Equilibrium Test

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Equilibrium Test - 34

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Question 1
1 / -0

When strong base $$(NaOH)$$ is added to the weak acid (acetic acid, $${ CH }_{ 3 }COOH$$), then dissociation of acetic acid increases; this effect is known as:

A
Common ion effect

B
Reverse ion effect

C
Saltation effect

D
Solubility effect

Solution

$$CH_3COOH\rightleftharpoons CH_3COO^{-}+H^+$$
When $$NaOH$$ is added $$H^+$$ combines with $$OH^{-}$$ so concentration of $$H^+$$ decreases so equilibrium will shift towards right.
$${ CH }_{ 3 }COOH+NaOH\longrightarrow { CH }_{ 3 }COONa+{ H }_{ 2 }O\quad $$
Ionization of acetic acid will increase with the progress of its neutralization. This effect is called reverse ion effect.
Question 2
1 / -0

For the equilibrium, $$ H_2 + I_2 \leftrightharpoons 2HI , $$ which of the following will effect the equilibrium constant?

A
Pressure change

B
Concentration change

C
Catalyst

D
Temperature change

Solution

Equilibrium constant depends only upon temperature and nature of reaction but independent of initial concentration of reactants, presence of catalyst and pressure.
Question 3
1 / -0

Which of the following acid / base pairs act as natural buffers in living systems?

A
$$H_2CO_3 / HCO_3{^-} $$

B
$$ {H_2PO_4} {^-} / HPO_4{^{2-}} $$

C
Histidine+ / histidine

D
All of these

Solution

$$H_2CO_3/HCO_3^- ; H_2PO_4^-/HPO_4^{2-} ; Histidine^+ / Histidine$$ all the buffers acts as natural buffers in living system.
Question 4
1 / -0

$${ H }_{ 2 }(g)+{ I }_{ 2 }(g)\rightleftharpoons 2HI(g)$$

The equilibrium constant for the given reaction is $$64$$. If the volume of the container is reduced to half of the original volume, the value of the equilibrium constant will be:

A
$$16$$

B
$$32$$

C
$$64$$

D
$$128$$

Solution

$${ H }_{ 2 }(g)+{ I }_{ 2 }(g)\longrightarrow 2HI(g)$$

for this reaction, $$\Delta { n }_{ g }=0$$

$$\therefore$$ The reaction and its equilibrium constant is not affected by a change in volume. Moreover, the equilibrium constant depends only on temperature.

Hence, the correct option is $$\text{C}$$
Question 5
1 / -0

The degree of dissociation of water at $${ 25 }^{ o }C$$ is $$1.8\times { 10 }^{ -7 }$$% and density is $$1.0g{ cm }^{ -3 }$$. The ionic constant for water is:

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

B
$$2.0\times { 10 }^{ -16 }$$

C
$$1.0\times { 10 }^{ -16 }$$

D
$$1.0\times { 10 }^{ -8 }$$

Solution

For the equilibrium:
$$H_2O\rightleftharpoons H^++OH^-$$
given $$\alpha =1.8\times 10^{ -7 }\times 1/100=1.8\times 10^{ -9 }$$
Since $$\alpha$$<<<1, $$[H_2O]$$ is not much affected.
therefore concentrations at equilibrium are as:
$$[H_2O]=\frac {1000\times 1}{18} \approx 55.5\ M$$
$$[H^+]=\alpha \times 55.5 \approx 1\times 10^{ -7 } $$
$$[OH^-]=\alpha \times 55.5 \approx 1\times 10^{ -7 } $$
Ionic constant of water: $$K=\frac {[OH^-][H^+]}{[H_2O]}$$
on substitution we get
$$K=\frac {10^{ -7 } \times 10^{ -7 }}{55.5}$$
$$K=\frac {10^{ -7 } \times 10^{ -7 }}{55.5}$$
$$K\approx 2\times 10^{ -16 }$$
Therefore option B is correct.
Question 6
1 / -0

Which of the following is not a buffer solution ?

A
$$CH_3COOH + CH_3COONa$$

B
$$H_3BO_3+Na_3BO_3$$

C
$$HClO_4 + NaClO_4$$

D
$$NH_4OH+ (NH_4)_2SO_4$$

Solution

Correct option: C
Hint: Buffer is a mixture of weak acid or base and its salt.
Explanation
A buffer solution is a mixture of a weak acid or a weak base and its salt.
In the given question, $$C{H_3}COOH$$ and $${H_3}B{O_3}$$ are weak acids whereas $$N{H_4}OH$$ is a weak base. $$HCl{O_4}$$ is a strong acid, its mixture with its salt cannot be a buffer solution according to the definition of buffer solution.

Hence, mixture of $${\textbf{HCl}}{{\textbf{O}}_{\textbf{4}}}$$ and $${\textbf{NaCl}}{{\textbf{O}}_{\textbf{4}}}$$ is not a buffer solution.
Question 7
1 / -0

What do you mean by buffer solution?

A
Buffer solution have no $$pH$$

B
Its $$pH$$ changes very little when a small amount of acid or base is added to the it

C
Its $$pH$$ changes very largely when a small amount of acid or base is added to the it

D
All solutions are buffer

Solution

It's $$pH$$ changes very little when a small amount of strong acid or base is added to it.
Buffer solutions are used as a means of keeping $$pH$$ at a nearly constant value in a wide variety of chemical applications. In nature, there are many systems that use buffering for $$pH$$ regulation. For example, the bicarbonate buffering system is used to regulate the $$pH$$ of blood.
Question 8
1 / -0

The pH of a $$0.1M$$ aqueous solution of a weak acid ($$HA$$) is $$3$$. Its degree of dissociation is

A
$$1$$%

B
$$10$$%

C
$$50$$%

D
$$25$$%

Solution

$$\left[ { H }^{ + } \right] =C.\alpha ={ 10 }^{ -3 }$$

$$0.1\times \alpha ={ 10 }^{ -3 }$$

$$\alpha ={ 10 }^{ -2 }= 10^{-2}\times100=1 \% $$
Question 9
1 / -0

To prepare a buffer solution of $$pH = 4.04$$, amount of Barium acetate to be added to $$100 mL$$ of $$0.1$$ M acetic acid solution [ $$pK_0(CH_3COO^-) =9.26$$] is:

A
$$0.05$$ mole

B
$$0.025$$ mole

C
$$0.1$$ mole

D
$$0.005$$ mole

Solution

For the buffer solution of $$ CH_3COOH $$ & $$ CH_3COO^- $$
$$ pH = pK_a + log \frac { [ CH_3COOH]}{[CH_3COO^-]} $$
$$ \Rightarrow $$ m. moles of $$ CH_COO^- = 50 $$
$$ \Rightarrow 4.04 = 4.74 + log \frac { ( 100 \times 0.1 ) }{ m. \quad moles \quad of \quad CH_3COO^-} $$
$$ \therefore $$ moles of $$ (CH_3COO)_2 $$ Ba required = 0.025
Question 10
1 / -0

If some $$He$$ gas is introduced into the equilibrium $${ PCl }_{ 5(g) }\rightleftharpoons { PCl }_{ 5(g) }+{ Cl }_{ 2(g) }$$ at constant pressure and temperature then equilibrium constant of reaction:

A
Increase

B
Decrease

C
Unchange

D
Nothing can be said

Solution

Given : If some
He gas is introduced into the equilibrium PCl5(g)⇌PCl5(g)+Cl2(g) at constant pressure and temperature then equilibrium constant of reaction:
Solution :
Kc will be unchanged because it depends on temperature only. I am giving you some more detail. Consider the dissociation equilibrium of $$PCl_5$$ : $$PCl_5$$ (g) ⇌ $$PCl_3$$ (g) + $$CL_2$$ (g) According to law of chemical equilibrium , equilibrium constant (Kc) : Kc = [$$PCl_3$$ ] [$$CL_2$$][$$PCl_5$$] . If an inert gas is added at constant volume then there is no effect on the dissociation of a compound as only the total pressure of the system increases and there is no change in the partial pressures of the gases. Hence , there will be not effect on the state of equilibrium.

The Correct Opt = C

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Equilibrium Test - 34

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Equilibrium Test - 34

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Question 1

Common ion effect

Reverse ion effect

Saltation effect

Solubility effect

Solution

Question 2

Pressure change

Concentration change

Catalyst

Temperature change

Solution

Question 3

$$H_2CO_3 / HCO_3{^-} $$

$$ {H_2PO_4} {^-} / HPO_4{^{2-}} $$

Histidine+ / histidine

All of these

Solution

Question 4

$$16$$

$$32$$

$$64$$

$$128$$

Solution

Question 5

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

$$2.0\times { 10 }^{ -16 }$$

$$1.0\times { 10 }^{ -16 }$$

$$1.0\times { 10 }^{ -8 }$$

Solution

Question 6

$$CH_3COOH + CH_3COONa$$

$$H_3BO_3+Na_3BO_3$$

$$HClO_4 + NaClO_4$$

$$NH_4OH+ (NH_4)_2SO_4$$

Solution

Question 7

Buffer solution have no $$pH$$

Its $$pH$$ changes very little when a small amount of acid or base is added to the it

Its $$pH$$ changes very largely when a small amount of acid or base is added to the it

All solutions are buffer

Solution

Question 8

$$1$$%

$$10$$%

$$50$$%

$$25$$%

Solution

Question 9

$$0.05$$ mole

$$0.025$$ mole

$$0.1$$ mole

$$0.005$$ mole

Solution

Question 10

Increase

Decrease

Unchange

Nothing can be said

Solution

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