Class 11 Chemistry

Equilibrium

NCERT Solution

Part 4

Question 46: The ionization constant of acetic acid is 1.74 × 10-5. Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of acetate ion in the solution and its pH.

Answer: `K_a=([CH_3CO\OH^-][H^+])/([CH_3CO\OH])`

`=([H^+]^2)/([CH_3CO\OH])`

Or, `[H^+]=sqrt(K_a[CH_3CO\OH])`

`=sqrt((1.74xx10^(-5)(5xx10^(-2))=9.33xx10^(-4)` M

[CH3COO-] = [H+] = 9.33 × 10-4 M

pH = - log (9.33 × 10-4)

= 4 - 0.9699 = 4 – 0.97 = 3.03

Question 47: It has been found that the pH of a 0.01 M solution of an organic acid is 4.15. Calculate the concentration of the anion, the ionization constant of acid and it pKa.

Answer: HA ⇌ H+

pH = - log [H+]

Or, log [H+] = - 4.16 = 5.85

Or, [H+] = 7.08 × 10-5 M

[A-] = [H+] = 7.08 × 10-5 M

`K_a=([H^+][A^-])/([HA])`

`=((7.08xx10^(-5))^2)/(10^(-2))=5.0xx10^(-7)`

pKa = - log Ka

= - log (5.0 × 10-7 = 7 – 0.699 = 6.301

Quesiton 48: Assuming complete dissociation, calculate the pH of the following solutions: (a) 0.003 M HCl (b) 0.005 M NaOH (c) 0.002 M Hbr (d) 0.002 M KOH

Answer: HCl + aq → H+ + Cl-

So, [H+ = [HCl] = 3 × 10-3 M

pH = - log (3 × 10-3) = 2.52

NaOH + aq → Na+ + OH-

So, [OH-] = 5 × 10-3 M

[H+] = `(10^(-14))/(5xx10^(-3))=2xx10^(-12)` M

pH = - log (2 × 10-12 = 11.70

HBr + aq → H+ + Br-

So, [H+] = 2 × 10-3 M

pH = - log (2 × 10-3) = 2.70

KOH + aq = K+ + OH-

So, [OH-] 2 × 10-3 M

[H+] = `(10^(-14))/(2xx10^(-3))=5xx10^(-12)`

pH = - log ( 5 × 10-12 = 11.30

Question 49: Calculate the pH of the following solutions:

(a) 2 g of TlOH dissolved in water to give 2 L of solution.

Answer: Molar conc of TlOH

`=2/(204+16+1)xx1/2=4.52xx10^(-3)` M

[OH- = [TlOH] = 4.52 × 10-3 M

[H+ = `(10^(-14))/(4.52xx10^(-3))=2.21xx10^(-12)` M

pH = - log (2.21 × 10-12) = 12 – 0.3424 = 11.66

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

Answer: Molar conc of Ca(OH)2

`=(0.3)/(40+34)xx1/(0.5)=8.11xx10^(-3)` M

Ca(OH)2 → Ca2+ + 2OH-

[OH-] = 2[Ca(OH)2]

= 2 × 8.11 × 10-3 = 16.22 × 10-3 M

pOH = - log (16.22 × 10-3) = 3 – 1.2101 = 1.79

pH = 14 – 1.79 = 12.21

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

Answer: Molar conc of NaOH

`=(0.3)/(40)xx1/(0.2)=3.75xx10^(-2)` M

[OH- = 3.75 × 10-2 M

pOH = - log (3.75 × 10-2) = 2 – 0.0574 = 1.43

pH = 14 – 1.43 = 12.57

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

Answer: M1V1 = M2V2

So, 13.6 M × 1 mL = M2 × 1000 mL

Or, M2 = 1.36 × 10-2 M

[H+] = [HCl]= 1/36 × 10-2 M

pH = - log (1.36 × 10-2)

= 2 – 0.1335 = 1.87

Question 50: The degree of ionization of a 0.1 M bromoacetic acid solution is 0.132. Calculate the pH of the solution and the pKa of bromoacetic acid.

Answer:

CH2(Br)COOHCH2(Br)COO-H+

Initial concC00
Equil concC – Cα

`K_a=(Cα^2)/(C(1-α))`

`Cα^2=0.1xx0.132^2=1.74xx10^(-3)`

pKa = - log (1.74 × 10-3 = 3 – 0.2405 = 2.76

[H+] = Cα = 0.1 × 0.132 = 1.32 × 10-2 M

pH = - log (1.32 × 10-2) = 2 – 0.1206 = 1.88

Question 51: The pH of 0.005 M codeine (C18H21NO3) solution is 9.95. Calculate its ionization constant and pKb.

Answer: Cod + H2 ⇌ Cod H+ + OH-

pH = 9.95

So, pOH = 14 – 9.95 = 4.05

It means that – log[OH-1] = 4.05

Or, log[OH-] = - 4.05 =5.95

Or [OH-] = 8.913 × 10-5 M

Now, `K_b=([Co\d H^+][OH^-])/([Co\d])=([OH^-]^2)/([Co\d])`

`=((8.91xx10^(-5))^2)/(5xx1o^(-3))=1.588xx10^(-6)`

pKb = - log (1.588 × 10-6) = 6 – 0.1987 = 5.8

Question 52: What is the pH of 0.001 M aniline solution? The ionization constant of aniline is 4.27 × 10-10. Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline.

Answer: Given, Kb = 4.27 × 10-10 and C = 0.0-01 M

We know: `K_a=Cα^2`

Or, 4.27 × 10-10 = 0.001 × α2

Or, α2 = 4270 × 10-10

Or, α = 6.53 × 10-4

So, [Anion] = Cα

= 0.01 × 6.53 × 10-4 = 0.65 × 10-5

So, pOH = - log(0.65 × 10-5) = 6.187

So, pH = 14 – 6.187 = 7.813

Now, Ka × Kb = Kw

Or, `K_a=(10^(-14))/(4.27xx10^(-10))=2.34xx10^(-10)`

This is the ionization constant of conjugate acid of aniline

Question 53: Calculate the degree of ionization of 0.05 M acetic acid if its pKa value is 4.74. How is the degree of dissociation affected when its solution also contains (a) 0.01 M (b) 0.1 M in HCl?

Answer: Given, C = 0.05, pKa = 4.74

pKa = - log (Ka)

Or Ka = 1.82 × 10-5

`K_a=Cα^2`

Or, `α=sqrt((K_a)/c)`

`=sqrt((1.82xx10^(-5))/(5xx10^(-2)))=1.908xx10^(-2)`

When HCl is added to the solution, concentration of H+ increases. So, the equilibrium shifts in backward direction.

Case 1: When 0.01 M HCl is taken

Let assume that amount of acetic acid dissociated after addition of HCl is x then:

CH3COOHH+CH3COO-
Initial conc0.05 M00
Dissociation conc0.05 – x ∼ 0.050.01 + x ∼ 0.01x

`K_a=([CH_3CO\O^-][H^+])/([CH_3CO\OH])`

Or, `K_a=(0.01x)/(0.05)`

Or, `x=(1.82xx10^(-5)xx0.05)/(0.01)`

= 1.82 × 10-3 × 0.05 M

Now, α = (Amount of acid dissociated) ÷ (Amount of acid taken)

`=(1.82xx10^(-5)xx0.05)/(0.05)=1.82xx10^(-5)`

Case II: When 0.1 M HCl is taken

`K_a=(0.1x)/(0.05)`

Or, `x=1.82xx10^(-5)xx(0.05)/(0.1)`

Or, `x=1.82xx10^(-4)xx0.05` M

So, `α=(1.82xx10^(-4)xx0.05)/(0.05)=1.82xx10^(-4)`

Question 54: The ionization constant of dimethylamine is 5.4 × 10-4. Calculate its degree of ionization in its 0.02 M solution. What percentage of dimethylamine is ionized if the solution is also 0.1 M in NaOH?

Answer: Given, Kb = 5.4 × 10-4

C = 0.02 M

α `=sqrt((K_b)/c)`

`=sqrt((5.410^(-4))/(0.02))=0.1643`

If 0.1 M of NaOH is added to the solution it undergoes complete ionization by virtue of being a strong base.

NaOHNa+OH-

0.1 M0.1 M

And,

(CH3)2NHH2(CH3)2NH2+OH-

0.02 – x ∼ 0.02 Mxx + 0.01 ∼ 0.1 M

Or, `K_b=([(CH_3)_2NH_2^+][OH^-])/([(CH_3)_2NH])`

Or, `5.4xx10^(-4)=(0.1x)/(0.02)`

Or, `x = 0.0054`

This means that in the presence of 0.1 M NaOH, 0.54% of dimethylamine is dissociated.

Question 55: Calculate the hydrogen ion concentration in the following biological fluids whose pH are given below.

(a) Human muscle fluid: 6.83

Answer: pH = 6.83 = - log [H+

Or, [H+ = 1.48 × 10-7 M

(b) Human stomach fluid: 1.2

Answer: pH = 1.2 = - log [H+]

Or, [H+ = 6.3 × 10-2 M

(c) Human blood: 7.38

Answer: pH = 7.38 = - log [H+]

Or, [H+] = 4.17 × 10-8 M

(d) Human saliva: 6.4

Answer: pH = 6.4 = - log {H+]

Or, [H+] = 3.98 × 10-7

Question 56: The pH of milk, black coffee, tomato juice, lemon juice and egg white are 6.8, 5.0, 4.2, 2.2 and 7.8 respectively. Calculate corresponding hydrogen ion concentration in each.

Answer: pH of milk = 6.8 = - log [H+]

Or, [H+ = 1.5 × 10-7 M

pH of black coffee = 5.0 = - log [H+]

Or, [H+ = 10-5 M

pH of tomato juice = 4.2 = - log [H+]

Or, [H+ = 6.31 × 10-5 M

pH of lemon juice = 2.2 = - log [H+]

Or, [H+] = 6.31 × 10-3 M

pH of egg white = 7.8 = - log [H+]

Or, [H+] = 1.5 × 10-8 M

Question 57: If 0.561 g of KOH is dissolved in water to give 200 mL of solution at 298 K, calculate the concentrations of potassium, hydrogen and hydroxyl ions. What is its pH?

Answer: [KOHaq] `=(0.561)/(1/5)=2.805` g L-1

`=2.805xx1/(56.11)=0.05` M

KOHaq → Kaq+ + OHaq-

[OH-] = [K+] = 0.05

[H+] `=(K_W)/([OH^+])`

`=(10^(-14))/(0.05)=2xx10^(-11)` M

So, pH = 12.70

Question 58: The solubility of Sr(OH)2 at 298 K is 19.23 g/L of solution. Calculate the concentrations of strontium and hydroxyl ions and the pH of the solution.

Answer: Solubility of Sr(OH)2 = 19.23 g L-1

So, concentration of Sr(OH)2 `=(19.23)/(121.63)=1.581` M

Ionization reaction can be given by following equation:

Sr(OH)2 (aq) → ⇌ Sr2+ (aq) + 2(OH-) (aq)

This means: [Sr2+] = 0.1581 M

[OH-] = 2 × 0.1581 = 0.3126 M

Now, KN = [OH-]{H+]

So, [H+] `=(10^(-4))/(0.3126)=3.2xx10^(-14)`

So, pH = 13.50

Question 59: The ionization constant of propanoic acid is 1.32 × 10-5. Calculate the degree of ionization of the acid in its 0.05 M solution and also its pH. What will be its degree of ionization if the solution is 0.01 M in HCl also?

Answer: Let us assume that degree of ionization of propanoic acid is α

HAH2H3O+A -
0.05 – α ∼ 0.050.05α0.05α

`K_a=([H_3O^+][A^-])/([HA])`

`=((0.5α)^2)/(0.05)=0.05&alpha^2`

Or, `α=sqrt((K_a)/(0.05)=1.63xx10^(-2)`

So, [H3O+] = 0.05α

= 0.05 × 1.63 × 10-2 = Kb × 1.5 × 10-4 M

So, pH = 3.09

Let us assume that degree of ionization in presence of 1 M HCl is α'

Then, [H3O+] = 0.01

[A- = 0.05α'

[HA] = 0.05

So, `K_a=(0.01xx0.5α')/(0.05)`

= 1.32 × 10-5 = 0.01α'

Or, α' = 1.3 × 10-5

Question 60: The pH of 0.1 M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.

Answer: Given, C = 0.1 M,

pH = 2.34 = - log [H+]

Or, [H+ = 4.5 × 10-3

Now, [H+ = Cα

Or, 4.5 × 10-3 = 0.1α

Or, `α=(4.5xx10^(-3))/(0.1)`

= 4.5 × 10-2 = 0.045

Now, Kα = Cα2

= 0.1 × (45 × 10-3)2

= 2.02 × 10-4