Equilibrium Test

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Equilibrium Test - 32

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Weekly Quiz Competition

Question 1
1 / -0

Which pair will show common ion effect?

A
$$Ba{ Cl }_{ 2 }+Ba{ \left( { NO }_{ 3 } \right) }_{ 2 }$$

B
$$NaCl+HCl$$

C
$${ NH }_{ 4 }OH+{ NH }_{ 4 }Cl$$

D
$$AgCN+KCN$$
Solution
Correct Answer: Options C and D.

Explanation:
- Under the influence of a common ion, the solubility of one electrolyte is decreased. This is known as the common ion effect.
- Due to the common ion effect, the disassociation reaction of the weak species starts to occur in the opposite direction.
- Common ion effect can only take place if at least one electrolyte is weak and there is a common ion present among the species.
- If both species are strong, then they are $$100\%$$ disassociated. Thus, no equilibrium is set up rather $$100\%$$ dissociation takes place.
1. $$BaCl_{2}$$ and $$Ba(NO_{3})_{2}$$
Both have $$Ba^{2+}$$ as a common ion.
$$BaCl_{2}$$ is a salt of $$Ba(OH)_{2}$$ and $$HCl$$. $$Ba(NO_{3})_{2}$$ is a salt of $$Ba(OH)_{2}$$ and $$HNO_{3}$$. Since all of these are strong acids and bases, these two species don’t exhibit the common ion effect.

2. $$NaCl$$ and $$HCl$$
Both have $$Cl^{-}$$ as a common ion.
$$NaCl$$ is a salt of $$Na(OH)$$ and $$HCl$$. $$HCl$$ is a strong acid itself. Since all of these are strong acids and bases, these two species don’t exhibit the common ion effect.

3. $$NH_{4}OH$$ and $$NH_{4}Cl$$
Both have $$NH_{4}^{+}$$ as a common ion.
$$NH_{4}OH$$ and $$NH_{4}Cl$$ are both weak species since $$NH_{4}^{+}$$ is a weak species. Therefore, these two can exhibit the common ion effect.

4. $$AgCN$$ and $$KCN$$
Both have $$CN^{-}$$ as a common ion.
$$KCN$$ is a salt of a weak acid and strong base and $$AgCN$$ is a sparingly soluble salt, they have $$CN^-$$ ion in common. So they will show the common ion effect.

Hence, Options $$C$$ and $$D$$ will show the common ion effect.
Question 2
1 / -0

A monoprotic acid in $$1.00M$$ solution is $$0.001$$% ionised. The dissociation constant of acid is:

A
$${ \alpha }^{ 2 }C+\alpha K-K=0$$

B
$${ \alpha }^{ 2 }C-\alpha K-K=0$$

C
$${ \alpha }^{ 2 }C-\alpha K+K=0$$

D
$${ \alpha }^{ 2 }C+\alpha K+K=0$$

Solution

The given solution is slightly ionized, which indicates it is a weak electrolyte. For weak electrolytes, we can use Ostwald's dilution law.
According to Ostwald's dilution law

$$K_a=\cfrac { { \alpha }^{ 2 }C }{ \left( 1-\alpha \right) } $$

Where, $$K_a$$ = Dissociation constant of a weak acid.
$$\alpha$$ = Degree of dissociation,
$$C$$ = Concentration of weak acid.

On solving the above formula for weak electrolyte we get.

$$\alpha^2C+\alpha K-K=0$$

Hence, the correct option is $$\text{A}$$
Question 3
1 / -0

When $${ NH }_{ 4 }Cl$$ is added to $${ NH }_{ 4 }OH$$ solution, the dissociation of ammonium hydroxide is reduced. It is due to:

A
common ion effect

B
hydrolysis

C
oxidation

D
reduction

Solution

When $$NH_4Cl$$ is added to $$NH_4OH$$ solution, concentration of $$NH_4^{+}$$ ions increases so the equilibrium shift towards left.So the dissociation of ammonium hydroxide is reduced.
Question 4
1 / -0

The ionisation constant of acetic acid is $$1.8\times { 10 }^{ -5 }$$. The concentration at which it will be dissociated to $$2$$% is:

A
$$1M$$

B
$$0.045M$$

C
$$0.018M$$

D
$$0.45M$$

Solution

When acetic acid is dissolved in water, it partially dissociates (2%).
Ionisation constant $$ K_a=1.87\times 10^{-5}$$
$$K_a=\cfrac{C\alpha \times C\alpha}{C(1-\alpha)}$$

We assume $$\alpha\sim 0\Rightarrow 1-\alpha\sim 1$$

$$\Rightarrow 1.8\times 10^{-5}=\cfrac{C\alpha^2}{1}={C\times 0.02\times 0.02}$$

$$\Rightarrow C = \cfrac{1.8\times 10^{-5}}{0.02^2}$$

$$=0.045M$$

Correct answer is option-B.
Question 5
1 / -0

One litre of water contains $${ 10 }^{ -7 }$$ mole of $${H}^{+}$$ ions. Degree of ionisation of water is:

A
$$1.8\times { 10 }^{ -7 }$$

B
$$0.8\times { 10 }^{ -9 }$$

C
$$5. 4\times { 10 }^{ -9 }$$

D
$$5 . 4\times { 10 }^{ -7 }$$

Solution

$$H_2O⇌H^{ + } + OH^{ - }\quad $$
$$C$$ $$0$$ $$0$$
$$C(1-\alpha )$$ $$C\alpha $$ $$C\alpha $$

$$C\alpha=10^{-7}$$

[H2O] =$$\dfrac{ 1000}{18}$$= 55.55 M

C= 55.5M

So $$\alpha=18\times10^{-10}$$

In percentage $$\alpha=1.8\times10^{-7}$$%
Question 6
1 / -0

The following reaction is known to occur in the human body:
$${ CO }_{ 2 }+{ H }_{ 2 }O\rightleftharpoons { H }_{ 2 }{ CO }_{ 3 }\rightleftharpoons { H }^{ + }+H{ CO }_{ 3 }^{ - }$$
If $${CO}_{2}$$ escapes from the system

A
pH will decrease

B
$${ H }^{ + }$$ ion concentration will decrease

C
$${ H }_{ 2 }{ CO }_{ 3 }$$ concentration will be unaltered

D
The forward reaction will be promoted

Solution

If $$CO_2$$ concentration will decrease then $$H_2CO_3$$ dissociates more towards left to give $$CO_2$$ and concentration of $$H^{+}$$ will decrease.
Question 7
1 / -0

The following equilibrium exists in aqueous solution:

$${ CH }_{ 3 }COOH\rightleftharpoons { H }^{ + }+{ CH }_{ 3 }CO{ O }^{ - }$$

If dilute $$HCl$$ is added:

A
The equilibrium constant will increase

B
The equilibrium constant will decrease

C
Acetate ion concentration will increase

D
Acetate ion concentration will decrease

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 8
1 / -0

The dissociation constants of two acids $$ H{ A }_{ 1 }$$ and $$H{ A }_{ 2 }$$ are $$3.0\times { 10 }^{ -4 }$$ and $$1.8\times { 10 }^{ -5 }$$ respectively. The relative strengths of the acids will be:

A
$$1:4$$

B
$$4:1$$

C
$$1:16$$

D
$$16:1$$

Solution

The dissociation constants of $$HA_1$$ and $$HA_2$$ are $$3\times10^{-4} $$ and $$1.8\times10^{-5}.$$

The strength of an acid is directly proportional to square root of dissociation constants of acids. So relative strength of the given acids are:
$$\dfrac { { (Acidic\ Strength })_{ HA_1 } }{ { (Acidic\ Strength })_{ HA_2} } =\dfrac { \sqrt { { (Dissociation\ Constant })_{ HA_1} } }{ \sqrt { { (Dissociation\ Constant })_{ HA_2 } } } $$

Relative Acidic Strength= $$\dfrac { { (Acidic Strength })_{ HA_1 } }{ { (Acidic Strength })_{ HA_2 } } =\dfrac { \sqrt { 3.0\times { 10 }^{ -4 } } }{ \sqrt { 1.8\times { 10 }^{ -5 } } } =4.08=4(approx)$$

relative strengths of acids will be $$4:1.$$
Question 9
1 / -0

A certain weak acid has dissociation constant of $$1.0\times { 10 }^{ -4 }$$. The equilibrium constant for its reaction with a strong base is:

A
$$1.0\times { 10 }^{ -4 }$$

B
$$1.0\times { 10 }^{ -10 }$$

C
$$1.0\times { 10 }^{ 10 }$$

D
$$1.0\times { 10 }^{ -14 }$$

Solution

Hence, the correct option is $$\text{C}$$
Question 10
1 / -0

The pH of a neutral water is $$6.5$$. Then the temperature of water:

A
Is $${ 25 }^{ o }C$$

B
Is more than $${ 25 }^{ o }C$$

C
Is less than $${ 25 }^{ o }C$$

D
Can be more or less than $${ 25 }^{ o }C$$

E
Cannot be predicted

Solution

The ionization increases with temperature. So the amount of $$H^+$$ active in solution will increase with temperature, the pH should decrease.
When there is decrease in pH of neutral water, it is normally assumed that the temperature is increased from room temperature $$25^0C.$$
So when the pH of neutral water is 6.5, the temperature should greater than $$25^0C.$$

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Equilibrium Test - 32

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Equilibrium Test - 32

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Question 1

$$Ba{ Cl }_{ 2 }+Ba{ \left( { NO }_{ 3 } \right) }_{ 2 }$$

$$NaCl+HCl$$

$${ NH }_{ 4 }OH+{ NH }_{ 4 }Cl$$

$$AgCN+KCN$$

Solution

Question 2

$${ \alpha }^{ 2 }C+\alpha K-K=0$$

$${ \alpha }^{ 2 }C-\alpha K-K=0$$

$${ \alpha }^{ 2 }C-\alpha K+K=0$$

$${ \alpha }^{ 2 }C+\alpha K+K=0$$

Solution

Question 3

common ion effect

hydrolysis

oxidation

reduction

Solution

Question 4

$$1M$$

$$0.045M$$

$$0.018M$$

$$0.45M$$

Solution

Question 5

$$1.8\times { 10 }^{ -7 }$$

$$0.8\times { 10 }^{ -9 }$$

$$5. 4\times { 10 }^{ -9 }$$

$$5 . 4\times { 10 }^{ -7 }$$

Solution

Question 6

pH will decrease

$${ H }^{ + }$$ ion concentration will decrease

$${ H }_{ 2 }{ CO }_{ 3 }$$ concentration will be unaltered

The forward reaction will be promoted

Solution

Question 7

The equilibrium constant will increase

The equilibrium constant will decrease

Acetate ion concentration will increase

Acetate ion concentration will decrease

Solution

Question 8

$$1:4$$

$$4:1$$

$$1:16$$

$$16:1$$

Solution

Question 9

$$1.0\times { 10 }^{ -4 }$$

$$1.0\times { 10 }^{ -10 }$$

$$1.0\times { 10 }^{ 10 }$$

$$1.0\times { 10 }^{ -14 }$$

Solution

Question 10

Is $${ 25 }^{ o }C$$

Is more than $${ 25 }^{ o }C$$

Is less than $${ 25 }^{ o }C$$

Can be more or less than $${ 25 }^{ o }C$$

Cannot be predicted

Solution

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