Ionic Equlibrium Test

Result

Ionic Equlibrium Test - 8

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Weekly Quiz Competition

Question 1
4 / -1

The compound whose 0.1 M solution is basic is?

A
Ammonium acetate

B
Ammonium chloride

C
Ammonium sulphate

D
Sodium acetate

Solution

Salt of weak acid & strong Base
CH₃COONa → CH₃COO^– + Na⁺ CH₃COO^–+ H₂O CH₃COOH + OH^–
Basic Solution
Question 2
4 / -1

Which of the following solution will have pH close to 1.0 ?

A
100 ml of M/100 HCl + 100 ml of M/10 NaOH

B
55 ml of M/10 HCl + 45 ml of M/10 NaOH

C
10 ml of M/10 HCl + 90 ml of M/10 NaOH

D
75 ml of M/5 HCl + 25 ml of M/5 NaOH

Solution

75mL of M/5HCl + 25mL of M/5 NaOH
Meq of HCl = 15 (∴ Mol. wt. and Eq. wt. of HCl or NaOH are same)
Meq of NaOH = 5 Hence, in solution 10meq of HCl remains in 100mL solution.
So, concentration of HCl in mixture = 10/100M = M/10
In M/10 HCl, [H⁺] = 10^-1 M
pH = -log[H⁺] = -log10^-1 = 1
Question 3
4 / -1

The ≈ pH of the neutralisation point of 0.1 N ammonium hydroxide with 0.1 N HCl is

A
1

B
6

C
7

D
9

Solution

Salt formed : NH₄Cl = 0.1 N
Solution will be slightly acidic due to Hydrolysis
Question 4
4 / -1

If equilibrium constant of
CH₃COOH + H₂O CH₃COO^- + H₃O⁺
Is 1.8 × 10^-5, equilibrium constant for
CH₃COOH + OH^- CH₃COO^- + H₂O is

A
1.8 × 10^-9

B
1.8 × 10⁹

C
5.55 × 10^-9

D
5.55 × 10¹⁰

Solution

Ka = 1.8 × 10^–5
CH₃COO^– + H₂O CH₃COOH + OH^-
Given reaction is reverse of above
Question 5
4 / -1

If 40 ml of 0.2 M KOH is added to 160 ml of 0.1 M HCOOH [K_a = 2 × 10^-4]. The pOH of the resulting solution is

A
3.4

B
3.7

C
7

D
10.3

Solution

m. equivalent of KOH = 8
m. equivalent of HCOOH = 16
Remaining m. eq. (HCOOH) = 8
Formed m. eq. (HCOOK) = 8
⇒ Acidic Buffer
pH = pKa = 4 – log2
= 3.7
pOH = 10.3
Question 6
4 / -1

A solution with pH 2.0 is more acidic than the one with pH 6.0 by a factor of :

A
3

B
4

C
3000

D
10,000

Solution

[H⁺]₁ = 10^–2 ; [H⁺]₂ = 10^–6
Question 7
4 / -1

The first and second dissociation constants of an acid H₂A are 1.0 × 10^-5 and 5.0 × 10^-10 respectively.
The overall dissociation constant of the acid will be:

A
5.0×10^-5

B
5.0×10¹⁵

C
5.0×10^-15

D
0.2×10⁵

Solution
Question 8
4 / -1

An aqueous solution contains 0.01 M RNH₂ (K_b= 2 × 10^-6) & 10^-4 M NaOH.
The concentration of OH^- is nearly:

A
2.414 ×10^-4 M

B
10^-4M

C
1.414×10^-4 M

D
2×10^-4M

Solution

x² + 10^–4 x – 2 × 10^–8 = 0
x = 10^–4
[OH–] = x + 10^–4
= 2 × 10^–4
Question 9
4 / -1

What volume of 0.2 M NH₄Cl solution should be added to 100 ml of 0.1 M NH₄OH solution to produce a buffer solution of pH = 8.7 ?
Given : pK_b of NH₄OH = 4.7 ; log 2 = 0.3

A
50 ml

B
100 ml

C
200 ml

D
none of these

Solution

pH = 8.7 ⇒ pOH = 5.3
Basic Buffer

If volume of salt = V ml
Question 10
4 / -1

The pK_a of a weak acid, HA, is 4.80. The pK_b of a weak base, BOH, is 4.78. The pH of an aqueous solution of the corresponding salt, BA, will be:

A
8.58

B
4.79

C
7.01

D
9.22

Solution

Salt of weak acid & weak base