Equilibrium Test

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Equilibrium Test - 9

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Weekly Quiz Competition

Question 1
1 / -0

The minimum volume of water required to dissolve $$0.1$$g lead(II) chloride to get a saturated solution ($$K_{sp}$$ of $$PbCl_2=3.2\times 10^{-8}$$; atomic mass of Pb$$=207u$$) is:

A
$$0.18$$ L

B
$$17.98$$ L

C
$$1.798$$ L

D
$$0.36$$ L

Solution

$$K{sp} = 3.2\times 10^{-8}$$

$$PbCl_2 \rightleftharpoons Pb^{2+} + 2Cl^-$$

$$K_{sp} = (S)(2S)^2$$

$$K_{sp} = 4S^3$$

$$s^3 = \dfrac{3.2\times 10^{-8}}{4}$$

$$s^3 = 8\times 10^{-9}$$

$$s = 2\times 10^{-3}$$

$$Molar\ mass\ of\ PbCl_2 = 278 g/mol$$

By using Molarity formula,

$$\therefore 2\times 10^{-3}=\dfrac{0.1}{{278}\times{x}}$$

$$\therefore x = \dfrac{0.1}{278\times 2\times 10^{-3}}$$

$$=\dfrac{100}{278\times 2}$$

$$x = 0.18L$$
Question 2
1 / -0

The gas phase reaction $$2{NO}_{2}(g)\rightarrow {N}_{2}{O}_{4}(g)$$ is an exothermic reaction. The decomposition of $${N}_{2}{O}_{4}$$, in equilibrium mixutre of $${NO}_{2}(g)$$ and $${N}_{2}{O}_{4}(g)$$, can be increased by:

A
increasing the pressure

B
addition of an inert gas at constant pressure

C
lowering the temperaure

D
addition of an inert gas at constant volume

Solution

Reaction at equilibrium

$$N_2O_4\rightleftharpoons 2NO_2$$

(1) According to Le chatelier's principle: Increasing the pressure on a gas reaction shifts the position of equilibrium towards the side with fewer molecules.
So, it will move in backward direction which leads to formation of $$N_2O_4 $$from $$NO_2$$.

(2) Addition of an inert gas at constant pressure will increase volume and equilibrium shifts towards more number of molecules i.e. there will be decomposition.

(3) Decomposition of $$N_2O_4$$ is endothermic. So, the reaction will move in forward reaction when the temperature is increased.
Question 3
1 / -0

In some solutions, the concentration of $$H_3O^+$$ remains constant even when small amounts of strong acid or strong base are added to them. These solutions are known as:

A
colloidal solutions

B
buffer solutions

C
true solutions

D
ideal solutions

Solution

In some solutions, the concentration of $$H_3O^+$$ remains constant even when small amounts of strong acid or strong base are added to them. These solutions are known as buffer solutions.
Question 4
1 / -0

At $$90^oC$$, pure water $$[H_3O^{\oplus}]$$ as $$10^{-6} mol\, L^{-1}$$ what is the value of $$K_w$$ at $$90^oC$$ ?

A
$$10^{-6}$$

B
$$10^{-12}$$

C
$$10^{-14}$$

D
$$10^{-8}$$

Solution

As we know, $$K_w = [H^+][OH^-]$$

At $$90^oC$$, $$[H^+] = [OH^-] = 10^{-6}$$

So $$K_w = 10^{-12}$$

Hence,option B is correct.
Question 5
1 / -0

An acidic buffer solution can be prepared by mixing the solutions of:

A
sodium acetate and acetic acid

B
ammonium chloride and ammonium hydroxide

C
sulphuric acid and sodium sulphate

D
sodium chloride and sodium hydroxide

Solution

A solution that resists change in pH value upon the addition of a small amount of strong acid or base (less than 1 %) or when a solution is diluted is called buffer solution.

Basic buffer solution: A basic buffer solution consists of a mixture of a weak base and its salt with a strong acid. The best-known example is a mixture of $$NH_4OH$$ and $$NH_4 Cl.$$

Acidic buffer solution: An acidic buffer solution consists of a solution of a weak acid and its salt with a strong base. The best-known example is a mixture of a solution of acetic acid and its salt with a strong base i.e. sodium acetate.

Hence, option $$A$$ is correct.
Question 6
1 / -0

Addition of $${NH}_{4}Cl$$ does not effect the $$pH$$ of solution of $${NH}_{4}OH$$.

A
True

B
False

C
Ambigous

D
None

Solution

Adding a common ion prevents the weak acid or weak base from ionizing as much as it would without the added common ion. The common ion effect suppresses the ionization of a weak acid by adding more of an ion that is a product of this equilibrium.

So addition of $${NH}_{4}Cl$$ reduces dissociation of $${NH}_{4}OH$$ and because of that its pH will decreases.
Question 7
1 / -0

The initial rate of hydrolysis methyl acetate ($$1\ M$$) by a weak acid ($$HA,\ 1M$$) is $$1/100^{th}$$ of that of a strong acid ($$HX,\ 1M$$), at $$25^{o}C$$. The $$K_{a}$$of $$HA$$ is:

A
1 x 10$$^{-4}$$

B
1 x 10$$^{-5}$$

C
1 x 10$$^{-6}$$

D
1 x 10$$^{-3}$$

Solution

The rate of the hydrolysis of methyl acetate is directly proportional to the hydrogen ion concentration.
In case of weak acid, the equilibrium constant is given by the expression $$K_a=\dfrac {[H^+][A^-]}{[HA]}=\dfrac {[H^+]^2}{[HA]}$$
Hence, $$[H^+]=\sqrt {K_a[HA]}$$

Hence, the expression for the ratio of the rate of the hydrolysis of methyl acetate in presence of weak acid and strong acid becomes $$\dfrac {r_{weak acid}} {r_{strong acid}}=\dfrac {\sqrt {K_a[HA]}}{[HA]}=\dfrac {1} {100}$$

Substitute $$[HA]=1 M$$

Hence, $$\dfrac {\sqrt {K_a \times 1}}{1}=\dfrac {1} {100}$$
$$K_a=1 \times 10^{-4}$$
Question 8
1 / -0

For the reversible reaction:

$$N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)+heat$$

The equilibrium shifts in forward direction:

A
by increasing the concentration of $$NH_3(g)$$

B
by decreasing the pressure

C
by decreasing the concentrations of $$N_2(g)$$ and $$H_2(g)$$

D
by increasing pressure and decreasing temperature

Solution

For the reversible reaction:
$$N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)+heat$$ The equilibrium shifts in forward direction by increasing pressure and decreasing temperature.

According to Le-Chatalier principle when equilibrium is disturbed by a change, the system will try to nullify the effect of change to restore the equilibrium.

Thus, when pressure is increased, the equilibrium will shift in a direction in which there is a decrease in the number of moles of gaseous substances. This will nullify the effect of increased pressure. This happens in the forward direction.

Similarly for an exothermic reaction, when the temperature is decreased, the equilibrium shifts in the forward direction.
Question 9
1 / -0

Which one of the following conditions will favour maximum formation of the product in the reaction,
$$A_{2}(g) + B_{2}(g) \rightleftharpoons X_{2}(g), \ \ \Delta_{r}H = -X\ kJ$$?

A
High temperature and high pressure

B
Low temperature and high pressure

C
High temperature and low pressure

D
Low temperature and low pressure

Solution

For the reaction, $$\displaystyle A_{2}(g) + B_{2}(g) \rightleftharpoons X_{2}(g) \ \ , \ \ \Delta_{r}H = -X\ kJ$$ suggest exothermic nature for forward reaction.

Exothermic reaction has negative value for $$\displaystyle \Delta_{r}H $$.

Exothermic reaction is favored at low temperature.

The forward reaction has $$\displaystyle \Delta n =1 -(1+1)=-1$$. Such reaction is favored at high pressure.
Question 10
1 / -0

Which will make basic buffer?

A
$$50mL$$ of $$0.1$$ M $$NaOH+25mL$$ of $$0.1M\,CH_3COOH$$

B
$$100mL$$ of $$0.1$$ M$$CH_3COOH+100mL$$ of $$0.1M\,NaOH$$

C
$$100mL$$ of $$0.1M\,HCl+200mL$$ of $$0.1M\,NH_4OH$$

D
$$100mL$$ of $$0.1M\,HCl+100mL$$ of $$0.1$$M\,NaOH$$

Solution

1) $$\begin{matrix} CH_ 3COOH+ & NaOH & \rightarrow & CH_ 3COONa+ & H\_ 2O \\ 25mL\times 0.1M=2.5mmol & \quad 50mL\times 0.1M=5mmol0 & & 0 & \\ 0 & 2.5mmol & & 2.5mmol & \end{matrix}$$
This is basic solution due to $$NaOH$$. This is not basic buffer.
2) $$\begin{matrix} CH_ 3COOH+ & NaOH & \rightarrow & CH_ 3COONa+ & H_ 2O \\ 100mL\times 0.1M=10mmol & \quad 100mL\times 0.1M=10mmol & & 0 & \\ 0 & 0 & & 10mmol & \end{matrix}$$
Hydrolysis of salt takes place. This is not buffer.
3) $$\begin{matrix} HCl & NH_{ 4 }OH & \rightarrow & NH_ 4Cl+ & H_ 2O \\ 100mL\times 0.1M=10mmol & \quad 200mL\times 0.1M=20mmol & & 0 & \\ 0 & 10 & & 10mmol & \end{matrix}$$
This is basic buffer.
4)$$\begin{matrix} HCl & NaOH & \rightarrow & NaCl+ & H_{ 2 }O \\ 100mL\times 0.1M=10mmol & \quad 100mL\times 0.1M=20mmol & & 0 & \\ 0 & 0 & & 10mmol & \end{matrix}$$
Neutral solution.

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Re-Attempt Weekly Quiz Competition

Equilibrium Test - 9

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Equilibrium Test - 9

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Question 1

$$0.18$$ L

$$17.98$$ L

$$1.798$$ L

$$0.36$$ L

Solution

Question 2

increasing the pressure

addition of an inert gas at constant pressure

lowering the temperaure

addition of an inert gas at constant volume

Solution

Question 3

colloidal solutions

buffer solutions

true solutions

ideal solutions

Solution

Question 4

$$10^{-6}$$

$$10^{-12}$$

$$10^{-14}$$

$$10^{-8}$$

Solution

Question 5

sodium acetate and acetic acid

ammonium chloride and ammonium hydroxide

sulphuric acid and sodium sulphate

sodium chloride and sodium hydroxide

Solution

Question 6

True

False

Ambigous

None

Solution

Question 7

1 x 10$$^{-4}$$

1 x 10$$^{-5}$$

1 x 10$$^{-6}$$

1 x 10$$^{-3}$$

Solution

Question 8

by increasing the concentration of $$NH_3(g)$$

by decreasing the pressure

by decreasing the concentrations of $$N_2(g)$$ and $$H_2(g)$$

by increasing pressure and decreasing temperature

Solution

Question 9

High temperature and high pressure

Low temperature and high pressure

High temperature and low pressure

Low temperature and low pressure

Solution

Question 10

$$50mL$$ of $$0.1$$ M $$NaOH+25mL$$ of $$0.1M\,CH_3COOH$$

$$100mL$$ of $$0.1$$ M$$CH_3COOH+100mL$$ of $$0.1M\,NaOH$$

$$100mL$$ of $$0.1M\,HCl+200mL$$ of $$0.1M\,NH_4OH$$

$$100mL$$ of $$0.1M\,HCl+100mL$$ of $$0.1$$M\,NaOH$$

Solution

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