Practice Start Test

Ionic Equilibrium

NEET Chemistry · 99 questions · Page 6 of 10 · Click an option or "Show Solution" to reveal answer

Q51
The rapid change of pH near the stoichiometric point of an acid-base titration is the basis of indicator detection. pH of the solution is related to ratio of the concentrations of the conjugate acid (HIn) and base (In −- ) forms of the indicator by the expression
A log⁡[In−][HIn]=pKIn−pH\log {{\left[ {I{n^ - }} \right]} \over {\left[ {HIn} \right]}} = p{K_{In}} - pH
B log⁡[HIn][In−]=pKIn−pH\log {{\left[ {HIn} \right]} \over {\left[ {I{n^ - }} \right]}} = p{K_{In}} - pH
C log⁡[HIn][In−]=pH−pKIn\log {{\left[ {HIn} \right]} \over {\left[ {I{n^ - }} \right]}} = pH - p{K_{In}}
D log⁡[In−][HIn]=pH−pKIn\log {{\left[ {I{n^ - }} \right]} \over {\left[ {HIn} \right]}} = pH - p{K_{In}}
Correct Answer
Option D
Solution

For an acid-base indicator H In ⇌ H + + In - K in =

[H+][In−][Hin]{{\left[ {{H^ + }} \right]\left[ {I{n^ - }} \right]} \over {\left[ {{H_{in}}} \right]}}

⇒\Rightarrow

[H+]=Kin[In−][Hin]\left[ {{H^ + }} \right] = {{{K_{in}}\left[ {I{n^ - }} \right]} \over {\left[ {{H_{in}}} \right]}}

Take – log on both sides

−log⁡[H+]=−log⁡(Kin[In−][Hin])- \log \left[ {{H^ + }} \right] = - \log \left( {{{{K_{in}}\left[ {I{n^ - }} \right]} \over {\left[ {{H_{in}}} \right]}}} \right)

⇒\Rightarrow pH = –log K In +

log⁡[In−][Hin]\log {{\left[ {I{n^ - }} \right]} \over {\left[ {{H_{in}}} \right]}}

⇒\Rightarrow pH = pK In +

log⁡[In−][Hin]\log {{\left[ {I{n^ - }} \right]} \over {\left[ {{H_{in}}} \right]}}

⇒\Rightarrow

log⁡[In−][Hin]\log {{\left[ {I{n^ - }} \right]} \over {\left[ {{H_{in}}} \right]}}

= pH - pK In

Q52
The solubility product of a sparingly soluble salt AX 2 is 3.2 × \times 10 −-11 . Its solubility (in moles/L) is
A 5.6 × \times 10 −-6
B 3.1 × \times 10 −-4
C 2 × \times 10 −-4
D 4 × \times 10 −-4
Correct Answer
Option C
Solution

<table class=tg> <tbody><tr> <th class=tg-9wq8>AX<sub>2</sub></th> <th class=tg-9wq8>⇌</th> <th class=tg-c3ow>A<sup>2+</sup></th> <th class=tg-c3ow>+</th> <th class=tg-c3ow>2X<sup>-</sup></th> </tr> <tr> <td class=tg-9wq8>s</td> <td class=tg-9wq8></td> <td class=tg-c3ow>s</td> <td class=tg-c3ow></td> <td class=tg-c3ow>2s</td> </tr> </tbody></table> <br><br>K<sub>sp</sub> = = [A<sup>2+</sup>] [X<sup>–</sup> ]<sup>2</sup> = s × (2s)<sup>2</sup> = 4s<sup>3</sup> <br><br>⇒\Rightarrow 3.2 ×\times 10<sup>−-11</sup> = 4s<sup>3</sup> <br><br>⇒\Rightarrow s<sup>3</sup> = 8 × 10<sup>–12</sup> <br><br>⇒\Rightarrow s = 2 × 10<sup>–4</sup> mol L<sup>–1</sup>

Q53
The solubility product of AgI at 25 o C is 1.0 × \times 10 −-16 mol 2 L −-2 . The solubility of AgI in 10 −-4 N solution of KI at 25 o C is approximately (in mol L −-1
A 1.0 × \times 10 −-16
B 1.0 × \times 10 −-12
C 1.0 × \times 10 −-10
D 1.0 × \times 10 −-8
Correct Answer
Option B
Solution

<table class=tg> <tbody><tr> <th class=tg-9wq8>AgI</th> <th class=tg-9wq8>⇌</th> <th class=tg-c3ow>Ag<sup>+</sup></th> <th class=tg-c3ow>+</th> <th class=tg-c3ow>I<sup>-</sup></th> </tr> <tr> <td class=tg-9wq8>s</td> <td class=tg-9wq8></td> <td class=tg-c3ow>s</td> <td class=tg-c3ow></td> <td class=tg-c3ow>s</td> </tr> </tbody></table> <br><br>K<sub>sp</sub> = s<sup>2</sup> <br><br>⇒\Rightarrow 1.0 × 10<sup>–16</sup> = s<sup>2</sup> <br><br>⇒\Rightarrow s = 1.0 × 10<sup>–8</sup> mol L<sup>–1</sup> <br><br>∴\therefore [Ag<sup>+</sup>] = 1.0 × 10<sup>–8</sup> mol L<sup>–1</sup> <br><br>Also, in 10<sup>–4</sup> N KI solution,<br><br> [I<sup>–1</sup>] = (10<sup>–4</sup> + 1.0 × 10<sup>–8</sup>) mol L<sup>–1</sup> <br><br>⇒\Rightarrow [I<sup>–1</sup>] = (10<sup>–4</sup>) mol L<sup>–1</sup> <br><br>[As 1.0 × 10<sup>–8</sup> mol L<sup>–1</sup> << 1.0 × 10<sup>–4</sup> mol L<sup>–1</sup>] <br><br>∴\therefore K<sub>sp</sub> of AgI = [Ag<sup>+</sup>][I<sup>–</sup>] <br><br>= (1.0 × 10<sup>–8</sup>)(10<sup>–4</sup>) <br><br>= 1.0 × 10<sup>–12</sup> mol L<sup>–1</sup>

Q54
Solution of 0.1 N NH 4 OH and 0.1 N NH 4 Cl has pH 9.25. Then find out pK b of NH 4 OH
A 9.25
B 4.75
C 3.75
D 8.25
Correct Answer
Option B
Solution

NH 4 OH and NH 4 Cl constitute to form a basic buffer. pOH = pK b + log

[Salt][Base]{{\left[ {Salt} \right]} \over {\left[ {Base} \right]}}

We know, pOH+ pH = 14 or pOH = 14 – pH ∴\therefore 14 - pH -

log⁡[Salt][Base]\log {{\left[ {Salt} \right]} \over {\left[ {Base} \right]}}

= pK b ⇒\Rightarrow 14 - 9.25 -

log⁡0.10.1\log {{0.1} \over {0.1}}

= pK b ⇒\Rightarrow 14 – 9.25 – 0 = pK b ⇒\Rightarrow pK b = 4.75

Q55
Solubility of MX 2 type electrolytes is 0.5 × \times 10 −-4 mole/lit., then find out K sp of electrolytes.
A 5 × \times 10 −-12
B 25 × \times 10 −-10
C 1 × \times 10 −-13
D 5 × \times 10 −-13
Correct Answer
Option D
Solution

<table class=tg> <tbody><tr> <th class=tg-9wq8>MX<sub>2</sub></th> <th class=tg-9wq8>⇌</th> <th class=tg-c3ow>Ag<sup>2+</sup></th> <th class=tg-c3ow>+</th> <th class=tg-c3ow>2X<sup>-</sup></th> </tr> <tr> <td class=tg-9wq8>s</td> <td class=tg-9wq8></td> <td class=tg-c3ow>s</td> <td class=tg-c3ow></td> <td class=tg-c3ow>2s</td> </tr> </tbody></table> <br><br>K<sub>sp</sub> = [M<sup>2+</sup>] [X<sup>–</sup>]<sup>2</sup> = (S)(2S)<sup>2</sup> = 4S<sup>3</sup> <br><br>⇒\Rightarrow K<sub>sp</sub> = 4(0.5 × 10<sup>–4</sup>)<sup>3</sup> = 5 × 10<sup>–13</sup>

Q56
Which has highest pH?
A CH 3 COOK
B Na 2 CO 3
C NH 4 Cl
D NaNO 3
Correct Answer
Option B
Solution

Na 2 CO 3 is a salt of weak acid H 2 CO 3 and strong base NaOH, therefore, its aqueous solution will be basic hence has pH more than 7.

Q57
Solubility of M 2 S salt is 3.5 × \times 10 −-6 then find out solubility product.
A 1.7 × \times 10 −-6
B 1.7 × \times 10 −-16
C 1.7 × \times 10 −-18
D 1.7 × \times 10 −-12
Correct Answer
Option B
Solution

M 2 S ⇌ 2M + + S 2– K sp = [M + ] 2 [S 2– ] = (2s) 2 (s) = 4s 3 K sp = 4(3.5 × 10 –6 ) 3 = 1.7 × 10 –16

Q58
In HS −- , I −- , R −- NH 2 , NH 3 order of proton accepting tendency will be
A I −- > NH 3 > R −- NH 2 > HS −-
B NH 3 > R −- NH 2 > HS −- > I −-
C R −- NH 2 > NH 3 > HS −- > I −-
D HS −- > R −- NH 2 > NH 3 > I −-
Correct Answer
Option C
Solution

Strong base has higher tendency to accept the proton.

Increasing order of base and hence the order of accepting tendency of proton is R −- NH 2 > NH 3 > HS −- > I −-

Q59
Ionisation constant of CH 3 COOH is 1.7 × \times 10 −-5 and concentration of H + ions is 3.4 × \times 10 −-4 . Then find out initial concentration of CH 3 COOH molecules.
A 3.4 × \times 10 −-4
B 3.4 × \times 10 −-3
C 6.8 × \times 10 −-4
D 6.8 × \times 10 −-3
Correct Answer
Option D
Solution

CH 3 COOH ⇌ CH 3 COO – + H + K c =

[CH3COO−][H+][CH3COOH]{{{\left[ {C{H_3}CO{O^ - }} \right]\left[ {{H^ + }} \right]} \over {\left[ {C{H_3}COOH} \right]}}}

⇒\Rightarrow [CH 3 COOH] =

3.4×10−4×3.4×10−41.7×10−5{{{3.4 \times {{10}^{ - 4}} \times 3.4 \times {{10}^{ - 4}}} \over {1.7 \times {{10}^{ - 5}}}}}

= 6.8 × 10 –3

Q60
Correct relation between dissociation constants of a dibasic acid is
A K a 1 = K a 2
B K a 1 > K a 2
C K a 1 < K a 2
D K a 1 = 1Ka2{1 \over {{K_{{a_2}}}}}
Correct Answer
Option B
Solution

In polyprotic acids the loss of second proton occurs much less readily than the first.

Usually the K a values for successive loss of protons from these acids differ by at least a factor of 10 –3 .

∴\therefore K a 1 > K a 2

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