Ionic Equilibrium Test

Result

Ionic Equilibrium Test - 7

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Weekly Quiz Competition

Question 1
1 / -0

A weak base B(OH)₂ has dissociation constant 10^-8 . The equilibrium constant for its reaction with excess of strong acid will be

A
10²⁰

B
10^-6

C
10^-20

D
10^-16

Solution

B(OH)₂ + 2H⁺ ⇌ B²⁺ + 2H₂O

K = 1/K_h

K_h for hydrolysis of B²⁺
Question 2
1 / -0

A weak monobasic acid (0.1 M) has a pH of 3 at a particular temperature (25°C). When this acid is neutralized by strong base (NaOH), what is the value of equilibrium constant at equivalent point at 25°C?

A
10^-5

B
10^-4

C
10^-7

D
10^-14

Solution

Let, the weak acid be HA

At equivalence point,
Question 3
1 / -0

The concentration of [OH^– ] left in a solution after mixing 50 ml of 0.2 M ACl₂ with 50 ml 0.4 m NaOH. [K_sp of A[OH]₂ = 1.2 × 10^–11]

A
2.8 × 10^–4 m/L

B
1.4 × 10^–4 m/L

C
2 × 10^–2 M

D
4 × 10^–2 M

Solution

[OH^–] = 2.8 × 10^–4
Question 4
1 / -0

A weak acid HA after treatment with 12 ml of 0.1 M strong base BOH has a pH of 5. At the end point, the volume of same base required is 20 ml, then K_a of acid is

A
1.5 × 10^–5

B
8.12 × 10^–6

C
1.8 × 10^–6

D
8.2 × 10^–5

Solution

Eq. of acid = Eq. of base

= 20 × 0.1 = 2

HA + BOH → BA + H₂O

⇒ K_a = antilog 4.8 = 1.5 × 10^–5
Question 5
1 / -0

The H⁺ ion concentration in 0.001 M CH₃COOH (K_a = 1.8 ×10^-5 ) is 1.34 ×10^-4 g ion/L. The H⁺ ion concentration if 0.164 g of CH₃COONa is added to a litre of 0.001 M CH₃COOH will be

A
9 ×10^-6

B
18 ×10^-6

C
4.5 ×10^-6

D
5 ×10^-6

Solution
Question 6
1 / -0

The pH at which an acid indicator with K_a = 1 × 10⁻⁵ changes colours when indicator concentration is 1 × 10⁻⁵ M is

A
4

B
5

C
6

D
3

Solution

For acid indicator HI_n,

when [I_n⁻ ] = [HI_n] indicator changes colour.

∴ KI_n = [H⁺] = 1 × 10⁻⁵

∴ pH = 5
Question 7
1 / -0

If K_b for fluoride ion at 25°C is 1.48 × 10⁻¹¹, the ionization constant of hydrofluoric acid in water at this temperature is

A
1.7 × 10⁻⁵

B
3.52× 10⁻³

C
6.75 × 10⁻⁴

D
5.38 × 10⁻²

Solution

K_a × K_w = K_ω ⇒ K_a

= 6.75 × 10⁻⁴
Question 8
1 / -0

If 50 ml of 0.2 M KOH is added to 40 ml of 0.5 M HCOOH, the pH of the resulting solution is (K_a = 1.8 × 10⁻⁴ )

A
3.75

B
5.6

C
7.5

D
3.4

Solution

Meq. of KOH = 50 × 0.2 = 10

Meq. of HCOOH = 40 × 0.2 = 20

∴ Meq. of KOH left = 10 Meq. of HCOOK formed = 10
Question 9
1 / -0

If the degree of ionization of water be 1.8 × 10^-9 at 298 K. Its ionization constant will be

A
1.8 × 10^-16

B
1 × 10^-14

C
1 × 10^-16

D
1.67 × 10^-14

Solution
Question 10
1 / -0

When a solution of benzoic acid was titrated with NaOH, the pH of the solution, when half of the acid was neutralized, will be 4.2. Dissociation constant of the acid is

A
6.31 × 10^-5

B
3.2 × 10^–5

C
8.7 × 10^–8

D
6.42 × 10^–4

Solution