Equilibrium Test

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Equilibrium Test - 10

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Weekly Quiz Competition

Question 1
1 / -0

Which one of the following pairs of solution is not an acidic buffer?

A
$$H_2CO_3$$ and $$Na_2CO_3$$

B
$$H_3PO_4$$ and $$Na_3PO_4$$

C
$$HClO_4$$ and $$NaClO_4$$

D
$$CH_2COOH$$ and $$CH_3COONa$$

Solution

An acidic buffer includes weak acid and its salt with a strong base.
$$HClO_4$$ and $$NaClO_4$$ contains strong acid $$HClO_4$$ and its salt with strong base $$NaOH$$.
$$HClO_4$$ and $$NaClO_4$$ is not an acidic buffer.
Hence, option C is correct.
Question 2
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At $$90^o C$$, pure water has $$[H^{+}]=10^{-6}M$$. The value of $$K_w$$ at $$90^o C$$ is __________.

A
$$10^{-6}$$

B
$$10^{-8}$$

C
$$10^{-12}$$

D
$$10^{-14}$$

Solution

At $$90^o C$$, pure water has $$[H^{+}]=10^{-6}M$$.

In pure water, $$[H_3O^{+}] = [OH^{+}]$$

The value of $$K_w$$ at $$90^o C,$$ $$K_w = [H_3O^+][OH^-]=10^{-6} \times 10^{-6} =10^{-12}$$
Question 3
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Which of the following according to Le-Chatelier's principle is correct?

A
Increase in temperature favours the endothermic reaction

B
Increase in temperature favours the exothermic reaction

C
Increase in pressure shifts the equilibrium in that side in which number of gaseous moles increases

D
All of the above are true

Solution

According to Le-Chatelier's principle increase in temperature favours the endothermic reaction while a decrease in temperature favour the exothermic reaction. Increase in pressure shifts the equilibrium in that side in which the number of moles of gaseous species decreases.

Note: An endothermic reaction is accompanied by the absorption of heat whereas an exothermic reaction is accompanied by the evolution of heat.

When the temperature is increased for an endothermic reaction, the added heat is absorbed during the forward reaction. Total amount of heat in the system thus decreases and the temperature decreases. This nullifies the effect of increase in temperature. Hence, the increase in temperature favours the endothermic reaction.
Question 4
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The degree of dissociation of $$P{ Cl }_{ 5 }\left( \alpha \right) $$ obeying the equilibrium $$P{ Cl }_{ 5 }\rightleftharpoons P{ Cl }_{ 3 }+{ Cl }_{ 2 }$$ is related to the equilibrium pressure by

A
$$\alpha \propto \cfrac { 1 }{ { P }^{ 4 } } $$

B
$$\alpha \propto \cfrac { 1 }{ \sqrt { P } } $$

C
$$\alpha \propto \cfrac { 1 }{ { P }^{ 2 } } $$

D
$$\alpha \propto P$$

Solution

$$P{ Cl }_{ 5 }\rightleftharpoons P{ Cl }_{ 3 }+{ Cl }_{ 2 }$$
$$ \displaystyle \alpha =$$ the degree of dissociation of $$ \displaystyle P{ Cl }_{ 5 }$$
$$ \displaystyle P = $$ total equilibrium pressure.
Suppose initially, 1 mole of $$ \displaystyle P{ Cl }_{ 5 }$$ is present.
$$ \displaystyle   \alpha $$ moles of $$ \displaystyle P{ Cl }_{ 5 }$$ will dissociate to form $$ \displaystyle \alpha $$ moles of $$ \displaystyle P{ Cl }_{ 3 }$$ and $$ \displaystyle \alpha $$ moles of $$ \displaystyle { Cl }_{ 2 }$$.
$$ \displaystyle (1- \alpha) $$ moles of $$ \displaystyle P{ Cl }_{ 5 }$$ will remain at equilibrium.
Total number of moles $$ \displaystyle = (1 - \alpha) + \alpha+ \alpha =(1+\alpha)$$

Mole fraction of $$ \displaystyle P{ Cl }_{ 5 } = \dfrac {(1 - \alpha)}{(1 + \alpha)}$$
Mole fraction of $$ \displaystyle P{ Cl }_{ 3 } = \dfrac {(   \alpha)}{(1 + \alpha)}$$
Mole fraction of $$ \displaystyle { Cl }_{ 2 } = \dfrac {( \alpha)}{(1 + \alpha)}$$

Partial pressure of $$ \displaystyle P{ Cl }_{ 5 } = \dfrac {(1 - \alpha)}{(1 + \alpha)}P$$
Partial pressure of $$ \displaystyle P{ Cl }_{ 3 } = \dfrac {(   \alpha)}{(1 + \alpha)}P$$
Partial pressure of $$ \displaystyle { Cl }_{ 2 } = \dfrac {( \alpha)}{(1 + \alpha)}P$$

$$ { K }_{ p }= \dfrac {P_{PCl3}P_{Cl2}}{P_{PCl5}}$$
$$\therefore { K }_{ p }=\cfrac { \cfrac { \alpha }{ 1+\alpha } P\times \cfrac { \alpha }{

1+\alpha } P }{ \cfrac { 1-\alpha }{ 1+\alpha } P }$$
$$\therefore { K }_{ p } =\cfrac { { \alpha }^{ 2 }P }{ 1-{

\alpha }^{ 2 } } $$
Assume $$1-{ \alpha }^{ 2 }=1$$ as the degree of dissociation is small.
$${ K }_{ p }={ \alpha }^{ 2 }P$$

$$\therefore \alpha =\sqrt { \cfrac { { K }_{ p } }{ P } } $$
$$ \displaystyle \alpha \propto \cfrac { 1 }{ \sqrt { P } }$$
Question 5
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In a chemical system, $$A(s)\rightleftharpoons 2B(g)+3C(g)$$, if partial pressure of $$C$$ is doubled, then partial pressure of $$B$$ will be:

A
$$2\sqrt { 2 } $$ times the original value

B
$$\cfrac { 1 }{ 2 } $$ times the original value

C
$$2$$ times the original value

D
$$\cfrac { 1 }{ 2\sqrt { 2 } } $$ times the original value

Solution

In a chemical system $$A(s)\rightleftharpoons 2B(g)+3C(g)$$.

As per the initial conditions,
$${ K }_{ p }={ p }_{ { B }_{ 1 } }^{ 2 }\times { p }_{ { C }_{ 1 } }^{ 3 }\quad $$.....(1)

But $$p_{ { C }_{ 2 }}=2 \times p_{{ C }_{ 1 }} $$ as partial pressure of $$C$$ is doubled.

As per the final conditions,
$${ K }_{ p }={ p }_{ { B }_{ 2 } }^{ 2 }\times { \left(2 { p }_{ { C }_{ 1 } }^{ } \right) }^{ 3 }\quad $$.....(2)

Substitute this in equation (3).

$$\therefore { p }_{ { B }_{ 1 } }^{ 2 }\times { p }_{ { C }_{ 1 } }^{ 3 } ={ p }_{ { B }_{ 2 } }^{ 2 } \times (2 \times p_{{ C }_{ 1 }} ) ^{ 3 }\quad $$

$$\therefore { p }_{ { B }_{ 1 } }^{ 2 }\times { p }_{ { C }_{ 1 } }^{ 3 } ={ p }_{ { B }_{ 2 } }^{ 2 }\times 8{ p }_{ { C }_{ 1 } }^{ 3 }\quad $$

$$\therefore { p }_{ { B }_{ 1 } }^{ 2 } ={ p }_{ { B }_{ 2 } }^{ 2 }\times 8 $$

$$\cfrac { { p }_{ { B }_{ 1 } }^{ 2} }{ { 8 } } ={ p }_{ { B }_{ 2 } }^{ 2 }$$

$$\cfrac { { p }_{ { B }_{ 1 } }^{ } }{ \sqrt { 8 } } ={ p }_{ { B }_{ 2 } }^{ }$$

$$\cfrac { { p }_{ { B }_{ 1 } }^{ } }{ 2\sqrt { 2 } } ={ p }_{ { B }_{ 2 } }^{ }$$

if partial pressure of $$C$$ is doubled, then partial pressure of $$B$$ will be $$\cfrac { 1 }{ 2\sqrt { 2 } } $$ times the original value.
Question 6
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In a sample of pure water, which one of the following statements is always true at all conditions of temperature and pressure.

A
$$[OH^-] = 1.0\times10^{-14} M $$

B
$$[H_3O^+] = 1.0\times10^{-7} M $$

C
$$[OH^-] = 1.0\times10^{-7} M $$

D
$$[H_3O^+]=[OH^-]$$
Question 7
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Buffer Solution is prepared by mixing _________________.

A
Weak acid and its salt of strong base

B
Strong acid + its salt of strong base

C
Weak acid + its salt of weak base

D
Strong base + its salts of strong acid

Solution

A solution that resists change in pH value upon addition of a small amount of strong acid or base (less than 1 %) or when the solution is diluted is called buffer solution.
An acidic buffer solution consists of a $$\text{solution of a weak acid and its salt with a strong base. }$$while basic buffer solution consists of a mixture of a weak base and its salt with strong acid.
Question 8
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Which of the following mixtures in aqueous solution of equimolar concentration acts as a buffer solution?

A
$$H NO_{3} + Na OH$$

B
$$H_{2}SO_{4} + K OH$$

C
$$NH_{4}OH(excess) + H Cl$$

D
$$CH_{3}COO H + Na OH (excess)$$

Solution

A mixture of ammonium hydroxide and $$HCl$$ react to form ammonium chloride. This also contains unreacted ammonium hydroxide.

$$NH_4OH+HCl \rightarrow NH_4Cl + H_2O$$

Thus, the resulting mixture of $$(NH_4OH+NH_4Cl)$$ is a basic buffer solution. It contains a mixture of weak base ammonium hydroxide and its salt (ammonium chloride) with strong acid $$(HCl)$$.

It is used in a qualitative analysis of group $$III$$ radicals.

A mixture of $$HNO_3$$ and $$NaOH$$ is a mixture of strong acid and strong base similarly a mixture of $$H_2SO_4$$ and $$KOH$$ is also a mixture of strong acid and strong base, thus they do not form buffer solution.

Similarly, in option D mixture will form a strong base and salt of strong base weak acid hence it will not form a buffer solution.

Hence, option C is correct.
Question 9
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$$H_{2}O$$ is a:

A
Proton acceptor

B
Proton donor

C
Both proton donor and proton acceptor

D
Neither proton donor, nor proton acceptor

Solution

Hint:
According to Bronsted-Lowry's theory, the compounds that have tendency to donate$$ \left(\mathrm{H}^{+}\right) $$ ion are acids and the compounds that have tendency to accept $$\left(\mathrm{H}^{+}\right)$$ ion are bases.
Explanation:
A compound such as water $$ \left(\mathrm{H}_{2} \mathrm{O}\right) $$ has many interesting properties. Water molecules can accept a proton to act as Bronsted-Lowry bases in certain circumstances. The following is an example of $$ \mathrm{HCl} $$ dissolving in water:
$$\mathrm{HCl}+\mathrm{H}_{2} \mathrm{O}_{(\ell)} \rightarrow \mathrm{H}_{3} \mathrm{O}_{\text {(aq) }}^{+}+\mathrm{Cl}_{(\mathrm{aq})}^{-}$$
The another possibility is water can act like a Bronsted-Lowry acid by donating a proton. Water donates a proton to a proton-accepting amide ion in the presence of ammonia, resulting in the following product:
$$\mathrm{H}_{2} \mathrm{O}_{(\ell)}+\mathrm{NH}_{2(a q)}^{-} \rightarrow \mathrm{OH}_{(a q)}^{-}+N H_{3(a q)}$$
Therefore, depending on the circumstances, $$ \left(\mathrm{H}_{2} \mathrm{O}\right) $$ can either act as a Bronsted-Lowry acid or a Bronsted-Lowry base. There are other substances that can react either as acids or bases in certain circumstances, but water is the most common example. It is called an amphiprotic compound when it either donates or accepts a proton, depending on the situation.
Final Answer: Option (C)
Question 10
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A solution which maintains constant pH when small amounts of acid or alkali are added is known as ____.

A
Indicator

B
Buffer

C
Amphoteric

D
Neutral

Solution

A buffer solution is one which resists changes in pH when small quantities of an acid or an alkali are added to it.

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Equilibrium Test - 10

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Question 1

$$H_2CO_3$$ and $$Na_2CO_3$$

$$H_3PO_4$$ and $$Na_3PO_4$$

$$HClO_4$$ and $$NaClO_4$$

$$CH_2COOH$$ and $$CH_3COONa$$

Solution

Question 2

$$10^{-6}$$

$$10^{-8}$$

$$10^{-12}$$

$$10^{-14}$$

Solution

Question 3

Increase in temperature favours the endothermic reaction

Increase in temperature favours the exothermic reaction

Increase in pressure shifts the equilibrium in that side in which number of gaseous moles increases

All of the above are true

Solution

Question 4

$$\alpha \propto \cfrac { 1 }{ { P }^{ 4 } } $$

$$\alpha \propto \cfrac { 1 }{ \sqrt { P } } $$

$$\alpha \propto \cfrac { 1 }{ { P }^{ 2 } } $$

$$\alpha \propto P$$

Solution

Question 5

$$2\sqrt { 2 } $$ times the original value

$$\cfrac { 1 }{ 2 } $$ times the original value

$$2$$ times the original value

$$\cfrac { 1 }{ 2\sqrt { 2 } } $$ times the original value

Solution

Question 6

$$[OH^-] = 1.0\times10^{-14} M $$

$$[H_3O^+] = 1.0\times10^{-7} M $$

$$[OH^-] = 1.0\times10^{-7} M $$

$$[H_3O^+]=[OH^-]$$

Question 7

Weak acid and its salt of strong base

Strong acid + its salt of strong base

Weak acid + its salt of weak base

Strong base + its salts of strong acid

Solution

Question 8

$$H NO_{3} + Na OH$$

$$H_{2}SO_{4} + K OH$$

$$NH_{4}OH(excess) + H Cl$$

$$CH_{3}COO H + Na OH (excess)$$

Solution

Question 9

Proton acceptor

Proton donor

Both proton donor and proton acceptor

Neither proton donor, nor proton acceptor

Solution

Question 10

Indicator

Buffer

Amphoteric

Neutral

Solution

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