=

We know that,

K_p = K_c[RT]^Δn

Where Δn is number of gas molecule in products minus number of gas molecule in reactant in balanced chemical equation.

(i)

2NOCl (g) ⇌2NO (g) + Cl₂ (g)

Number of gaseous species on product side is 3 (2 NO and 1 Cl₂ ) and on reactant side is 2 (2 NOCl )

Thus, Δn = 3 - 2 = 1

T = 500 K

R = 0.0821 L atm K^-1

Given, K_p = 1.8 × 10^-2atm

∴ K_c =

=

= 4.4 × 10 �^-4mol L^-1

(ii)

CaCO₃ (s) ⇌CaO(s) + CO₂(g)

Number of gaseous species on product side is 1 (1 CO₂) and on reactant side is 0.

Thus, Δn = 1 – 0 = 1

T = 1073 K

K_p = 167 atm

K_c =

=

= 1.9 mol L^-1

Equilibrium constant for forward reaction is

K_c^fwd =

For reverse reaction, Equilibrium constant is given as,

K_c^rev =

= 1/K_c^fwd

= 1/6.3×10¹⁴

= 1.59 × 10^-15

[(Equilibrium constant for reverse reaction) = 1/(equilibrium constant of forward reaction)

i.e, K_c^rev = ]

Concentration of solid or pure liquid,

[substance] =

=

=

Since density and molar mass doesn’t vary, concentration of the substance remains constant.

Pure liquids and solid have a constant concentration. Composition of solids remain same throughout a reaction even though amount may change.

Similarly, pure liquids have same composition (since it has only one substance)

Equilibrium constant of reaction

2N₂ (g) + O₂ (g) ⇌2N₂O (g)

is given as,

K_c =

= 2.0 × 10^–37

Initial concentration of substances,

[N₂O] = 0

[N₂] = number of mole of N₂/Volume of container

= 0.482/10 = 0.0482

[O₂] = number of mole of O₂/Volume of container

= 0.933/10 = 0.0933

2N₂ (g) + O₂ (g) ⇌2N₂O (g)

where x is concentration of O₂ that has reacted.

Thus, equilibrium constant,

K_c =

=

Since value of equilibrium constant is small, the amount of O₂ and N₂ reacted is very less and thus can be neglected.

∴ K_c =

= 2 × 10^-37

i.e., (2x)² = (2 × 10^-37) ×(0.0482)²× 0.0933

2x= 4.335 × 10^-41

x = 2.168 × 10^-41

Thus, Equilibrium concentrations of substances are,

[N₂] = 0.0482 – 4.335 × 10^-41 = 0.0482 mol L^-1

[O₂] = 0.0933 – 2.168 × 10^-41 = 0.0933 mol L^-1

[N₂O] = 4.335 × 10^-41mol L^-1

The chemical reaction is,

Initial amount:

Let x amount of Br₂ be reacted to attain equilibrium. Then amount of substance left at equilibrium,

2NO (g) + Br₂ (g) ⇌2NOBr (g)

Equilibrium amount: 0.087 – 2x 0.0437 – x 2x

Given, equilibrium concentration of NOBr is 0.0518,

i.e, 2x = 0.0518

x = 0.0259

Thus,

Amount of NO at equilibrium = 0.087 - 2 × 0.0259

= 0.0352 mole

Amount of Br₂ at equilibrium = 0.0437 – 0.0259

= 0.0178 mole

We know that,

K_p = K_c[RT]^Δn

Where Δn is number of gas molecule in products minus number of gas molecule in reactant in balanced chemical equation.

For reaction,

2SO₂(g) + O₂(g) ⇌2SO₃ (g)

Δn = 2 – 3 = -1

T = 450 K

K_p= 2.0 × 10¹⁰ bar

K_c =

On putting the values in the above equation, we get

Partial pressure of HI = 0.04 atm

sum of partial pressures = total pressure,

thus,

P(H₂)+P(I₂)= P – P(HI)

Where,

P is total pressure = 0.2 atm

P(H₂) is Partial pressure of H₂ = x

P(I₂) is Partial pressure of I₂= x

P(HI) is Partial pressure of HI = 0.04 atm

∴ x + x = 0.2 – 0.04

= 0.16

i.e, x = 0.08 atm

K_p=

= = 4

Concentration of substance,

[substance] =

Thus, concentrations of given substances are,

[N₂] = 1.57/20 mol L^-1

[H₂] = 1.92/20 mol L^-1

[NH₃] = 8.13/20 mol L^-1

For a chemical change,

aA + bB ⇌ cC + dD

Reaction quotient Q_c for the reaction is,

Q_c =

Where [substance] is concentration of substance at a particular time.

For given reaction

N₂ (g) + 3H₂ (g)⇌2NH₃ (g)

Reaction quotient is,

Q_c = =

Q_c >K_c . Therefore, the reaction proceeds in reverse direction in order to attain equilibrium. (Equilibrium constant is reaction coefficient at equilibrium state)

For a chemical change,

aA + bB ⇌ cC + dD

Equilibrium constant K_c for the reaction,

K_c =

In question, expression for equilibrium constant is given as,

K_c =

Comparing the power of each concentration term we can find the chemical reaction.

The reaction here is given by,

4NO + 6H₂O ⇌ 4NH₃ + 5O₂

Initially there was 1 mole of H₂O and CO in a 10 L flask.

At equilibrium, 40% of H₂O has reacted thus amount of H₂O remaining is 0.6 mole.

The amount of CO reacted is also 0.4 mole.

Amount of H₂ and CO₂ formed is also 0.4 mole.

H₂O (g)+ CO (g)⇌H₂ (g) + CO₂ (g)

H₂O (g) + CO (g) ⇌H₂ (g) + CO₂ (g)

Equilibrium constant for the reaction is,

K_c =

=

= 4/9

For the reaction,

H₂ (g) + I₂ (g) ⇌2HI (g)

Equilibrium constant is given to be 54.8.

i.e, K_c =

Let x be equilibrium concentration of H₂ at equilibrium. Then equilibrium concentration of I₂ will also be x. Since, initially there was only HI present, equal amount of H₂ and I₂ will be produced.

At equilibrium:

K_c =

= = 54.8

x =

= 0.0675

Therefore, concentration of H₂ and I₂ is 0.0675 mol L^-1

For the given reaction, let’s assume that the molar concentration of I₂ and Cl₂ at equilibrium is x mol/L.

The given reaction is

2ICl (g) ⇌ I₂ (g) + Cl₂ (g)

Initial concentration 0.78M 0 0

Equilibrium concentration (0.78-2x)M x x

Equilibrium constant (K_c) is defined as a number that expresses the relationship between the amounts of products and reactants present at equilibrium in a reversible chemical reaction at a given temperature.

Here, K_c = 0.14

For the given reaction,

Where, [I₂]=concentration of I₂ in mol/L at equilibrium

[Cl₂]=concentration of Cl₂ in mol/L at equilibrium

[ICl]=concentration of ICl in mol/L at equilibrium

X = 1.67

Therefore, at equilibrium,

[H₂]=[I₂]=0.167 M

[ICl]= 0.78 - (2x0.167) = 0.446 M

For the given reaction, lets assume that p atm is the pressure exerted by ethene and hydrogen gas at equilibrium.

Therefore, the given reaction is:

C₂H₆ (g) ⇌ C₂H₄ (g) + H₂ (g)

Equilibrium pressure constant (K_p) is defined as a number that expresses the relationship between the partial pressures of products and reactants present at equilibrium in a reversible chemical reaction at a given temperature.

Here. K_p= 0.04 atm

For the given reaction,

Where, = partial pressure pf C₂H₄ in atom at equilibrium

= partial pressure of H₂ in atom at equilibrium

= partial pressure of C₂H₆ in atom at equilibrium

p²=0.16-0.04p

p²+0.04p-0.16=0

Now,

(Taking positive value)

p=0.38

Therefore at equilibrium, [C₂H₆]= 4 - p = 4 - 0.38

=3.62 atom

(i) The reaction quotient (Q) measures the relative amounts of products and reactants present during a reaction at a particular point in time. It also indicates the direction in which the reaction is going to proceed.

For the given reaction of ethanol and acetic acid that results in the formation of ethyl acetate and water, the reaction quotient is-

CH₃COOH (l)+C₂H₅OH (l)⇌CH₃COOC₂H₅(l)+H₂O (l)

Therefore,

(ii) To find if equilibrium has been reached, we need to calculate the Reaction quotient (Q_c) and compare it with the Equilibrium constant (K_c).

We know that,

Since the value of the Reaction quotient (Q_c) is less than that of the Equilibrium constant (K_c), equilibrium has not been reached yet. The process of reaction is still taking place to form the products.

For the given reaction, let’s assume that the initial molar concentration of PCl₅ is x mol/L.

The given reaction is:

Moles of PCl₅ decomposed during the reaction=Moles of PCl₃ formed=Moles of Cl₂ formed=(x-0.05) mol/L

Equilibrium constant (K_c) is defined as a number that expresses the relationship between the amounts of products and reactants present at equilibrium in a reversible chemical reaction at a given temperature.

Here, K_c = 0.0083

For the given reaction,

Where, [PCl₃] = concentration of PCl₃ in mol/L at equilibrium

[Cl₂] = concentration of Cl₂ in mol/L at equilibrium

[PCl₅] = concentration of PCl₅ in mol/L at equilibrium

(x-0.05)² = 0.000415

(x-0.05) = 0.0203

x = 0.07 mol/L

Therefore, [PCl₃] = (x-0.05) = 0.07-0.05 = 0.02 mol/L

[Cl₂] = (x-0.05) = 0.07-0.05 = 0.02 mol/L

The given reaction is:

The reaction quotient (Q_p) measures the partial pressures of products and reactants present during a reaction at a particular point in time. It also indicates the direction in which the reaction is going to proceed.

Here,

And K_p=0.265

Since Q_p>K_p, it indicates that the reaction will move in the backward direction to attain equilibrium. This means that the partial pressure of CO₂ will decrease and the partial pressure of CO will increase by the same amounts. If the partial pressure of CO₂ will decrease by p atm, the partial pressure of CO will increase by p atm.

Hence, and

Equilibrium pressure constant (K_p) is defined as a number that expresses the relationship between the partial pressures of products and reactants present at equilibrium in a reversible chemical reaction at a given temperature.

Therefore,

0.371 + 0.265p = 0.8 – p

1.265p = 0.429

p = 0.339 atm

The given reaction is:

The reaction quotient (Q) measures the relative amounts of products and reactants present during a reaction at a particular point in time. It also indicates the direction in which the reaction is going to proceed.

Here,

Where, [NH₃]=concentration of NH₃ in mol/L at a particular instant

[N₂]=concentration of N₂ in mol/L at a particular instant

[H₂]=concentration of H₂ in mol/L at a particular instant

Therefore,

And K_c=0.061, where K_c is the equilibrium constant.

Since, Q_c<K_c, the reaction is not in equilibrium and will proceed in the forward direction in order to attain equilibrium.

For the given reaction, let’s assume that x mol/L of BrCl decompose in order to attain equilibrium.

The given reaction is

Equilibrium constant (K_c) is defined as a number that expresses the relationship between the amounts of products and reactants present at equilibrium in a reversible chemical reaction at a given temperature.

Here, K_c = 32

For the given reaction,

Where, [Br₂]=concentration of Br₂ in mol/L at equilibrium

[Cl₂]=concentration of Cl₂ in mol/L at equilibrium

[BrCl]=concentration of BrCl in mol/L at equilibrium

Molar concentration of BrCl at equilibrium = (0.0033 mol/L – 0.003 mol/L) = 0.0003 mol/L

Let us assume that the total mass of the gaseous mixture is 100g

Therefore, mass of CO = 90.55g

And mass of CO₂ = (100-90.55) = 9.45g

Number of moles of CO,

Numb we of moles of CO₂,

Partial pressure of CO,

Partial pressure of CO₂,

Equilibrium pressure constant (K_p) is defined as a number that expresses the relationship between the partial pressures of products and reactants present at equilibrium in a reversible chemical reaction at a given temperature.

Therefore,

Also, we know that,

Where K_c = Equilibrium constant

K_p = 14.19 atm

R = 0.081 L atm/K mol

T = 1127 K

Δng = Sum of stochiometric coefficients of products – Sum of stochiometric coefficients of reactants = 2-1 = 1

For the given reaction,

NO (g) + 1/2 O₂ (g) ⇌ NO₂ (g)

Given:

Standard free energy change of the formation of the product (NO₂)

⇒Δ_fG^° (NO₂) = 52.0 kJ/mol

Standard free energy change of the formation of one reactant (NO)

⇒Δ_fG^° (NO) = 87.0 kJ/mol

Standard free energy change of the formation of second reactant(O₂)

⇒Δ_fG^° (O₂) = 0 kJ/mol

a) Calculation of ΔG^°

ΔG^° = Difference in free energy of the reaction when all the reactants and products are in the standard state (1 atmospheric pressure and 298K) Hence,

ΔG^° = ∑ Δ_fG^° (Products) - ∑ Δ_fG^° (Reactants)

⇒ΔG^° = Δ_fG^° (NO₂) – [Δ_fG^° (NO) + 1/2 Δ_fG^° (O₂)]

⇒ΔG^° = 52.0 – [87.0 + 1/2 x 0]

⇒ΔG^° = -35.0 kJ mol^-1

b) Equilibrium constant for the formation of NO₂ from NO and O₂ at 298K

From standard free change of a reaction, we know that,

ΔG^° = -2.303 RT log K_c

Where, ΔG^° is the standard free change of a reaction, value of ΔG^° is -35000 (calculated in part(a))

R = gas constant, value of R is 8.314 J/mol-K

T= Absolute temperature, value of T is 298K (given)

K_c= Equilibrium constant

Using the formula, we write,

log K_c =

⇒log K_c =

⇒log K_c = 6.134

∴ K_c = antilog (6.134)

⇒K_c = 1.361 x 10⁶

When the pressure on the system is increased, the volume decreases proportionately. According to Le Chatelier’s principle, the equilibrium will shift in the direction in which there is a decrease in the number of moles, i.e., towards the direction in which there is a decrease in the volume. In general, an increase in pressure applied to the system at equilibrium favours the reaction in the direction which takes place with a decrease in a total number of moles and a decrease in the pressure favours the reaction in the direction which takes place with an increase in total number of moles. If there is no change in the number of moles of gases in a reaction, a pressure change does not affect the equilibrium.

Hence, according to Le Chatelier’s principle, if the pressure on the system is decreased, then the equilibrium shifts in the direction in which the number of moles of gases is more.

a) The number of moles of reaction products will increase. In the given reaction,

PCl₅ (g) ⇌ PCl₃ (g) + Cl₂ (g)

The number of moles of gaseous products is more than that of gaseous reactants. Thus, the reaction will proceed in the forward reaction. As a result, the number of moles of reaction products will increase.

b) The number of moles of reaction products will decrease. In the given reaction,

CaO (s) + CO₂ (g) ⇌ CaCO₃ (s)

The number of moles of gaseous products is less than that of

gaseous reactants. Thus, the reaction will proceed in the backward

reaction. As a result, the number of moles of reaction products will

decrease.

c) The number of moles of reaction products remains same.

In the given reaction,

3Fe (s) + 4H₂O (g) ⇌ Fe₃O₄ (s) + 4H₂ (g)

The number of moles of gaseous products equal to gaseous

reactants. Thus, the decreasing of pressure does not affect the

equilibrium. As a result, the number of moles of reaction products

will remain same.

Note: Le Chatelier’s principle states that “if a system at equilibrium is subjected to a change of concentration, pressure or temperature, the equilibrium shifts in a direction that tends to undo the effect of the change”

According to Le Chatelier’s principle, an increase in pressure applied to the system at equilibrium favours the reaction in the direction which takes place with a decrease in a total number of moles. If there is no change in the number of moles of gases in a reaction, a pressure change does not affect the equilibrium.

Hence, reactions affected will be those in which number of moles of products (n_p) is not equal to number of moles of reactants (n_p )

∴ n_p ≠ n_r. Hence, the reactions (i), (iii), (iv), (v) and (vi) will be affected. By applying Le Chatelier’s principle, we can predict the direction. In general,

⇒The reaction will go to the left if n_p > n_r

⇒The reaction will go tp the right if n_p < n_r

(i)COCl₂ (g) ⇌CO (g) + Cl₂ (g)

Number of moles of products (n_p)= 2

Number of moles of reactants (n_r) = 1

∴ n_p > n_r

According to Le Chatelier’s principle, the reaction will proceed in the backward reaction.

(ii) CH₄ (g) + 2S₂ (g) ⇌CS₂ (g) + 2H₂S (g)

Number of moles of products (n_p)= 3

Number of moles of reactants (n_r) = 3

∴ n_p = n_r

According to Le Chatelier’s principle, the reaction will not be affected by the pressure.

(iii) CO₂ (g) + C (s) ⇌ 2CO (g)

Number of moles of products (n_p)= 2

Number of moles of reactants (n_r) = 1

∴ n_p > n_r

According to Le Chatelier’s principle, the reaction will proceed in the backward direction.

(iv) 2H₂ (g) + CO (g) ⇌ CH₃OH (g)

Number of moles of products (n_p)= 1

Number of moles of reactants (n_r) = 3

∴ n_p < n_r

According to Le Chatelier’s principle, the reaction will proceed in the forward direction.

(v) CaCO₃ (s) ⇌CaO (s) + CO₂ (g)

Number of moles of products (n_p)= 2

Number of moles of reactants (n_r) = 1

∴ n_p > n_r

According to Le Chatelier’s principle, the reaction will proceed in the backward direction.

(vi) 4 NH₃ (g) + 5O₂ (g) ⇌ 4NO (g) + 6H₂O(g)

Number of moles of products (n_p)= 10

Number of moles of reactants (n_r) = 9

∴ n_p > n_r

According to Le Chatelier’s principle, the reaction will proceed in the backward direction.

Given:

For the given reaction,

H₂(g) + Br₂(g) ⇌ 2HBr(g)

The equilibrium constant(K_c) is 1.6 × 10⁵

Initial pressure of HBr = 10bar

Temperature = 1024K

For equilibrium constant in terms of pressure, we apply the formula given below:

K_p =

Where P_x and P_Y are the partial pressures of x and y(products)

P_A and P_B are the partial pressures of a and b (reactants)

As we know that the relationship between K_p and K_c is K_p=K_c (R/T)^Δn

Where Δn = no. of moles of products – no. of moles of reactants

Hence for the given reaction,

Δn = 0

∴ K_p = K_c

As the given reaction is a reversible reaction. Hence, we can write

2HBr(g) ⇌ H₂(g) + Br₂(g), ∴ K_c =

Initial 10bar

At

Equilibrium 10-P P/2 P/2

Using the above formula,

K_p =

K_p = = K_c

⇒ =

⇒ =

Taking square root of both sides, we get

⇒ =

⇒4.0 × 10² = 2(10-P)

⇒402 P = 20

⇒P =

⇒P = 4.97 × 10^-2 bar

Hence, at equilibrium

P_H2 = P_Br2 = P/2

⇒ = 2.48 × 10^-2 bar

P_HBr = 10-P ≈ 10 bar

Thus, the equilibrium pressure of H₂ and Br₂ is 2.48 × 10^-2 bar and HBr is 10 bar

a) For the given reaction,

CH₄ (g) + H₂O (g) ⇌ CO (g) + 3H₂ (g)

By using the formula,

K_p =

Where P_x and P_Y are the partial pressures of x and y(products)

P_A and P_B are the partial pressures of a and b (reactants)

Hence,

K_p =

b) (i) According to Le Chatelier’s principle, an increase in pressure applied to the system at equilibrium favours the reaction in the direction that takes place with a decrease in a total number of moles. Hence for the reaction,

CH₄ (g) + H₂O (g) ⇌ CO (g) + 3H₂ (g)

Number of moles of products (n_p)= 4

Number of moles of reactants (n_r) = 2

∴ n_p > n_r

According to Le Chatelier’s principle, the equilibrium will shift in the backward direction and value of K_p decreases.

(ii) If the temperature is raised, i.e., heat is applied to the system, then according to Le Chatelier’s principle, the equilibrium will shift to the side that absorbs heat, i.e., in the backward reaction. An endothermic reaction is favoured by low temperature.

Hence for the endothermic reaction,

CH₄ (g) + H₂O (g) ⇌ CO (g) + 3H₂ (g)

The equilibrium shifts in the forward direction and K_p increases.

(iii) By using a catalyst, equilibrium composition will not be disturbed but equilibrium will be attained quickly and value of K_p remains unaffected.

Note: Role of the catalyst is that it increases the rate of both forward and backward reactions equally; the equilibrium will be attained in less time, i.e., the same amount of product will be formed in less time. Catalyst does not effect equilibrium constant and heat of reaction.

According to Le Chatelier’s principle, if in a reaction in equilibrium, the concentration of any reactant is increased, the equilibrium shifts in the forward reaction. On the other hand, if the concentration of any product is increased, the equilibrium shifts in the backward direction. The reverse happens if the concentrations are decreased.

For the given reaction,

2H₂(g) + CO (g) ⇌CH₃OH (g)

a) If the addition of H₂ is taking place, it means the concentration of reactant is increasing. According to Le Chatelier’s principle, the equilibrium shifts in the forward direction.

b) If the addition of CH₃OH is taking place, it means the concentration of product is increasing. According to Le Chatelier’s principle, the equilibrium shifts in the backward direction.

c) If the removal of CO is taking place, it means the concentration of reactant is decreasing, the equilibrium shifts in the backward direction. According to Le Chatelier’s principle, the equilibrium shifts in the backward direction.

d) If the removal of CH₃OH is taking place, it means the concentration of product is decreasing, the equilibrium shifts in the forward direction. According to Le Chatelier’s principle, the equilibrium shifts in the forward direction.

Equilibrium constant(K_c) is the product of the molar concentrations of the product, each raised to the power equal to its stoichiometric coefficient divided by the product of the molar concentrations of the reactants, each raised to the power equal to its stoichiometric coefficient is constant at constant temperature.

∴ K_c = (Law of Chemical Equilibrium)

For the given decomposition reaction,

PCl₅ (g) ⇌ PCl₃ (g) + Cl₂ (g)

Given: The equilibrium constant (K_c) is 8.3 × 10^-3

a) From Law of Chemical Equilibrium, we know that

⇒K_c =

Where X and Y are the products and A and B are the reactants

Hence, for the given decomposition reaction

PCl₅ (g) ⇌ PCl₃ (g) + Cl₂ (g)

⇒K_c =

b) Reverse reaction: PCl₃ (g) + Cl₂ (g) ⇌ PCl₅ (g)

K_c^’ =

K_c = 8.3 × 10^-3 (given)

∴ K_c^’ =

⇒K_c^’ = 120.48

c) (i) If more PCl₅ is added in the reaction, no effect as K_C is constant at constant temperature.

(ii) If the pressure is increased, no effect as K_C is constant at constant temperature.

(iii) In an endothermic reaction, the value of K_c increases with an increase in temperature. Since the given reaction is an endothermic reaction, the value of K_c will increases with the increase of temperature.

The given reaction is CO (g) + H₂O (g) ⇌ CO₂ (g) + H₂ (g)

Given: Temperature = 400° C

Partial pressure of CO (P_CO) = Partial pressure of H₂O (P_H2O) = 4.0 bar

K_p = 10.1 bar

Formula: K_p =

Let the partial pressure of H₂ (P_H2) at equilibrium = P bar

By applying the formula given below:

K_p =

Where P_x and P_Y are the partial pressures of x and y(products)

P_A and P_B are the partial pressures of a and b (reactants)

Hence, K_p =

As K_p = 10.1 bar (given)

∴ 10.1 =

⇒10.1 =

By taking square roots of both the sides, we get

⇒√10.1 =

⇒0.3178 =

⇒12.712– 0.3178 P = P

⇒4.178 P = 12.712

⇒P = 3.042 bar

Thus, the partial pressure of H₂ (P_H2) at equilibrium = P bar = 3.042 bar.

The reaction given in (c) will have appreciable concentration of

reactants and products whose K_c are 1.8.

Explanation:

If the value of K_c lies between 1 × 10^-3 and 1 × 10³, a reaction has an appreciable concentration of reactants and products.

⇒In reaction (A),

Cl₂ (g) ⇌ 2Cl (g)

K_c= 5 ×10^–39 (given)

The value of K_c does not lie between 1× 10^-3 and 1× 10³ Hence, this reaction has not the appreciable concentration of reactants and products.

⇒In reaction (B),

Cl₂ (g) + 2NO (g) ⇌ 2NOCl (g)

K_c = 3.7 × 10⁸ (given)

The value of K_c does not lie between 1× 10^-3 and 1×10³ Hence, this reaction has not appreciable concentration of reactants and products.

⇒In reaction (B),

Cl₂ (g) + 2NO₂ (g) ⇌ 2NO₂Cl (g)

K_c = 1.8 (given)

The value of K_c lies between 1× 10^-3 and 1×10³.Hence, this reaction has appreciable concentration of reactants and products.

Note: If K_c > 10³, products predominate over reactants, i.e., if Kc is very large, the reaction proceeds nearly to completion.

If Kc < 10^–3, reactants predominate over products, i.e., if Kc is very small, the reaction proceeds rarely.

Given:

The reaction is 3O₂ (g) ⇌ 2O₃ (g)

The equilibrium constant of the given reaction = 2.0×10^–50

Temperature = 25° C

Equilibrium concentration of O₂ ([O₂]) = 1.6 ×10^–2 M

To find out the equilibrium concentration of O₃, we will apply the

Law of Chemical Equilibrium, i.e., K_c =

Where X and Y are the products and A and B are the reactants.

Hence, equilibrium constant (K_c) of the given reaction,

3O₂ (g) ⇌ 2O₃ (g)

K_c =

As K_c = 2.0×10^–50 (given) and [O₂] = 1.6 ×10^–2 M (given)

∴ 2.0×10^–50 =

⇒[O₃]² = 2.0×10^–50 × (1.6 × 10^-2 M)³

⇒[O₃]² = 8.192 × 10^-56

⇒[O₃] =

⇒[O₃] = 2.86 × 10^-28 M

Thus, the equilibrium concentration of O₃ is 2.86 × 10^-28M

Given:

The reaction given is CO(g) + 3H₂(g) ⇌ CH₄(g) + H₂O(g)

No. of moles of CO = 0.30 mol

No. of moles of H₂ = 0.10 mol

No. of moles of H₂O = 0.02 mol

The equilibrium constant of the given reaction (K_c) = 3.90

As we know that molar concentration = No. of moles / Volume of flask. Hence,

Molar concentration of [CO] = = 0.30 M

Molar concentration of [H₂] = = 0.10 M

Molar concentration of [H₂O] = = 0.02 M

To find out the concentration of CH₄, we will apply the formula

K_c =

Where X and Y are the products and A and B are the reactants.

Hence, for the given reaction,

CO(g) + 3H₂(g) ⇌ CH₄(g) + H₂O(g)

Equilibrium constant (K_c) =

As the K_c = 3.90 (given)

[CO] = 0.30 M

[H₂] = 0.10 M

[H₂O] = 0.02 M

∴ 3.90 =

⇒CH₄ = 0.0585 M = 5.85 × 10^-2 M

Thus, the concentration of CH₄ in the mixture is 5.85 × 10^-2 M

Consider the following acid-base reaction:

HCl+ H₂O ⇌ H₃O⁺+ Cl-

HCl is an acid because it donates a proton to water (H₂O) and H₂O is a base because it accepts a proton from H₂Oand hence is a base. Thus, there are two acid-base pairs in the reaction. These are HCl-Cl^- and H₃O⁺ -H₂O. These acid-base pairs are called conjugate acid-base pairs.

Lewis acid is defined as the substance (atom, ion or molecule) which is capable of accepting a pair of electrons. In other words, a Lewis acid is an electron pair acceptor.

⇒All cations are Lewis acids because they have a tendency to accept the pair of electrons.

⇒Molecules having a central atom with incomplete octet

Hence, H⁺ and being cations are Lewis acids and in BF_3,the boron is electron deficient and has an empty orbital, so it can accept a pair of electrons, making it a Lewis acid.

Bronsted acid: A bronsted acid is a substance that can donate a proton. A bronsted acid is proton-donor.

When an acid loses a proton, the residual part of it has a tendency to regain a proton. Therefore, it behaves as a conjugate base.

Conjugate base ⇌ Conjugate acid - H⁺ (loses proton)

Bronsted base: A bronsted base is a substance that can accept a proton from an acid. A bronsted base is proton-acceptor.

When a base gain a proton, the residual part of it has a tendency to accept the proton. Therefore, it behaves as conjugate acid.

Conjugate base ⇌ Conjugate acid + H⁺ (gains proton)

Bronsted acid: A bronsted acid is a substance that can donate a proton. A bronsted acid is proton-donor.

Bronsted base: A bronsted base is a substance that can accept a proton from an acid. A bronsted base is proton-acceptor.

When a base gain a proton, the residual part of it has a tendency to accept the proton. Therefore, it behaves as conjugate acid.

When an acid loses a proton, the residual part of it has a tendency to regain a proton. Therefore, it behaves as a conjugate base.

Conjugate base ⇌ Conjugate acid - H⁺

Conjugate base ⇌ Conjugate acid + H⁺

Lewis acid is defined as the substance (atom, ion or molecule) which is capable of accepting a pair of electrons.

⇒All cations are Lewis acids because they have a tendency to accept the pair of electrons. Example Ag+

⇒Molecules having a central atom with an incomplete octet. Example BF₃

Lewis base is defined as the substance which is capable of donating a pair of electrons.

⇒All anions are Lewis bases because they have a tendency to donate the pair of electrons. Example F^-

⇒Molecules in which one of the atoms has got at least one pair of electrons. Example NH₃,

a) OH^– can donate an electron pair. Hence, it is a Lewis base.

b) F^– can donate an electron pair. Hence it a Lewis base.

c) H⁺ can accept an electron pair. Hence, it is a Lewis acid.

d) BCl₃ is deficient in electron as its octet is incomplete. Hence, it can accept electron pair and is, therefore, a Lewis acid.

Given:

[H⁺] = 3.8 × 10^–3 M

To calculate of pH, we apply the formula

pH = -log[H⁺]

or pH = log

where pH is defined as the negative logarithm of the concentration (in mol per litre) of hydrogen ions which it contain or pH of the solution is the logarithm of the reciprocal of H⁺ ion concentration.

pH = -log [H⁺]

⇒pH = -log (3.8 × 10^–3)

⇒pH = -log 3.8 – log 10^-3

⇒pH = -log 3.8 – (-3 log 10)

⇒pH = -0.58 + 3

⇒pH = 2.42

Thus, pH value of soft drink is 2.42

Given:

pH= 3.76

To find out the concentration of hydrogen ion in it, we apply the formula:

pH = -log[H⁺]

where pH is defined as the negative logarithm of the concentration (in mol per litre) of hydrogen ions which it contains.

Using the formula, we write,

log[H⁺] = -pH

By taking antilog of both the sides, we get

[H⁺] = antilog [-pH]

⇒[H⁺] = antilog (-3.76)

⇒[H⁺] = 1.74 × 10^-4

Thus, the concentration of hydrogen ion in the sample of vinegar is 1.74 × 10^-4M

Given:

The ionization constant of HF = 6.8 × 10^–4

The ionization constant of HCOOH = 1.8 × 10^-4

The ionization constant of HCN = 4.8 × 10^-9

⇒Conjugate base of HF = F^-

⇒Conjugate base of HCOOH = HCOO^-

⇒Conjugate base of HCN = CN^-

To find out the ionization constants of the corresponding conjugate base, we apply the formula:

K_w = K_a × K_b

Where K_w is the ionic productof water.

K_a is the ionization constant of acid

K_b is the ionization constant of base

To calculate the ionization constant of conjugate base of HF, i.e., F^-,

Using the formula, we write

K_b =

As K_a of HF = 6.8 × 10^–4 (given)

K_w = 10^-14 (same value for every acid or base)

⇒K_b =

⇒K_b = 1.5 × 10^-11

To calculate the ionization constant of conjugate base of HCOOH, i.e., HCOO^-,

Using the formula, we write

K_b =

As K_a of HCOOH = 1.8 × 10^–4 (given)

K_w = 10^-14

⇒K_b =

⇒K_b = 5.6 × 10^-11

To calculate the ionization constant of conjugate base of HCN, i.e., CN^-,

Using the formula, we write

K_b =

As K_a of HCN = 4.8 × 10^–9 (given)

K_w = 10^-14

⇒K_b =

⇒K_b = 2.08 × 10^-6

Given:

The ionization constant of phenol (K_a) =1.0 × 10^–10

Concentration of phenolate ion (C₆H₅O ^-) in phenol = 0.05 M

Concentration of phenolate ion in sodium phenolate = 0.01 M

Ionization of phenol takes place,

∴ Using the formula,

K_a =

K_a =

As K_a = 1.0 × 10^–10 (given)

∴ 1.0 × 10^–10 =

⇒1.0 × 10^–10 =

As the value of ionization constant is very less, x will be very small. Thus, we can ignore x in the denominator.

∴ x² = 5 × 10^-12

⇒x = √ 5 × 10^-12

⇒x = 2.23 × 10^-6 M

Hence [H⁺] = [C₆H₅O^-] = 2.23 × 10^-6 M

Thus, the concentration of the phenolate ion is 2.23 × 10^-6 M

Let α be the degree of ionization of phenol in the presence of 0.01 M C₆H₅ONa

[C₆H₅OH] = 0.05 – 0.05α ≈ 0.05M

[C₆H₅O^-] = 0.01 + 0.05α ≈0.01 M

[H⁺] = 0.05α

K_a =

⇒K_a =

As K_a = 1.0 × 10^–10 (given)

∴ 1.0 × 10^–10

⇒1.0 × 10^–10 = 0.01α

⇒α = 1 × 10^-8

Thus, the degree of ionization is 1 × 10^-8

Given:

First ionization constant of H₂S= 9.1×10^-8

Second dissociation constant of H₂S =1.2×10^-13

[H₂S] = 0.1M

[HCl] = 0.1M

To calculate the [HS^–]

In the absence of 0.1M HCl

Let the concentration of HS^- ion be xM

Then, Using the formula:

K_a =

K_a1 =

As K_a1 = 9.1 × 10^–8 (given)

∴ 9.1×10^-8 =

⇒9.1×10^-8 =

⇒(9.1 × 10^-8) (0.1 – x) = x²

Taking 0.1 – x 0.1

⇒(9.1 × 10^-8) (0.1) = x²

⇒x = √ 9.1 × 10^-8

⇒x = 9.54 × 10^-5 M

Thus, in the absence of HCl, the concentration of HS^- ion is 9.54 × 10^-5 M

In the presence of 0.1 M HCl, let the concentration of HS^- ion be yM

Then at equilibrium

Then, using the formula:

K_a1 =

K_a1 =

As K_a1 = 9.1 × 10^–8

∴ 9.1×10^-8 =

⇒(9.1 × 10^-8)(0.1 - y) =

Taking 0.1 – y 0.1

0.1 + y≈ 0.1

∴ (9.1 × 10^-8)(0.1) = y(0.1)

⇒y = 9.1 × 10^-8 M

Thus, in the presence of 0.1M HCl, the concentration of HS^- ion is 9.1 × 10^-8 M

To calculate [S^2–]

In the absence of HCl

Let the concentration of S^2- ion be x M

HS^-↔ H⁺ + S^2–

[HS^-] = 9.54 × 10^-5M (from first ionization)

[H⁺] = 9.54 × 10^-5M (from first ionization)

[S^2-] = xM

By applying the formula:

K_a =

K_a1 =

As K_a2 = 1.2 × 10^–13 (given)

∴ 1.2 × 10^–13=

⇒x = 1.2 × 10^–13M

Thus, in the absence of HCl, the concentration of S^2- ion is 1.2 × 10^–13 M

Case 2: In the presence of 0.1 M HCl

Let the concentration of S^2- ion be x M

HS^-↔ H⁺ + S^2–

[HS^-] = 9.1 × 10^-8 (from first ionization, Case 2)

[H⁺] = 0.1 M(from HCl, Case 2)

[S^2-] = yM

By applying the formula:

K_a =

K_a2 =

As K_a2 = 1.2 × 10^–13 (given)

∴ 1.2 × 10^–13=

⇒10.92 × 10^-21 = 0.1 y

⇒y = 1.092 × 10^–19M

Thus, in the presence of 0.1 M HCl, the concentration of S^2- ion is 1.092 × 10^–19 M

Given:

Ionization constant of acetic acid = 1.74 × 10^–5

0.05 M solution

Ionization of CH₃COOH

CH₃COOH ⇒ CH₃COO^– + H⁺

By applying the formula,

K_a =

K_a =

At equilibrium [CH₃COO^– ]= [H⁺]

Hence,

K_a=

As K_a = 1.74 × 10^–5 (given)

[CH₃COOH] = 0.05 M

∴ 1.74 × 10^–5 =

⇒[H⁺]² = 1.74 × 10^–5 × 5 × 10^-2M

⇒[H⁺] =

⇒[H⁺] = 9.33 × 10^-4 M

As, [CH₃COO^–] = [H⁺]

Thus, the concentration of acetate ion is 9.33 × 10^-4M

To calculate pH, we apply the formula,

pH = -log[H⁺]

⇒pH = -log(9.33 × 10^-4)

⇒pH = -log 9.33 – log 10^-4

⇒pH = - log 9.33 – (-4) log 10

⇒pH = 4 - log 9.33

⇒pH = 4 – 0.9699

⇒pH = 3.03

Thus, its pH is 3.03.

Given:

pH of = 4.15

Concentration of HA = 0.01M

To calculate the concentration of the anion, we apply the formula:

pH = -log[H⁺]

To calculate the ionization constant of the acid, we apply the formula:

K_a =

To calculate the pK_a of an organic acid, we apply the formula:

pK_a = -log k_a

where K_a is the ionization constant of the acid

Let the HA be the acid

The ionization of organic acid HA is HA ⇌ H⁺ + A^-

pH = -log[H⁺]

We can also write,

log[H⁺] = -pH

As pH = 4.15 (given)

∴ log[H⁺] = -4.15

By taking antilog of both the sides, we get

[H⁺] = 7.08 × 10^-5

[A^-] = [H⁺] = 7.08 × 10^-5 M

Thus, the concentration of anion is 7.08 × 10^-5M

K_a =

K_a =

As [A^-] = [H⁺] = 7.08 × 10^-5 (calculated above)

[HA] = 10^-2 (given)

∴ K_a =

⇒K_a = 5.0 × 10^-7

Thus, the ionization constant of the acid is 5.0 × 10^-7

pK_a = -log k_a

As K_a = 5.0 × 10^-7

∴ pK_a = -log (5.0 × 10^-7)

⇒pK_a = - log 5 – (-7) log 10

⇒pK_a = 7 - log 5

⇒pK_a = 7 – 0.699

⇒pK_a = 6.301

Thus, pk_a is 6.301

Note: pK_a is a measure of acid strength. It depends on the identity and chemical properties of the acid. pH is a measure of [H⁺] in a solution. For acids, the smaller the pKa, the more acidic the substance is (the more easily a proton is lost, thus the lower the pH).

To calculate the pH, we apply the formula:

For acidic solution, pH = -log[H₃O⁺]

For basic solution, pOH = - log[OH^-]

Also, we use the given formula to calculate pH and pOH

pH + pOH = 14

A)0.003M HCl

Given:

[HCl] = 0.003 M

Ionization of HCl

HCl + H₂O ⇌ H₃O⁺ + Cl-

Since HCl is completely ionized(given)

[HCl] = [H₃O⁺]

⇒[H₃O⁺] = 0.003 M

By applying the formula,

pH = -log[H₃O⁺]

As [H₃O⁺] = 0.003 M

∴ pH = -log[0.003]

⇒pH = 2.52

Thus, the pH of the solution is 2.52

B) 0.005 M NaOH

Given:

[NaOH] = 0.005M

Ionization of NaOH,

NaOH ⇌ Na⁺ + OH-

Since NaOH is completely ionized (given)

[NaOH] = [OH^-]

⇒[OH^-] = 0.005 M

By applying the formula for basic solution:

pOH = -log[OH^-]

As [OH^-] = 0.005 M

∴ pOH = -log [0.005]

⇒pOH = 2.30

As pOH + pH =14

⇒pH = 14 - pOH

⇒pH = 14 – 2.30

⇒pH = 11.70

Thus, the pH of the solution is 11.70

c)0.002 M HBr

Given:

[HBr] = 0.002 M

Ionization of HBr

H₂O + HBr ⇌ H₃O⁺ + Br-

Since HBr is completely ionized(given)

[HBr] = [H₃O⁺]

⇒[H₃O⁺] = 0.002 M

By applying the formula (for acidic solution)

pH = -log[H₃O⁺]

As [H₃O⁺] = 0.002 M

∴ pH = -log[0.002]

⇒pH = 2.69

Thus, the pH of the solution is 2.69

d)0.002 M KOH

Given:

[KOH] = 0.002M

Ionization of KOH

KOH + ⇌ K⁺ + OH-

Since KOH is completely ionized (given)

[KOH] = [OH^-]

⇒[OH^-] = 0.002 M

By applying the formula for basic solution,

pOH = -log[OH^-]

As [OH^-] = 0.002 M

∴ pOH = -log[0.002]

⇒pOH = 2.69

As pOH + pH =14

⇒pH = 14 - pOH

⇒pH = 14 – 2.69

⇒pH = 11.31

Thus, the pH of the solution is 11.31

To find out the concentration, we apply the formula:

Concentration=

Where n= No. of moles, V= Volume

As we know that n= ()

∴ Concentration=

To calculate [H⁺], we apply the formula:

K_w = [H⁺] [OH^-]

Where K_w is the ionic product of water which is defined as the product of molar concentration of H⁺ and OH^- ions.

To calculate pH, we apply the formula:

pH = -log[H⁺]

a) For 2g of TiOH dissolved in water to give 2L of solution,

Given:

Mass of TiOH(m) = 2g

Volume(V) = 2l

Molar mass of TiOH(M) = 221 g/mol

To calculate concentration of TiOH, we apply the formula,

Concentration=

[TiOH] =

⇒[TiOH] = 4.52 × 10^-3 M

Now, the ionisation of TiOH

⇒TiOH↔ Ti⁺ + OH^-

TiOH is completely ionised

∴ [TiOH] = [OH^-] = 4.52 × 10^-3 M

To calculate [H⁺], we apply the formula

K_w = [H⁺] [OH^-]

As the value of K_w is taken as 10^-14

∴ [H⁺] =

⇒[H⁺] = 2.21 × 10^-12 M

To calculate pH, we apply the formula:

pH = -log[H⁺]

⇒pH = - log (2.21 × 10^-12)

⇒pH = -log 2.21 – (-12) log 10

⇒pH = 12 – log2.21

⇒pH = 12 – 0.3424

⇒pH = 11.65

Thus, the pH of the given solution is 11.65

b) For 0.3g of Ca(OH)₂ dissolved in water to give 500mL of solution,

Given:

Mass of Ca(OH)₂ (m) = 0.3g

Volume(V) = 500 ml = L = 0.5L

Molar mass of Ca(OH)₂ (M) = 40 × 2(16 + 1) = 74 g/mol

To calculate concentration of Ca(OH)₂, we apply the formula,

Concentration=

⇒[Ca(OH)₂] =

⇒[Ca(OH)₂] = 8.1 × 10^-3 M

Now, the ionisation of Ca(OH)₂

⇒Ca(OH)₂↔ Ca²⁺ + 2OH^-

Ca(OH)₂ is completely ionised

∴ [Ca(OH)₂] = 2[OH^-]

⇒2[OH^-]= 2 × 8.1 × 10^-3 M

⇒16 × 10^-3 M

To calculate pOH, we apply the formula:

pOH = -log[OH^-]

⇒pOH = -log (16 × 10^-3)

⇒pOH = -log 16 – (-3) log 10

⇒pOH = 3 -1.201

⇒pOH = 1.79

As we know that pH + pOH = 14

∴ pH = 14 – 1.79

⇒pH = 12.21

Thus, the pH of the given solution is 12.21

c)for 0.3g of NaOH dissolved in water to give 500mL of solution,

Given:

Mass of NaOH (m) = 0.3g

Volume(V) = 200 ml = L = 0.2L

Molar mass of NaOH (M) = 40 g/mol

To calculate concentration of NaOH, we apply the formula,

Concentration=

⇒[NaOH] =

⇒[NaOH] = 3.75 × 10^-2 M

Now, the ionisation of NaOH

⇒NaOH↔ Na + OH^-

NaOH is completely ionised

∴ [NaOH] = [OH^-]

⇒[OH^-]= 3.75 × 10^-2 M

To calculate pOH, we apply the formula:

pOH = -log[OH^-]

⇒pOH = -log (3.75 × 10^-2 ) M

⇒pOH = -log 3.75 – (-2) log 10

⇒pOH = 2-log 3.75

⇒pOH = 1.42

As we know that pH + pOH = 14

∴ pH = 14 – 1.42

⇒pH = 12.57

Thus, the pH of the given solution is 12.57

d)For 1mL of 13.6 M HCl diluted with water to give 1 L of solution:

Given:

M₁ = 13.6M

V₁ = 1ml

V₂ = 1L = 1000ml

∴ 13.6 × 1 mL = M₂ × 1000 mL

⇒M₂ = 1.36 × 10^-2

Ionization of HCl

HCl ⇌ H⁺ + Cl-

As it is completely ionized

∴ [H⁺] = [HCl]

⇒1.36 × 10^-2

To calculate pH, we apply the formula:

pH = -log[H⁺]

⇒pH = - log (1.36× 10^-2)

⇒pH = -log 1.36 – (-2) log 10

⇒pH = 2 – log 1.36

⇒pH = 2 – 0.1335

⇒pH = 1.8665

Thus, the pH of the solution is 1.87

Given:

Degree of ionization (α)= 0.132

Concentration(C) = 0.1 M

Ionization of bromoacetic acid

By applying the formula

K_a =

⇒K_a =

⇒K_a=

We can neglect C(1-α) as its value is very small

∴ K_a = Cα²

As α= 0.132, C = 0.1 M (Given)

Hence,

⇒K_a = 0.1× 0.132 × 0.132

⇒K_a = 1.74 × 10^-3

To calculate Pk_a , we apply the formula:

Pk_a = -logK_a

As K_a = 1.74 × 10^-3

∴ pk_a = -log ( 1.74 × 10^-3)

⇒Pk_a = -log 1.74 – (-3) log 10

⇒Pk_a = -0.2405 + 3

⇒Pk_a = 2.76

Thus, the pK_a of bromoacetic acid is 2.76

[H⁺] = Cα

⇒[H⁺] = 0.1 × 0.132

⇒[H⁺] = 1.32 × 10^-2

To calculate pH, we apply the formula:

pH= -log[H⁺]

⇒pH= -log (1.32 × 10^-2)

⇒pH= -log 1.32 + 2log 10

⇒pH = 1.88

Thus, the pH of the solution is 1.88

Given:

pH of Codeine(C₁₈H₂₁NO₃) solution = 9.95

[Cod] = 0.005M

Ionisation of Cod:

Cod + H₂O ↔ CodH⁺ + OH^-1

As we know that pH + pOH = 14

∴ 9.95 + pOH =14

⇒pOH = 14- 9.95

⇒pOH = 4.05

As we know that, pOH = -log[OH^-]

∴ 4.05 = -log[OH]

By taking antilog of both the sides, we get

Antilog 4.05 = -[OH^-]

⇒[OH^-] = 8.913 × 10^-5

As we know that, K_b(ionization constant)

⇒K_b=

Hence, for the reaction, Cod + H₂O ↔ CodH⁺ + OH^-1

K_b =

As the reaction is completely ionized,

∴ [CodH⁺] = [OH^-]

Hence, K_b=

As [OH^-] = 8.913 × 10^-5 (calculated)

[Cod] = 0.005M

K_b=

⇒K_b = 1.588 × 10^-6

Thus, the ionization constant of codeine is 1.588 × 10^-6

To calculate Pk_b we apply the formula:

Pk_a = -logK_b

As K_b = 1.74 × 10^-3

∴ pk_b = -log (1.588 × 10^-6)

⇒Pk_b = -log 1.588 – (-6) log 10

⇒Pk_b = 5.80

Thus, the Pk_b of codeine is 5.80

Given:

[C₆H₅NH₂] = 0.001M

Ionization constant of aniline (K_a) = 4.27 × 10^-10

Ionisation of aniline:

C₆H₅NH₂ + H₂O ↔ C₆H₅NH₂ + OH^-

K_a =

At equilibrium, []

∴ K_a =

As K_a = 4.27 × 10^-10 (given)

[C₆H₅NH₂] = 0.001M (given)

∴ 4.27 × 10^-10 = .

⇒[OH] = √(4.27 ×0^-10) (0.001)

⇒[OH] = 6.534 × 10^-7

As we know that, pOH = -log[OH^-]

∴ pOH = -log[OH]

⇒pOH = -log (6.534 × 10^-7)

⇒pOH = 6.18

As we know that pH + pOH = 14

∴ pH + 6.18 =14

⇒pH = 14- 6.18

⇒pH = 7.82

By applying the formula

K_b =

⇒K_b =

⇒K_a=

We can neglect C(1-α) as its value is very small

∴ K_b = Cα²

As C = 0.001 M (Given)

K_b = 4.27 × 10^-10 (given)

Hence,

⇒4.27 × 10^-10= 0.001× α²

⇒α = 6.53 × 10^-4

Thus, the degree of ionization aniline is 6.53 × 10^-4

As we know that pk_a + pK_b = 14 (for a conjugate acid and base)

∴ Pk_a = -logK_b

As K_b = 4.27 × 10^-10 (given)

∴ pk_a = -log ( 4.27 × 10^-10 )

⇒Pk_a= -log 4.27 – (-10) log 10

⇒Pk_a = -0.62 + 10

⇒Pk_a = 9.37

pk_a + pK_b = 14

⇒Pk_a = 14 – 9.37

⇒Pk_a = 4.63

As we know that, pk_a = -logK_a

-log K_a = 4.63

By taking antilog of both the sides, we get

K_a = antilog 4.63

⇒K_a = 2.4 × 10^-5

Thus, the ionization constant of the conjugate acid of aniline is 2.4 × 10^-5

Given:

Concentration of acetic acid = 0.05M

pK_a = 4.74

As we know that, pk_a = -logK_a

We can also write,

∴ pK_a = -logK_a

⇒4.74 = -logK_a

By taking antilog of both the sides, we get

Antilog -4.74 = K_a

⇒K_a = 1.82 × 10-5

By applying the formula, k_a = cα²

Where c is the concentration and α is the degree of ionization

We can also write,

α =√

K_a = 1.82 × 10-5(given)

C = 0.05M (given)

∴ α =√

⇒α = 1.908 × 10^-2

In the presence of HCl due to the high concentration of H+ , the dissociation equilibrium will shift in the backward direction i.e., dissociation of acetic acid will decrease.

a) In the presence of HCl, let x be the amount of acetic acid dissociated after the addition of HCl.

Ionization of acetic acid

We can assume, 0.05-x≈ 0.05

0.01+x ≈ 0.01

To calculate the degree of ionization, we apply the formula

α=

Where α =

K_a is the ionization constant

c is the concentration

K_a = 1.82 × 10-5(given)

C = 0.01M (given)

∴ α =

⇒α =1.82 × 10^-3

Thus, the degree of ionization in the presence of 0.01M HCl is 1.82 × 10^-3

b) In the presence of 0.1 M HCl, let y be the amount of acetic acid dissociated after the addition of HCl.

Ionization of acetic acid

We can assume, 0.05-x≈ 0.05

0.1+x ≈ 0.1

To calculate the degree of ionization, we apply the formula

α=

Where α =

K_a is the ionization constant

c is the concentration

K_a = 1.82 × 10-5(given)

C = 0.1M (given)

∴ α =

α =1.82 × 10^-4

Thus, the degree of ionization in the presence of 0.1M HCl is 1.82 × 10^-4

Given:

K_b = 5.4 × 10^–4.

c=0.02M

To calculate the degree of ionization, we apply the formula

α=

Where K_b is the ionization constant

c is the concentration

∴ α =√

⇒α =0.1643

Thus, the degree of ionization is 0.1643

Ionization of 0.1 M NaOH;

Since, (CH₃)₂ NH⁺₂]=x

[OH^– ]=x+0.1 ≈ 0.1 M

Using the formula,

K_b =

⇒K_b =

As K_b = 5.4 × 10^–4.(given)

[(CH₃)₂ NH] = 0.02 M (given)

∴ 5.4 × 10^–4 =

⇒x = 0.054

Thus, It means that in the presence of 0.1 M NaOH, 0.54% of dimethylamine will get dissociated.

a) Human muscle fluid 6.83:

⇒pH = 6.83

⇒pH = – log [H+]

⇒6.83 = – log [H+]

⇒[H⁺] =1.48 × 10^–7 M

(b) Human stomach fluid, 1.2:

⇒pH =1.2

⇒1.2 = – log [H⁺]

⇒[H⁺] = 0.063

(c) Human blood, 7.38:

⇒pH = 7.38

⇒pH = log [H⁺]

⇒[H⁺] = 4.17 × 10^–8 M

d)Human saliva, 6.4:

⇒pH = 6.4

⇒6.4 = – log [H+]

⇒[H+] = 3.98 × 10^–7

We can calculate the hydrogen ion concentration by applying the formula, pH = –log [H+]

(i) pH of milk = 6.8

Since, pH = –log [H+]

⇒6.8 = –log [H+]

⇒log [H+] = –6.8

By taking antilog of both the sides, we get

⇒[H+] = antilog (–6.8)

⇒[H⁺]= 1.5×19–7 M

(ii) pH of black coffee = 5.0

Since, pH = –log [H⁺]

⇒5.0 = –log [H⁺]

⇒log [H⁺] = –5.0

By taking antilog of both the sides, we get

⇒[H⁺] = antilog (–5.0)

⇒[H⁺] = 10^-5 M

(iii) pH of tomato juice = 4.2

Since, pH = –log [H⁺]

⇒4.2 = –log [H⁺]

⇒log [H⁺] = –4.2

By taking antilog of both the sides, we get

⇒[H⁺] = antilog (–4.2)

⇒[H⁺]= 6.31×10^-5 M

(iv) pH of lemon juice = 2.2

Since, pH = –log [H⁺]

⇒2.2 = –log [H⁺]

⇒log [H+] = –2.2

By taking antilog of both the sides, we get

⇒[H⁺] = antilog (–2.2)

⇒[H⁺]= 6.31×10^-3 M

(v) pH of egg white = 7.8

Since, pH = –log [H⁺]

⇒7.8 = –log [H⁺]

⇒log [H⁺] = –7.8

By taking antilog of both the sides, we get

⇒[H⁺] = antilog (–7.8)

⇒[H⁺]=1.58×10^-8 M

Given:

Mass of KOH (m) = 0.561g

Volume(V) = 200 ml = L = 0.2L

Molar mass of KOH (M) = 56 g/mol

To calculate concentration of KOH, we apply the formula,

Concentration=

⇒[KOH] =

⇒[KOH] = 0.05 M

Now, the ionisation of KOH

⇒KOH↔ K + OH^-

At equilibrium

∴ [K] = [OH^-] = 0.05 M

To calculate [H⁺], we apply the formula:

K_w = [H⁺] [OH^-]

Where K_w is the ionic product of water which is defined as the product of molar concentration of H⁺ and OH^- ions.

Using the formula, we write

[H⁺] =

As the value of K_w is taken as 10^-14

[OH^-]= 0.05 M (calculated)

∴ [H⁺] =

⇒[H⁺] = 2.0 × 10^-13

Thus, the concentration of potassium, hydrogen, and hydroxyl ions are 0.05, 0.05 and 2.0 × 10^-13 respectively

To calculate pH, we apply the formula:

pH = –log [H⁺]

⇒pH = - log (2.0 × 10^-13)

⇒pH = 13- log2

⇒pH = 12.69

Thus, the pH is 12.69

Given:

Solubility of Sr(OH)₂ = 19.23 g/L

Molar mass of Sr(OH)₂ = 87.6 + 34 = 121.6 g/mol

To calculate concentration of Sr(OH)₂, we apply the formula,

Concentration=

⇒[Sr(OH)₂] =

⇒[Sr(OH)₂] = 0.1581 M

Ionisation of Sr(OH)₂

Sr(OH)₂ ↔ Sr²⁺ + 2(OH^-)

As it is completely ionised,

∴ [Sr²⁺] = 0.1581 M

[OH^-] = 2× 0.1581 = 0.3126M

Thus, the concentration of strontium and hydroxyl ions are 0.1581 and 0.3126

Now,

K_w = [OH^-][H⁺]

Using the formula, we write

[H⁺] =

As the value of K_w is taken as 10^-14

[OH^-]= 0.3126 M (calculated)

∴ [H⁺] =

⇒[H⁺] = 3.2 × 10^-14 M

To calculate pH of the solution, we apply the formula,

pH= -log(H⁺)

⇒pH= -log (3.2 × 10^-14)

⇒pH = -log 3.2 – (-14) log 10

⇒pH= 13.49

Thus, the pH of the solution is 13.49

Given:

K_a = 1.32 × 10^–5

C = 0.005 M

Assuming α to be very small, applying formula directly, we have

α = √

⇒α = √

⇒α = 1.62 × 10^-2

Thus, the degree of ionization in 0.05M solution is 1.62 × 10^-2

Let the degree of ionization of propanoic acid be x

Ionization of propanoic acid

By applying the formula

K_a =

⇒K_a =

⇒K_a=

As K_a = 1.32 × 10^–5

, C = 0.1 M (given)

Hence,

x= 1.32 × 10^-3

Thus, the degree of ionization in 0.01M solution is 1.32 × 10^-3

Given:

pH = 2.34

Concentration of HCNO = 0.1M

Ionization of HCNO,

HCNO↔ H⁺ + CNO^-

As pH = -log[H⁺]

pH = 2.34 (given)

∴ log[H⁺] = -2.34 = -3.66

By taking antilog of both the side, we get

[H⁺] = antilog -3.66

⇒[H⁺] = 4.57 × 10^-3 M

At equilibrium, [H⁺] = [CNO^-] =4.57 × 10^-3 M

∴ Using the formula,

K_a =

⇒K_a =

⇒K_a =

⇒K_a = 2.09 × 10^-4

To calculate degree of ionization (α) applying formula directly, we have α = √

⇒α = √

⇒α = 0.0457

Thus, the ionization constant of the acid is 2.09 × 10^-4 and the degree of ionization is 0.00457

Given:

K_a of nitrous acid is 4.5 × 10^–4

[NO₂] = 0.04 M

To calculate the degree of hydrolysis, we apply the formula:

K_h = K_w / K_a

Where K_h is a degree of hydrolysis of a salt which is defined as the fraction of one mole of the salt which is hydrolyzed, when the equilibrium is attained. Hence for a salt (acid and base),

K_h = K_w / K_a

NaNO₂ is the salt of a strong base (NaOH) and a weak acid (HNO₂)

Using the formula, we write

K_h =

As K_w = 10^-14 (same value)

K_a = 4.5 × 10^–4 (given)

∴ K_h =

⇒K_h = 0.22 × 10^-10

Thus, the degree of hydrolysis is 0.22 × 10^-10

NO₂ + H₂O ↔ HNO₂ + OH^-

[NO₂] = 0.04 M (given)

Now, if x moles of the salt undergo hydrolysis, then the concentration of the various species will be:

[NO₂] = 0.04 M – x ≈0.04 M

[HNO₂] = x

[OH] = x

Using the formula,

K_h =

⇒K_h =

⇒K_h =

As K_h == 0.22 × 10^-10 (calculated)

∴ 0.22 × 10^-10 M× 0.04 M = x²

⇒X =

⇒X = 0.93 × 10^-5

As [OH] = x

⇒[OH^-] = 0.93 × 10^-5

Now,

K_w = [OH^-][H⁺]

Using the formula, we write

[H⁺] =

As the value of K_w is taken as 10^-14

⇒[OH^-]= 0.93 × 10^-5(calculated)

∴ [H⁺] =

⇒[H⁺] = 10.75 × 10^-9

To calculate pH of the solution, we apply the formula,

pH= -log(H⁺)

⇒pH= -log (10.75 × 10^-9)

⇒pH = -log 10.75 – (-9) log 10

⇒pH= 7.96

Thus, the pH of the solution is 7.96

Given:

pH = 3.44

[C₆H₅NCl] = 0.02M

As we know that,

pH = -log[H⁺]

3.44= -log[H⁺]

By taking antilog of both the sides, we get

Antilog 3.44= [H⁺]

[H⁺] = 3.63 × 10^-4

Ionisation of pyridinium hydrochloride [C₅H₅NCl]

C₆H₅NCl + aq ↔ C₅H₅NCl + H⁺

Using the formula,

K_h =

⇒K_h =

As it is completely ionized,

∴ [C₅H₅NCl] = [H⁺] = 3.63 × 10^-4 M

As [C₆H₅NCl] = 0.02M (given)

∴ K_h =

⇒K_h =

⇒K_h = 6.6 × 10^-6

To calculate ionization constant, we apply the formula:

K_h = K_w / K_a

Or K_a = K_h / K_w

As the value of K_w is taken as 10^-14

K_h = 6.6 × 10^-6 (given)

∴ K_a =

⇒K_a = 1.51 × 10^-9

Thus, the ionization constant of pyridine is 1.51 × 10^-9

Note: K_h is the degree of hydrolysis of a salt which is defined as the fraction of a mole of the salt which is hydrolyzed. When the equilibrium is attained. Hence for a salt (acid and base), K_h = K_w / K_a

(i) NaCl is neutral as it is salt of strong acid(HCl) and strong base(NaOH)

(ii)KBr is neutral as it is salt of strong acid(HBr) and strong base(KOH)

(iii) NaCN is basic as it is a salt of strong base(NaOH) and weak acid(HCN)

(iv) NH₄NO₃ is acidic as it a salt of strong acid(HNO₃) and weak base (NH₄OH)

(v) NaNO₂ is basic as it a salt of strong base(NaOH) and weak acid(HNO₂)

(vi) KF is basic as it is salt of strong base(KOH) and weak acid(HF)

Given:

Ionization constant of chloroacetic acid is 1.35 × 10^–3

C = 0.1M

Ionization of chloroacetic acid:

ClCH₂COOH ↔ ClCH₂COO^- + H⁺

Using the formula,

K_a =

⇒K_a =

As it is completely ionized,

∴ [ClCH₂COO^-] = [H⁺]

As [ClCH₂COOH] = 0.02M (given)

As K_a = 1.35× 10^-3 (given)

∴ 1.35× 10^-3=

^⇒[H⁺] = √0.02× 1.35× 10^-3

⇒[H⁺] = 1.16 × 10^-2

To calculate pH of the solution, we apply the formula,

pH= -log[H⁺]

⇒pH= -log (1.16 × 10^-2)

⇒pH = -log 1.16 – (-2) log 10

⇒pH= 1.94

Thus, the pH of 0.1 M acid is 1.94

Now, To calculate the pH of 0.1M sodium salt solution:

As we know that ClCH₂COONa is salt of weak acid(ClCH₂COOH) and a stong base(NaOH)

Hence, by applying the formula:

pH = -1/2 [logK_w + log K_a – log c]

Where K_w is the ionic product of water

K_a is the ionization constant

C is the concentration

As we know the value of K_w is 10^-14

K_a = 1.35 × 10^–3 (given)

C = 0.1 M (given)

∴ pH = -1/2 [log10^-14 + log (1.35 × 10^–3) + log 0.1]

⇒pH = -1/2 [-14 + (-3 + 0.1303) – (-1)]

⇒pH = 7+ 1.44+ 0.5

⇒pH = 7.94

Thus, the pH of 0.1 M sodium salt solution is 7.94

Given:

Ionic product of water(K_w) = 2.7 × 10^–14

As we know that K_w = [H⁺][OH^-]

For neutral water [H⁺]=[OH^-]

∴ K_w = [H⁺][H⁺]

⇒[H⁺]² = 2.7 × 10^–14

⇒[H⁺] =

⇒[H⁺] = 1.643 × 10^–7

Since, pH= -log[H⁺]

⇒pH = -log (1.643 × 10^–7)

⇒pH = -log 1.643 + 7log 10

⇒pH = 7- 0.1256

⇒pH = 6.78

Thus, the pH of the neutral water is 6.78

(a) for 10 mL of 0.2M Ca(OH)₂ and 25 mL of 0.1M HCl

10 ml of 0.2 M Ca(OH)₂= 10× 0.2 milimole = 2 milimoles

25 ml of 0.1 M HCl = 25× 0.1 milimole = 2.5 milimoles

Ca(OH)₂ + 2HCl ↔ CaCl₂ + 2H₂O

0.1 a mole of Ca(OH)₂ reacts with 2 millimole of HCl

∴ 2.5 mole of HCl reacts with 1.25 millimole of Ca(OH)₂

_∴ Ca(OH)₂ left = 1=1.25 = 0.75 milimole

Volume of reaction mixture = 10mL+ 25mL = 35mL

∴ Molarity of Ca(OH)₂in the mixture solution is:

Molarity =

Hence,

Molarity = = 0.0214 M

⇒[OH^-] = 2 × 0.0124 M

⇒[OH^-] = 0.0428 M

To caculate pOH, we use the formula:

pH = - log[OH^-]

⇒pH= - log (4.28× 10^-2)

⇒pH = -log 4.28 + 2log 10

⇒pH = 2 – 0.6314

⇒pH = 1.37

As we know that, pH + pOH =14

∴ pH = 14 – 1.37

⇒pH = 12.63

Hence, the pH is 12.62

(b) for 10 mL of 0.01M H₂SO₄ and 10 mL of 0.01M Ca(OH)₂

10 ml of 0.01 M H₂sO₄ = 10× 0.01 milimole = 0.1 milimole

10 ml of 0.01 M Ca(OH)₂ = 10× 0.01 milimole = 0.1 milimole

Ca(OH)₂ + H₂sO₄ = CaSO₄ + 2H₂O

1 mole of Ca(OH)₂ reacts with 1 mole of H₂sO₄

∴ 0.1 millimole of Ca(OH)₂ will react completely with 0.1 minimal of H₂SO₄. Hence solution will be neutral with pH= 7

(C) 10 mL of 0.1M H₂SO₄ and 10 mL of 0.1M KOH

10 ml of 0.01 M H₂sO₄ = 10× 0.01 milimole = 0.1 milimole

10 ml of 0.1 M KOH = 10× 0.1 milimole = 1 milimole

2KOH+ H₂sO₄ = K₂SO₄ + 2H₂O

I mole of KOHreacts with 0.5 milimole of H₂sO₄

_∴ H₂SO₄ left = 1=0.5 = 0.5 milimole

Volume of reaction mixture = 10+ 10 = 20 mL

∴ Molarity of H₂SO₄ in the mixture solution is:

Molarity =

Hence,

Molarity = = 2.5 × 10^-12 M

⇒[H⁺] = 2 × 2.5 × 10^-12

⇒[H⁺] = 5× 10^-2

To caculate pH, we use the formula:

pH = - log[H⁺]

⇒pH= - log (5× 10^-2)

⇒pH = -log 5 + 2log 10

⇒pH = 2 – 0.699

⇒pH = 1.3

Hence, the pH is 1.3

To determine the solubility, we apply the formula

K_sp = [A⁺][B^-]

Where K_sp is the solubility product which is equal to the product of ionic concentrations in a saturated solution.

a) Silver chromate

ionization of silver chromate:

Ag₂CrO₄ ↔ 2Ag⁺ + CrO₄^2-

As we know that,

Ks_p = [A⁺] [B^-]

Where A and B are the ions dissolved

In the above reaction,

⇒[A⁺] = [Ag⁺]

⇒[B^-] = [CrO₄^2-]

∴ K_sp = [Ag⁺]² + [CrO₄^2-]

As K_sp of Ag₂CrO₄ =1.1 × 10^-12 (given)

Let ‘s’ be the solubility of Ag₂CrO₄

[Ag⁺] = 2s

[CrO₄^2-] =s

∴ 1.1 × 10^–12 = (2s)² s

⇒1.1 × 10^-12 = 4(s)³

^⇒0.275 × 10^–12 = s³

⇒s =

⇒s = 0.65 × 10^-4

Thus, Molarity of Ag⁺ = 2s= 2× 0.65× 10^-4 = 1.30× 10^-4 M

Molarity of CrO₄²- = s = 0.65 × 10^-4 M

b) Barium chromate

ionization of barium chromate:

BaCrO₄ ↔ Ba²⁺ + CrO₄^2-

As we know that,

Ks_p = [A⁺] [B^-]

Where A and B are the ions dissolved

In the above reaction,

⇒[A⁺] = [Ba²⁺]

[B^-] = [CrO₄^2-]

∴ K_sp = [Ba²⁺] + [CrO₄^2-]

As K_sp of BaCrO₄ =1.2 × 10^-10 (given)

Let ‘s’ be the solubility of BaCrO₄

[Ba²⁺] =s

[CrO₄²-] =s

∴ 1.2 × 10^–10 = (s× s)

^⇒1.2 × 10^-10 = (s)²

⇒s =

⇒s = 1.09 × 10^-5

Thus, Molarity of Ba²⁺ and CrO₄^2- = s= 1.09× 10^-5M

c) Ferric hydroxide

ionization of ferric hydroxide

Fe(OH)₃ ↔ Fe³⁺ + 3OH^-

As we know that,

Ks_p = [A⁺] [B^-]

Where A and B are the ions dissolved

In the above reaction,

⇒[A⁺] = [Fe³⁺]

⇒[B^-] = [OH^-]

∴ K_sp = [Fe³⁺] + [OH^-]

As K_sp of Fe(OH)₃ =1.0 × 10^-38 (given)

Let ‘s’ be the solubility of Fe(OH)₃

[Fe³⁺] =s

[3OH^-] = 3s

∴ 1.0× 10^–38 = (s)(3s)³

^⇒1.0 × 10^-38 = (27s)⁴

^⇒0.37 × 10^–38 = s³

⇒s =

⇒s = 1.39 × 10^-10

Thus, Molarity of Fe³⁺ = s= 1.39× 10^-10 M

Molarity of OH^- = 3s = 4.17 × 10^-10M

d) Lead chloride

ionization of lead chloride:

PbCl₂ ↔ Pb²⁺ + 2Cl^-

As we know that,

Ks_p = [A⁺] [B^-]

Where A and B are the ions dissolved

In the above reaction,

⇒[A⁺] = [Pb²⁺]

⇒[B^-] = [Cl^-]

∴ K_sp = [Pb²⁺] + [Cl^-]²

As K_sp of PbCl₂ =1.6 × 10^-5 (given)

Let ‘s’ be the solubility of PbCl₂

[Pb²⁺] =s

[Cl^-] =2 s

∴ 1.6 × 10^–5 = (s)(2s)²

^⇒1.6 × 10^-5 = (4s)³

^⇒0.4 × 10^–5 = s³

⇒s =

⇒s = 1.58 × 10^-2

Thus, Molarity of Pb²⁺ = s= 1.58× 10^-2 M

Molarity of Cl^- = 2s = 3.16 × 10^-2 M

e) Mercurous Iodide

ionization of mercurous iodide:

Hg₂I₂ ↔ Hg₂²⁺ + 2I^-

As we know that,

Ks_p = [A⁺] [B^-]

Where A and B are the ions dissolved

In the above reaction,

⇒[A⁺] = [Hg₂²⁺]

⇒[B^-] = [I^-]²

∴ K_sp = [Hg₂²⁺] + [I^-]²

As K_sp of Hg₂I₂ = 4.5 × 10^–29

Let ‘s’ be the solubility of Hg₂I₂

[Hg₂²⁺] =s

[I^-] 2s

∴ 4.5 × 10^–29 = (s)(2s)²

⇒4.5 × 10^–29 = (4s)³

⇒4.5 × 10^–29 = s³

⇒s =

⇒s = 2.24× 10^-10

Thus, Molarity of Hg₂²⁺ = s= 2.24× 10^-10 M

Molarity of I- = 2s = 4.48 × 10^-10 M

Note: K_sp increases with increase in temperature.

⇒In a saturated solution, K_sp = [A⁺][B^-]

⇒In an unsaturated solution of AB, K_sp> [A⁺][B^-] means more solute can be dissolved.

⇒In a super saturated solution of AB, K_sp< [A⁺][B^-] means precipitation will start to occur

Given:

K_sp of Ag₂CrO₄ =1.1 × 10^–12

K_sp of AgBr = 5.0 × 10^–13

To find the ratio of molarities of Ag₂CrO₄ and AgBr saturated solutions, first, we will calculate their solubilities separately and then calculate the ratio. Hence,

Ionization of Ag₂CrO₄:

Ag₂CrO₄ ↔ 2Ag⁺ + CrO₄^2-

As we know that,

Ks_p = [A⁺] [B^-]

Where A and B are the ions dissolved

In the above reaction:

⇒[A⁺] = [Ag⁺]

⇒[B^-] = [CrO₄^2-]

∴ K_sp = [Ag⁺]² + [CrO₄^2-]

As K_sp of Ag₂CrO₄ =1.1 × 10^–12

Let ‘s’ be the solubility of Ag₂CrO₄

[Ag⁺] =2s

[CrO₄^2-] =s

∴ 1.1 × 10^–12 = (2s)²(s)

^⇒1.1 × 10^–12 = (4s)³

⇒s =

⇒s = 6.5× 10^-5

Ionization of AgBr:

AgBr↔ Ag⁺ + Br^-

As we know that,

Ks_p = [A⁺] [B^-]

Where A and B are the ions dissolved

In the above reaction,

⇒[A⁺] = [Ag⁺]

⇒[B^-] = [Br^-]

∴ K_sp = [Ag⁺] + [Br^-]

As K_sp of AgBr =5.0 × 10^–13

Let s’ be the solubility of AgBr

∴ 5.0 × 10^–13 = (s)(s)

^⇒1.1 × 10^–12 = (s)²

⇒s =

s’ = 7.07× 10^-7

Now the ratio of their solubilities

⇒ = = 9.91

Thus, the ratio of molarities of their saturated solutions is 9.91

Given:

Ks_p of cupric iodate = 7.4 × 10^–8

Equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate are mixed together.

Ionization of sodium iodate:

2NaIO₃ + CuCrO₄ Na₂CrO₄ + Cu(IO₃)₂

After mixing, [NaIO₃] = [IO₃^-] = = 10^-3M

[CuCrO₄] = [Cu²⁺] = = 10^-3M

For cupric iodate, the ionization is:

Cu(IO₃)_2↔ [Cu²⁺][IO₃^-]

As we know that,

Ks_p = [A⁺] [B^-]

Where A and B are the ions dissolved

In the above reaction,

⇒[A⁺] = [Cu²⁺]

⇒[B^-] = [IO₃^-]

∴ K_sp = [Cu²⁺][IO₃^-]

⇒K_sp = (10^-3)(10^-3)

⇒K_sp = 10^-9

As we know that precipitation only occur when Ks_p < [A⁺][B^-]

As ionic product 10^-9 is less than the K_sp (7.4 × 10^–8), hence no precipitation will occur.

Given:

The ionization constant of benzoic acid (K_a) is 6.46 × 10^–5

Ks_p for silver benzoate is 2.5 × 10^–13

pH = 3.19

Ionization of silver benzoate:

C₆H₅COOAg ↔ C₆H₅COO^- + Ag⁺

Solubility in water: Let solubility in water is x mol/l Then

[C₆H₅COO^- ] = [Ag⁺] = x mol/l

As we know that,

Ks_p = [A⁺] [B^-]

Where A and B are the ions dissolved

In the above reaction,

^⇒[A⁺] = C₆H₅COO^-

⇒[B^-] = Ag⁺

^⇒∴ K_sp = [C₆H₅COO^-][Ag⁺]

As K_sp = 2.5 × 10^–13 (given)

⇒2.5 × 10^–13 = x²

⇒x = 5 × 10^-7

Solubility in buffer of pH =3.19

As we know that,

pH = –log [H⁺]

∴ –log [H⁺] = 3.19

By taking antilog of both the sides, we get

⇒[H⁺] = antilog -3.19

⇒[H⁺] = 6.457 × 10^-4 M

C₆H₅COO^- ions combine with the H⁼ ions o from benzoic acid but [H⁺] remains almost constant because we have buffer solution. Now,

C₆H₅COOH ↔ C₆H₅COO^- + H⁺

As we know that,

K_a =

⇒K_a =

⇒ =

[H⁺] = 6.457 × 10^-4 M (calculated above)

K_a = 6.46 × 10^–5 (given)

∴ = = = 10

⇒[C₆H₅COOH] = 10 [C₆H₅COO^-]

Suppose solubility in the buffer solution is y mol/l.

Then as most of the benzoate ion is converted into benzoic acid molecules(which remains almost ionized) we have

Y = [Ag⁺] = [C₆H₅COO^-] + [C₆H₅COOH]

As [C₆H₅COOH] = 10 [C₆H₅COO^-] (calculated above)

∴ y = [C₆H₅COO^-] + 10 [C₆H₅COO^-]

⇒y = 11 [C₆H₅COO^-]

⇒[C₆H₅COO^-] =

As we know that,

K_sp = [C₆H₅COO^-][Ag⁺]

As K_sp = 2.5× 10^-13 (given0

[Ag⁺] = y, [C₆H₅COO^-] =

∴ 2.5× 10^-13 = × y

⇒Y² = 2.75 × 10^-12

⇒Y = √ 2.75 × 10^-12

⇒Y = 1.66 × 10^-6

Now,

= = = 3.32

Thus, the silver benzoate is 3.32 times more soluble in a buffer of pH compared to its solubility in pure water

Note: In case of weak acids, the solubility is more in the acidic solution than in water. The reason, in general, may be explained as below:

Taking example of C₆H₅COOAg, we have

C₆H₅COOAg ↔ C₆H₅COO^- + Ag⁺

In acidic solution, the anions (C₆H₅COO^- in the present case) undergo protonation in presence of acid. Thus, C₆H₅COO^- ions are removed. Hence equilibrium shifts forward producing more Ag⁺ ions. Alternatively, as C₆H₅COO^- are removed, Q_sp decreases. In order to maintain solubility product equilibrium (Q_sp= K_sp), Ag⁺ ion concentration must increase. Hence solubility is more.

Note: A buffer solution is defined as a solution which resists any change in its pH value even when small amount of acid or base added to it.

Given:

K_sp of FeS = 6.3 × 10^–18

Let the concentration of solution of FeSO₄ and Na₂Sis x mol/L

On mixing the equimolar (equal moles) solutions, the volume of the concentration become half

Thus, [FeSO₄] = [Na₂S] = M

∴ [Fe²⁺] = [S^2-] = M

Ionization of ferrous sulphide:

FeS Fe²⁺ +S^2-

As we know that,

Ks_p = [A⁺] [B^-]

Where A and B are the ions

In the above reaction,

^⇒[A⁺] = Fe²⁺

⇒[B^-] = S^2-

∴ K_sp = [Fe²⁺][S^2-]

As K_sp = 6.3 × 10^-18 (given)

⇒6.3×10^-18 =

⇒ =6.3×10^-18

Thus, x = 5.02×10^-9

As we know that precipitation only occur when Ks_p < [A⁺][B^-]

Thus, if the concentration of ferrous sulphate and sodium sulhide are equal to or less than that of 5.02×10^-9, then there no precipitation of ferrous sulphide takes place.

Given:

For calcium sulphate, K_sp is 9.1 × 10^–6

Ionization of calcium sulphate

CaSO₄Ca²⁺ + SO₄^2-

As we know that, K_sp = [A⁺] [B^-]

In the above reaction,

[A⁺] = Ca²⁺

[B^-] = SO₄^2-

∴ K_sp = [Ca²⁺] [SO₄^2-]

Let the solubility of calcium sulphate is x.

So, K_sp = x²

∴9.1×10^-6=x²

∴x=3.02×10^-3

As molar mass of CaSO₄ is :

⇒40 +32 + 4× 16

⇒40+32+64

⇒136g/mol.

Solubility of calcium sulphate in g/mol is

=3.02×10^-3×136

=0.41 g/L

Thus, for dissolving 0.41g, water required = 1L

∴ for dissolving 1g, water required = L

⇒2.44L

Thus, minimum volume of water required to dissolve 1 g of CaSO₄ is 2.44L

Given:

Concentration of sulphide ion [S^2-] = 1.0 × 10^–19 M

Volume of solution containing S^2- ion = 10 ml

Volume of metals salt solution added = 5ml

As 10ml of solution containing S^2- ion is mixed with 5ml of metals salt solution. Hence, after mixing volume will be 10ml + 5mL = 15ml

Now the concentration of sulphide ion is:

[S^2-] = 1× 10^-19 M × ml

⇒6.67 × 10^-20

The concentration of metals solution [M²⁺] is

⇒[Fe²⁺]= [Mn²⁺] = [Zn²⁺] = [Cd²⁺] = × 0.04 = 1.33 × 10^-2

As we know that precipitation will take place in the solution for which ionic product is greater than solubility product. K_sp < [A⁺][B^-]

Ionic product for each of these will be [M²⁺][S^2-]

⇒6.67 × 10^-20 ×1.33 × 10^-2

⇒8.87 × 10^-22

As this is greater than solubility product of ZnS and CdS, therefore ZnCl₂ and CdCl₂ solutions will be precipitated.