MCQ Questions for Class 11 Medical Chemistry Equilibrium Quiz 9

In order to prepare a buffer of

$pH = 8.26$ , the amount of

$(NH_4)_2 SO_4$ required to be mixed with one litre of

$0.1 (M) NH_3 (aq),$

$pK_b$ =

$4.74$ is:

0%
$1.0\ mole$
0%
$10.0\ mole$
0%
$0.50 \ mole$
0%
$5\ mole$

Explanation

$pOH={ pK }_{ b }+log\dfrac { \left[ salt \right] }{ \left[ base \right] }$

$\Rightarrow 14-pH=4.74+log\dfrac { \left[ salt \right] }{ 0.1 }$

$\Rightarrow 14-8.26=4.74+log\dfrac { \left[ salt \right] }{ 0.1 }$

$\Rightarrow \dfrac { \left[ salt \right] }{ 0.1 } ={ 10 }^{ 1 }=10$

$\Rightarrow \left[ salt \right] =1$

Since

$1$ mole of

${ \left( { NH }_{ 4 } \right) }_{ 2 }{ SO }_{ 4 }$ gives two moles of

${ NH }_{ 4 }^{ + }$ . So,

$0.5$ mole of

${ \left( { NH }_{ 4 } \right) }_{ 2 }{ SO }_{ 4 }$ in one litre will be needed

${ \left( { NH }_{ 4 } \right) }_{ 2 }{ SO }_{ 4 }\rightleftharpoons 2{ NH }_{ 4 }^{ + }+{ SO }_{ 4 }^{ 2- }$

$\therefore$ correct answer will be C.

In a buffer solution, the ratio of the concentration of

$NH_4CI$ and

$NH_4OH$ is 1 : 1 when it changes to 2 : 1, what will be the effect on pH of the buffer?

0%
Increase
0%
Decrease
0%
No change
0%
First increase then decrease

Explanation

in a basic buffer ,

$pOH=pK_{b}+log\dfrac{\left [ salt \right ]}{\left [ base \right ]}$

$\therefore$ increasing Base i.e.

$NH_{4}Cl$ will increase

$pOH$ and hence

$pH$ will decrease.

A buffer solution of pH = 5.15 is to be prepared from acetic acid (Ka =

$$1.80x10^5$$ ), sodium acetate and water. What must be the molar ratio of acetate ion to acetic acid in this solution to obtain this pH?

0%
$0.05 mole$
0%
$2.54 mole$
0%
$0.1 mole$
0%
$0.005 mole$

Explanation

$\begin{array}{l} \text { To create:-buffer of } p H=5.15 \text { from } \\ \text { acctic acid }\left(k a=1.8 \times 10^{-5}\right) \text {, sodium acetate } \\ \text { and water. } \\ \text { To find:- molar ratio of acetate ion to } \\ \text { acetic acid } \\ \text { According to Henderson- Hassel batch equation } \\ p H=p k a+\log \left[\frac{\text { salt }}{\text { acid }}\right]. \text { } \end{array}$

$\text { Here } p k a=-\log \left(1 \cdot 8 \times 10^{-5}\right)=4 \cdot 7447$

$\begin{aligned} \therefore & p H-p k a=\log _{10}\left[\frac{[a c e t a t e]}{[a c i d]}\right]\\ & 5 \cdot 15-4 \cdot 7447=\log _{10}\left[\frac{(a\text {cetate ion })}{\text { (acetic acid) }}\right.\end{aligned}$

$10^{0.3753}=\frac{\text { [acetate ion] }}{\text { [acetic acid] }}$

[acetate]

$\frac{\text { ion }]}{\text { [acetic acid] }}=2 \cdot 54$

So, the correct answer is B) 2.54

The total number of different kind of buffers obtained during the titration of

$H_{3}PO_{4}$ with NaOH are

0%
3
0%
1
0%
2
0%
0

Each of the following forms a true solution in water. Which of them will allow the least etectric current to flow through it?

0%
Alum
0%
Table saIt
0%
Cane sugar
0%
Mohr's salt

Which of the following solutions display common ion effect?

0%
$HCl+NaCl$
0%
${ H }_{ 2 }S + KH$
0%
$NaHS+HCl$
0%
${ H }_{ 2 }{ SO }_{ 4 }+{ K }_{ 2 }S$

Explanation

In common ion effect there should be common ions in the two or more chemicals. Here,

$HCl+NaCl$ shows common ion effect. Because

$Cl^-$ is common in both.

Which of the following is an example of weak electrolyte?

0%
$H_3BO_3$
0%
$H_2SO_4$
0%
$HNO_3$
0%
$HClO_3$

Explanation

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.

One liter of solution contains

${ 10 }^{ -5 }$ moles of

${ H }^{ + }$ ions at

${ 25 }^{ o }C$ .Percentage ionisation of water in solution is:

0%
$1.8\times { 10 }^{ -7 }\%$
0%
$1.8\times { 10 }^{ -5 }\%$
0%
$3.6\times { 10 }^{ -9 }\%$
0%
$1.8\times { 10 }^{ -11 }\%$

Explanation

Molarity of water=

$55.5$ mole

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

Let

$x$ be the degree of dissociation

Then

$55.5x= 10^{-5}$

$\Rightarrow x= 1.8 \times 10^{-7}$

$x= 1.8 \times 10^{-7}\times 100= 1.8 \times 10^{-5}$

Dissolution of salt participate in water shows:

0%
Particulate nature of matter.
0%
Continuous nature of matter.
0%
Particle size of salt is very big.
0%
All of these.

If the first dissociation of

$X{ \left( OH \right) }_{ 3 }$ is 100% where as second dissociation is 50% and third dissociation is negligible then the pH of

$4\times { 10 }^{ -3 }$

$M\quad X{ \left( OH \right) }_{ 3 }$ is :

0%
$11.78$
0%
$10.78$
0%
$2.5$
0%
$2.22$

Explanation

First dissociation

$X(OH)_3 \longrightarrow X(OH)^+_2+OH^-$
Second dissociation

$X(OH)^+_2 \longrightarrow X(OH)^{2+}+OH^-$
Total

$[OH^-]=4 \times 10^{-3} +2 \times 10^{-3}=6 \times 10^{-3}$

$pOH=3-\log 6=2.22$

$\therefore pH=11.78$

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

0%
5.44
0%
6.44
0%
6.0
0%
4.73

Explanation

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

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

$\Rightarrow pH= 5.74 + 0.6989$

$\therefore pH = 6.44$

Hence, option B is correct .

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

0%
${ (55.4\times { 10 }^{ 14 } })^{ -1 }$
0%
$1 \times { 10 }^{ -14 }$
0%
$\dfrac { 1\times { 10 }^{ -14 } }{ 18 }$
0%
$none\ of\ these$

Explanation

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

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

The values of dissociation constants of some acids (at

${ 25 }^{ 0 }C$ ) are as follows. Indicate which is the strongest acid in water?

0%
${ 1.4\times 10 }^{ -2 }$
0%
${ 1.6\times 10 }^{ -4 }$
0%
${ 4.4\times 10 }^{ -10 }$
0%
${ 4.3\times 10 }^{ -7 }$

Explanation

More is

$K_{a}$ lesser is

$pK_{a}(pK_{a}=-\log K_{a})$ more is acidic strength.

Option A is correct.

Ionic product of

$H_2O$ is

$10^{-14}$ . The

$H^+$ ion concentration in

$0.1\ M$

$NaOH$ solution is:

0%
$10^{-11}\ M$
0%
$10^{-13}\ M$
0%
$10^{-1}\ M$
0%
$10^{-4}\ M$

Explanation

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

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

Which among the following salts will give basic solution on hydrolysis?

0%
${NH}_{4}Cl$
0%
$Na{NO}_{3}$
0%
${K}_{2}{CO}_{3}$
0%
$KCl$

Explanation

Answer

$C: K_2CO_3$

$NH_4Cl\longrightarrow$ Acidic salt.

$NaNO_3\longrightarrow$ Neutral salt.

$K_2CO_3\longrightarrow$ Basic salt.

$K_2CO_3+2H_2O\longrightarrow \underset{strong-base}{2KOH}+H_2CO_3$

$KCl\longrightarrow$ Neutral salt.

The mass of acetic acid present in 500 ml of solution in which it is 1% ionised (

$Ka$ of

$CH_{3}COOH=1.8\times 10^{-5}$ )

0%
5.4 g
0%
12.6 g
0%
6.4 g
0%
10.8 g

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 :

0%
$60\%$
0%
$83.5\%$
0%
$85.3\%$
0%
$41.6\%$

Explanation

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 .

When 0.1 mole of an acid is added to 2 lit of a buffer solution, the pH of the buffer decreases by 0.The buffer capacity of the solution is:

0%
0.6
0%
0.4
0%
0.2
0%
0.1

Explanation

$0.1$ 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

$0.1$

What is

$[{ NH }_{ 4 }^{ + }]$ in a solution containing 0.02M

${ NH }_{ 3 }$ (

${ K }_{ b }={ 1.8\times 10 }^{ -5 }$ ) and 0.01M KOH?

0%
${ 1.8\times 10 }^{ -5 }$
0%
${ 9\times 10 }^{ -6 }$
0%
${ 3.6\times 10 }^{ -5 }$
0%
None of the above

Explanation

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

A buffer solution can be prepared from a mixture of:

0%
${ CH }_{ 3 }COONa$ and $HCl$ in 1:1 mole ratio
0%
${ CH }_{ 3 }COONa$ and $CH_3COOH$ in 1:1 mole ratio
0%
${ CH }_{ 3 }COONa$ and $HCl$ in 1:2 mole ratio
0%
${ NH }_{ 4 }Cl$ and $NaOH$ in 1:2 mole ratio

Which of the following has maximum

$pK_a$ :-

0%
$CH_2FCOOH$
0%
$CH_3ClCOOH$
0%
$CH_3COOH$
0%
$HCOOH$

Explanation

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.

The solubility of CaF

$_2$ (K

$_{sp} = 5.3\times 10^{-9}$ ) in

$0.1$ M solution of NaF would be : (Assume no reaction of cation/anion) .

0%
$5.3 \times 10^{-10}$ M
0%
$5.3 \times 10^{-8}$ M
0%
$5.3 \times 10^{-7}$ M
0%
$5.3 \times 10^{-11}$ M

Explanation

$(C)\ 5.3\times 10^{-7}m$

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

$K_{sp}=[Ca^{}2+][F^-]^2=S(S+0.1)^2=S\times 0.1^2=5.3\times 10^{-9}$

Note:

$S<<0.1$ so,

$S+ 0.1 \approx 0.1$

$\Rightarrow S=5.3\times 10^{-7}\ M$

Four grams of

$NaOH$ solid are dissolved in just enough water to make

$1$ litre of solution. What is the

$[H^+]$ of the solution?

0%
$10^{-2}$ moles/litre
0%
$10^{-1}$ moles/litre
0%
$10^{-12}$ moles/litre
0%
$10^{-13}$ moles/litre

Explanation

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,

$[H^+]=10^{-13}\ moles/litre$

$K_a$ for

$HF$ is

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

$K_b$ for the fluoride ion.

0%
$3.5\times 10^{-4}$
0%
$1.0\times 10^{-7}$
0%
$2.9\times 10^{-11}$
0%
$1.0\times 10^{-14}$

Explanation

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

$K_b$ 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

$(C)$

Salt which on heating produces a mixture of two gases is :

0%
$NaNO_3$
0%
$Bi(NO_3)_3$
0%
$Pb(NO_3)_2$
0%
$NH_4NO_3$

Explanation

$Pb\left ( NO_{3} \right )_{2}$ on heating it gives a brown coloured gas

$NO_{2}$ and

$O_{2}$ gas.

Equation is as follows:

$2Pb\left ( NO_{3} \right )_{2}\overset{673K}{\rightarrow} 4NO_{2}+ 2PbO +O_{2}$

Here it is a mixture of two gas i.e. nitrogen-dioxide and oxygen gas.

Hence option C is the correct answer

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:

0%
$0.8\times 10^{-4}$
0%
$1.79\times 10^{-5}$
0%
$0.182\times 10^{-4}$
0%
$1\times 10^{-4}$

The unit of ionic product of water

$K_w$ are

0%
$Mol^{-1}L^{-1}$
0%
$Mol^{-2}L^{-2}$
0%
$Mol^{-2}L^{-1}$
0%
$Mol^{2}L^{-2}$

Explanation

as,

$K_w=[H^+][OH^−]$

$Units \ of\ K_w=(MolL^{−1})(MolL^{−1})$

$=Mol^2L^{−2}$

thus, D is the correct answer .

Oxidation number of

$Cl$ in

${ CaOCl }_{ 2 }$ (bleaching powder) is:

0%
zero, since it contains ${ Cl }_{ 2 }$
0%
$-1$ , since it contains ${ Cl }^{ - }$
0%
$-1$ , since it contains ${ ClO }^{ - }$
0%
$+1$ and $-1$ since it contains ${ ClO }^{ - }$ and ${ Cl }^{ - }$

Explanation

$CaOCl_2\to Ca^{2+}+OCl^-+Cl^-$

In bleaching powder,

$CaOCl_2,$ the two

$Cl$ atoms are in different oxidation states i.e., one

$Cl$ atom having oxidation number of

$-1$ and the other as

$OCl^-$ having oxidation number of

$+1$ .

Therefore,

$+1$ and

$-1$ since it contains

$ClO^-$ and

$Cl^-.$

Ans is option

$D.$

Solution of

$FeCl_3$ and

$H_2O$ is :

0%
Basic
0%
Acidic
0%
Neutral
0%
None of these

Explanation

The reaction between ferric chloride (

$FeCl_{3}$ ) and water is given by the following reaction,

$FeCl_{3} +H_{2}O \rightarrow Al(OH)_{3} +3 HCl$

Since hydrochloric acid is one of the side product of the reaction so the solution will be acidic in nature.

The number of

$H^+$ ions in

$1\ cc$ of a solution of

$pH = 13$ is:

0%
$6.023 \times 10^7$
0%
$1 \times 10^{-13}$
0%
$6.023 \times 10^{13}$
0%
$1 \times 10^{16}$

The concentration of

$\left[ {{H^ + }} \right]$ and concentration of

$\left[ {{OH^ + }} \right]$ of a

$0.1\,M$ aqueous solution of

$2\%$ ionised weak acid is [ionic product of water

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

0%
$2 \times {10^{ - 3}}M$ and $5 \times {10^{ - 12}}M$
0%
$1 \times {10^{ - 3}}M$ and $3 \times {10^{ - 11}}M$
0%
$0.02 \times {10^{ - 3}}M$ and $5 \times {10^{ - 11}}M$
0%
$3 \times {10^{ - 2}}M$ and $4 \times {10^{ - 13}}M$

Explanation

$HA$

$\rightleftharpoons$

$H^{+} + A^{-}$
t=0:

$c$
t=t:

$c(1 - \alpha)$

$c \alpha$

$\therefore$

$[H^{+}] = c \alpha = 0.1 \times 0.02 = 2 \times 10^{-3}$

Hence

$[OH^{-}] = \dfrac{10^{-14}}{2 \times 10^{-3}} = 5 \times 10^{-12}$

Hence, Option "B" is the correct answer.

Dissociation constant of

$0.002 M$ acotic add collection having conduct of

$8 \times 10^{-5} scm^{-1}$ &

$\mathring {A}$

$400 s cm^2 mole^{-1}$ is

0%
$1.7 \times 10^{-4}$
0%
$2.1 \times 184$
0%
$2.9 \times 10^{-4}$
0%
$1.7 \times 10^{-3}$

Addition of sodium acetate solution to acetic acid causes the following change-

0%
pH increases
0%
pH decreases
0%
pH remains unchanged
0%
pH becomes 7

Explanation

Dissociation of

$CH_{3}COOH$ is suppressed by the addition of sodium acetate

$( CH_{3}COONa)$ due to common ion

$(CH_{3}COO^{-})$ effect. The

$[H^+]$ decreases raising the pH of the acid solution.

Hence, pH will increase.

Therefore, option A is correct.

Which pair can not act as buffer?

0%
$CH_3COOH + CH_3COONa$
0%
Borax + $H_3BO_3$
0%
$HCl + NH_4OH$
0%
$NH_4Cl + HCl$

Explanation

$NH_4Cl$ and

$HCl$ can not behave as buffer

$NH_4Cl$ is formed from the reaction of

$NH_3$ , a weak base, and

$HCl$ , a strong acid.

The common ion effects is shown by which of the following sets of solutions:-

0%
$BaCl_{2}+Ba(NO_{3})_{2}$
0%
$NaCl+HCl$
0%
$NH_{4}OH+NH_{4}Cl$
0%
$None$

Electrolytic conduction is due to the movement of

0%
molecules
0%
atoms
0%
ions
0%
electrons

Explanation

Electrolytic conduction involves the

$\textbf{movement of ions}$ throughout a pure liquid or solution. The major difference between them is that one involves the movement of electrons and the other involves the movement of ions.

Option (C) is correct.

Chlorides are mineral salts, and cannot be affected by

0%
Chemical Actions
0%
Biological Action
0%
Temperature
0%
Turbidity

What

$[H^+]$ must be maintained in a saturated solution to ppt.

$Pb^{+2}$ but not

$Zn^{+2}$ from a solution in which each ion is present at a concentration of

$0.01\ M$ ?

Given,

$K_{sp}$ for

$H_2S=1.1\times 10^{-22}$ ,

$K_{sp}$ for

$ZnS=10^{-21}$ .

0%
$3.3\times 10^{-4}$
0%
$3.3\times 10^{-2}$
0%
$6.6\times 10^{-2}$
0%
$6.6\times 10^{-4}$

Which of the following reacts with water?

0%
$CHCl_{3}$
0%
$CCl_{4}$
0%
$CCl_{3}CHO$
0%
$CH_{2}ClCH_{2}Cl$

Explanation

Hydrocarbons and its halogen derivatives are insoluble or immiscible in water and hence it does not react with water.
only $CCl_3CHO$ reacts with water to form soluble hydrate which was used to be a hypnotic and seductive agent.
Trichloroacetaldehyde $CCl_3CHO$ reacts with water to form a very stable product $CCl_3CH(OH)_2$ which is a hydrate of trichloracetaldehyde.
It is also called chloral hydrate.
It is stable due to intramolecular hydrogen bonding in its butterfly-like structure.
$CH_3Cl, CCl_4 ,$ and $CH_2ClCH_2Cl$ do not react with water.

Option C is correct.

1921054_1338518_ans_0bbc8e47c9a0405688de0d3a83c56fee.png

$C+O_{2}\rightarrow CO_{2} +Heat$

0%
Combination reaction
0%
Decomposition reaction
0%
Both $(1)$ and $(2)$
0%
none of these

Explanation

The reaction between carbon and oxygen to form carbon dioxide can be classified as a combination reaction as carbon directly combines with oxygen to form carbon dioxide

$(CO_2)$ and gives off heat and light.

Decomposition reaction means breaking of a compound into simpler compounds or molecules on the application of heat or electricity. It is the reverse of a combination reaction.

Hence, option (A) is correct.

Which of the following salts is insoluble in water at room temperature but soluble in boiling water?

0%
$CaCl_2$
0%
$BaCl_2$
0%
$SrCl_2$
0%
$PbCl_2$

Explanation

Chlorides of Group(II) halides are soluble in both hot and cold water, whereas, Chlorides of

$\mathrm{{Pb}^{+2}, {Ag}^+ \ \& \ {{Hg}_2}^{+2}}$ are insoluble in water at room temperature but soluble in boiling water.

Hence, Option "D" is the correct answer.

Two elements have electronegativity of 1.2 and 3.Bond formed between them would be ________________.

0%
predominantely ionic
0%
predominantely covalent
0%
co-ordinate bond
0%
metallic bond

Basic strength of

$NH_{4}OH$ in presence of

$NH_{4}CI$

0%
Increase
0%
Remains unchanged
0%
Decreases
0%
Some times increases or sometimes decrease

The

$pH$ of rain water is approximately_____.

0%
6.5
0%
7.5
0%
5.6
0%
7.0

Explanation

Rainwater becomes acidic because gases present in the environment are dissolved so its

$pH$ will be less than 7.

$pH$ of rainwater is approximate

$5.6$ .

Solubility of

$MX_{ 2 }$ type electrolytes is

$0.5\times 10^{ -4 } mol/L$ , then find out

${ K }_{ sp }$ of electrolytes.

0%
$5\times 10^{ -12 }$
0%
$25\times 10^{ -10 }$
0%
$1\times 10^{ -13 }$
0%
$5\times 10^{ -13 }$

Explanation

For

$MX_{2}$ type of electrolyte

$MX_{2}\rightleftharpoons M^{+2} + 2X^{-}$ . So if the solubility is

$S$ then the solubility product would be

$4S^{3}$

Here, solubility S=

$0.5 \times 10^{-4}mol/L$

So, solubility product

$K_{sp}=4\times (0.5 \times10^{-4})^{3}$

$5 \times 10^{-13}$ So the correct option is D

Hydrated cobalt(II) chloride decomposes on heating.
The equation for the reaction is

$CoCl_{2}.6H_{2}O CoCl_{2} + 6H_{2}O$
The reaction is reversed by adding water.
Which row describes the colour change and the type of reaction for the reverse reaction?

0%
Colour change - blue to pink, type of reaction - endothermic
0%
Colour change - blue to pink, type of reaction - exothermic
0%
Colour change - pink to blue, type of reaction - endothermic
0%
Colour change - pink to blue, type of reaction - exothermic

Explanation

$\text{Hydrated cobalt(II) chloride when added to water its colour change from blue to pink. }$

$\text{This is a exothermic reaction.}$

The number of hydronium ions in

$1 ml$ of an aqueous solution of

$pH$

$12.0$ at

$25^{\circ}C$ is

0%
$0.01$
0%
$10^{-12}$
0%
$6.02 \times 10^{8}$
0%
$6.02 \times 10^{11}$

Predict the direction in which the following equilibrium will shift backward.

$\underset{(g)}{N_2} + \underset{(g)}{3H_2} \Leftrightarrow \underset{(g)}{2NH_3 \,\, \Delta H} = -100 \, Joule$

0%
increasing Temperature
0%
increasing pressure
0%
addition of catalyst
0%
on addition of $NH_{3}$

Which of the following pair will show common ion effect?

0%
HCN +KOH
0%
HBr +AgBr
0%
HCOOH + $CH_3COOH$
0%
KOH +AgOH

$90^oC$ the concentration of

$H_3O^+$ in pure water is

$10^{-4}$

$mol. Lit^-$ . What is the value of

$K_W$ at this temperature ?

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

Explanation

For pure water,

$[H^+]=[OH^−]$

and

$K_w=[H^+][OH^−]$

$=[H^+]^2$

$=[10^{−6}]^2=10^{−12}$

CBSE Questions for Class 11 Medical Chemistry Equilibrium Quiz 9 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Equilibrium Quiz 9 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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