Equilibrium Test

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Equilibrium Test - 38

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Question 1
1 / -0

The pH of a solution which is 0.1 M in HA and 0.5 M in NaA. $${K_a}$$ for HA is $$1.8 \times {10^{ - 6}}$$

A
5.44

B
6.44

C
6.0

D
4.73

Solution

$$ pH = pKa + log\frac{[salt]}{[acid]} $$

$$ pH = -log(1.8 \times 10^{-6}) + log(\dfrac{0.5}{0.1}) $$

$$\Rightarrow pH= 5.74 + 0.6989$$

$$ \therefore pH = 6.44$$

Hence, option B is correct .
Question 2
1 / -0

Ionic product of $$H_2O$$ is $$10^{-14}$$. The $$H^+$$ ion concentration in $$0.1\ M$$ $$NaOH$$ solution is:

A
$$10^{-11}\ M$$

B
$$10^{-13}\ M$$

C
$$10^{-1}\ M$$

D
$$10^{-4}\ M$$

Solution

Solution:- (B) $${10}^{-13} \; M$$
As we know that,
$${K}_{2} = \left[ {H}^{+} \right] \left[ {OH}^{-} \right]$$
Given:-
$${K}_{w} = {10}^{-14}$$
$$\left[ {OH}^{-} \right] = 0.1 = {10}^{-1}$$
$$\therefore \left[ {H}^{+} \right] = \cfrac{{10}^{-14}}{{10}^{-1}} = {10}^{-13}$$
Hence the $${H}^{+}$$ ion concentration will be $${10}^{-13} \; M$$.
Question 3
1 / -0

The mass of acetic acid present in 500 ml of solution in which it is 1% ionised ($$Ka$$ of $$CH_{3}COOH=1.8\times 10^{-5}$$)

A
5.4 g

B
12.6 g

C
6.4 g

D
10.8 g
Question 4
1 / -0

The value of observed and calculate molecular weights of silver nitrate are $$92.64$$ and $$170$$ respectively.

The degree of dissociation of silver nitrate will be :

A
$$60\%$$

B
$$83.5\%$$

C
$$85.3\%$$

D
$$41.6\%$$

Solution

Given ,

observed molecular weights of silver nitrate = 92.64 g

calculate molecular weights of silver nitrate = 170 g

i for $$AgNO_{3} = \frac{normal-mol.wt}{observed -mol.wt} $$

$$ \Rightarrow 1 + \alpha $$

putting values , we get

$$ \alpha = 0.835 = $$83%

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

$$A+B\rightleftharpoons C+D$$.If initially the concentration of A and B both equal but at equilibrium,concentration of D will be twice of that of A,then what will be the equilibrium constant of the reaction?

A
$$\cfrac {4}{9}$$

B
$$\cfrac {9}{4}$$

C
$$\cfrac {1}{9}$$

D
$$4$$

Solution

Option $$D$$ is the correct answer which is $$4$$

Consider the reaction is $$A+BC\rightleftharpoons D$$

initially $$\begin{array}{l} \begin{array} { *{ 20 }{ c } }a & a \end{array} \\ a-x \end{array}$$ $$\begin{array}{l} \begin{array} { *{ 20 }{ c } }0 & 0 \end{array} \\ \begin{array} { *{ 20 }{ c } }x & x \end{array} \end{array}$$

we have,

$$\begin{array}{l} \left[ D \right] =2\left[ A \right] \\ x=2\left( { a-x } \right) \\ \Rightarrow x=2a-2x \\ x=\frac { { 2a } }{ 3 } \end{array}$$

$$\begin{array}{l} { k_{ eq } }=\frac { { \left[ C \right] \left[ D \right] } }{ { \left[ A \right] \left[ B \right] } } \\ =\frac { { \left( { \frac { { 2a } }{ 3 } } \right) \left( { \frac { { 2a } }{ 3 } } \right) } }{ { \left( { a-\frac { { 2a } }{ 3 } } \right) \left( { a-\frac { { 2a } }{ 3 } } \right) } } =\frac { { { { \left( { \frac { { 2a } }{ 3 } } \right) }^{ 2 } } } }{ { { { \left( { \frac { a }{ 3 } } \right) }^{ 2 } } } } ={ 2^{ 2 } } \\ =4 \end{array}$$
Question 6
1 / -0

If 0.1 M of a weak acid is taken and its percentage of the degree of ionization is $$1.34\%$$, then its ionization constant will be:

A
$$0.8\times 10^{-4}$$

B
$$1.79\times 10^{-5}$$

C
$$0.182\times 10^{-4}$$

D
$$1\times 10^{-4}$$

Solution
Question 7
1 / -0

What is $$[{ NH }_{ 4 }^{ + }]$$ in a solution containing 0.02M $${ NH }_{ 3 }$$ ($${ K }_{ b }={ 1.8\times 10 }^{ -5 }$$) and 0.01M KOH?

A
$${ 1.8\times 10 }^{ -5 }$$

B
$${ 9\times 10 }^{ -6 }$$

C
$${ 3.6\times 10 }^{ -5 }$$

D
None of the above

Solution

Solution:
$$NH_4OH \rightleftharpoons NH_4^+ +OH^−$$

$$KOH\to K^+ + OH^-$$

$$K_b=\dfrac{[NH^+_4][OH^−]^2}{[NH_4OH]}$$

$$1.8×10^{−5}=\dfrac{[X][10^{−2}]}{2×10^{−2}}$$

$$\Rightarrow x=3.6×10^{−5}$$

Hence the correct option: C
Question 8
1 / -0

Four grams of $$NaOH$$ solid are dissolved in just enough water to make $$1$$ litre of solution. What is the$$[H^+]$$ of the solution?

A
$$10^{-2}$$ moles/litre

B
$$10^{-1}$$ moles/litre

C
$$10^{-12}$$ moles/litre

D
$$10^{-13}$$ moles/litre

Solution

Moles of $$NaOH=\dfrac{4}{40}=0.1$$

$$NaOH$$ dissolves into water, thus dissociating into its respective ions as

$$NaOH(aq.)\rightleftharpoons Na^+(aq.)+Cl^-(aq.)$$

Hence, $$[OH^-]=\dfrac{0.1}{1}=0.1\ moles/litre$$

Since, $$[H^+][OH^-]=10^{-14}\ moles/litre$$

Therefore, $$[H^+]=10^{-13}\ moles/litre$$
Question 9
1 / -0

$$K_a$$ for $$HF$$ is $$3.5\times 10^{-4}$$. Calculate $$K_b$$ for the fluoride ion.

A
$$3.5\times 10^{-4}$$

B
$$1.0\times 10^{-7}$$

C
$$2.9\times 10^{-11}$$

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

Solution

$$K_a = 3.5 \times 10^{-4}$$

$$K_b$$ for fluoride ion wills be :

$$K_b= \dfrac{K_w}{K_a}=\dfrac{1.0 \times 10^{-14}}{3.5 \times 10^{-4}}=2.9 \times 10^{-11}$$

$$\therefore $$ correct option is $$(C)$$
Question 10
1 / -0

$$C(s)+H_2O(g) \rightleftharpoons CO(g)+H_2(g)$$; $$ \Delta H < 0$$.

The above equilibrium will proceed in forward direction when _________.

A
it is subjected to high pressure

B
it is subjected to high temperature

C
inert gas (Argon) is added at constant pressure

D
carbon (solid) is added

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Equilibrium Test - 38

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Equilibrium Test - 38

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Question 1

5.44

6.44

6.0

4.73

Solution

Question 2

$$10^{-11}\ M$$

$$10^{-13}\ M$$

$$10^{-1}\ M$$

$$10^{-4}\ M$$

Solution

Question 3

5.4 g

12.6 g

6.4 g

10.8 g

Question 4

$$60\%$$

$$83.5\%$$

$$85.3\%$$

$$41.6\%$$

Solution

Question 5

$$\cfrac {4}{9}$$

$$\cfrac {9}{4}$$

$$\cfrac {1}{9}$$

$$4$$

Solution

Question 6

$$0.8\times 10^{-4}$$

$$1.79\times 10^{-5}$$

$$0.182\times 10^{-4}$$

$$1\times 10^{-4}$$

Solution

Question 7

$${ 1.8\times 10 }^{ -5 }$$

$${ 9\times 10 }^{ -6 }$$

$${ 3.6\times 10 }^{ -5 }$$

None of the above

Solution

Question 8

$$10^{-2}$$ moles/litre

$$10^{-1}$$ moles/litre

$$10^{-12}$$ moles/litre

$$10^{-13}$$ moles/litre

Solution

Question 9

$$3.5\times 10^{-4}$$

$$1.0\times 10^{-7}$$

$$2.9\times 10^{-11}$$

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

Solution

Question 10

it is subjected to high pressure

it is subjected to high temperature

inert gas (Argon) is added at constant pressure

carbon (solid) is added

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