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Ionic Equilibrium

JEE Chemistry · 82 questions · Page 1 of 9 · Click an option or "Show Solution" to reveal answer

Q1
Which of the following is a Lewis acid?
A PH3
B B(CH3)3
C NaH
D NF3
Correct Answer
Option B
Solution

As B can accept electron pair, so it is Lewis Acid.

Q2
The strength of an aqueous NaOH solution is most accurately determined by titrating : (Note : consider that an appropriate indicator is used)
A Aq. NaOH in a pipette and aqueous oxalic acid in a burette
B Aq. NaOH in a burette and aqueous oxalic acid in a conical flask
C Aq. NaOH in a volumetric flask and concentrated H2SO4 in a conical flask
D Aq. NaOH in a burette and concentrated H2SO4 in a conical flask
Correct Answer
Option B
Solution

Aq. NaOH in a burette and aqueous oxalic acid in a conical flask.

Q3
Hydrogen ion concentration in mol / L in a solution of pH = 5.4 will be :
A 3.98 ×\times 10-6
B 3.68 ×\times 10-6
C 3.88 ×\times 106
D 3.98 ×\times 108
Correct Answer
Option A
Solution
pH=−log⁡[H+]=log⁡1[H+];pH = - \log \left[ {{H^ + }} \right] = \log {1 \over {\left[ {{H^ + }} \right]}};
                      5.4=log⁡1[H+]\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,5.4 = \log {1 \over {\left[ {{H^ + }} \right]}}

On solving,

[H+]=3.98×10−6\left[ {{H^ + }} \right] = 3.98 \times {10^{ - 6}}
Q4
When rain is accompanied by a thunderstorm, the collected rain water will have a pH value :
A slightly higher than that when the thunderstorm is not there
B uninfluenced by occurence of thunderstorm
C which depends on the amount of dust in air
D slightly lower than that of rain water without thunderstorm
Correct Answer
Option D
Solution

The rain water after thunderstorm contains dissolved acid and therefore the

pHpH

is less than rain water without thunderstorm.

Q5
Given below are two statements. Statement I : In the titration between strong acid and weak base methyl orange is suitable as an indicator. Statement II : For titration of acetic acid with NaOH phenolphthalein is not a suitable indicator. In the light of the above statements, choose the most appropriate answer from the options given below :
A Statement I is false but Statement II is true
B Statement I is true but Statement II is false
C Both Statement I and Statement II are true
D Both Statement I and Statement II are false
Correct Answer
Option B
Solution

Titration curve for strong acid and weak base initially a buffer of weak base and conjugate acid is : Formed, thus pH falls slowly and after equivalence point, so the pH falls sharply so methyl orange, having pH range of 3.2 to 4.4 will weak as indicator.

So, statement I is correct.

Titration curve for weak acid and strong base (NaOH) Initially weak acid will form a buffer so pH increases slowly but after equivalence point.

It rises sharply covering range of phenolphthalein so it will be suitable indicator so statement II is false.

Q6
A solution is 0.1 M in Cl−- and 0.001 M in CrO42−_4^{2 - }. Solid AgNO3 is gradually added to it. Assuming that the addition does not change in volume and Ksp(AgCl) = 1.7 ×\times 10−-10 M2 and Ksp(Ag2CrO4) = 1.9 ×\times 10−-12 M3. Select correct statement from the following :
A AgCl precipitates first because its Ksp is high.
B Ag2CrO4 precipitates first as its Ksp is low.
C Ag2CrO4 precipitates first because the amount of Ag+ needed is low.
D AgCl will precipitate first as the amount of Ag+ needed to precipitate is low.
Correct Answer
Option D
Solution

Conc. of Cl−- = 0.1 M = 10−-1 M Conc. of CrO

42−_4^{2 - }

= 0.001 M = 10−-3 M Ksp(AgCl) = [Ag+][Cl−-] [Ag+]AgCl =

1.7×10−1010−1=1.7×10−9{{1.7 \times {{10}^{ - 10}}} \over {{{10}^{ - 1}}}} = 1.7 \times {10^{ - 9}}
Ksp(Ag2CrO4)=[Ag+]2[CrO42−]{K_{sp}}(A{g_2}Cr{O_4}) = {[A{g^ + }]^2}[CrO_4^{2 - }]
[Ag+]=1.9×10−1210−3=19×10−4[A{g^ + }] = \sqrt {{{1.9 \times {{10}^{ - 12}}} \over {{{10}^{ - 3}}}}} = \sqrt {19} \times {10^{ - 4}}

∴\therefore AgCl will be precipitate first

Q7
The solubility product of a salt having general formula MX2, in water is: 4 ×\times 10-12 . The concentration of M2+ ions in the aqueous solution of the salt is :
A 2.0 ×\times 10-6 M
B 4.0 ×\times 10-10 M
C 1.0 ×\times 10-4 M
D 1.6 ×\times 10-4 M
Correct Answer
Option C
Solution
MX2 ⇌  M2+s          +  2X−2sM{X_2}\,\rightleftharpoons\,\,\mathop {{M^{ 2 + }}}\limits_{s\,\,\,\,\,\,\,\,} \,\, + \,\,\mathop {2{X^ - }}\limits_{2s}

Where

ss

is the solubility of

MX2M{X_2}

then

Ksp={K_{sp}} =
s×(2s)2s \times {\left( {2s} \right)^2}

=

4s34{s^3}

; ⇒\Rightarrow

4×10−12=4s3;4 \times {10^{ - 12}} = 4{s^3};

⇒\Rightarrow

s=1×10−4s = 1 \times {10^{ - 4}}
[M2+]=s\left[ {{M^{ 2+ }}} \right] = s
=1×10−4= 1 \times 10^ {- 4}
Q8
Species acting as both Bronsted acid and base is :
A (HSO4)-1
B Na2CO3
C NH3
D OH-1
Correct Answer
Option A
Solution
(HSO4)−   {\left( {HS{O_4}} \right)^ - }\,\,\,

can accept and donate a proton

(HSO4)−+H+→H2SO4   {\left( {HS{O_4}} \right)^ - } + {H^ + } \to {H_2}S{O_4}\,\,\,

(acting as base)

(HSO4)−−H+→H2SO42−.   {\left( {HS{O_4}} \right)^ - } - {H^ + } \to {H_2}S{O_4}^{2 - }.\,\,\,

(acting as acid)

Q9
What is the conjugate base of OH-?
A O2
B H2O
C O-
D O-2
Correct Answer
Option D
Solution

Conjugate acid-base pair differ by only one proton.

OH−→H++O2−  O{H^ - } \to {H^ + } + {O^2}^ - \,\,

Conjugate base of

OH−O{H^ - }

is

O2−{O^{2 - }}
Q10
The first and second dissociation constants of an acid H2A are 1.0 ×\times 10−5 and 5.0 ×\times 10−10 respectively. The overall dissociation constant of the acid will be :
A 5.0 ×\times 10−5
B 5.0 ×\times 1015
C 5.0 ×\times 10−15
D 5.0 ×\times 105
Correct Answer
Option C
Solution
H2A ⇌ H+  +  HA−{H_2}A\,\rightleftharpoons\,{H^ + }\,\, + \,\,H{A^ - }

∴\therefore

   K1=1.0×10−5=[H+][HA−][H2A]\,\,\,{K_1} = 1.0 \times {10^{ - 5}} = {{\left[ {{H^ + }} \right]\left[ {H{A^ - }} \right]} \over {\left[ {{H_2}A} \right]}}

(Given)

HA−→H++A−H{A^ - } \to {H^ + } + {A^ - }

∴\therefore

   K2=5.0×10−10\,\,\,{K_2} = 5.0 \times {10^{ - 10}}
=[H+][A−−][HA−]= {{\left[ {{H^ + }} \right]\left[ {{A^{ - - }}} \right]} \over {\left[ {H{A^ - }} \right]}}

(Given)

K=[H+]2[A2−][H2A]K = {{{{\left[ {{H^ + }} \right]}^2}\left[ {{A^{2 - }}} \right]} \over {\left[ {{H_2}A} \right]}}
=K1×K2= {K_1} \times {K_2}
=(1.0×10−5)×(5×10−10)= \left( {1.0 \times {{10}^{ - 5}}} \right) \times \left( {5 \times {{10}^{ - 10}}} \right)
=5×10−15= 5 \times {10^{ - 15}}
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