Chapter 7 Equilibrium

Vapour pressure is the pressure exerted by the vapour in equilibrium with the liquid at a fixed temperature.

(a) If the volume of the container is suddenly increased, then the vapour pressure would decrease initially. This is because the amount of vapour remains the same, but the volume increases suddenly. As a result, the same amount of vapour is distributed in a larger volume.

(b) Since the temperature is constant, the rate of evaporation also remains constant. When the volume of the container is increased, the density of the vapour phase decreases. As a result, the rate of collisions of the vapour particles also decreases. Hence, the rate of condensation decreases initially.

(c) When equilibrium is restored finally, the rate of evaporation becomes equal to the rate of condensation. In this case, only the volume changes while the temperature remains constant. The vapour pressure depends on temperature and not on volume. Hence, the final vapour pressure will be equal to the original vapour pressure of the system.

The equilibrium constant (K_c) for the give reaction is:

K_c   =   [SO₃]2  /  [SO₂]2 [O₂]

      =  (1.90)² M²  / (0.60)² (0.8321) M³

      = 12.239 M^-1 Approx.

Hence, K_c for the equilibrium is 12.239 M^-1

 

Partial pressure of I atoms,

p_I  =  p_total x 40/100

    =  10⁵ x 40/100 

    =  4 x 10⁴ Pa 

Partial pressure of I₂ molecules,

p_I 2   =  p_total x 60/100

       =  10⁵ x 60/100 

       =  6 x 10⁴ Pa 

Now, for the given reaction,

Kp  = (pI)²  / pI₂

     =  (4 x 10⁴ ) ² Pa²  / 6 x 10⁴ Pa

     =  2.67 x 10⁴ Pa

 

 

 

(a) The relation between K_p and K_c is given as:

K_p = K_c (RT)^Δn

(a) Here, Δn = 3 - 2 = 1

R = 0.0831 barLmol^-1K^-1

T = 500 K

Kp = 1.8 x 10^-2

Now, K_p = K_c (RT)^Δn

⇒ K_c  = 1.8 x 10^-2  /  (0.0831x500)

         = 4.33 x 10^-4 (approx.)

 

(b) Here, Δn = 2 - 1 = 1

R = 0.0831 barLmol^-1K^-1

T = 1073 K

K^p = 167

Now, K_p = K_c (RT)^Δn

⇒ K_c  = 167 / (0.0831x1073)

         = 1.87 (approx.)

It is given that K_c for the forward reaction is 6.3 × 10¹⁴  

Then, K_c  for the reverse reaction will be,

K^’_c  = 1 / K_c

     = 1 /  6.3 × 10¹⁴  

     = 1.59 x 10^-15

: For a pure substance (both solids and liquids),

Pure Substance  = Number of moles  /  Volume

                         = Mass/Molecular Mass / Volume

                         = Mass /  Volume x Molecular Mass

                         = Density /  Molecular Mass

Now, the molecular mass and density (at a particular temperature) of a pure substance is always fixed and is accounted for the equilibrium constant. Therefore, the values of pure substances are not mentioned in the equilibrium constant expression.

Let the concentration of N₂O at equilibrium be x. The given reaction is:

                                2N₂ (g)         +        O₂ (g)        ↔       2N₂O (g)

Intial Concentration:   0.482 mol            0.933 mol                  0 mol

At equilibrium           (0.482-x) mol        (0.933 -x) mol            x mol

Therefore, at equilibrium, in the 10 L vessel:

N₂ = 0.482-x / 10

O₂ = 0.933-x/2  / 10

N₂O = x / 10

The value of equilibrium constant i.e.K_c = 2.0 × 10^-37 is very small. Therefore, the amount of N₂ and O₂ reacted is also very small. Thus, x can be neglected from the expressions of molar concentrations of N₂ and O₂.

Then,

N2  =  0.482/10 = 0.0482 molL-1 and O2 = 0.933/10 = 0.0933 molL-1

Now,

The given reaction is:

2NO (g)         +           Br₂ (g)       ↔          2NOBr (g)

2 mol                         1 mol                        2 mol    

Now, 2 mol of NOBr are formed from 2 mol of NO. Therefore, 0.0518 mol of NOBr are formed from 0.0518 mol of NO.

Again, 2 mol of NOBr are formed from 1 mol of Br.

Therefore, 0.0518 mol of NOBr are formed from 0.0518/2 mol of Br, or 0.0259 mol of NO.

The amount of NO and Br present initially is as follows:

[NO] = 0.087 mol [Br₂] = 0.0437 mol

Therefore, the amount of NO present at equilibrium is: [NO] = 0.087 - 0.0518 = 0.0352 mol

And, the amount of Br present at equilibrium is: [Br₂] = 0.0437-0.0259 = 0.0178 mol

For the given reaction,

Δn = 2 - 3 =-1

T = 450 K

R = 0.0831 bar L bar K^-1 mol^-1

K_p = 2.0 × 10¹⁰ bar ^-1

We know that,

K_p =  K_c (RT)Δn

The initial concentration of HI is 0.2 atm. At equilibrium, it has a partial pressure of 0.04 atm. Therefore, a decrease in the pressure of HI is 0.2 - 0.04 = 0.16. The given reaction is:

                                  2HI (g)           ↔             H₂ (g)       +        I₂ (g)

Intial concentration       0.2 atm                           0                         0

At equilibrium              0.04 atm                       0.16/2                   2.15/2

Therefore,

 

Therefore, Hence, the value of K_p for the given equilibrium is 4.0.

The given reaction is:

N₂ (g) + 3H₂ (g) ↔ 2NH₃ (g)

The given concetration of various species is:

N₂  = 1.57/20 molL^-1

H₂ = 1.92/20 molL^-1

NH₃ =  8.13/20 molL^-1

Now, reaction quotient Qc is:

Qc  = (8.13/20)²  /  (1.57/20) (1.92/20)³

        = 2.4 x 10^-3

Since,Qc ≠  Kc the reaction mixture is not at equilibrium.

Again, Qc >  Kc. Hence, the reaction will proceed in the reverse direction.

The balanced chemical equation corresponding to the given expression can be written as:

4NO(g)  + 6H₂O(g)   ↔   4NH_3(g)  + 5O_2(g)

The given reaction is:

                                  H₂O (g)   +  CO (g)  ↔   H₂ (g)  +  CO₂ (g)

Initial Concentration:     1/10 M       1/10 M           0              0

At equilibrium              1-0.4/10 M  1-0.4/10 M  0.4/10M 0.4/10M

                                = 0.06M       0.06M         0.04M     0.04M

Therefore, the equilibrium constant for the reaction,

 

It is given that equilibrium constant K_c for the reaction

H₂ (g) + I₂ (g) ↔ 2HI (g)  is 54.8.

 

Therefore, at equilibrium, the equilibrium constant K'_cfor the reaction

2HI (g) ↔ H₂ (g) + I₂ (g)    will be  1/54.8

 

HI  =  0.5 molL^-1

Let the concentrations of hydrogen and iodine at equilibrium be x molL^-1 .

Hence, at equilibrium,

The given reaction is:
                                 2ICl (g)  ↔  I₂ (g) + Cl₂ (g)
Initial conc.                0.78 M           0           0
At equilibrium         (0.78 - 2x) M     x M        x M

Now we can write, Kc  =  I₂  x Cl₂   /  (ICl)²  

Let p be the pressure exerted by ethene and hydrogen gas (each) at equilibrium.

Now, according to the reaction,

                  C₂H₆ (g) ↔ C₂H₄ (g) + H₂ (g)

Initial conc.   4.0atm            0           0
At equilibrium 4.0-p             p            p

We can write,

Hence, at equilibrium,

C2H6  - 4  - p  = 4 -.038

= 3.62 atm

(i) Reaction quotient,

(ii) Let the volume of the reaction mixture be V. Also, here we will consider that water is a solvent and is present in excess. The given reaction is:

Therefore, equilibrium constant for the given reaction is:

Let the concentrations of both PCl₃ and Cl₂ at equilibrium be x molL^-1. The given reaction is:

                            PCl₅ (g)         ↔    PCl₃ (g)   +   Cl₂(g)

at equilibrium       0.5x10^-1molL^-1         xmolL^-1     xmolL^-1

it is given that the value of equilibrium constant, Kc is 8.3x10^–3

Now we can write the expression for equilibrium as:

The given reaction is:

                               N₂ (g)     +   3H₂ (g)  ↔  2NH₃ (g)

at a particular time: 3.0molL^-1  2.0 molL^-1     0.5molL^-1

Now, we know that,

Qc =  [NH₃]² / [N₂][H₂]³

      =  (0.5)²  / (3.0)(2.0)³

      = 0.0104

It is given that K_c = 0.061

Since,Qc ≠  Kc, the reaction mixture is not at equilibrium.

Again, Qc <  Kc, the reaction will proceed in the forward direction to reach equilibrium.

Let the amount of bromine and chlorine formed at equilibrium be x. The given reaction is:

                      2BrCl (g) ↔ Br₂ (g) + Cl₂ (g)

Initial Conc.     3.3x10^-3          0            0

at equilibrium   3.3x10^-3 -2x    x            x

Now, we can write,

Kc = [Br₂][Cl₂]  /  [BrCl]²

⇒  (x) x (x) /  (3.3x10^-3 -2x)²  = 32

⇒  x  /  (3.3x10^-3 -2x)  = 5.66

⇒ x  =  18.678x10^-3 - 11.32x

⇒ x  + 11.32x = 18.678x10^-3

⇒ 12.32x = 18.678x10^-3

⇒ x = 1.5 x 10^-3

Therefore, at equilibrium,

[BrCl] =  3.3x10^-3  -  (2 x 1.5 x 10^-3)

         = 3.3x10^-3  -   3.0x10^-3

            = 0.3  x 10^-3

           = 3.0 x 10^-4 mol L^-1

Let the total mass of the gaseous mixture be 100 g.

Mass of CO = 90.55 g

And, mass of CO₂ = (100 - 90.55) = 9.45 g

Now, number of moles of CO, n_co = 90.55/28 = 3.234 mol

Number of moles of CO₂,  n_co2 = 9.45/44 = 0.215 mol

Partial pressure of CO,

For the given reaction,

Δn = 2 - 1 = 1

we know that,

Kp  =  Kc  (RT)^Δn

⇒ 14.19 = Kc  (0.082 x 1127)¹

⇒ Kc = 14.19 / 0.082 x 1127

= 0.154 (arrpox.)

(a) For the given reaction,

ΔG° = ΔG°( Products) - ΔG°( Reactants)

ΔG° = 52.0 -{87.0 + 0} = -35.0 kJ mol^-1

(b) We know that,

ΔG° = RT log K_c

ΔG° = 2.303 RT log K_c

Hence, the equilibrium constant for the given reaction K_c is 1.36 × 106

(a) The number of moles of reaction products will increase. According to Le Chatelier's principle, if pressure is decreased, then the equilibrium shifts in the direction in which the number of moles of gases is more. In the given reaction, the number of moles of gaseous products is more than that of gaseous reactants. Thus, the reaction will proceed in the forward direction. As a result, the number of moles of reaction products will increase.

(b) The number of moles of reaction products will decrease.

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

The reactions given in (i), (iii), (iv), (v), and (vi) will get affected by increasing the pressure.

The reaction given in (iv) will proceed in the forward direction because the number of moles of gaseous reactants is more than that of gaseous products.

The reactions given in (i), (iii), (v), and (vi) will shift in the backward direction because the number of moles of gaseous reactants is less than that of gaseous products.

Given,k_p for the reaction i.e., H₂(g) + Br₂(g) ↔ 2HBr(g)  is 1.6 ×10⁵

Therefore, for the reaction

2HBr(g) ↔ H₂(g) + Br₂(g)

the equilibrium constant will be,

K'_p = 1/K_p

     = 1/1.6 ×10⁵

      = 6.25x10^-6

Now, let p be the pressure of both H₂ and Br₂ at equilibrium.

                    2HBr(g) ↔ H₂(g) + Br₂(g)

Initial Conc.      10             0         0

at equilibrium 10-2p           p         p

Now, we can write,

(a) For the given reaction,

 

(b) (i) According to Le Chatelier's principle, the equilibrium will shift in the backward direction.

(ii) According to Le Chatelier's principle, as the reaction is endothermic, the equilibrium will shift in the forward direction.

(iii) The equilibrium of the reaction is not affected by the presence of a catalyst. A catalyst only increases the rate of a reaction. Thus, equilibrium will be attained quickly.

(a) According to Le Chatelier's principle, on addition of H₂, the equilibrium of the given reaction will shift in the forward direction.

(b) On addition of CH₃OH, the equilibrium will shift in the backward direction.

(c) On removing CO, the equilibrium will shift in the backward direction.

(d) On removing CH₃OH, the equilibrium will shift in the forward direction.

(c) (i) Kc would remain the same because in this case, the temperature remains the same.

(ii) Kc is constant at constant temperature. Thus, in this case, Kc would not change.

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

Let the partial pressure of both carbon dioxide and hydrogen gas be p. The given reaction is:

                        CO (g)   +    H₂O (g)    ↔    CO₂ (g)    +   H₂ (g)

Initial Conc.      4.0 bar          4.0 bar             0                   0

At equilibrium    4.0-p           4.0-p                 p                   p

It is given that K_p  =  10.1

Now,

Hence, at equilibrium, the partial pressure of H₂ will be 3.04 bar.

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

Thus, the reaction given in (c) will have appreciable concentration of reactants and products.

The given reaction is:

3O₂ (g) ↔ 2O₃ (g)

Then, Kc =  [O₃ (g)]²  /  [O₂ (g)]³

It is given that Kc =  2.0 ×10^–50  and O₂ (g) = 1.6 ×10^–2

Then we have,

2.0 ×10^–50  =  [O₃ (g)]²   / [1.6 ×10^–2 ]³

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

⇒ [O₃ (g)]²   = 8.192 x 10^-56

⇒ O₃ (g)     =   2.86x10^-28 M

Hence, the concentration of O₂ (g)  = 2.86x10^-28 M

Let the concentration of methane at equilibrium be x.

                         CO(g)  +  3H₂(g)  ↔  CH₄(g) +  H₂O(g)

At equilibrium    0.3/1 M    0.1/1 M       x             0.02/1 M

It is given that K_c= 3.90.

Therefore,

Kc  =  [CH₄(g)] [ H₂O(g)]  /  [ CO(g) ] [H₂(g)]³

⇒  [x] [0.02]  / (0.3)  (0.1)³  = 3.90

⇒ x  = (3.90) (0.3)  (0.1)³  / [0.02]

⇒ x = 0.00117 / 0.02

      = 0.0585 M

      = 5.85 x 10^-2

Hence, the concentration of CH₄ at equilibrium is 5.85 × 10^-2 M.

A conjugate acid-base pair is a pair that differs only by one proton. The conjugate acid-base for the given species is mentioned in the table below.

Species	Conjugate acid-base
HNO2	NO-2 (base)
CN–	HCN (Acid)
HClO4	ClO-4 (base)
F –	HF (Acid)
OH–	H2O(Acid) / O2-  (base)
CO2–3	HCO–3 (Acid)
S-	HS- (Acid)

Lewis acids are those acids which can accept a pair of electrons. For example, BF₃, H⁺, and NH⁺₄ are Lewis acids.

The table below lists the conjugate bases for the given Bronsted acids.

Bronsted acid	Conjugate base
HF	F-
H2SO4	HSO-4
HCO3	CO2-3

 

The table below lists the conjugate acids for the given Brönsted bases.

Bronsted base	Conjugate acid
NH–2	NH3
NH3	NH+4
HCOO–	HCOOH

 

The table below lists the conjugate acids and conjugate bases for the given species.

Species	Conjugate acid	Conjugate base
H2O	H3O+	OH-
HCO–3	H2CO3	CO2-3
HSO-4	H2SO4	SO2-4
NH3	NH+4	NH-2

 

(a) OH^- is a Lewis base since it can donate its lone pair of electrons.

(b) F^{- i}s a Lewis base since it can donate a pair of electrons.

(c) H⁺ is a Lewis acid since it can accept a pair of electrons.

(d) BCl₃ is a Lewis acid since it can accept a pair of electrons.

Given that:

[H⁺]  =  3.8 × 10^–3

∴ pH value of soft drink

= -log [H⁺] 

=  -log [3.8 × 10^–3] 

= -log [3.8]  - log[10^–3]

= -log [3.8]  +  3

= -0.58 + 3

=  2.42

Given that:

pH = 3.76

It is known that,

pH = -log [H⁺]

⇒  log [H⁺]  = -pH

⇒ [H⁺] = antilog (-pH)

          = antilog(-3.76)

          = 1.74x10^-4 M

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

It is known that,

K_b  =  K_w / K_a

Given

K_a of HF = 6.8 × 10^-4

Hence, K_b of its conjugate base F-  =  K_w / K_a

= 10^-14 / 6.8 × 10^-4

= 1.5 x 10^-11

Given, K_a of HCOOH = 1.8 × 10^-4

Hence, K_b of its conjugate base HCOO^-  = K_w / K_a

= 10^-14 / 1.8 × 10^-4

= 5.6x10^-11

Given, K_a of HCN = 4.8 × 10^-9

Hence, Kb of its conjugate base CN^- = K_w / K_a

= 10^-14 / 4.8 × 10^-9

= 2.8 x 10^-6

Ionization of phenol:

                    C₆H₅OH + H₂O  ↔  C₆H₅O^- +  H₃O⁺

Initial Conc.          0.05                    0              0

At equilibrium    0.05-x                    x               x

Ka   =  [C₆H₅O^- x  H₃O⁺]   /  C₆H₅OH

Ka   =  x x  / 0.05-x

Ka   =  x²  / 0.05-x

As the value of the ionization constant is very less, x will be very small. Thus,

we can ignore x in the denominator

Now, let α be the degree of ionization of phenol in the presence of 0.01 M C₆H₅ONa.

 

(i) To calculate the concentration of HS^- ion:

Case I (in the absence of HCl):

Let the concentration of HS^- be x M.

             H₂S  ↔  H⁺  +  HS^-

C_i          0.1         0         0

C_f       0.1-x         x         x

Then K_a1  = [H⁺ ] [ HS^-]  /  H₂S

9.1 × 10^–8  = xx / 0.1-x

(9.1 × 10^–8)  (0.1-x) = x²

Taking 0.1 - x M ; 0.1M, we have

(9.1 × 10^–8)  (0.1) = x²

9.1 x 10^-9  = x²

Case II (in the presence of HCl):

In the presence of 0.1 M of HCl, let [HS^-] be y M.

Then,       H₂S  ↔  H⁺  +  HS^-

C_i          0.1          0         0

C_f        0.1-y         y         y

Also,   HCI    ↔   H⁺   +   CI^-

                         0.1       0.1

Now,  K_a1  =  [H⁺ ] [ HS^-]  /  H₂S

K_a1   =   [y] [0.1+y] / [0.1-y]

9.1 × 10^–8  =  y x 0.1  / 0.1              (∵ 0.1-y; 0.1M) (and 0.1+y; 0.1M)

9.1 × 10^–8  =  y

⇒ [ HS^-]  = 9.1 × 10^–8

 

(ii) To calculate the concentration of [S^2-]:

Case I (in the absence of 0.1 M HCl):

HS^-    ↔   H⁺   + S^2-

HS^-    =  9.54 x 10^-5 M  (From first ionization, case I)

Let S^2- be X.

Also, [H⁺]  =  9.54 x 10^-5 M  (From first ionization, case I)

K_a2  =  (9.54 x 10^-5) (X)  / (9.54 x 10^-5)

       1.2x10-13  =  X  = S^2-

Case II (in the presence of 0.1 M HCl):

Again, let the concentration of HS^- be X' M.

 

1) CH₃COOH    ↔    CH₃COO^-  + H⁺     K_a = 1.74x10^-5

2) H₂O + H₂O     ↔  H₃O⁺   +   OH^-        K_w  = 1.0x10^-14

Since K_a > K_w

         CH₃COOH  + H₂O  ↔     CH₃COO^-  +    H₃O⁺

Ci  =  0.05                                     0                   0

       0.05- 0.05α                     0.05α             0.05α  

k_a =  0.05α x   0.05α  /   0.05- 0.05α  

= 0.05α x   0.05α  /   0.05 (1-α)

= 0.05α²  /  (1-α)

⇒ 1.74x10^-5  =  0.05α²  /  (1-α)

⇒ 1.74x10^-5   -  1.74x10^-5 α  =  0.05α² 

⇒ 0.05α²  +    1.74x10^-5 α  -  1.74x10^-5 =0

D =  b² - 4ac

   = (1.74x10^-5 )² - 4 (0.05) (1.74x10^-5)

   = 3.02x10^-25  + 0.348x10^-5

Method 2:

Degree of Dissociation,

CH₃COOH   ↔   CH₃COO^- + H⁺

Thus, concentration of CH₃COO^- = c.α 

= 0.05x1.86x10^-2

= 0.93x10^-2

=.00093M

Since [oAc-] = [H⁺]

[H⁺] = .00093 = 0.093x10-2

pH = -log[H⁺]

     =  -log (0.93x10^-2)

∴ pH = 3.03

Hence, the concentration of acetate ion in the solution is 0.00093 M and its Ph is 3.03.

 

Let the organic acid be HA.

⇒ HA ↔ H⁺  + A^-

Concentration of HA = 0.01 M

pH = 4.15

-log[H⁺] = 4.15

[H⁺]  = 7.08x10^-5

Now, K_a = [H⁺] [A^-]  / [HA] 

[H⁺]  = [A^-]  = 7.08x10^-5

[HA]  = 0.01

Then,

K_a =  7.08x10^-5 x 7.08x10^-5  /  0.01

K_a =  5.01x10^-7

pK_a = -log K_a

        =  -log(5.01x10^-7)

pK_a = 6.3001

 

(i) 0.003MHCl:

H₂O   +   HCl  ↔  H3O⁺   +   Cl^-

Since HCl is completely ionized,

[H₃O⁺] = [ HCl]

⇒ [H₃O⁺] = = 0.003

Now

pH = -log [H₃O⁺]  = -log (0.003)

= 2.52

Hence, the pH of the solution is 2.52.

 

(b) 0.005 M NaOH

NaOH_(aq)  ↔ Na⁺_(aq) + HO^-_(aq)

[NaOH] =  [ HO^-]

⇒  [ HO^-]  = 0.05

pOH =  -log[ HO^-]  = -log (0.05)

= 2.30

∴ pH  = 14 - 2.30 = 11.70

Hence, the pH of the solution is 11.70.

 

(c) 0.002 M HBr

HBr  + H₂O  ↔  H₃O⁺ + Br^-

[HBr] = [H₃O⁺]

⇒  [H₃O⁺] = 0.002

∴ pH = -log [H₃O⁺]  = -log (0.002)

= 2.69

Hence, the pH of the solution is 2.69.

 

(d) 0.002 M KOH

KOH_(aq)  ↔  K⁺_(aq)  + OH^-_(aq)

[OH^-]  =  [KOH]

⇒ [OH^-] = 0.002

Now pOH  = -log[OH^-]  = -log (0.002)

= 2.69

∴ pH = 14-2.69  =  11.31

Hence, the pH of the solution is 11.31.

For 2g of TlOH dissolved in water to give 2 L of solution:

[TIOH(aq)] = 2/2 g/L

= 2/2 x 1/221 M

= 1/221 M

TIOH_(aq)  →  TI⁺_(aq)  +  OH^-_(aq)

OH^-_(aq)  =  TIOH_(aq) = 1/221M

Kw  =  [H⁺] [OH^-]

10^-14 = [H⁺] [1/221]

[H⁺] = 221x10^-14

⇒ pH  =  -log [H⁺]  = -log ( 221x10^-14)

= 11.65

 

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

Ca(OH)₂  →   Ca²⁺ + 2OH^-

[Ca(OH)₂]  = 0.3x1000/500 = 0.6M

OH^-_(aq) = 2 x [Ca(OH)_2(aq)] = 2 x 0.6 = 1.2M

[H+]  = Kw  /  OH^-_(aq)

= 10^-14/1.2 M

= 0.833 x 10^-14

pH  =  -log(0.833 x 10^-14)

= -log(8.33 x 10^-13)

= (-0.902 + 13)

= 12.098

 

(c) For 0.3 g of NaOH dissolved in water to give 200 mL of solution:

NaOH →  Na ⁺_(aq)  + OH^-_(aq)

[NaOH] = 0.3 x 1000/200 = 1.5M

[OH^-_(aq)] = 1.5M

Then [H+]  =  10^-14 / 1.5

= 6.66 x 10^-13

pH  =  -log ( 6.66 x 10^-13)

= 12.18

 

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

13.6 x 1 mL       =   M₂ x 1000 mL

(Before dilution)   (after dilution)

13.6 x 10^-3 =  M₂ x 1L

M₂  =  1.36 x 10^-2

[H⁺] = 1.36 × 10^-2

pH = - log (1.36 × 10^-2)

= (- 0.1335 + 2)

= 1.866 = 1.87

Degree of ionization, α = 0.132

Concentration, c = 0.1 M

Thus, the concentration of H₃O⁺ = c.α

= 0.1 × 0.132

= 0.0132

pH = -log [H⁺]

      = -log (0.0132)

     = 1.879 : 1.88

Now,

K_a  =  Cα²

      =  0.1 x (0.132)²

 K_a = 0.0017

pK_α  = 2.75

c = 0.005

pH = 9.95

pOH = 4.05

pH = -log (4.05)

4.05 =  - log [OH^-]

[OH^-] = 8.91x 10^-5

cα = 8.91x 10^-5

α = 8.91x 10^-5  /  5 x 10^-3  =  1.782  x 10^-2

Thus , K_b  =  cα²

= 0.005 x (1.782)² x 10^-4

= 0.005 x 3.1755 x 10^-4

= 0.0158 x 10^-4

K_b  =  1.58 x 10^-6

PK_b  =  -logK_b 

= -log (1.58 x 10^-6)

= 5.80

 

K_b  =   4.27 x 10^-10

c  =  0.001M

pH  = ?

α  =  ?

K_b  =  cα²

4.27 x 10^-10  = 0.001 x α²

4270 x 10^-10 = α²

α = 65.34 x 10^-4

Then (anion) = cα  = 0.001 x 65.34 x 10^-4

= 0.65 x 10^-5

pOH = -log ( 0.65 x 10^-5)

= 6.187

pH  =  7.813

Now

K_a x K_b = Kw

 ∴ 4.27 x 10^-10 x Ka = Kw

K_a  =  10^-14 / 4.27 x 10^-10

= 2.34 x 10^-5

Thus, the ionization constant of the conjugate acid of aniline is 2.34 × 10^-5. 

c =  0.05 M

pK_a  =  4.74

pK_a  = -log (K_a)

K_a  =  1.82 x 10^-5

When HCl is added to the solution, the concentration of H⁺ ions will increase. Therefore, the equilibrium will shift in the backward direction i.e., dissociation of acetic acid will decrease.

 

Case I: When 0.01 M HCl is taken.

Let x be the amount of acetic acid dissociated after the addition of HCl.

                               CH₃COOH     ↔     H⁺    +   CH₃COO^-

Initial Conc.             0.05M                   0                0

After dissociation   0.05-x            0.01+x              x

As the dissociation of a very small amount of acetic acid will take place, the values i.e., 0.05 - x and 0.01 + x can be taken as 0.05 and 0.01 respectively.

K_a  =  [CH₃COO^-] [ H⁺]  / [CH₃COOH]

∴ K_a  = (0.01)  (x) / (0.05)

x  =   1.82 x 10^-5  x(multiply) 0.05 /  0.01

x  =  1.82 x 10^-3  x(multiply) 0.05

Now

α =  Amount of acid dissociated /  amount of acid taken

= 1.82 x 10^-3  x 0.05  / 0.05

= 1.82 x 10^-3

 

Case II: When 0.1 M HCl is taken.

Let the amount of acetic acid dissociated in this case be X. As we have done in the first case, the concentrations of various species involved in the reaction are:

[CH₃COOH]  = 0.05 - X  : 0.05M

[CH₃COO^-] =  X

[ H⁺]   =   0.1+X ; 0.1M

K_a  =  [CH₃COO^-] [ H⁺]  / [CH₃COOH]

∴ K_a  = (0.1)  (X) / (0.05)

x  =   1.82 x 10^-5  x(multiply) 0.05 /  0.1

x  =  1.82 x 10^-4  x(multiply) 0.05

Now

α =  Amount of acid dissociated /  amount of acid taken

= 1.82 x 10^-4  x 0.05  / 0.05

= 1.82 x 10^-4

Kb  =  5.4x10-4

c =  0.02M

Now, if 0.1 M of NaOH is added to the solution, then NaOH (being a strong base) undergoes complete ionization.

NaOH_(aq)     ↔   Na⁺_(aq)   +  OH^- _(aq)

                          0.1M             0.1M

and

(CH₃)₂ NH  + H₂O   ↔   (CH₃)₂ NH⁺₂  +  O^-H

0.02-x                                      x                 x

Then (CH₃)₂ NH⁺₂   = x

[OH^-]  =  x + 0.1 ; 0.1

⇒ K_b = [(CH₃)₂ NH⁺₂ ] [OH^-]   /   [CH₃)₂ NH]

5.4x10-4  =  x x 0.1 / 0.02

 x  =  0.0054

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 x 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 = - log [H⁺]

∴ [H⁺] = 4.17 x 10^-8 M

 

(d) Human saliva, 6.4:

pH = 6.4

6.4 = - log [H⁺]

[H⁺] = 3.98 x 10^-7

The hydrogen ion concentration in the given substances can be calculated by using the given relation:

pH = -log [H⁺]

(i) pH of milk = 6.8

Since, pH = -log [H⁺]

6.8 = -log [H⁺]

log [H⁺] = -6.8

[H⁺] = anitlog(-6.8)

= 1.5x10-7M

 

(ii) pH of black coffee = 5.0

Since, pH = -log [H⁺]

5.0 = -log [H⁺]

log [H⁺] = -5.0

[H⁺] = anitlog(-5.0)

= 10^-5M

 

(iii) pH of tomato juice = 4.2

Since, pH =-log [H⁺]

4.2 =-log [H⁺]

log [H⁺] = -4.2

[H⁺] = anitlog(-4.2)

= 6.31x10^-5M

 

(iv) pH of lemon juice = 2.2

Since, pH = -log [H⁺]

2.2 = -log [H⁺]

log [H⁺] = -2.2

[H⁺] = anitlog(-2.2)

=6.31x10^-3M

 

(v) pH of egg white = 7.8

Since, pH = -log [H⁺]

7.8 = -log[H⁺]

log[H⁺] = -7.8

[H⁺] = anitlog(-7.8)

= 1.58x10^-8M

[KOH_(aq)] = 0.561 / (1/5)g/L

= 2.805 g/L

= 2.805 x 1/56.11 M

= 0.05M

KOH_(aq)  → K⁺_(aq)  +  OH^-_(aq)

[OH^-]  = 0.05M =  [K⁺]

[H^-] [H⁺]  =  K_w

[H⁺] K_w / [OH^-]

= 10^-14 / 0.05 = 2x10^-13 M

∴ pH = 12.70

 

Solubility of Sr(OH)₂ = 19.23 g/L

Then, concentration of Sr(OH)₂

= 19.23 / 121.63 M

= 0.1581 M

Sr(OH)_2(aq) →  Sr²⁺_(aq) + 2 (OH^-)_(aq)

∴Sr²⁺  = 0.1581M

[OH^-] =  2 x 0.1581M = 0.3126 M

Now

Kw  =  [OH^-] [H⁺]

10^-14 / 0.3126  = [H⁺]

⇒ [H⁺] =  3.2 x 10^-14

∴ pH = 13.495 = 13.50

Let the degree of ionization of propanoic acid be α.

Then, representing propionic acid as HA, we have:

HA     +    H₂O      ↔      H₃O⁺     +  A^-

(0.05-0.0α) ≈ 0.05       0.05α     0.05α

K_a  =  [H₃O⁺]  [A^-]  / [HA]

=  (0.05α)(0.05α)  /  0.05

= 0.05 α²

Then, [H₃O⁺]  = 0.05α = 0.05 x 1.63 x 10^-2  =  Kb . 15 x 10^-4 M

∴ pH  =  3.09

In the presence of 0.1M of HCl, let α' be the degree of ionization.

Then, [H₃O⁺]  = 0.01

[A^-]  =  0.05α'

[HA]  =  0.05

Ka  =  0.01 x 0.05α'  / 0.05

⇒ 1.32 x 10^-5 =  0.1 α'

α' =  1.32 x 10^-3

c = 0.1 M

pH = 2.34

-log [H⁺] =  pH

-log [H⁺] = 2.34

[H⁺]  = 4.5 x 10^-3

also

[H⁺]  = cα

4.5 x 10^-3  =   0.1  α

α =  4.5 x 10^-3  /  0.1

α  = 45 x 10^-3 

Then,

K_a  =  cα²

= 0.1 (45 x 10^-3)²

= 202.5 x  10^-6

= 2.02 x 10^-4

 

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

NO^-₂  +  H₂O  ↔   HNO₂  +  OH^-

K_h  =  [ HNO₂ ] [ OH^-]  / [NO^-₂]

⇒  K_w / k_a  =  10^-14 /  4.5 x 10^-4  = 0.22 x 10^-10

Now, If x moles of the salt undergo hydrolysis, then the concentration of various species present in the solution will be:

[NO^-₂ ]  = 0.04 - x ; 0.04

[ HNO₂ ]  =  x

[ OH^-]  =  x

K_h  =  x² / 0.04 =  0.22 x 10^-10

x²  =  0.0088 x 10^-10

x  =  0.093 x 10^-5

∴ [ OH^-]  =  0.093 x 10^-5M

[H₃O⁺]  =  10^-14 /  0.093 x 10^-5  = 10.75 x 10^-9M

⇒  pH  = -log(10.75 x 10^-9)

= 7.96

Therefore, degree of hydrolysis

= x / 0.04  =  (0.093 x 10^-5 ) / 0.04 =  2.325 x 10^-5

 

Given,  pH  =  3.44

We know that

pH  =  -log [H⁺]

∴ [H⁺]  =  3.63x10^-4

Then K_h  = ( 3.63x10^-4 )² / 0.02         (∵ Concentration is 0.02M)

⇒ K_h  =  6.6 x 10^-6

Now , K_h  =  K_w / K_a

⇒  K_a  =  K_w / K_h

= 10^-14 / 6.6 x 10^-6

= 1.51 x 10^-9

 

Ionic product,

K_w   =  [H⁺] [ OH^-] 

Let [H⁺]   =  x.

Since [H⁺]  =  [ OH^-] , K_w  =  x²

⇒ K_w at 310K is 2.7x 10^-14

∴   2.7x 10^-14 = x²

⇒ x  =  1.64 x 10^-7

⇒ [H⁺]   = 1.64 x 10^-7

⇒ pH  = -log [H⁺] 

= -log (1.64 x 10^-7)

= 6.78

Hence, the pH of neutral water is 6.78.

CaSO_4(s)   ↔   Ca ²⁺ _(aq) + SO^2-_4(aq)

K_sp   =  [ Ca ²⁺ ] [ SO^2- ]

Let the solubility of CaSO₄ be s.

Then, K_sp   = s²

9.1 x 10^-6  = s²

s  =  3.02 x 10^-3 mol/L

Molecular mass of CaSO₄ = 136 g/mol

Solubility of  CaSO₄ in gram/L = 3.02 × 10^-3 × 136 = 0.41 g/L

This means that we need 1L of water to dissolve 0.41g of CaSO₄

Therefore, to dissolve 1g of CaSO₄ we require  = 1/0.41 L = 2.44 Lof water.