Test: Common Ion Effect


10 Questions MCQ Test Chemistry Class 11 | Test: Common Ion Effect


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QUESTION: 1

The following equilibrium exists in aqueous solution

CH3COOH  CH3COO- + H+

If dilute HCl is added:

Solution:

If dilute HCl is added then concentration of H+ ions will increase so according to Lechatelier's principle equilibrium will shift in backward direction. So acetate ion concentration will decrease.

QUESTION: 2

Consider KA, the potassium salt of a weak acid, HA. If a sample of KA is dissolved in water, with no other substances added, which of the following statements is true (approximately)?

Solution:

The correct answer is option C
[HA] = [OH–]
The reaction is: A– + H2O ⇌ HA + OH–

QUESTION: 3

Which of the following is an example of basic buffer?

Solution:

The correct answer is Option B.

NH4Cl is a strong electrolyte hence, dissociates completely, NH4OH is a weak electrolyte. Its dissociation is further suppressed by common ion NH4+ provided by NH4Cl in the solution.
NH4Claq  ⇌ NH4aq+ +Claq−
NH4OHaq ⇌ NH4aq++ OHaq−
This solution acts as a basic buffer and maintains its pH around 9.25. It is capable of resisting the change in pH in addition to a small amount of acid or alkali.
To Explain:
If a small amount of HCl is added to this solution, H+
ions of HCl get neutralized by OH− ions already present and more of NH4
​OH molecules get ionized to compensate for the loss of OH− ions. Thus pH practically, remains constant.
If a small amount of NaOH is added to this solution, OH− ions of NaOH combine with NH4+ ions already present in large numbers forming weakly ionized NH4OH. Thus, pH of solution remains practically unchanged.

QUESTION: 4

 If the value of the solubility product for AgBr is 4.0 x 10-12 at 25°C, calculate the solubility of AgBr(s) in water.

Solution:

Ksp = 4.0 x 10-12 
For AB type salt, Solubility = Ksp1/2
= (4.0 x 10-12)1/2
= 2 x 10-6

QUESTION: 5

A mixture of CH3COOH and CH3COONa behaves as

Solution:

Mixtures of CH3COOH and CH3COONa are called acidic buffer. Since the constitution of this buffer is a weak acid and its constituting salt, it is called Acidic buffer.

QUESTION: 6

Role of NH4Cl in qualitative analysis of third group cations

Solution:

The correct answer is option D
Common ion effect is observed when a solution of weak electrolyte is mixed with a solution of strong electrolyte, which provides an ion common to that provided by weak electrolyte.
The NH4OH is a weak base and it does not ionise completely. Thus, due to presence of common ion NH4+ in NH4Cl, it suppresses the ionisation of weak base NH4OH in order to decrease the OH- concentration so that higher group cations will not get precipitated.
Thus the pair NH4​OH+NH4​Cl shows a common ion effect. 
Ammonium chloride suppresses the ionization of ammonium hydroxide
.

QUESTION: 7

When the value of Kc is very small then

Solution:
QUESTION: 8

A mixture of NH4Cl and NH4OH shows no change in pH upon addition small amount of HCl. This is because it is

Solution:

A mixture of weak base and its salt with a strong acid serves as an basic buffer which resists changes in pH upon addition of small amount of acid or base.

 

 

QUESTION: 9

 We eat a variety of foods still pH of our blood does not change every time. The reason is

Solution:

The correct answer is option C
pH of blood remains constant because of the buffer system present in the blood. Acid-base buffers confer resistance to a change in the pH of a solution when hydrogen ions (protons) or hydroxide ions are added or removed. An acid-base buffer typically consists of a weak acid, and its conjugate base (salt).It used to neutralized the extra added protons or OH- in blood.The buffer for maintaining acid-base balance in the blood is the carbonic-acid-bicarbonate buffer.So pH of blood remains even after eating spicy food.

QUESTION: 10

 For HF, pKa = 3.45. What is the pH of an aqueous buffer solution that is 0.1M HF (aq) and 0.300 M KF (aq)?

Solution:

Given, pKa = 3.45
Concentration of HF = 0.1 M, concentration of KF = 0.300 M
For acidic buffer;
pH = pKa + log [salt of weak acid]/[weak acid]
= 3.45 + log0.3/0.1
= 3.45 + 0.48
= 3.93

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