MCQ Questions for Class 11 Medical Chemistry Equilibrium Quiz 7

What is rate?

0%
Change / Time
0%
Time / Concentration
0%
Time / Change
0%
Rate / Time
0%
Concentration / Time

Explanation

The rate is the change of concentration with time.
Rate

$\displaystyle = \dfrac {d}{dt}C$

Assertion:

$NaCl(aq)$ is an electrolyte.
Reason: It forms ions in solution.

0%
Both Assertion and Reason are true and Reason is the correct explanation of Assertion
0%
Both Assertion and Reason are true but Reason is not the correct explanation of Assertion
0%
Assertion is true but Reason is false
0%
Assertion is false but Reason is true
0%
Both Assertion and Reason are false

Explanation

pure water is a very good insulator, but if we add sodium chloride we create ions to form an electrolytic solution.

$NaCl(s) + H_2O(l) \longrightarrow Na^+(aq) + Cl^-(aq) + H_2O(l)$

Therefore, $NaCl$ is an electrolyte.

Statement I: Water makes a good buffer.
Statement II: A good buffer will resist changes in pH.

0%
Statement 1 and Statement 2 are correct and Statement 2 is the correct explanation of Statement 1
0%
Both the Statement 1 and Statement 2 are correct and Statement 2 is NOT the correct explanation of Statement 1
0%
Statement 1 is correct but Statement 2 is not correct
0%
Statement 1 is not correct but Statement 2 is correct
0%
Both the Statement 1 and Statement 2 are not correct.

The Haber process is used for producing ammonia from nitrogen and hydrogen. This reaction could be forced to produce more ammonia by :

0%
Increasing the reaction pressure
0%
Decreasing the reaction pressure
0%
Adding a catalyst
0%
Both b and c
0%
None of the above

The hydrogen sulfate or bisulfate ion,

$H{SO}_{4}^{-}$ , can act as either an acid or a base in water solution. In which of the following equations does

$H{SO}_{4}^{-}$ act as an acid?

0%
$H{SO}_{4}^{-}+{H}_{2}O\rightarrow {H}_{2}{SO}_{4}+{OH}^{-}$
0%
$H{SO}_{4}^{-}+{H}_{3}{O}^{+}\rightarrow {SO}_{3}+2{H}_{2}O$
0%
$H{SO}_{4}^{-}+{OH}^{-}\rightarrow {H}_{2}{SO}_{4}+{O}^{2-}$
0%
$H{SO}_{4}^{-}+{H}_{2}O\rightarrow {SO}_{4}^{2-}+{H}_{3}{O}^{+}$
0%
None of these

Explanation

Option A:

${ H{ SO }_{ 4 } }^{ - }$ acts as base as it accepts

${ H }^{ + }$ ion

Option B:

${ H{ SO }_{ 4 } }^{ - }$ acts as base as it gives away

${ OH }^{ - }$ ion

Option C:

${ H{ SO }_{ 4 } }^{ - }$ acts as base as it accepts

${ H }^{ + }$ ion

${ H{ SO }_{ 4 } }^{ - }$ acts as acid as it gives away

${ H }^{ + }$ ion

Hence, D is the answer.

What does a buffer do?

0%
It keeps the $pH$ of a solution from changing very much.
0%
It maintains the equilibrium of a reaction.
0%
It maintains the amount of acid in a solution
0%
It maintains the amount of base in a solution
0%
It keeps the $pH$ of a solution from changing at all.

Explanation

A buffer keeps the

$pH$ of a solution from changing very much.
A buffer is a solution that contains equal amounts of a weak acid and its conjugate base, or a weak base and its conjugate acid.
Such solution resists the change in

$pH$ when a small amount of an acid or a base is added.
For example, a mixture of acetic acid and sodium acetate acts as an acid buffer solution. Similarly, a mixture of ammonium hydroxide and ammonium chloride is a basic buffer solution.

It accepts a proton. It is called as:

0%
a Bronsted acid
0%
a Bronsted base
0%
a strong acid
0%
a weak base

Explanation

A Bronsted acid can accept a proton

$(H^+)$ . For example a basic salt, such as

$Na^{\oplus}H^ {\circleddash}$ , can take up a

$H^+$ from

$H_2O$ to form

$OH^{\circleddash}$ ions.

$F^ {\circleddash}(aq)+H_2O(l) \rightleftharpoons HF (aq)+OH^ {\oplus}(aq)$

The type of bonding present in a substance can be determined from experimental data.
Based on the information in the table below, which of the following substances is MOST likely to represent the identity of the solid?

Melting Point	Density	Conductivity	Conductivity in Aqueous Solution
$801^{\circ}C$	$2.17\ g/cc$	None	Excellent

0%
Platinum, $Pt$
0%
Sodium chloride, $NaCl$
0%
Glucose, $C_{6}H_{12}O_{6}$
0%
Quartz, $SiO_{2}$

Explanation

$(B)$ Sodium chloride,

$NaCl$

$Reason$ : Sodium chloride is an ionic compound with the chemical formula

$NaCl$ , in which sodium(cation)

$Na^{+}$ and chloride(anion)

$Cl^{-}$ ions are in the ratio

$1 : 1$ .It is commonly known as table salt, common salt or halite. Seawater is the major source of this salt.

Some properties of sodium chloride are as follows -

$1)$ It is easily soluble in water and partially or insoluble in other liquids.

$2)$ They are white crystals with a desity of

$2.17 g/cc$ , which do not have an odor but possess a taste.

$3)$ In its aqueous state

$NaCl$ acts as a good conductor of electricity due to the free movement of ions.

$4)$ It has a melting point of

$801^{o} C$ and a boilng point of

$1,413 ^{o} C$ .

$Uses$ :

$1)$ This salt is used in glass production.

$2)$ It is widely used in food industries as a food preservative and as a flavour enhancer.

$Ca{SO}_{4}$ is somewhat soluble in water.

$I$ . When

${H}_{2}{SO}_{4}$ is added to a solution of

$Ca{SO}_{4}$ , the solubility of the

$Ca{SO}_{4}$ will be increased.

$II$ . The addition of

${H}_{2}{SO}_{4}$ will lower the

$pH$ of the solution.

0%
Statement $I$ is true, Statement $II$ is true
0%
Statement $I$ is true, Statement $II$ is false
0%
Statement $I$ is false, Statement $II$ is true
0%
Statement $I$ is false, Statement $II$ is false

Explanation

When

$H_2SO_4$ is added to a

$CaSO_4$ solution, due to common ion effect, the solubility of

$CaSO_4$ decreases. Further, as the acid concentration increases, the

$pH$ of the solution lowers due to addition of

$H_2SO_4$ .

Hence, statement

$I$ is false, statement

$II$ is true.

The colour of bleaching powder is :

0%
green
0%
pale yellow
0%
blue
0%
white

Addition of water to this solution will not change

$[H_3 O^+]$ .

0%
Chemical pH indicator
0%
Acid/base buffer
0%
Anhydrous solution
0%
Hypotonic solution

Explanation

Addition of water to

$X$ solution does not change the

$pH$ of the solution, which means the concentration of the species present changes in such a way that the

$pH$ remains the same. This is possible for acid/base buffer.

$pH=pk_a+\log \cfrac{[salt]}{[acid]}$

Since the volume changes the same for both salt and acid, the ratio

$\cfrac{[salt]}{[acid]}$ remains the same and hence

$pH$ also remains the same. Same in the case of basic buffer solution.

Which of the following items are not bleached by bleaching powder?

0%
Straw
0%
Silk
0%
Ivory
0%
None of the above

For each pair of chemicals below, the amounts shown are dissolved, together, in water to make 1.0 L of solution. Which mixture will form a buffer solution?

0%
0.2 mol of $\displaystyle HCl$ and 0.1 mol of $\displaystyle { K }_{ 2 }{ SO }_{ 3 }$
0%
0.2 mol of $\displaystyle NaOH$ and 0.4 mol of $\displaystyle HF$
0%
0.1 mol of $\displaystyle HBr$ and 0.1 mol of $\displaystyle Ba{ \left( OH \right) }_{ 2 }$
0%
0.4 mol of $\displaystyle HCl$ and 0.2 mol of $\displaystyle { NH }_{ 3 }$
0%
0.2 mol of $\displaystyle NaOH$ and 0.4 mol of $\displaystyle HCl$

Explanation

A buffer solution is an aqueous solution consisting of a mixture of an acid and its conjugate base. Its

$pH$ changes very little when a small amount of strong acid or base is added to it. Hence solution of

$NaOH /HF$ is a buffer solution.

Bleaching powder loses ___________ by the action of dilute acids in excess.

0%
oxygen
0%
calcium
0%
chlorine
0%
water

Explanation

Reaction of bleaching powder with

$dil.HCl$ :

$CaOCl_2 + 2HCl \rightarrow CaCl_2 + Cl_2 + H_2O$

On long standing, bleaching powder undergoes auto-oxidation into:

0%
$Ca(ClO_3)_2, CaCl_2$
0%
$CaO, CaCl_2$
0%
$CaClO, CaClO_3$
0%
$CaClO_2, CaCl_2$

Explanation

On long standing, bleaching powder undergoes auto-oxidation into calcium chlorate

$Ca(ClO_3)_2$ and

$CaCl_2$ .

$6CaOCl_2 \rightarrow 6 Ca(ClO_3)_2 + 5 CaCl_2$ .
The bleaching powder is

$CaOCl_2$

Bleaching powder loses oxygen in the presence of:

0%
calcium chloride
0%
cobalt chloride
0%
sodium chloride
0%
none of the above

Explanation

Bleaching powder loses oxygen in the presence of cobalt chloride catalyst.

$2CaOCl_2 \xrightarrow {CoCl_2} 2CaCl_2 + O_2$

State whether the given statement is true or false.

Ideal solutions doesn't obey Roult's law.

0%
True
0%
False

Which of the following solutions show positive deviation from Raoult's law?

0%
Acetone + Ethanol
0%
Acetone + Benzene
0%
Water + Methanol
0%
All of the above

Which of the following compounds will be dissociated into its constituent ions?

0%
Aqueous $KCl$
0%
Solid $KCl$
0%
Aqueous $MgCl_2$
0%
Solid $MgCl_2$

Explanation

When electric current is passed through an electrolyte in its molten or aqueous state, the electrolyte breaks up into its constituent ions.
When aqueous

$KCl$ undergoes electrolysis, the ions

$K^+$ and

$Cl^-$ are formed where potassium transfers its one valence electron to Chlorine. So potassium becomes positively charged and chlorine becomes negatively charged.
When aqueous

$MgCl_2$ undergoes electrolysis, the ions

$Mg^{2+}$ and two

$Cl^-$ ions are formed where magnesium transfers its two valence electron to chlorine. Thus magnesium gets two positive charges and each chlorine gets one negative charge.

An example for a neutral buffer is :

0%
ammonium hydroxide and ammonium chloride
0%
acetic acid and sodium acetate
0%
acetic acid and ammonium hydroxide
0%
citric acid and sodium citrate

Which of the following is false of a binary solution showing false deviation from Roult's law?

0%
$P_A < X_AP_A^o$
0%
$\Delta H_{mix} < 0$
0%
$\Delta V_{mix} < 0$
0%
None of the above

Dissociation events in

$H_2O \rightarrow H^+ +OH^-$ is termed as ionization because :

0%
the electron is initially shared between both atoms, thus the dissociation event into ions involves the transfer of an electron from one atom to the other
0%
it does not involve any electron transfer
0%
charged species formed
0%
no charge species formed

Explanation

$H_2O\longrightarrow H^++OH^-$

This dissociation is termed as ionization because initially there is a polar covalent compound

$H_2O$ .

This dissociation leads to the formation of ions

$H^+$ and

$OH^-$

This dissociation involves a transfer of an electron from one atom to another leading to the formation of charged species.

When HCl gas is passed through a saturated solution of common salt, pure NaCl is precipitated because __________.

0%
HCl is highly ionised in solution
0%
HCl is highly soluble in water
0%
The solubility product of NaCl is lowered by HCl
0%
The ionic product of $[Na^+] [Cl^-]$ exceeds the solubility product of NaCl

Explanation

As the ionic product of

$[Na^+][Cl^-]$ exceeds the solubility product, NaCl is precipitated as the precipitation occurs only when ionic product exceeds solubility product.

Name of the organic acid present in tomato is :

0%
tartaric acid
0%
malic acid
0%
lactic acid
0%
oxalic acid

Sulfuric acid in air causes:

0%
hair loss
0%
acid precipitation
0%
skin problems
0%
respiratory problems

Which of the following statement is correct?

0%
Sodium hydroxide turns blue litmus red
0%
Nitric acid turns red litmus blue
0%
Sodium hydroxide turns red litmus blue
0%
Nitric acid does not change blue litmus to red

Which two sets of reactants best represents the amphoteric character of

$Zn{ \left( OH \right) }_{ 2 }$ ?
Set I

$Zn{ \left( OH \right) }_{ 2 }\left( s \right)$ and

$OH\left( aq \right)$
Set II

$Zn{ \left( OH \right) }_{ 2 }\left( s \right)$ and

${ H }_{ 2 }O\left( I \right)$
Set III

$Zn{ \left( OH \right) }_{ 2 }\left( s \right)$ and

${ H }^{ + }\left( aq \right)$
Set IV

$Zn{ \left( OH \right) }_{ 2 }\left( s \right)$ and

$N{ H }_{ 3 }\left( aq \right)$

0%
III and II
0%
I and III
0%
IV and I
0%
II and IV

Explanation

The amphoteric character of

$Zn{ \left( OH \right) }_{ 2 }$ is represented by I and III.

$Zn(OH)_2+2OH^-\rightarrow ZnO_2^{2-}+2H_2O$

$Zn(OH)_2+2H^+\rightarrow Zn^{2+}+2H_2O$

A chemical which turns red litmus to blue is:

0%
acidic in nature
0%
basic in nature
0%
neutral in nature
0%
can be acidic or basic in nature

Explanation

Litmus paper is used to find out the acidic or basic nature of a solution. Following observations takes place when an acidic or basic solution interact with litmus paper.

$\left.\begin{matrix} Acid & \rightarrow & Blue\, to \, Red \\ Base & \rightarrow & Red \, to \, Blue \end{matrix}\right\}$ Litmus paper colour change

Thus, the chemical compound would be basic in nature.

How many grams of ammonium chloride should be dissolved in 500 mL of water to have a solution of pH 4.5?

[

$\displaystyle K_b$ for ammonium hydroxide is

$\displaystyle1.8 \times 10^{-5}$ ]

0%
25.22
0%
48.15
0%
53.5
0%
90

Explanation

$\displaystyle [H^+]=10^{-pH}=10^{-4.5}=3.162 \times 10^{-5} mol/L$

$\displaystyle [H^+]=\sqrt {\dfrac {K_wC}{K_b}}$

$\displaystyle C = \dfrac {[H^+]^2K_b}{K_w}$

$\displaystyle C = \dfrac {3.162 \times 10^{-5} \times 3.162 \times 10^{-5} \times 1.8 \times 10^{-5}}{1 \times 10^{-14}}=1.8 \: mol/L$
Number of moles of ammonium chloride present in 500 mL

$\displaystyle = \dfrac {1.8 \times 500}{1000}=0.9$ moles.
The molar mass of ammonium chloride is 53.5 g/mol. The mass of ammonium chloride will be

$\displaystyle 0.90 \times 53.5 = 48.15$ g

The process in which metal surface is made inactive is called:

0%
passivation
0%
galvanizing
0%
corrosion
0%
pickling

Ammonia is basic in nature. Hence it turns:

0%
red litmus to blue
0%
blue litmus to red
0%
white litmus to red
0%
white litmus to blue

Explanation

Red litmus turns blue under basic or alkaline conditions.

Ammonia is basic in nature hence it will turn red litmus to blue.

Hence, option

$A$ is correct.

Degree of ionisation does not depend on:

0%
nature of the solvent
0%
nature of the electrolyte
0%
dilution
0%
molecular mass of the electrolyte

Explanation

Degree of ionization

$(\alpha)$ depends on-

1) Concentration of solute

2) Temperature

3) Nature of electrolysis

4) Nature of solvent

5) Dilution

$\alpha$ " does not depend on molecular mass of electrolyte.

Curd taste sour because this is a/an :

0%
base
0%
salt
0%
acid
0%
liquid

Explanation

The bacteria Lactobacillus (present in a small quantity) when added to a large amount of milk, converts lactose to lactic acid, which makes the curd sour in taste.

Hence the option

$C$ is correct.

When a blue litmus paper is dipped in a solution, it remains blue. What can be concluded about this solution?

0%
Th solution is neutral.
0%
The solution is basic or neutral.
0%
The solution is acidic or neutral.
0%
None of the above.

The number of ions given by one molecule of

$K_{4}Fe(CN)_{6}$ after complete dissociation is____________.

0%
$5$
0%
$11$
0%
$2$
0%
$10$

Explanation

$K_4Fe(CN)_6\longrightarrow 4K^++[Fe(CN)_6]^{-4}$

Total number of ions on complete dissociation=

$5$

As ions in co-ordination sphere will not dissociate so

$Fe^{2+}$ and

$CN^-$ ions will not dissociate into their respective ions.

Which of the following will not function as a buffer solution?

$(i)$

$NaCl$ and

$NaOH$

$(ii)$

$NaOH$ and

${NH}_{4}OH$

$(iii)$

${CH}_{3}COO{NH}_{4}$ and

$HCl$

$(iv)$ Borax and boric acid

0%
$(i),(ii),(iii)$
0%
$(ii),(iii),(iv)$
0%
$(i),(iii),(iv)$
0%
$(i),(ii),(iii),(iv)$

Explanation

A buffer solution is a salt of either weak acid and its salt with a strong base (

$AcOH + NaOAc$ ) or a weak base and its salt with a strong acid (

$NH_3+NH_4Cl$ ).

(i)

$NaCl$ and

$NaOH$ consists of salt of strong acid and strong base.

(ii)

$NaOH$ and

$NH_4OH$ consists of a weak base and strong base.

(iii)

$CH_3COONH_4$ and

$HCl$ consists of strong acid and a salt.

(iv) Borax and boric acid is an acid buffer solution. It consists of a weak acid (boric acid) and it's salt (borax) with a strong base.

Hence, option A is correct.

The state of equilibrium refers to:

0%
State of rest
0%
Dynamic state
0%
Stationary state
0%
State of inertness

Acid turns blue litmus to:

0%
White
0%
Yellow
0%
Red
0%
Green

A monoprotic acid in

$1.00M$ solution is

$0.001$ % ionised. The dissociation constant of acid is:

0%
$1\times { 10 }^{ -3 }$
0%
$1\times { 10 }^{ -6 }$
0%
$1\times { 10 }^{ -8 }$
0%
$1\times { 10 }^{ -10 }$

Explanation

According to Ostwald's dilution law

$K=\cfrac { { \alpha }^{ 2 }C }{ \left( 1-\alpha \right) }$

As degree of ionisation is very less than 1 so above equation reduces to

$K=C\alpha^2$

$K=1(10^{-5})^2$

$K=10^{-10}$

Which pair will show common ion effect?

0%
$Ba{ Cl }_{ 2 }+Ba{ \left( { NO }_{ 3 } \right) }_{ 2 }$
0%
$NaCl+HCl$
0%
${ NH }_{ 4 }OH+{ NH }_{ 4 }Cl$
0%
$AgCN+KCN$

Explanation

Correct Answer: Options C and D.

Explanation:

Under the influence of a common ion, the solubility of one electrolyte is decreased. This is known as the common ion effect.
Due to the common ion effect, the disassociation reaction of the weak species starts to occur in the opposite direction.
Common ion effect can only take place if at least one electrolyte is weak and there is a common ion present among the species.
If both species are strong, then they are $100\%$ disassociated. Thus, no equilibrium is set up rather $100\%$ dissociation takes place.

1. $BaCl_{2}$ and $Ba(NO_{3})_{2}$
Both have $Ba^{2+}$ as a common ion.
$BaCl_{2}$ is a salt of $Ba(OH)_{2}$ and $HCl$ . $Ba(NO_{3})_{2}$ is a salt of $Ba(OH)_{2}$ and $HNO_{3}$ . Since all of these are strong acids and bases, these two species don’t exhibit the common ion effect.

2. $NaCl$ and $HCl$
Both have $Cl^{-}$ as a common ion.
$NaCl$ is a salt of $Na(OH)$ and $HCl$ . $HCl$ is a strong acid itself. Since all of these are strong acids and bases, these two species don’t exhibit the common ion effect.

3. $NH_{4}OH$ and $NH_{4}Cl$
Both have $NH_{4}^{+}$ as a common ion.
$NH_{4}OH$ and $NH_{4}Cl$ are both weak species since $NH_{4}^{+}$ is a weak species. Therefore, these two can exhibit the common ion effect.

4. $AgCN$ and $KCN$
Both have $CN^{-}$ as a common ion.
$KCN$ is a salt of a weak acid and strong base and $AgCN$ is a sparingly soluble salt, they have $CN^-$ ion in common. So they will show the common ion effect.

Hence, Options $C$ and $D$ will show the common ion effect.

When

${ NH }_{ 4 }Cl$ is added to

${ NH }_{ 4 }OH$ solution, the dissociation of ammonium hydroxide is reduced. It is due to:

0%
common ion effect
0%
hydrolysis
0%
oxidation
0%
reduction

Explanation

When

$NH_4Cl$ is added to

$NH_4OH$ solution, concentration of

$NH_4^{+}$ ions increases so the equilibrium shift towards left.So the dissociation of ammonium hydroxide is reduced.

A monoprotic acid in

$1.00M$ solution is

$30$ % ionised. The dissociation constant of acid is related degree dissociation as:

0%
${ \alpha }^{ 2 }C+\alpha K-K=0$
0%
${ \alpha }^{ 2 }C-\alpha K-K=0$
0%
${ \alpha }^{ 2 }C-\alpha K+K=0$
0%
${ \alpha }^{ 2 }C+\alpha K+K=0$

Explanation

The given solution is slightly ionized, which indicates it is a weak electrolyte. For weak electrolytes, we can use Ostwald's dilution law.

According to Ostwald's dilution law

$K_a=\cfrac { { \alpha }^{ 2 }C }{ \left( 1-\alpha \right) }$

Where,

$K_a$ = Dissociation constant of a weak acid.

$\alpha$ = Degree of dissociation,

$C$ = Concentration of weak acid.

On solving the above formula for weak electrolyte we get.

$\alpha^2C+\alpha K-K=0$

Hence, the correct option is

$\text{A}$

The ionisation constant of acetic acid is

$1.8\times { 10 }^{ -5 }$ . The concentration at which it will be dissociated to

$2$ % is:

0%
$1M$
0%
$0.045M$
0%
$0.018M$
0%
$0.45M$

Explanation

When acetic acid is dissolved in water, it partially dissociates (2%).

Ionisation constant

$K_a=1.87\times 10^{-5}$

$K_a=\cfrac{C\alpha \times C\alpha}{C(1-\alpha)}$

We assume

$\alpha\sim 0\Rightarrow 1-\alpha\sim 1$

$\Rightarrow 1.8\times 10^{-5}=\cfrac{C\alpha^2}{1}={C\times 0.02\times 0.02}$

$\Rightarrow C = \cfrac{1.8\times 10^{-5}}{0.02^2}$

$=0.045M$

Correct answer is option-B.

One litre of water contains

${ 10 }^{ -7 }$ mole of

${H}^{+}$ ions. Degree of ionisation of water is:

0%
$5 . 4\times { 10 }^{ -7 }$
0%
$1.8\times { 10 }^{ -7 }$
0%
$0.8\times { 10 }^{ -9 }$
0%
$5. 4\times { 10 }^{ -9 }$

Explanation

$H_2O⇌H^{ + } + OH^{ - }\quad$

$C$

$0$

$C(1-\alpha )$

$C\alpha$

$C\alpha=10^{-7}$

[H2O] =

$\dfrac{ 1000}{18}$ = 55.55 M

C= 55.5M

$\alpha=18\times10^{-10}$

In percentage

$\alpha=1.8\times10^{-7}$ %

Which one does not give a buffer solution?

0%
Ammonia and sodium hydroxide in water
0%
Sodium acetate and acetic acid in water
0%
Ammonia and ammonium chloride in water
0%
Sodium acetate and hydrochloric acid in water

A certain weak acid has dissociation constant of

$1.0\times { 10 }^{ -4 }$ . The equilibrium constant for its reaction with a strong base is:

0%
$1.0\times { 10 }^{ -4 }$
0%
$1.0\times { 10 }^{ -10 }$
0%
$1.0\times { 10 }^{ 10 }$
0%
$1.0\times { 10 }^{ -14 }$

Explanation

Hence, the correct option is

$\text{C}$
1469734_662618_ans_3a9d73ac3bfc4d6eb544299c30b21ecc.png

The pH of a neutral water is

$6.5$ . Then the temperature of water:

0%
Is ${ 25 }^{ o }C$
0%
Is more than ${ 25 }^{ o }C$
0%
Is less than ${ 25 }^{ o }C$
0%
Can be more or less than ${ 25 }^{ o }C$
0%
Cannot be predicted

Explanation

The ionization increases with temperature. So the amount of

$H^+$ active in solution will increase with temperature, the pH should decrease.

When there is decrease in pH of neutral water, it is normally assumed that the temperature is increased from room temperature

$25^0C.$

So when the pH of neutral water is 6.5, the temperature should greater than

$25^0C.$

The dissociation constants of two acids

$H{ A }_{ 1 }$ and

$H{ A }_{ 2 }$ are

$3.0\times { 10 }^{ -4 }$ and

$1.8\times { 10 }^{ -5 }$ respectively. The relative strengths of the acids will be:

0%
$1:4$
0%
$4:1$
0%
$1:16$
0%
$16:1$

Explanation

The dissociation constants of

$HA_1$ and

$HA_2$ are

$3\times10^{-4}$ and

$1.8\times10^{-5}.$

The strength of an acid is directly proportional to square root of dissociation constants of acids. So relative strength of the given acids are:

$\dfrac { { (Acidic\ Strength })_{ HA_1 } }{ { (Acidic\ Strength })_{ HA_2} } =\dfrac { \sqrt { { (Dissociation\ Constant })_{ HA_1} } }{ \sqrt { { (Dissociation\ Constant })_{ HA_2 } } }$

Relative Acidic Strength=

$\dfrac { { (Acidic Strength })_{ HA_1 } }{ { (Acidic Strength })_{ HA_2 } } =\dfrac { \sqrt { 3.0\times { 10 }^{ -4 } } }{ \sqrt { 1.8\times { 10 }^{ -5 } } } =4.08=4(approx)$

relative strengths of acids will be

$4:1.$

The

$pH$ of an aqueous solution of a

$0.1M$ solution of a weak monoprotic acid which is

$1$ % ionised is:

0%
$1$
0%
$2$
0%
$3$
0%
$11$

Explanation

Concentration of solution

$= C = 0.1M$

Dissociation is

$\alpha=0.01$

After dissociation, the concentration of

$H^+$ ions will be

$C\alpha=0.1\times 0.01=10^{-3}=[H^+]$

We know the fomrula for pH as

$pH=-log[H^+]=-log(10^{-3})=+3$

Which of the following statements is correct?

0%
$p{ K }_{ w }$ increases with increase of temperature
0%
$p{ K }_{ w }$ decreases with increase of temperature
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$p{ K }_{ w }=14$ at all temperatures
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$p{ K }_{ w }=pH$ at all temperatures

Explanation

As temperature increases dissociation will increase so number of

$H^{+}$ and

$OH^{-}$ will increase.As

$K_w=[H^{+}][OH^{-}]$ so

$K_w$ will increase so

$pK_w$ will decrease with increase in temperature.

CBSE Questions for Class 11 Medical Chemistry Equilibrium Quiz 7 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Equilibrium Quiz 7 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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