Equilibrium Test

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Equilibrium Test - 37

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following will not show common ion effect on addition of $$HCl$$ ?

A
$$CH_3COOH$$

B
$$H_2S$$

C
$$C_6H_5COOH$$

D
$$H_2SO_4$$

Solution
Question 2
1 / -0

For preparing a buffer solution of $$pH=7.0$$, which buffer system you will choose?

A
$${ H }_{ 3 }{ PO }_{ 4 },{ H }_{ 2 }{ PO }_{ 4 }^{ - }$$

B
$${ H }_{ 2 }{ PO }_{ 4 }^{ - }, { HPO }_{ 4 }^{ 2- }$$

C
$${ HPO }_{ 4 }^{ 2- },{ PO }_{ 4 }^{ 3- }$$

D
$${ H }_{ 3 }{ PO }_{ 4 },{ PO }_{ 4 }^{ 3- }$$

Solution

Using $${H_2PO_4}^-$$, $${HPO_4}^{-2}$$, a buffer solution of $$pH=7$$ is prepared.
The equilibrium reaction involved in this solution will be dissociation of $${H_2PO_4}^{-}$$ in water.
$${H_2PO_4}^- + H_2O \rightleftharpoons {HPO_4}^{2-}+{H_3O}^+$$
For phosphate buffer, $$p{K_a}$$ value of $${H_2PO_4}^-$$ is equal to $$7.2$$ so that buffer system is suitable for $$pH$$ range $$7.2 \pm 1$$ or from $$6.2$$ to $$8.2$$
Question 3
1 / -0

Which of the following is an example of weak electrolyte?

A
$$H_3BO_3$$

B
$$H_2SO_4$$

C
$$HNO_3$$

D
$$HClO_3$$

Solution

Boric acid is very weak acid since it dose not give $$H^+$$ , instead it takes $$OH^-$$ from $$H_2O$$ and become $$B{ \left( OH \right) }_{ 4 }^{ - }$$ and Proton is lost from $$H_2O$$ not from boric acid.
Question 4
1 / -0

A monoprotic acid in $$100\ M$$ solution is $$0.001\%$$ ionized. The dissociation constant of this acid is:

A
$$1.0\ \times 10^{-3}$$

B
$$1.0\ \times 10^{3}$$

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

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

Solution

$$HA\leftrightharpoons H^{ + }+A^{ - }$$

% ionization $$=0.001$$ %
$$[HA]=100\ M$$

Ionization percent $$=\dfrac { [{ H }^{ + }] }{ [HA] } \times 100$$

$$0.001=\dfrac { [{ H }^{ + }]\times 100 }{ 100 } \\ [{ H }^{ + }]=0.001\\ K_{ a }=\dfrac { [{ H }^{ + }][{ A }^{ - }] }{ [HA] }$$

$$=\dfrac { 0.001\times 0.001 }{ 100 }$$

$$ =1\times 10^{ -8 }$$
Question 5
1 / -0

Ionization constant $$({ K }_{ a })$$ for three weak monobasic acids $$HA$$, $$HB$$ and $$HD$$ are $${ 10 }^{ -3 },{ 10 }^{ -7 }$$ and $${ 10 }^{ -9 }$$ ,respectively at $${ 25 }^{ o }C$$. Which of the following is correct prediction?

A
$${ (pH) }_{ NaA\ }>\ ({ pH) }_{ NaB }$$

B
$${ (pH) }_{ NaA }<\ ({ pH) }_{ NaB }$$

C
$${ (pH) }_{ NaA }> ({ pH) }_{ NaD }$$

D
$${ (pH) }_{ NaB }=7$$

Solution

$${ pH }=p{ K }_{ a }+\log _{ 10 }{ (\cfrac { [salt] }{ [Acid] } ) } $$ {Henderson Hassel batch equation}
$$\therefore { (pH) }_{ NaA }<{ (pH) }_{ NaB }$$
Question 6
1 / -0

One liter of solution contains $${ 10 }^{ -5 }$$ moles of $${ H }^{ + }$$ ions at $$ { 25 }^{ o }C$$.Percentage ionisation of water in solution is:

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

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

C
$$3.6\times { 10 }^{ -9 }\%$$

D
$$1.8\times { 10 }^{ -11 }\%$$

Solution

Molarity of water= $$55.5$$ mole
$$H_2O \rightleftharpoons H^++OH^-$$
Let $$x$$ be the degree of dissociation
Then $$55.5x= 10^{-5}$$
$$\Rightarrow x= 1.8 \times 10^{-7}$$
% $$x= 1.8 \times 10^{-7}\times 100= 1.8 \times 10^{-5}$$
Question 7
1 / -0

When 0.1 mole of an acid is added to 2 lit of a buffer solution, the pH of the buffer decreases by 0.5. The buffer capacity of the solution is:

A
0.6

B
0.4

C
0.2

D
0.1

Solution

$$0.1$$ moles of acid is added to $$2L$$ of buffer solution. $$pH$$ of buffer decreases by $$0.5$$.

We know, Buffer capacity $$=\dfrac { No.\quad of\quad moles\quad of\quad acid\quad or\quad base\quad added\quad to\quad 1\quad litre\quad of\quad solution }{ change\quad in\quad pH } $$

$$\phi =\dfrac { \dfrac { 0.1 }{ 2 } }{ 0.5 } $$

$$\phi =\dfrac { 1 }{ 10 } $$

$$\phi =0.1$$

buffer capacity of solution is $$0.1$$
Question 8
1 / -0

Which of the following has maximum $$pK_a$$ :-

A
$$CH_2FCOOH$$

B
$$CH_3ClCOOH$$

C
$$CH_3COOH$$

D
$$HCOOH$$

Solution

The correct answer is $$(A)$$.
Because,
$$C{H_2}FCOOH$$ is the strongest acid in the following:
Hence,
Fluorine is an electron withdrawing group the acidic strength increases.
Question 9
1 / -0

At $$25^{o}C$$, the dissociation constant for pure water is given by :-

A
$${ (55.4\times { 10 }^{ 14 } })^{ -1 }$$

B
$$ 1 \times { 10 }^{ -14 }$$

C
$$\dfrac { 1\times { 10 }^{ -14 } }{ 18 } $$

D
$$none\ of\ these$$

Solution

Pure water contains ions. There is self-ionisation constant of water, $${ K }_{ w }$$, which has no units. It is the product of the hydronium ions times the hydroxide ions,
$${ K }_{ w }=\left[ { H }_{ 3 }{ O }^{ + } \right] \left[ { OH }^{ - } \right] $$
the ions are equal in concentrations, which experiments have shown to be $$1.0\times { 10 }^{ -7 }$$ for each one at $${ 25 }^{ 0 }C$$
$$\Rightarrow { K }_{ w }=\left( { 10 }^{ -7 } \right) \left( { 10 }^{ -7 } \right) $$
$${ K }_{ w }=1.0\times { 10 }^{ -14 }$$
$$\Rightarrow$$ At $${ 25 }^{ 0 }C$$, the dissociation constant for pure water is given by $$1.0\times { 10 }^{ -14 }$$.
Question 10
1 / -0

Which of the following is a biodegradable polymer?

A
Cellulose

B
PVC

C
Nylon-6

D
Polythene

Solution

The biodegradable polymer is cellulose because many types of microorganisms are known to biodegrade cellulose. Cellulose is the main constituent of plant cell walls which make the leaves, branches strong and hard. Bacteria and fungi are of particular interest because they are the most widely available degrading microorganisms.

Hence, option A is correct.

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Equilibrium Test - 37

Result

Equilibrium Test - 37

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Question 1

$$CH_3COOH$$

$$H_2S$$

$$C_6H_5COOH$$

$$H_2SO_4$$

Solution

Question 2

$${ H }_{ 3 }{ PO }_{ 4 },{ H }_{ 2 }{ PO }_{ 4 }^{ - }$$

$${ H }_{ 2 }{ PO }_{ 4 }^{ - }, { HPO }_{ 4 }^{ 2- }$$

$${ HPO }_{ 4 }^{ 2- },{ PO }_{ 4 }^{ 3- }$$

$${ H }_{ 3 }{ PO }_{ 4 },{ PO }_{ 4 }^{ 3- }$$

Solution

Question 3

$$H_3BO_3$$

$$H_2SO_4$$

$$HNO_3$$

$$HClO_3$$

Solution

Question 4

$$1.0\ \times 10^{-3}$$

$$1.0\ \times 10^{3}$$

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

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

Solution

Question 5

$${ (pH) }_{ NaA\ }>\ ({ pH) }_{ NaB }$$

$${ (pH) }_{ NaA }<\ ({ pH) }_{ NaB }$$

$${ (pH) }_{ NaA }> ({ pH) }_{ NaD }$$

$${ (pH) }_{ NaB }=7$$

Solution

Question 6

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

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

$$3.6\times { 10 }^{ -9 }\%$$

$$1.8\times { 10 }^{ -11 }\%$$

Solution

Question 7

0.6

0.4

0.2

0.1

Solution

Question 8

$$CH_2FCOOH$$

$$CH_3ClCOOH$$

$$CH_3COOH$$

$$HCOOH$$

Solution

Question 9

$${ (55.4\times { 10 }^{ 14 } })^{ -1 }$$

$$ 1 \times { 10 }^{ -14 }$$

$$\dfrac { 1\times { 10 }^{ -14 } }{ 18 } $$

$$none\ of\ these$$

Solution

Question 10

Cellulose

PVC

Nylon-6

Polythene

Solution

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