MCQ Questions for Class 11 Medical Chemistry Equilibrium Quiz 4

A saturated salt water solution was heated and allowed to cool without adding any more salt in it.

Which of the following will take place?

0%
Some salt appears to settle at the bottom.
0%
Some more salt can be dissolved now.
0%
No change takes place.
0%
Both $A$ or $B$

Ionic solutions are good conductors of Electricity.

0%
True
0%
False

Dilute sulphuric acid splits into:

0%
oxygen ions and hydrogen ions.
0%
oxygen ions, hydrogen ions and sulphur ions.
0%
hydrogen ions, oxygen ions and sulphate ions.
0%
hydrogen ions and sulphate ions.

Explanation

In a solution, dilute sulphuric acid splits into ions by the following reaction :-

$H_2SO_4 \rightarrow 2H^+ + SO_4^{2-}$

Thus dilute sulphuric acid splits into two hydrogen atoms (

$2H^+$ ) and a sulphate ion

$SO_4^{2-}$ .

Hence, the correct answer is option

$(D)$ .

Solubility of

$AgCN$ is maximum in :

0%
acidic buffer solution
0%
basic buffer solution
0%
in pure water
0%
equal in all solution

Explanation

Reaction:

$AgCN+HA\rightarrow HCN$ due to this reaction, reaction shifts forward and solubility of AgCN increases.

So AgCN has maximum solubility in buffer solution.

Available chlorine is liberated from bleaching powder when it :

0%
is heated
0%
reacts with acid
0%
reacts with ${H_2 O}$
0%
reacts with ${CO_2}$

Explanation

Reaction is:

$Ca(OCl)_2+ 4HCl \rightarrow CaCl_2 + 2H_2O + 2Cl_2$

$Ca(OCl)_2+H_2O \rightarrow Ca(OH)_2 + Cl_2$

$Ca(OCl)_2+ CO_2\rightarrow CaCO_3 +Cl_2$

In water, or aqueous solutions of

$HCl$ or

$H_{2}SO_{4},$ proton exists as:

0%
$H_{3}O^{+}$
0%
$H(H_{2}O)_{4}^{+}$
0%
$H(H_{2}O)_{n}^{+}$
0%
$all\ correct$

Explanation

In water, or aqueous solutions of

$HCl$ or

$H_{2}SO_{4},$ proton exists as either

$H_{3}O^{+}$ or

$H(H_{2}O)_{2}^{+}$ or

$H(H_{2}O)_{3}^{+}$ .

Thus, proton does not exist as free proton

$H^+$ but it is solvated with one, two or three molecules of water.

Option D is correct.

$0.10\space M$ solution of

$HF$ is

$8.0\mbox{%}$ ionized. What is the

$K_a$ ?

0%
$6.4\times10^{-4}$
0%
$8.8\times10^{-4}$
0%
$6.95\times10^{-4}$
0%
$7.6\times10^{-4}$

Explanation

As we know,

$C\alpha^2 = K_a = 0.1\times(0.08)^2 = 6.4\times10^{-4}$ .

Chloride which is insoluble in water is :

0%
sodium chloride
0%
potassium chloride
0%
mercurous chloride
0%
calcium chloride

$pH$ of following solution is not affected by dilution:

0%
$0.01\ M\ CH_{3}COONH_{4}$
0%
$0.01\ M\ NaH_{2}PO_{4}$
0%
$0.01\ M\ NaHCO_{3}$
0%
All of these

Does

$Mg(OH)_2$ react with sodium hydroxide?

0%
True
0%
False

Fear or excitement generally causes one to breathe rapidly and it results in the decrease of concentration of

${CO}_{2}$ in blood. In what way, it will change the

$pH$ of blood?

0%
$pH$ will increase
0%
$pH$ will decrease
0%
No change
0%
$pH$ will be $7$

Explanation

As we know,

$CO_{2}$ is acidic in nature. So, a decrease in its concentration should increase the

$pH$ . But, blood contains serum which has proteins in it. This acts as a buffer solution. We already know that buffer solutions resist a change in the

$pH$ even on addition of a small amount of acid or small amount of base or dilution, etc.

Therefore, the

$pH$ of the blood remains unaffected.

Hence,option C is correct.

${90}^{o}C$ pure water has

$[{H}_{3}{O}^{+}]={10}^{-6}$

$mol/litre$ . The value of

${K}_{w}$ at

${90}^{o}C$ is

0%
${10}^{-6}$
0%
${10}^{-12}$
0%
${10}^{-8}$
0%
${10}^{-14}$

Explanation

For pure water,

$[{H}^{+}]=[{OH}^{-}]$
and

${K}_{w}=[{H}^{+}][{OH}^{-}]={ \left[ { H }^{ + } \right] }^{ 2 }={ \left[ { 10 }^{ -6 } \right] }^{ 2 }={10}^{-12}$

Calcium phosphate is present in tooth enamel, its nature is:

0%
basic
0%
acidic
0%
neutral
0%
none of the above

Addition of large quantity of Iron salt will coagulate blood.

0%
True
0%
False

Which of the following is acidic in nature?

0%
Lime juice
0%
Human blood
0%
Lime water
0%
Antacid

Explanation

Human blood, lime water and antacid(sodium bicarbonate) are alkaline in nature which have

$pH$ greater than

$7$ while lime juice which has

$pH$ less than

$7$ is acidic in nature.

Hence, the correct option is A.

If a salt of weak acid or base is added to a solution of its acid or base respectively, the:

0%
dissociation of acid or base is diminished
0%
the $pH$ of the solution in case of acid increases and in case of base decreases
0%
mixing of two leads for precipitation
0%
none of the above

Explanation

Due to common ion effect, if a salt of weak acid or base is added to a solution of its acid or base respectively, the dissociation of acid or base is diminished.

As a result, concentration of hydrogen ions or hydroxide ion will change and pH of solution increases in case of acid and decreases in case of base.

$pH = -log [ H^+]$

To protect tooth decay we are advised to brush our teeth regularly. The nature of the tooth paste commonly used is:

0%
acidic
0%
neutral
0%
basic
0%
corrosive

Explanation

Correct answer: Option $C$

Hint: Toothpaste contains calcium carbonate.

Explanation:

Decay of teeth starts when

$pH$ of mouth is below

$5.5$ .
5

Toothpaste contains basic

$CaCO_3$ , which keeps the

$pH$ of our mouth high, which prevents the growth of bacteria that thrive in an acidic medium.

Also, it prevents tooth decay by protecting the

$(Ca)_3(PO_4)_2$ which would dissociate into hydrogen phosphate in an acidic medium,

$NaCl + H_2SO_4\overset{\Delta}{\longrightarrow}HCl \ +$

0%
$NaHSO_4$
0%
$Na_2SO4$
0%
$Na_2SO_3$
0%
none of the above

Explanation

Reacting sodium chloride and sulphuric acid at elevated temperature leads to production of sodium bisulphate and hydrogen chloride. Little amount of sodium sulphate is also formed which reacts with the sulphuric acid to yield sodium bisulphate or sodium hydrogen sulphate.

$NaCl\left( s \right) +{ H }_{ 2 }{ SO }_{ 4 }\rightarrow { NaHSO }_{ 4 }\left( s \right) +HCl$

In the dissociation of

$NH_4OH$ , if excess if

$NH_4Cl$ is added before adding

$NH_4OH$ , the concentration of:

0%
$NH_4^+$ ions increases and $OH^-$ ions decreases
0%
both $NH_4^+$ ions and $OH^-$ ions increases
0%
$NH_4^+$ ions decreases and $OH^-$ ions increases
0%
both $NH_4^+$ ions and $OH^-$ ions decreases

Explanation

From the law of Mass action, the dissociation of

$NH_4OH$ takes place and we have,

$\dfrac{[NH_4^+][OH^-]}{[NH_4OH]} =K$
Ammonium chloride, a strong electrolyte, ionises almost completely as follows:

$NH_4Cl \leftrightarrow NH_4^+ + Cl^-$
So, if excess of

$NH_4Cl$ is added before adding

$NH_4OH$ , the concentration of

$NH_4^+$ ions is increased and consequently the concentration of

$OH^-$ ions is decreased.

Assertion: Due to common ion effect, the solubility of

$HgI_2$ is expected to be less in an aqueous solution of KI than in water. But

$HgI_2$ dissolves in an aqueous solution of KI to form a clear solution.
Reason:

$I^{\circleddash}$ ion is highly polarisable.

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Assertion is incorrect but Reason is correct
0%
Both Assertion and Reason are incorrect

Explanation

Due to common ion effect, the solubility of

$HgI_2$ is expected to be less in an aqueous solution of KI than in water as

$HgI_2 + KI \longrightarrow K_2[HgI_4]$ .
since

$I^{\circleddash}$ ion is large sized and therefore is highly polarisable.
But (R) is not the correct explanation of (A)

Acids do not show acidic behaviour in:

0%
water
0%
absence of water
0%
basic solution
0%
none of these

Explanation

Any acid reacts with water forming hydronium ions which are responsible for giving it acidic behaviour and increasing the concentration of hydronium ions which are formed from water giving hydrogen ions and hydroxide ions. Acidity is given by the

$pH$ of the solution which increases on the addition of an acid to it.

Assertion: On addition of

$NH_4Cl$ to

$NH_4OH, pH$ decreases but remains greater than 7.
Reason: Addition of

$\overset{\circleddash}{N}H_4$ ion decreases ionisation of

$NH_4$ , thus [OH] decreases and also pH decreases.

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Assertion is incorrect but Reason is correct
0%
Both Assertion and Reason are incorrect

Explanation

Adding a common ion prevents the weak acid or weak base from ionizing as much as it would without the added common ion. The common ion effect suppresses the ionization of a weak acid by adding more of an ion that is a product of this equilibrium.

So addition of

$CH_3COONa$ to

$CH_3COOH$ increases the pH of solution but addition of

$NH_4Cl$ to

$NH_4OH$ decreases the pH of solution.

Assertion:

$pH$ value of

$HCN$ solution decreases when

$NaCN$ is added to it
Reason:

$NaCN$ provides a common ion

$CN^{\circleddash}$ to

$HCN$

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Assertion is incorrect but Reason is correct
0%
Both Assertion and Reason are incorrect

Explanation

$(A)$ is wrong because the addition of

$NaCN$ to

$HCN$ , due to common ion

$(CN^{\circleddash})$ , the degree of dissociation of

$HCN$ is suppressed and hence less

$[H^{\oplus}]$ and increase in

$pH.$

A farmer treats the soil of his field with lime when the soil has:

0%
acidic nature
0%
basic nature
0%
neutral nature
0%
None of the above

Which of the following reagents will give white precipitate with the aqueous solution of sulphurous acid?

0%
$BaCI_2$
0%
$HCl$
0%
$NaCl$
0%
$KCl$

Explanation

Barium chloride reacts with sulphurous acid to form white precipitate of barium sulphite.

$\displaystyle BaCl_2 + H_2SO_3 \rightarrow BaSO_3 \downarrow + 2HCl$

Which of the following will supress the ionisation of acetic acid in aqueous solution?

0%
$NaCl$
0%
$HCI$
0%
$KCI$
0%
Unpredicatble

Explanation

So addition of

$CH_3COONa$ to

$CH_3COOH$ increases the

$pH$ of solution but addition of

$NH_4Cl$ to

$NH_4OH$ decreases the

$pH$ of solution.

Here, the addition of

$HCl$ acid will suppress the ionisation of acetic acid.

In the third group of qualitative analysis, the precipatitating reagent is

$NH_4Cl / NH_4OH$ . The function of

$NH_4Cl$ is to

0%
increase the ionisation of $NH_4OH$
0%
supress the ionisation of $NH_4OH$
0%
convert the ions of group third into their respective chlorides
0%
stabilise the hydroxides of group $III$ cations

Explanation

So, addition of

$CH_3COONa$ to

$CH_3COOH$ increases the pH of solution but addition of

$NH_4Cl$ to

$NH_4OH$ decreases the pH of solution.

So, the function of

$NH_4Cl$ to suppress the ionisation of

$NH_4OH$ and to reduce conentration of hydroxide ion in solution.

Which of the following statements(s) is(are) correct?

0%
The pH of $1.0 \times 10^{-8} M$ solution of HCI is 8
0%
The conjugate base of $H_2PO_4^{\circleddash}$ is $HPO_4^{2-}$
0%
Autoprotolysis constant of water increases with temperature
0%
When a solution of weak monoprotic acid is titrated against a strong base, at half-neutralisation, point $pH = (1/2) pK_a$

Explanation

The following statements B and C are correct.
(A) The pH of

$1.0 \times 10^{-8} M$ solution of HCI cannot be 8 as it is not an alkaline solution which has pH greater than 7.
(B) The conjugate base of

$H_2PO_4^{\circleddash}$ is

$HPO_4^{2-}$ . The acid and the conjugate base differ by 1 proton only.
(C) Autoprotolysis constant of water increases with temperature as the autoionization of water increases with temperature.
(D) When a solution of weak monoprotic acid is titrated against a strong base, at half-neutralisation, point

$pH = pK_a$

The solubility of

$AgI$ in

$NaI$ solutions is less than that in pure water because:

0%
$AgI$ forms complex with $NaI$
0%
of common ion effect
0%
solubility product of $AgI$ is less than that of $NaI$
0%
the temperature of the solution decreases

Explanation

$AgI \rightleftharpoons Ag^+ + I^-$

$NaCl \rightarrow Na^+ + I^-$

Sodium iodide is a strong electrolyte and is completely dissociated. This increases the iodide ion concentration in solution and suppresses the ionization of

$AgI$ . Hence, the solubility of

$AgI$ decreases.

This is called common ion effect.

Aqueous solutions of

$HNO_3, CH_3COOH$ and

$CH_3COOK$ of identical concentrations are given. The pair(s) of the solution which may form a buffer upon mixing is(are):

0%
$NaOH$ and $CH_3COOH$
0%
$HNO_3$ and $CH_3COOK$
0%
$CH_3COOH$ and $CH_3COOK$
0%
$HNO_3+ CH_3COOH$

Explanation

An acidic buffer solution contains a weak acid and its salt with strong base.
The following pairs will form buffer solutions.
(A)

$NaOH$ and

$CH_3COOH$ react to form

$CH_3COONa$ .

$NaOH + CH_3COOH \rightarrow CH_3COONa + H_2O$
A mixture

$CH_3COOH + CH_3COONa$ is a buffer solution.
(B)

$HNO_3$ and

$CH_3COOK$ react to form

$CH_3COOH$ .

$HNO_3 + CH_3COOK \rightarrow CH_3COOH + KNO_3$
A mixture

$CH_3COOH + CH_3COONa$ is a buffer solution.
(C)

$CH_3COOH$ and

$CH_3COOK$ is a buffer solution.

$H_3PO_4 \rightleftharpoons H^{\oplus}+H_2PO_4^{\circleddash}; K_{a_{1}} :$

$H_3PO_4^{\circleddash} \rightleftharpoons H^{\oplus}+HPO_4^{2-}; K_{a_{2}}:$

$HPO_4^{2-} \rightleftharpoons H^{\oplus}+PO_4^{3-}; K_{a_{3}}:$
Mark out the incorrect statements

0%
$K_{a_{1}} > K_{a_{2}} > K_{a_{3}}$
0%
$pH(H_2PO_4^{\circleddash})=\displaystyle \frac{pK_{a_1}+pK_{a_2}}{2}$
0%
Both $H_3PO_4$ and $H_2PO_4^{\circleddash}$ are more acidic than $HPO_4^{2-}$
0%
Only $HPO_4^{2-}$ is amphiprotic anion in the solution

Explanation

After losing one

$H^+$ ion, the tendency to lose another

$H^+$ ion decreases due to the existing charge on the molecule and incomplete octet.
Hence,

$K_{a_{1}} > K_{a_{2}} > K_{a_{3}}$ , which implies Both

$H_3PO_4$ and

$H_2PO_4^{\circleddash}$ are more acidic than

$HPO_4^{2-}$ .
Amphiprotic molecules are those which can either donate or accept a proton (H+)

$HPO_4^{2-}$ and

$H_2PO_4^{-}$ are amphiprotic anion in the solution.
Further, the Ka values cannot be directly related.

The

$pH$ of a dilute solution of acetic acid was found to be

$4.3$ The addition of a small crystal of sodium acetate will cause

$pH$ to:

0%
become less than $4.3$
0%
become more than $4.3$
0%
remain equal to $4.3$
0%
unpredictable

Explanation

Due to this common ion effect, when we add sodium acetate dissociation of acetic acid decreases and solution will have less number of hydrogen ion and so, pH increases. (as

$pH = -log [H^+]$ )

A buffer solution can be prepared from a mixture of

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

Explanation

Hint: A buffer solution can be prepared by mixing a weak acid or base with the salt of a strong base or strong acid.

Explanation:

Sodium acetate and acetic acid in water can form buffer solution because sodium acetate is a salt of strong acid $NaOH$ and weak acid acetic acid.

$CH_3COONa+CH_3COOH$

Sodium acetate and hydrochloric acid in water cannot form a buffer solution because $HCl$ is a strong acid.
Ammonia and ammonium chloride in water can from buffer solution because ammonia is a weak base and ammonium chloride is a salt of ammonia and strong acid $HCl$ .

$NH_4Cl+NH_3$

Ammonia and sodium hydroxide in water cannot form buffer solutions because both are bases.

Final Answer: A buffer solution can be prepared from a mixture of Sodium acetate and acetic acid in water and Ammonia and ammonium chloride in water.

Correct options $A$ and $C$

$pOH$ of water is

$7.0$ at

$298 K$ . If water is heated to

$350 K$ , which of the following should be true?

0%
$pOH$ will decrease
0%
$pOH$ will increase
0%
$pOH$ will remain seven
0%
Concentration of $H^{\oplus}$ ions will increae but that of $\overset{\circleddash}{O}H$ will decrease.

Explanation

As we increase the temp. more water molecules will dissociate and the solution will have more no of hydrogen ion as well as hydroxide ion.

As a result, both

$pH$ and

$pOH$ of solution will decrease.

$55^0 C$ , autoprotolysis constant of water is

$4 \times {10}^{-14}$ . If a given sample of water has a

$pH$ of 6.9, then it is:

0%
acidic
0%
basic
0%
neutral
0%
explosive

$pOH$ of

${H}_{2}O$ is

$7.0$ at

$298\ K$ . If water is heated at

$350\ K$ , which of the following should be true?

0%
$pOH$ will decrease
0%
$pOH$ will increase
0%
$pOH$ will remain $7.0$
0%
Concentration of ${H}^{+}$ ions will increase but that of $O{H}^{-}$ will decrease

Explanation

$K_w = [H^+][OH^-]$

$\left[ { H }^{ + } \right] =\left[ O{ H }^{ - } \right] ={ 10 }^{ -7 }M$ for a neutral solution at

$25^o C$ only.

After that as we increase temp. more no of water molecules dissociate and conc. of both ions increases.

$pH$ and

$pOH$ both decreases around

$350\:K$ .

Calculate :

${K}_{b}$ for

$B{({OH}_{4})}^{-}$ given

${K}_{a}(B{(OH)}_{3})=6\times {10}^{-10}$

0%
$8.3\times {10}^{-6}$
0%
$1.66\times {10}^{-5}$
0%
$3.3\times {10}^{-5}$
0%
None of these

Explanation

As we know,

${K}_{a}\times {K}_{b}={10}^{-14}$ and

Given,

${K}_{a}(B{(OH)}_{3})=6\times {10}^{-10}$

Therefore,

${K}_{b}$ =

$1.66\times {10}^{-5}$

The solubility of

$Ca{ F }_{ 2 } \left( { K }_{ sp } = 3.4 \times { 10 }^{ -11 } \right)$ in

$0.1 M$ solution of

$NaF$ would be

0%
$3.4\times { 10 }^{ -12 }M$
0%
$3.4\times { 10 }^{ -10 }M$
0%
$3.4\times { 10 }^{ -9 }M$
0%
$3.4\times { 10 }^{ -13 }M$

Explanation

$CaF_2 \rightarrow Ca^{+2} + 2F^-$

$S$

$0.1$

$CaF_2: { K }_{ sp } = 3.4 \times { 10 }^{ -11 } = [Ca^{+2}][F^-]^2$

$S = \displaystyle \frac {3.4 \times { 10 }^{ -11 }}{(0.1)^2} = 3.4 \times { 10 }^{ -9 }$

Addition of

$HCl$ will not suppress the ionization of:

0%
acetic acid
0%
benzoic acid
0%
${H}_{2}S$
0%
sulphuric acid

Explanation

Any acid weaker than

$HCl$ will be suppressed by

$HCl$ . Hence, among the given options, only sulphuric acid is an acid with comparable acidity strength to

$HCl$ . The same can also be verified using

$K_a$ values from the data.

When solid

$KCl$ is added to a saturated solution of

$AgCI$ in

$H_2O$

0%
Nothing happens
0%
Solubility of $AgCl$ decreases
0%
Solubility of $AgCl$ increases
0%
Solubility product of $AgCl$ increases

Explanation

$AgCl \rightleftharpoons Ag^+ + Cl^-$

$KCl \rightarrow K^+ + Cl^-$

Potassium chloride is a strong electrolyte and is completely dissociated. This increases the chloride ion concentration in solution and suppresses the ionization of

$AgCl$ . Hence, the solubility of

$AgCl$ decreases.

Buffer solutions can be prepared from mixtures of:

0%
$HCl$ and $NaCl$
0%
$NaH_2PO_4$ and $Na_2HPO_4$
0%
$CH_3COOH + NaCl$
0%
$NH_3OH+NH_3$

Explanation

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.

Basic buffer solution: A basic buffer solution consists of a mixture of a weak base and its salt with strong acid.

Acidic buffer solution: An acidic buffer solution consists of a solution of a weak acid and its salt with a strong base.

So among given options,

$NaH_2PO_4 + Na_2HPO_4$ will form a buffer solution.

Which of the following expressions is/are not true?

0%
$\left[ { H }^{ + } \right] =\left[ O{ H }^{ - } \right] =\sqrt { { K }_{ w } }$ for a neutral solution at all temperatures
0%
$\left[ { H }^{ + } \right] >\sqrt { { K }_{ w } } \& \left[ O{ H }^{ - } \right] <\sqrt { { K }_{ w } }$ for an acidic solution
0%
$\left[ { H }^{ + } \right] <\sqrt { { K }_{ w } } \& \left[ O{ H }^{ - } \right] >\sqrt { { K }_{ w } }$ for an alkaline solution
0%
$\left[ { H }^{ + } \right] =\left[ O{ H }^{ - } \right] ={ 10 }^{ -7 }M$ for a neutral solution at all temperatures

Explanation

$K_w = [H^+][OH^-]$

For neutral solution,

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

For acidic solution,

$[H^+] > [OH^-]$

For basic solution,

$[H^+] < [OH^-]$

$\left[ { H }^{ + } \right] =\left[ O{ H }^{ - } \right] =\sqrt { { K }_{ w } }$ for a neutral solution at all temperatures

$\left[ { H }^{ + } \right] >\sqrt { { K }_{ w } } \& \left[ O{ H }^{ - } \right] <\sqrt { { K }_{ w } }$ for an acidic solution

$\left[ { H }^{ + } \right] <\sqrt { { K }_{ w } } \& \left[ O{ H }^{ - } \right] >\sqrt { { K }_{ w } }$ for an alkaline solution

$\left[ { H }^{ + } \right] =\left[ O{ H }^{ - } \right] ={ 10 }^{ -7 }M$ for a neutral solution at

$25^o C$ only. After that as we increase temp. more no of water molecules dissociate and conc. of both ions increases.

$90^oC$ , pure water has

$[H_3O^{\oplus}]=10^{-6.7}\, mol\, L^{-1}$ . What is the value of

$K_w$ at

$90^oC$ ?

0%
$10^{-6}$
0%
$10^{-12}$
0%
$10^{-13.4}$
0%
$10^{-6.7}$

Explanation

$90^oC$ , pure water has

$[H_3O^{\oplus}]=10^{-6.7}\, mol\, L^{-1}$ and

$[OH^-] = 10^{-6.7}$

$K_w = [H^+][OH^-] = 10^{-6.7}\times10^{-6.7} = 10^{-13.4}$

Determine

$[{OH}^{-}]$ of a

$0.050\ M$ solution of ammonia to which has been added sufficient

${NH}_{4}Cl$ to make the total

$[{NH}_{4}^{+}]$ equal to

$0.100 M$ .

$[{K}_{b({NH}_{3})}=1.8\times {10}^{-5}]$

0%
$[{OH}^{-}]=9.0\times {10}^{-6}$
0%
$[{OH}^{-}]=9.0\times {10}^{-8}$
0%
$[{OH}^{-}]=9.0\times {10}^{-2}$
0%
$[{OH}^{-}]=9.0\times {10}^{-9}$

Explanation

${NH}_{4}Cl\longrightarrow {NH}_{4}^{+}+{Cl}^{-}$

${NH}_{4}OH\longrightarrow {NH}_{4}^{+}+{OH}^{-}$

${K}_{b}=\cfrac { \left[ { NH }_{ 4 }^{ + } \right] \left[ OH \right] }{ \left[ { NH }_{ 4 }OH \right] }$

$[{NH}_{4}^{+}]=$ is due to salt because

${NH}_{4}OH$ ionise less amount due to common ions effect

$1.8\times {10}^{-5}=\cfrac{0.1\times [{OH}^{-}]}{0.05}$

$9\times {10}^{-6}=[{OH}^{-}]$

The ionic product of water ______ if a few drops of acid or base are added to it.

0%
increases
0%
decreases
0%
remains the same
0%
can not predict

In aqueous solution, the hydronium ion is further hydrated to species like:

0%
${H_5 O_2^{+}}$
0%
${H_7 O_3^{+}}$
0%
${H_8 O_3^{+}}$
0%
${H_9 O_4^{+}}$

Assertion: A solution of sodium acetate and ammonium acetate can act as a buffer.
Reason: A buffer solution consists of a mixture of a weak acid and its conjugate base or vice versa.

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Both Assertion and Reason are true and Reason is the correct explanation of Assertion
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Assertion is true but Reason is false
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Explanation

A buffer is a solution that can resist pH change upon the addition of an acidic or basic components. It is able to neutralize small amounts of added acid or base, thus maintaining the pH of the solution relatively stable. A buffer must consist of a weak conjugate acid-base pair, meaning either a. a weak acid and its conjugate base, or b. a weak base and its conjugate acid. For example, the following could function as buffers when together in solution:
Acetic acid (weak organic acid w/ formula

$CH_3COOH$ ) and a salt containing its conjugate base, the acetate anion (

$CH_3COO^-$ ), such as sodium acetate (

$CH_3COONa$ )

Assertion: A buffer has a definite pH value which changes, on keeping it or on diluting it.

Reason: A buffer solution is a mixture consisting of a weak acid and its conjugate base or vice versa.

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

Assertion: AgCl is less soluble in aqueous sodium chloride solution than in pure water.

Reason: AgCl dissociates completely and more rapidly than NaCl.

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

Explanation

$AgCl \rightleftharpoons Ag^+ + Cl^-$

$NaCl \rightarrow Na^+ + Cl^-$
Sodium chloride is a strong electrolyte and is completely dissociated. This increases the chloride ion concentration in solution and suppresses the ionization of

$AgCl$ . Hence, the solubility of

$AgCl$ decreases.

In the presence of a common ion (incapable of forming complex ion), the solubility of salt _______ in solution.

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increases
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decreases
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remains the same
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cannot predict

Explanation

$AB \rightarrow A^+ + B^-$

$BC \rightarrow B^+ + C^-$
Since

$B^+$ is incapable of forming a complex salt it tends to decrease the solubility by Le-Chatelier's principle.

CBSE Questions for Class 11 Medical Chemistry Equilibrium Quiz 4 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Equilibrium Quiz 4 - MCQExams.com

0:0:1

Practice Class 11 Medical Chemistry Quiz Questions and Answers

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