Ionic Equilibrium — Page 5 | JEE Chemistry

Q41

pKa of a weak acid (HA) and pKb of a weak base (BOH) are 3.2 and 3.4, respectively. The pH of their salt (AB) solution is :

A 6.9

B 7.0

C 1.0

D 7.2

Correct Answer

Option A

Solution

The salt (AB) given is a salt is of weak acid and weak base.

Hence its pH can be calculated by the formula $\therefore$ pH = 7 +

{1 \over 2}p{K_a} - {1 \over 2}p{K_b}

= 7 +

{1 \over 2}\left( {3.2} \right) - {1 \over 2}\left( {3.4} \right)

= 6.9

Q42

Arrange the following solutions in the decreasing order of pOH : (A) 0.01 M HCl (B) 0.01 M NaOH (C) 0.01 M CH3COONa (D) 0.01 M NaCl

A (B) > (C) > (D) > (A)

B (A) > (D) > (C) > (B)

C (A) > (C) > (D) > (B)

D (B) > (D) > (C) > (A)

Correct Answer

Option B

Solution

(A) 10–2 M HCl $\Rightarrow$ [H+] = 10–2 M $\Rightarrow$ pH = 2 $\Rightarrow$ pOH = 14 - 2 = 12 (B) 10–2 M NaOH $\Rightarrow$ [OH–] = 10–2 M $\Rightarrow$ pOH = 2 (C) 10–2 M CH3COO–Na+ $\Rightarrow$ [OH+] > 10–7 $\Rightarrow$ pOH < 7 (D) 10–2 M NaCl $\Rightarrow$ Neutral pOH = 7 Order of pOH value A > D > C > B

Q43

In a sautrated solution of the sparingly soluble strong electrolyte AgIO3 (Molecular mass = 283) the equilibrium which sets in is AgIO3(s)

\leftrightharpoons

Ag+(aq) +

IO_3^-

If the solubility product constant Ksp of AgIO3 at a given temperature is 1.0

\times

10−8, what is the mass of AgIO3 contained in 100 ml of its saturated solution?

A 28.3 × 10−2 g

B 2.83 × 10−3 g

C 1.0 × 10−7 g

D 1.0 × 10−4 g

Correct Answer

Option B

Solution

Let

\,\,\,s = \,\,\,

solubility

Ag{\rm I}{O_3}\,\rightleftharpoons\,\mathop {A{g^ + }}\limits_s \,\,\mathop {{\rm I}{O_3}^ - }\limits_s

{K_{sp}} = \left[ {A{g^ + }} \right]\left[ {{\rm I}{O_3}^ - } \right]

= s \times s = {s^2}

Given

\,\,\,{K_{sp}} = 1 \times {10^{ - 8}}

$\therefore$

\,\,\,s = \sqrt {{K_{sp}}} = \sqrt {1 \times {{10}^{ - 8}}}

= 1.0 \times {10^4}\,\,mol/lit

= 1.0 \times {10^{ - 4}} \times 283\,g/lit

(as Molecular mass of

Ag\,{\rm I}{O_3} = 283

)

= {{1.0 \times {{10}^{ - 4}} \times 283 \times 100} \over {1000}}gm/100ml

= 2.83 \times {10^{ - 3}}\,gm/100\,ml

Q44

The pKa of a weak acid, HA, is 4.80. The pKb of a weak base, BOH, is 4.78. The pH of an aqueous solution of the corresponding salt, BA, will be

A 9.58

B 4.79

C 7.01

D 9.22

Correct Answer

Option C

Solution

In aqueous solution

BA

(salt) hydrolyses to give

BA\,\, + \,\,{H_2}O\,\rightleftharpoons\,BOH\,\, + \,\,HA

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,

Base acid Now

pH

is given by

pH = {1 \over 2}p{K_w} + {1 \over 2}pKa - {1 \over 2}p{K_b}

Substituting given values, we get

pH = {1 \over 2}\left( {14 + 4.80 - 4.78} \right) = 7.01

Q45

An alkali is titrated against an acid with methyl orange as indicator, which of the following is a correct combination?

A .tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; overflow:hidden;padding:10px 5px;word-break:normal;} .tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;} .tg .tg-baqh{text-align:center;vertical-align:top} .tg .tg-amwm{font-weight:bold;text-align:center;vertical-align:top} Base Acid End point Weak Strong Colourless to pink

B Base Acid End point Strong Strong Pinkish red to yellow

C Base Acid End point Weak Strong Yellow to pinkish red

D Base Acid End point Strong Strong Pink to colourless

Correct Answer

Option C

Solution

Methyl orange is a pH indicator frequently used in titrations because of its clear and distinct colour change.

Under acidic conditions, it is red or pinkish red and under alkaline conditions, it turns yellow.

Let's consider each of the options : Option A : When a weak base is titrated against a strong acid, the solution at the end point is slightly acidic.

However, the color change from colorless to pink is not associated with methyl orange.

Option B : When a strong base is titrated against a strong acid, the solution is neutral at the end point.

Methyl orange changes color from pinkish-red in acidic conditions to yellow in basic conditions, so this color change doesn't match the description.

Option C : When a weak base is titrated against a strong acid, the solution is slightly acidic at the end point.

Methyl orange would change from yellow in the basic solution to red in the acidic solution at the end point.

This corresponds to a "yellow to pinkish-red" color change.

Option D : When a strong base is titrated against a strong acid, the solution is neutral at the end point.

The color change described here is "pink to colorless" which does not correspond with the characteristic changes of methyl orange.

Therefore, the correct answer is Option C: Weak base, Strong acid, Yellow to pinkish-red.

Q46

Solubility product of silver bromide is 5.0

\times

10–13. The quantity of potassium bromide (molar mass taken as 120g of mol–1) to be added to 1 litre of 0.05 M solution of silver nitrate to start the precipitation of AgBr is :

A 1.2

\times

10–10 g

B 1.2

\times

10–9 g

C 6.2

\times

10–5 g

D 5.0

\times

10–8 g

Correct Answer

Option B

Solution

AgBr\,\rightleftharpoons\,A{g^ + } + B{r^ - }

{K_{sp}} = \left[ {A{g^ + }} \right]\left[ {B{r^ - }} \right]

For precipitation to occur Ionic product

>

Solubility product

\left[ {B{r^ - }} \right] = {{{K_{sp}}} \over {\left[ {A{g^ + }} \right]}} = {{5 \times {{10}^{ - 13}}} \over {0.05}} = {10^{ - 11}}

i.e., precipitation just starts when

{10^{ - 11}}\,

moles of

KBr

is added to

1\ell \,AgN{O_3}\,

solution $\therefore$

\,\,\,

Number of moles of

\,\,\,B{r^ - }\,\,

needed from

KBr = {10^{ - 11}}

$\therefore$

\,\,\,

Mass of

KBr = {10^{ - 11}} \times 120

= 1.2 \times {10^{ - 9}}g

Q47

The minimum volume of water required to dissolve 0.1 g lead (II) chloride to get a saturated solution (Ksp of PbCl2 = 3.2

\times

10-8 atomic mass of Pb = 207 u ) is :

A 0.36 L

B 17.98 L

C 0.18 L

D 1.798 L

Correct Answer

Option C

Solution

Given, Ksp of PbCl2 = 3.2 $\times$ 10 $-$ 8 .tg .tg PbCl2 $\rightleftharpoons$ Pb2+ 2Cl Initially 1 0 0 At equilibrium 1-s s 2s Ksp = [s] [2s]2

\therefore\,\,\,\,

Ksp = 4s3

\therefore\,\,\,\,

4s3 = 3.2 $\times$ 10 $-$ 8 $\Rightarrow$

\,\,\,\,

s3 = 8 $\times$ 10 $-$ 9 $\Rightarrow$

\,\,\,\,

s = 2 $\times$ 10 $-$ 3 M Solubility =

{{no.\,\,of\,moles\,\,of\,\,pbc{l_2}\,} \over {volume\,(in\,\,litre)}}

\therefore\,\,\,\,

2 $\times$ 10 $-$ 3 =

{{0.1} \over {278}}

$\times$

{1 \over V}

$\Rightarrow$

\,\,\,\,

V =

{{0.1} \over {278}}

$\times$

{{{{10}^3}} \over 2}

= 0.18 L

Q48

The Ksp for bismuth sulphide (Bi2S3) is 1.08

\times

10

-

73. The solubility of Bi2S3 in mol L

-

1 at 298 K is :

A 1.0

\times

10

-

15

B 2.7

\times

10

-

12

C 3.2

\times

10

-

10

D 4.2

\times

10

-

8

Correct Answer

Option A

Solution

Ksp = (2s)2(3s)3 = 108s5 108s5 = 108 × 10–75 s = 1.0 × 10–15 mol/L

Q49

Let the solubility of an aqueous solution of Mg(OH)2 be x then its Ksp is :

A 4x3

B 108x5

C 27x4

D 9x

Correct Answer

Option A

Solution

Mg{\left( {OH} \right)_2} \to \mathop {\left[ {M{g^{2 + }}} \right]}\limits_x + 2\mathop {\left[ {O{H^ - }} \right]}\limits_{2x}

{K_{sp}} = \left[ {Mg} \right]{\left[ {OH} \right]^2}

= \left[ x \right]{\left[ {2x} \right]^2}

= x.4{x^2} = 4{x^3}.

Q50

Class XII students were asked to prepare one litre of buffer solution of

\mathrm{pH} \,8.26

by their Chemistry teacher: The amount of ammonium chloride to be dissolved by the student in

0.2\, \mathrm{M}

ammonia solution to make one litre of the buffer is : (Given:

\mathrm{pK}_{\mathrm{b}}\left(\mathrm{NH}_{3}\right)=4.74

Molar mass of

\mathrm{NH}_{3}=17 \mathrm{~g} \mathrm{~mol}^{-1}

Molar mass of

\mathrm{NH}_{4} \mathrm{Cl}=53.5 \mathrm{~g} \mathrm{~mol}^{-1}

)

A 53.5 g

B 72.3 g

C 107.0 g

D 126.0 g

Correct Answer

Option C

Solution

For basic Buffer,

\mathrm{pOH}=\mathrm{pK}_{\mathrm{b}}+\log \frac{[\text{ salt }]}{[\text{ Base }]}

\mathrm{pOH}=14-8.26=5.74

5.74=4.74+\log \frac{\left[\mathrm{NH}_{4} \mathrm{Cl}\right]}{0.2}

\left[\mathrm{NH}_{4} \mathrm{Cl}\right]=2 \mathrm{M}

Moles of

\mathrm{NH}_{4} \mathrm{Cl}=2 \times 1=2

moles Weight of

\mathrm{NH}_{4} \mathrm{Cl}=2 \times 53.5=107 \mathrm{~g}