MCQ Questions for Class 11 Medical Chemistry Equilibrium Quiz 8

The reaction,

$Pb(OH{ ) }_{ 2 }+HN{ O }_{ 3 }\rightarrow Pb(OH)N{ O }_{ 3 }+{ H }_{ 2 }O$ shows that

$Pb(OH)N{ O }_{ 3 }$ is

0%
acidic salt
0%
basic salt
0%
base
0%
acid

A buffer solution is used in:

0%
preparation of potash alum
0%
the removal of ${ PO }_{ 4 }^{ 3- }$ ions
0%
increasing the pH value of a solution
0%
precipitation of $Cr{ \left( OH \right) }_{ 3 }$ from $$Cr{ Cl }_{ 3 }$p

Explanation

$PO_4^{3-}$ removal columns are designed to remove

$PO_4^{3-}$ from a buffer solution. For example, the

$PO_4^{3-}$ concentration can be reduced from

$10mM$ to lower than

$0.001mM$ . The principle of phosphate removal is based on interactions between

$PO_4^{3-}$ and column resin. The

$PO_4^{3-}$ stays on the column. Other buffer components including regular buffer salts and non-charged molecules such as sugar and glycerol do not bind to the resin, therefore they stay in the sample solution.

Which of the following expressions is not true?

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

Explanation

For a neutral solution at

$25^0C$

$pH$ and

$pOH$ are both equal to 7 so

$[H^{+}]=[OH^{-}]=\sqrt{K_w}$ .

For an acidic solution

$pH<7$ so

$[H^{+}]>\sqrt{K_w}$ . and

$[OH^{-}]<\sqrt{K_w}$ .As the temperature increases dissociation will increase so

$pH$ of neutral solution will decrease.So

$[H^{+}]$ is not equal to

$[OH^{-}]$ or

$\sqrt{K_w}$ at all temperatures.

$1M$

$NaCl$ and

$1M$

$HCl$ are present in an aqueous solution. The solution is:

0%
Not a buffer solution with $pH< 7$
0%
Not a buffer solution with $pH> 7$
0%
A buffer solution with $pH< 7$
0%
A buffer solution with $pH= 7$

Explanation

$HCl$ is strong acid so it cannot make a buffer and

$HCl$ with

$NaCl$ have

$pH<7$ .

Which of the following will function as buffer?

0%
$NaCl+NaOH$
0%
Borax $+$ boric acid
0%
$Na{ H }_{ 2 }{ PO }_{ 4 }+NaH{ PO }_{ 4 }$
0%
${ NH }_{ 4 }Cl+{ NH }_{ 4 }OH$

Explanation

$NaCl+NaOH$ is (salt of strong acid and strong base+strong base).So it will not function as buffer.

Borax(Salt of weak acid and strong base)+boric acid(weak acid) so they will function as buffer.

$Na_2HPO_4+NaHPO_4$ will not function as buffer.

$NH_4Cl$ (Salt of strong acid and weak base)+

$NH_4OH$ (salt of weak base) so they will function as buffer.

When strong base

$(NaOH)$ is added to the weak acid (acetic acid,

${ CH }_{ 3 }COOH$ ), then dissociation of acetic acid increases; this effect is known as:

0%
Common ion effect
0%
Reverse ion effect
0%
Saltation effect
0%
Solubility effect

Explanation

$CH_3COOH\rightleftharpoons CH_3COO^{-}+H^+$

When

$NaOH$ is added

$H^+$ combines with

$OH^{-}$ so concentration of

$H^+$ decreases so equilibrium will shift towards right.

${ CH }_{ 3 }COOH+NaOH\longrightarrow { CH }_{ 3 }COONa+{ H }_{ 2 }O\quad$
Ionization of acetic acid will increase with the progress of its neutralization. This effect is called reverse ion effect.

The weak acid,

$HA$ has a

${K}_{a}$ of

$1.00\times { 10 }^{ -5 }$ . If

$0.1$ mol of this acid is dissolved in one litre of water, the percentage of acid dissociated at equilibrium is closet to:

0%
$1$ %
0%
$99.9$ %
0%
$0.1$ %
0%
$99$ %

Explanation

0.1 mole of acid is dissolved in 1 litre of water means

$[HA]=0.1M$

Let '

$\alpha$ ' be degree of dissociation,

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

$0.1$

$0.1(1-\alpha)$

$0.1\alpha$

${ K }_{ a }=\cfrac { \left[ { H }^{ + } \right] \left[ { A }^{ - } \right] }{ \left[ HA \right] } =\cfrac { { 0.1 }^{ 2 }{ \alpha }^{ 2 } }{ 0.1(1-\alpha) }$

Let

$\alpha<<1$ so

$1-\alpha=1$

$K_a=0.1\alpha^2=10^{-5}$

$\alpha=10^{-2}$

% of acid dissociated=

$1$ %

Aqueous solution of

$H{ NO }_{ 3 },KOH,{ CH }_{ 3 }COOH$ and

${CH}_{3}COONa$ of identical concentrations are provided. The pair(s) of solutions which form a buffer upon mixing is (are):

0%
$H{ NO }_{ 3 }$ and ${ CH }_{ 3 }COOH$
0%
$KOH$ and ${ CH }_{ 3 }COONa$
0%
$H{ NO }_{ 3 }$ and ${ CH }_{ 3 }COONa$
0%
${ CH }_{ 3 }COOH$ and ${ CH }_{ 3 }COONa$

If concentration of two acids are some, their relative strengths can be compared by:

0%
${ \alpha }_{ 1 }/{ \alpha }_{ 2 }$
0%
$K_{ 1 }/K_{ 2 }$
0%
${ \left[ { H }^{ + } \right] }_{ 1 }/{ \left[ { H }^{ + } \right] }_{ 2 }$
0%
$\sqrt { K_{ 1 }/K_{ 2 } }$

Explanation

Relative strength of two acids can be compared by their degree of dissociation.

$HA\rightleftharpoons H^++A^-$

$C$

$C-C\alpha$

$C\alpha$

If concentration of two acids are same so their relative strength can be compared by their

$[H^{+}]$ concentration.

$K_a=C\alpha^2$

$\alpha=(K_a/C)^{0.5}$

If concentration of two acids are same so their relative strength can be compared by square root of their dissociation constants.

For dissociation constant

$(K)$ and ionic product

$({K}_{w})$ of water which is correct?

0%
$K>{ K }_{ w }$
0%
${ K }_{ w }>K$
0%
${ K }_{ w }=K$
0%
None of these

Explanation

The value of ionisation constant of water is $1.8 \times 10^{-16} \text{and} \dfrac{1}{55.55}$ times that of $K_w$ , the ionic product of water.

So $K_w>K$ .

Hence option B is correct.

Which mixture forms a buffer when dissolved in

$1\ L$ of water?

0%
$0.2\ mol\ NaOH+0.2\ mol \ HBr$
0%
$0.2\ mol \ NaCl+0.3\ mol\ HCl$
0%
$0.4\ mol\ H{NO}_{2}+0.2\ mol\ NaOH$
0%
$0.5\ mol\ {NH}_{3}+0.5\ mol\ HCl$

Explanation

(C)

$0.4 mol HNO_2 + 0.2 mol NaOH$ will form a buffer as it is a mixture of weak acid and its conjugate base.

Which of the following statements is/are correct about the ionic product of water?

0%
$K$ (equilibrium constant of water) $< { K }_{ w }$ (ionic product of water)
0%
$pK>p{ K }_{ w }$
0%
At $300K$ , ${ K }_{ w }$ of water becomes ${10}^{-12}$
0%
Ionic product of water at ${ 25 }^{ o }C$ is ${ 10 }^{ -14 }$

Explanation

Ionic product of water(

$K_w$ ) is always greater than equilibrium constant of water(

$K$ ).So

$pK_a=-\log(K)$ is always greater than

$pK_w=-\log(K_w)$ .Ionic product of water at

$25^{0}C$ is

$10^{-14}$ .As temperature increases there will increase in dissociation of water so concentration of

$H^{+}$ and

$OH^{-}$ increases so

$K_w$ increases and it becomes

$10^{-12}$ .

Which of the following is not a buffer solution ?

0%
$CH_3COOH + CH_3COONa$
0%
$H_3BO_3+Na_3BO_3$
0%
$HClO_4 + NaClO_4$
0%
$NH_4OH+ (NH_4)_2SO_4$

Explanation

Correct option: C

Hint: Buffer is a mixture of weak acid or base and its salt.

Explanation

A buffer solution is a mixture of a weak acid or a weak base and its salt.

In the given question, $C{H_3}COOH$ and ${H_3}B{O_3}$ are weak acids whereas $N{H_4}OH$ is a weak base. $HCl{O_4}$ is a strong acid, its mixture with its salt cannot be a buffer solution according to the definition of buffer solution.

Hence, mixture of ${\textbf{HCl}}{{\textbf{O}}_{\textbf{4}}}$ and ${\textbf{NaCl}}{{\textbf{O}}_{\textbf{4}}}$ is not a buffer solution.

The solubility of

${ A }_{ 2 }{ X }_{ 5 }$ is

$x\ mol\ { dm }^{ -3 }$ . Its solubility product is:

0%
$36{ x }^{ 6 }$
0%
$64\times { 10 }^{ 4 }{ x }^{ 7 }$
0%
$126{ x }^{ 7 }$
0%
$1.25\times { 10 }^{ 4 }{ x }^{ 7 }$

Explanation

Let the solubility be

$S.$

${ As }_{ 2 }X_{ 3 }\leftrightharpoons 2As^{ 5+ }+5X^{ 2- }\\ \quad \quad \quad \quad 2S\quad \quad \quad 5S$

Solubility product is

$K_{ sp }=[As^{ 5+ }]^{ 2 }\times [S^{ 2- }]^{ 5 }\\$

Let

$[{ As }_{ 2 }S_{ 5 }]=S,[As^{ 5+ }]=2S, [S^{ 2- }]=5S\\ Ksp=(2S)^{ 2 }\times (5S)^{ 5 }=4S^{ 2 }\times 3125S^{ 5 }=12500S^{ 7 }$

The solution of blue vitriol in water is acidic because:

0%
$CuSO_4$ reacts with water
0%
$Cu^{2+}$ reacts with water
0%
$SO^{2-}_4$ reacts with water
0%
$CuSO_4$ removes $OH^-$ ions from water

The degree of dissociation of water at

${ 25 }^{ o }C$ is

$1.8\times { 10 }^{ -7 }$ % and density is

$1.0g{ cm }^{ -3 }$ . The ionic constant for water is:

0%
$1.0\times { 10 }^{ -14 }$
0%
$2.0\times { 10 }^{ -16 }$
0%
$1.0\times { 10 }^{ -16 }$
0%
$1.0\times { 10 }^{ -8 }$

Explanation

For the equilibrium:

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

given

$\alpha =1.8\times 10^{ -7 }\times 1/100=1.8\times 10^{ -9 }$

Since

$\alpha$ <<<1,

$[H_2O]$ is not much affected.

therefore concentrations at equilibrium are as:

$[H_2O]=\frac {1000\times 1}{18} \approx 55.5\ M$

$[H^+]=\alpha \times 55.5 \approx 1\times 10^{ -7 }$

$[OH^-]=\alpha \times 55.5 \approx 1\times 10^{ -7 }$

Ionic constant of water:

$K=\frac {[OH^-][H^+]}{[H_2O]}$

on substitution we get

$K=\frac {10^{ -7 } \times 10^{ -7 }}{55.5}$

$K\approx 2\times 10^{ -16 }$

Therefore option B is correct.

What do you mean by buffer solution?

0%
Buffer solution have no $pH$
0%
Its $pH$ changes very little when a small amount of acid or base is added to the it
0%
Its $pH$ changes very largely when a small amount of acid or base is added to the it
0%
All solutions are buffer

Explanation

It's

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

Buffer solutions are used as a means of keeping

$pH$ at a nearly constant value in a wide variety of chemical applications. In nature, there are many systems that use buffering for

$pH$ regulation. For example, the bicarbonate buffering system is used to regulate the

$pH$ of blood.

The pH of a

$0.1M$ aqueous solution of a weak acid (

$HA$ ) is

$3$ . Its degree of dissociation is

0%
$1$ %
0%
$10$ %
0%
$50$ %
0%
$25$ %

Explanation

$\left[ { H }^{ + } \right] =C.\alpha ={ 10 }^{ -3 }$

$0.1\times \alpha ={ 10 }^{ -3 }$

$\alpha ={ 10 }^{ -2 }= 10^{-2}\times100=1 \%$

To prepare a buffer solution of

$pH = 4.04$ , amount of Barium acetate to be added to

$100 mL$ of

$0.1$ M acetic acid solution [

$pK_0(CH_3COO^-) =9.26$ ] is:

0%
$0.05$ mole
0%
$0.025$ mole
0%
$0.1$ mole
0%
$0.005$ mole

Explanation

For the buffer solution of

$CH_3COOH$ &

$CH_3COO^-$

$pH = pK_a + log \frac { [ CH_3COOH]}{[CH_3COO^-]}$

$\Rightarrow$ m. moles of

$CH_COO^- = 50$

$\Rightarrow 4.04 = 4.74 + log \frac { ( 100 \times 0.1 ) }{ m. \quad moles \quad of \quad CH_3COO^-}$

$\therefore$ moles of

$(CH_3COO)_2$ Ba required = 0.025

Assuming each salt to be 90% dissociated, which of the following will have highest boiling point.

0%
Decimolar $Al_2(SO_4)_3$
0%
Decimolar $BaCl_2$
0%
Decimolar $Na_2SO_4$
0%
A solution obtained by mixing equal volumes of (B) and (C)

Explanation

$Al_2(SO_4)_3$ dissociate to give

$5$ ions as it is a pure crystalline solid having monoclinic end-centered crystal structure.

$\alpha = 90 \% = 0.9$

Boiling point depends on number of particles . More the number of particles, more the boiling point.

The concentration of

$[H^{+}]$ and concentration of

$[OH^{-}]$ of a

$0.1\ M$ aqueous solution of

$2$ M ionised weak acid is: [ionic product of water

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

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

Explanation

Degree of ionization,

$\alpha = \cfrac{{\left[ {{H^ + }} \right] \times 100}}{{\left[ {HA} \right]}}$

$\left[ {{H^ + }} \right] = \cfrac{{\alpha \times \left[ {HA} \right]}}{{100}}$

$= \cfrac{{2 \times 0.1}}{{100}}$

$\left[ {{H^ + }} \right] = 2 \times {10^{ - 3}}M$ .

${K_W} = \left[ {{H^ + }} \right]\left[ {O{H^ - }} \right]$

${10^{ - 14}} = 2 \times {10^{ - 3}} \times \left[ {O{H^ - }} \right]$

$\left[ {O{H^ - }} \right] = \frac{{{{10}^{ - 14}}}}{{2 \times {{10}^{ - 3}}}}$

$\left[ {O{H^ - }} \right] = 5 \times {10^{ - 12}}M$ .

How many grams of

$CO_2$ gas is dissolved in a

$1$ lt bottle of carbonated water if the manufacturer uses a pressure of

$2.4$ atmosphere in the bottling process at

$25^o$ C? Given

$K_H$ of

$CO_2$ water

$=29.76$ atm/mole/l at

$25^o$ C.

0%
$3.52$
0%
$4.2$
0%
$3.1$
0%
$2.5$

Explanation

According to Hennry's Law,

$p=CK_h$

$p=2.4 atm$

$K_h=29.76\cfrac{atm}{mol/L}$

$C=\cfrac{p}{K_h}=\cfrac{2.4}{29.76}=0.08\cfrac{mol}{L}$

$0.08 mol$ of

$CO_2$ are present in

$1L$ of solution.

Grams of

$CO_2=(moles)(molar\quad mass)=(0.08)(44)=3.52$

Buffer solution

$A$ of a weak monoprotic acid and its sodium salt in the concentration ratio

$x : y$ has

$pH = (pH)$ ,. Buffer solution

$B$ of the same acid and its sodium salt in the concentration ratio

$y : x$ has

$pH = (pH)_{2}$ . If

$(pH)_{2} - (pH)_{1} = 1$ unit and

$(pH)_{1} + (pH)_{2} = 9.5\ units$ , then:

0%
$\dfrac {x}{y} = 3.162$
0%
$pK_{a} = 4.75$
0%
$\dfrac {x}{y} = 2.36$
0%
$pK_{a} = 5.25$

Explanation

$(pH)_1= pK_a+\log \cfrac {y}{x}$

$\Rightarrow (pH)_2=pK_a+\log \cfrac {x}{y}$

$\Rightarrow (pH)_2-(pH)_1=1=\log \cfrac {x}{y}-\log \cfrac {y}{x}= 2 \log \cfrac {x}{y}$

$\therefore \log \cfrac {x}{y}=\cfrac {1}{2} \Rightarrow \cfrac {x}{y}= 3.162$

$(pH)_1+(pH)_2=9.5=2pK_a+\log \cfrac {y}{x}\times \cfrac {x}{y}=2pK_a+0$

$\therefore pK_a=4.75$

To prepare a buffer of pH 8.26 amount of

$({ NH }_{ 4 }{ ) }_{ 2 }{ SO }_{ 4 }$ to be added to 500 mL of 0.01 M

${ NH }_{ 4 }OH$ solution is :

$[pK_{ a }({ NH }_{ 4 }^{ + })\ =9.26]$

0%
0.05 mole
0%
0.025 mole
0%
0.10 mole
0%
0.005 mole

Explanation

Let

$(NH_4)_2SO_4$ to be added to 500 mL solution = x mol

$[NH^{+}_4]=4xM$

$pH=8.26$

$pOH=14−8.26=5.74$

$pK_b=14−9.26=4.74$

$pOH=pK_b+log\dfrac{[NH^{+}_4]}{[NH_4OH]}$

$5.74=4.74+log\dfrac{4x}{0.01}$

$log4x\times 10^2 = 1$

$4x\times 10^2=1$

$x=\dfrac{1}{40}\ mol = 0.025\ mol$

A buffer solution made up of

$BOH$ and

$BCl$ of total molarity 0.29 M has

$pH = 9.6$ and

${ K }_{ b }=1.8\ \times \ { 10 }^{ -5 }$ . Concentration of salt and base respectively is:

0%
0.09 M and 0.2 M
0%
0.2 M and 0.09 M
0%
0.1 M and 0.19 M
0%
0.19 M and 0.1 M

Explanation

$pH+pOH=14$

$\therefore\ pOH=14-9.6=4.4$

${for\ basic\ buffer}=p{OH}=p{K_b}+\log \dfrac {[salt]}{[Base]}$

$4.4=-\log (1.8\times 10^{-5})+\log \dfrac {[salt]}{[Base]}$

$4.4=4.74+\log \dfrac {[salt]}{[Base]}$

$-0.34=\log \dfrac {[salt]}{[Base]}$

${[Base] =2.18\ [salt]}$

Now,

Total molarity

$= [salt] +[Base]= 0.29\ M$

$[salt]\ 3.18=0.29\ M$

$salt=0.09\ M$

$\therefore \ Base =0.29-0.09=0.2\ M$

For preparing a buffer solution of

$pH=7.0$ , which buffer system you will choose:

0%
$H_{3}PO_{4},H_{2}PO_4{^-}$
0%
$H_{2}PO_4{^-},\ HPO^{2-}_{4}$
0%
$HPO^{2-}_{4},PO^{3-}_{4}$
0%
$H_{3}PO_{4},PO^{3-}_{4}$

Explanation

Solution:- (B)

${{H}_{2}P{O}_{4}}^{-}, {HP{O}_{4}}^{2-}$

${{H}_{2}P{O}_{4}}^{-}$ is a weak acid, thus it's conjugate base will be a strong base and thus will form a buffer of

$pH$ approx

$7$ .

The pH of 0.2 M solution of acid HQ =Then the value of

$K_{a}$ for HQ will be

0%
$10^{-7}$
0%
$5\times10^{-6}$
0%
$5\times10^{-3}$
0%
$3\times10^{-3}$

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

$H_3PO_4$ with

$NaOH$ are:

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

Explanation

There will be 3 buffer solution they are

$P{ O }_{ 4 }^{ 3- }$ - with

$NaOH$

$HP{ O }_{ 4 }^{ 2- }$ with

$NaOH$

$HP{ O }_{ 4 }^{ - }$ with

$NaOH$

At a temperature under high pressure

$K_w(H_2O) \, = \, 10^{10}$ , a solution of pH 5.4 is said to be:

0%
Acidic
0%
Basic
0%
Neutral
0%
Amphitoric

At what temperature liquid water will be in equilibrium with water vapour?

$\Delta H_{vap} = 40.73 \ kJ \ mol^{-1}$ ,

$\Delta S_{vap} = 0.109 \ kJ \ K^{-1} \ mol^{-1}$ ,

0%
282.4 K
0%
373.6 K
0%
100 K
0%
400 K

Explanation

The given reaction is :-

${ H }_{ 2 }O\left( l \right) \rightleftharpoons { H }_{ 2 }O\left( g \right)$

Now, at equilibrium,

$\triangle G=0$

Now, from fibb's free energy change equation

$\triangle G=\triangle H-T\triangle S$

$\Rightarrow \quad \triangle H=T\triangle S$

$\Rightarrow \quad T=\dfrac { \triangle H }{ \triangle S } =\dfrac { 40.73 }{ 0.1079 } =373.6K$

$\left( \because \triangle Hrap=40.73KJ/mol,\quad \triangle Srap=0.109KJ{ K }^{ -1 }{ mol }^{ -1 } \right)$

Several blocks of magnesium are fixed to the bottom of a ship to

0%
Prevent puncture by undersea rocks
0%
Keeps away the sharks
0%
Prevent action of water and salts
0%
Make the ship lighter

What will be the value of pH of

$0.01 mol\ dm^{-3} CH_3COOH (K_a = 1.74 \times 10^{-5})$ ?

0%
$3.4$
0%
$3.6$
0%
$3.9$
0%
$3.0$

Explanation

$\underset{C(1-\alpha)}{CH_3COOH}+H_2O\longrightarrow \underset{C\alpha}{H_3O}+\underset{C\alpha}{CH_3COO}$

$K_a=C\alpha^2\Rightarrow \alpha=\sqrt{\dfrac{Ka}{C}}$

$[H_3O^+]=\sqrt[C]{\dfrac{Ka}{c}}=\sqrt{Ka.C}$

$=\sqrt{0.01\times 1.74\times 10^{-5}}$

$\sqrt{1.74\times 10^{-7}}=4,17\times 10^{-4}$

$PH_2-\log(4.17\times 10^{-4})=3.38\simeq 3.4$

Choose the correct answer from the alternatives given.

The human stomach produces acid

$X$ which helps in the digestion of food. Acid

$X$ is _______.

0%
acetic acid
0%
methanoic acid
0%
hydrochloric acid
0%
citric acid

Explanation

The human stomach produces hydrochloric acid. It helps our body to break down, digest, and absorb nutrients such as protein. It also eliminates bacteria and viruses in the stomach, protecting your body from infection. Low levels of hydrochloric acid can have a profound impact on the body's ability to properly digest and absorb nutrients.

So, acid

$X$ is hydrochloric acid

$(HCl)$ .

Thus, option C is correct.

An acid

$HA$ ionizes as,

$HA\rightleftharpoons H^++A^-$ . The pH of

$1.0$ M solution is

$5$ . Its dissociation constant would be:

0%
$1\times 10^{-10}$
0%
$5$
0%
$5\times 10^{-8}$
0%
$1\times 10^{-5}$

Explanation

$HA\rightleftharpoons H^+ +A^-$

Initial

$\quad 1m\quad 0\quad 0$

equilibrium

$1-x\quad x\quad x$

given

$\left\{pH=5\right\}$

$5=-\log [H^+]$

$[H^+]=10^{-5}$

$\left\{x=10^{-5}\right\}$

$Ka=\dfrac {[H^+] [A^-]}{[HA]}$

$Ka=\dfrac {(10^{-5} (10^{-5}))}{1-10^{-5}}$

$Ka=1\times 10^{-10} \left\{(1-10^{-5}\approx 1)\right\}$

At 373 K, steam and water are in equilibrium and

$\Delta H$ = 40.98 kJ

$mol^{-1}$ .What will be

$\Delta S$ fro conversion of water into steam?

$H_2O_{(l)} \rightarrow H_2O_{(g)}$

0%
109.8 J $K^{-1}mol^{-1}$
0%
31 J $K^{-1}mol^{-1}$
0%
21.98 J $K^{-1}mol^{-1}$
0%
326 J $K^{-1}mol^{-1}$

Explanation

The given reaction is :-

${ H }_{ 2 }O\left( l \right) \longrightarrow { H }_{ 2 }O\left( g \right)$

So, we have

$\triangle Svap=\dfrac { \triangle Hvap }{ Tb }$

$=\dfrac { 40.98\times 1000 }{ 373 } =109.8{ JK }^{ -1 }{ mol }^{ -1 }$

Which one of the following mixture does not act as a buffer solution?

0%
Boric acid and borax
0%
Sodium phosphate & disodium hydrogen phosphate
0%
Sodium propionate and propionic acid
0%
Sodium acetate and sodium propionate

Which of the following graphs are correct for the give reaction

${H}_{2}(g)+{CO}_{2}(g)\rightleftharpoons {H}_{2}(g)+CO(g)$
Assume initially only

${H}_{2}$ and

${CO}_{2}$ are present

Explanation

The equation will be-

${ H }_{ 2 }(g)+{ CO }_{ 2 }(g)\rightleftharpoons { H }_{ 2 }O(g)+CO$

The correct option is D

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

$HCl$ ?

0%
$CH_3COOH$
0%
$H_2S$
0%
$C_6H_5COOH$
0%
$H_2SO_4$

The equivalent conductance of

$M/32$ solution of monobasic acid is

$8.0$ mho

$cm^2$ and at infinite dilution is

$400$ mho

$cm^2$ . The dissociation constant of this acid is:

0%
$1.25\times 10^{-5}$
0%
$1.25\times 10^{-6}$
0%
$6.25\times 10^{-4}$
0%
$1.25\times 10^{-4}$

Explanation

Degree of dissociation,

$\alpha=\frac{\lambda ^{c}}{\lambda ^{infinite}}$

$\alpha=\frac{8}{400} = 2\times 10^{^{-2}}$

$Ka = \frac{C\alpha ^{2}}{1-\alpha}$

$C\alpha ^{2}$ ( 1>>

$\alpha ^{2}$ )

$1.25\times 10^{-5}$

For preparing a buffer solution of

$pH=7.0$ , which buffer system you will choose?

0%
${ H }_{ 3 }{ PO }_{ 4 },{ H }_{ 2 }{ PO }_{ 4 }^{ - }$
0%
${ H }_{ 2 }{ PO }_{ 4 }^{ - }, { HPO }_{ 4 }^{ 2- }$
0%
${ HPO }_{ 4 }^{ 2- },{ PO }_{ 4 }^{ 3- }$
0%
${ H }_{ 3 }{ PO }_{ 4 },{ PO }_{ 4 }^{ 3- }$

Explanation

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

$8.2$

A buffer solution with

$pH =9$ is to be prepared by mixing

$NH_4CI$ and

$NH_4OH$ . Calculate the number of moles of

$NH_4CI$ that should be added to one litre pf

$1.0 M$

$NH_4OH$ .

$[K_b = 1.8 \times 10^{-5}]$

0%
$3.4$
0%
$2.6$
0%
$1.5$
0%
$1.8$

Explanation

Given

$pH=9, pOH=14-9=5$

$pOH=pK_b+\log \cfrac{[salt]}{[base]}$

$5=-\log(1.8\times 10^{-5})+\log\cfrac{[salt]}{1.0}=4.745+\log[salt]$

$\Longrightarrow [salt]=1.79\simeq 1.8$

$1.8$ No. of moles of

$NH_4CI$ to be added to

$1L$ of

$1.0M$

$NH_4OH.$

Which of the following solutions when added to

$1L$ of a

$0.1\ M$

${ CH }_{ 3 }COOH$ solution will cause no change in either the degree of dissociation of

${ CH }_{ 3 }COOH$ or the pH of the solution

$[Ka = 1.6\ \times\ 10^{-5}$ for

${ CH }_{ 3 }COOH]$

0%
$0.6\ mM$ $HCOOH (K_{a}=8\ \times 10^{-4})$
0%
$0.1\ M\ CH_{3}COONa$
0%
$0.1$ of $2mM\ HCI$
0%
$0.1 M$ ${ CH }_{ 3 }COOH$

Find the percentage of ionisation of 0.2 M acetic acid solution, whose dissociation constant is

$1 .8 \times 10^{-5}$ .

0%
0.198
0%
0.290
0%
0.950
0%
none of these

${ H }_{ 3 }{ PO }_{ 4 }$ is a tribasic acid and one of its salt is

${ NaH }_{ 2 }{ PO }_{ 4 }$ What volume of

$1M\quad NaOH$ solution should be added to

$12g\ { NaH }_{ 2 }{ PO }_{ 4 }$ to convert it into

${ Na }_{ 3 }{ PO }_{ 4 }$ ? (

$at.wt$ of

$P=31$ )

0%
$100\ ml$
0%
$200\ ml$
0%
$80\ ml$
0%
$300\ ml$

Explanation

$Na{ H }_{ 2 }{ PO }_{ 4 }+NaOH\rightarrow { Na }_{ 2 }{ HPO }_{ 4 }+{ H }_{ 2 }O\quad \quad -(i)$

Moles

$=\cfrac { 12 }{ 120 }$

$\cfrac { 12 }{ 120 }$

${ Na }_{ 2 }{ HPO }_{ 4 }+NaOH\rightarrow { Na }_{ 3 }{ PO }_{ 4 }+{ H }_{ 2 }O\quad \quad \quad -(ii)$

Moles

$=\cfrac { 12 }{ 120 }$

$\cfrac { 12 }{ 120 }$

Moles of

$NaOH$ required

$=\cfrac { 12 }{ 120 } +\cfrac { 12 }{ 120 }$

$=0.2$ moles

Now,

$Mularity=\cfrac { Moles }{ Vol.of\quad sol\quad in\quad ltrs }$

$\therefore$ Volume of

$NaOH$ solution

$=\cfrac { 0.2 }{ 1 }$

$=0.2$ litres

$=200$ ml

A monoprotic acid in

$100\ M$ solution is

$0.001\%$ ionized. The dissociation constant of this acid is:

0%
$1.0\ \times 10^{-3}$
0%
$1.0\ \times 10^{3}$
0%
$1.0\ \times 10^{-8}$
0%
$1.0\ \times 10^{-10}$

Explanation

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

% ionization

$=0.001$ %

$[HA]=100\ M$

Ionization percent

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

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

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

$=1\times 10^{ -8 }$

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?

0%
${ (pH) }_{ NaA\ }>\ ({ pH) }_{ NaB }$
0%
${ (pH) }_{ NaA }<\ ({ pH) }_{ NaB }$
0%
${ (pH) }_{ NaA }> ({ pH) }_{ NaD }$
0%
${ (pH) }_{ NaB }=7$

Explanation

${ pH }=p{ K }_{ a }+\log _{ 10 }{ (\cfrac { [salt] }{ [Acid] } ) }$ {Henderson Hassel batch equation}

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

$2HI \rightleftharpoons H_2 + I_2$

The equilibrium constant of the above reaction is

$6.4$ at

$300$ K. If

$0.25$ mole each of

$H_2$ and

$I_2$ are added to the system, the equilibrium constant will be:

0%
1.6
0%
3.2
0%
0.8
0%
6.4

Strong electrolyte of the following is?

0%
$01M$ $HAc$
0%
$0.1M$ $HCl$
0%
$0.1M$ $KCl$
0%
$0.1M$ $NaCl$

Statement (A): For a liquid-vapour equilibrium the system must be a closed system.

Statement (B): For the equilibrium

${ H }_{ 2 }{ O }_{ \left( s \right) }\rightleftharpoons { H }_{ 2 }{ O }_{ \left( 1 \right) }$ , High temperature favours the formation of liquid

Statement (C): For the equilibrium

${ NH }_{ 4 }{ CI }_{ \left( s \right) }\rightleftharpoons { NH }_{ 4 }{ CI }_{ \left( g \right) }-x$ K cal low temperature favours the formation of solid

${ NH }_{ 4 }CI$

0%
All the statements are correct
0%
All the statements are incorrect
0%
A, B are correct and C is incorrect
0%
A, B are incorrect and C is correct

100 mL of 20.8%

$BaCl_2$ solution and 50 mL of 9.8%

$H_2SO_4$ solution will form

$BaSO_4$

$(Ba=137, Cl=35.5, S=32, H=1, O=16)$

$BaCl_2+H_2SO_4\rightarrow BaSO_4+2HCl$

0%
23.3 g
0%
11.65 g
0%
30.6 g
0%
None of these

Explanation

$100ml$ of

$20.8$ %

$BaCl_2$ solution=

$20.8g$

$BaCl_2$

$50ml$ of

$9.8$ %

$H_2SO_4$ solution=

$4.9g$

$H_2SO_4$

Reaction:

$BaCl_2+H_2SO_4\longrightarrow BaSO_4\downarrow +2HCl$

$208 g mol^{-1}$

$98 g mol^{-1}$

$233 g mol^{-1}$

$\therefore 98g$

$H_2SO_4$ reacts with

$208g$

$BaCl_2$

$4.9g$

$H_2SO_4$ reacts with

$\cfrac {208}{98}\times 4.9=10.4 g$

$BaCl_2$

$98g$

$H_2SO_4$ will produce

$233g$

$BaSO_4$

$\therefore 4.9g$

$H_2SO_4$ will produce=

$\cfrac {233}{98}\times 4.9=11.65g$

$BaSO_4$

CBSE Questions for Class 11 Medical Chemistry Equilibrium Quiz 8 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Equilibrium Quiz 8 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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