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Ionic Equilibrium

NEET Chemistry · 99 questions · Page 3 of 10 · Click an option or "Show Solution" to reveal answer

Q21
MY and NY 3 , two nearly insoluble salts, have the same K sp values of 6.2 × \times 10 −-13 at room temperature. Which statement would be true in regard to MY and NY 3 ?
A The salts MY and NY 3 are more soluble in 0.5 M KY than in pure water.
B The addition of the salt of KY to solution of MY and NY 3 will have no effect on their solubilities.
C The molar solubilities of MY and NY 3 in water are identical.
D The molar solubility of MY in water is less than that of NY 3 .
Correct Answer
Option D
Solution

MY ⇌ M + + Y – s s s K sp = s.s = s 2 ⇒\Rightarrow s =

6.2×10−13\sqrt {6.2 \times {{10}^{ - 13}}}

= 7.87 × 10 –7 mol L –1 NY 3 ⇌ N + + 3Y – s s 3s K sp = s.(3s) 3 = 27s 4 6.2 × 10 –13 = 27s 4 ⇒\Rightarrow s = 3.89 × 10 –4 mol L –1 Hence, molar solubility of MY in water is less than that of NY 3 .

Q22
What is the pH of the resulting solution when equal volumes of 0.1 M NaOH and 0.01 M HCl are mixed?
A 2.0
B 7.0
C 1.04
D 12.65
Correct Answer
Option D
Solution

HCl + NaOH →\to NaCl + H 2 O Initial 0.01 0.1 0 0 Final 0 0.09 0.01 0.01 As equal volumes of HCl and NaOH are added so the volume of resulting solution becomes double and the concentration of the solution becomes half.

∴\therefore [OH - ] =

0.092{{0.09} \over 2}

= 0.045 M ∴\therefore pOH = – log[OH – ] = –log [0.045] = 1.35 ∴\therefore pH = 14 – pOH = 14 – 1.35 = 12.65

Q23
Which one of the following pairs of solution is not an acidic buffer ?
A CH 3 COOH and CH 3 COONa
B H 2 CO 3 and Na 2 CO 3
C H 3 PO 4 and Na 3 PO 4
D HClO 4 and NaClO 4
Correct Answer
Option D
Solution

An acidic buffer is a mixture of a weak acid and its salt with a strong base.

Among CH 3 COOH, H 2 CO 3 , H 3 PO 4 and HClO 4 , the HClO 4 is a strong acid while all other are weak acid thus, HClO 4 and NaClO 4 does not constitute to form an acidic buffer.

Q24
The K sp of Ag 2 CrO 4 , AgCl, AgBr and Agl are respectively, 1.1 × \times 10 −-12 , 1.8 × \times 10 −-10 , 5.0 × \times 10 −-13 , 8.3 × \times 10 −-17 . Which one of the following salts will precipitate last if AgNO 3 solution is added to the solution containing equal moles of NaCl, NaBr, Nal and Na 2 CrO 4 ?
A AgBr
B Ag 2 CrO 4
C Agl
D AgCl
Correct Answer
Option B
Solution

From the Ksp values of the given salts calculate the solubility values.

Salt having highest solubility will precipitate at last. <table class=tg> <tbody><tr> <th class=tg-nrix>AgCrO<sub>4</sub></th> <th class=tg-nrix>⇌</th> <th class=tg-baqh>2Ag<sup>+</sup></th> <th class=tg-baqh>+</th> <th class=tg-baqh>CrO<sub>4</sub><sup>2-</sup></th> </tr> <tr> <td class=tg-nrix>s</td> <td class=tg-nrix></td> <td class=tg-baqh>2s</td> <td class=tg-baqh></td> <td class=tg-baqh>s</td> </tr> </tbody></table> <br><br>K<sub>sp</sub> = (2s)<sup>2</sup>(s) = 1.1 × 10<sup>–12</sup> <br><br>⇒\Rightarrow s = 0.65 × 10<sup>–4</sup> <br><br> <table class=tg> <tbody><tr> <th class=tg-nrix>AgCl</th> <th class=tg-nrix>⇌</th> <th class=tg-baqh>Ag<sup>+</sup></th> <th class=tg-baqh>+</th> <th class=tg-baqh>Cl<sup>-</sup></th> </tr> <tr> <td class=tg-nrix>s</td> <td class=tg-nrix></td> <td class=tg-baqh>s</td> <td class=tg-baqh></td> <td class=tg-baqh>s</td> </tr> </tbody></table> <br><br>K<sub>sp</sub> = s × s <br><br>⇒\Rightarrow 1.8 × 10<sup>–10</sup> = s <sup>2</sup> <br><br>⇒\Rightarrow s = 1.34 × 10<sup>–5</sup> <br><br> <table class=tg> <tbody><tr> <th class=tg-nrix>AgBr</th> <th class=tg-nrix>⇌</th> <th class=tg-baqh>Ag<sup>+</sup></th> <th class=tg-baqh>+</th> <th class=tg-baqh>Br<sup>-</sup></th> </tr> <tr> <td class=tg-nrix>s</td> <td class=tg-nrix></td> <td class=tg-baqh>s</td> <td class=tg-baqh></td> <td class=tg-baqh>s</td> </tr> </tbody></table> <br><br>K<sub>sp</sub> = s × s <br><br>⇒\Rightarrow 5 × 10<sup>–13</sup> = s<sup>2</sup> <br><br>⇒\Rightarrow s = 0.71 × 10<sup>–6</sup> <br><br> <table class=tg> <tbody><tr> <th class=tg-nrix>AgI</th> <th class=tg-nrix>⇌</th> <th class=tg-baqh>Ag<sup>+</sup></th> <th class=tg-baqh>+</th> <th class=tg-baqh>I<sup>-</sup></th> </tr> <tr> <td class=tg-nrix>s</td> <td class=tg-nrix></td> <td class=tg-baqh>s</td> <td class=tg-baqh></td> <td class=tg-baqh>s</td> </tr> </tbody></table> <br><br>K<sub>sp</sub> = s × s <br><br>⇒\Rightarrow 8.3 × 10<sup>–17</sup> = s<sup>2</sup> <br><br>⇒\Rightarrow s = 0.9 × 10<sup>–8</sup> <br><br>∴\therefore Solubility of Ag<sub>2</sub>CrO<sub>4</sub> is maximum so, it will precipitate at last.

Q25
Which of the following salts will give highest pH in water?
A KCl
B NaCl
C Na 2 CO 3
D CuSO 4
Correct Answer
Option C
Solution

Na 2 CO 3 is salt of strong base, NaOH and weak acid, H 2 CO 3 hence the pH value of the solution will be high.

Q26
The dissociation constant of weak acid is 1 × \times 10 −-4 . In order to prepare a buffer solution with a pH = 5, the [Salt]/[Acid] ratio should be
A 4 : 5
B 10 : 1
C 5 : 4
D 1 : 10
Correct Answer
Option B
Solution

Given, K a = 1 × 10 –4 pK a = – log (1× 10 –4 ) = 4 pH = pK a + log

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

⇒\Rightarrow 5 = 4 + log

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

⇒\Rightarrow log

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

= 1 ⇒\Rightarrow

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

= 10 = 10 : 1

Q27
At 100 o C the K w of water is 55 times its value at 25 o C. What will be the pH of neutral solution? (log 55 = 1.74)
A 7.00
B 7.87
C 5.13
D 6.13
Correct Answer
Option D
Solution

K w at 25 o C = 1 × 10 –14 At 25ºC K w = [H + ] [OH – ] = 10 –14 At 100°C (given) K w = [H + ] [OH – ] = 55 × 10 –14 for a neutral solution [H + ] = [OH – ] [H + ] 2 = 55 × 10 –14 or [H + ] = (55 × 10–14) 1/2 pH = – log [H + ] On taking log on both side – log [H + ] = –log (55 × 10 –14 ) 1/2 ⇒\Rightarrow pH =

12{1 \over 2}

- log55 - 14log10 ⇒\Rightarrow pH = 6.13

Q28
The values of K sp of CaCO 3 and CaC 2 O 4 are 4.7 × \times 10 −-9 and 1.3 × \times −-9 respectively at 25 o C. If the mixture of these two is washed with water, what is the concentration of Ca 2+ ions in water ?
A 5.831 × \times 10 −-5 M
B 6.856 × \times 10 −-5 M
C 3.606 × \times 10 −-5 M
D 7.746 × \times 10 −-5 M
Correct Answer
Option D
Solution

<table class=tg> <tbody><tr> <th class=tg-9wq8>CaCO<sub>3</sub></th> <th class=tg-9wq8>→\to</th> <th class=tg-c3ow>Ca<sup>2+</sup></th> <th class=tg-c3ow>+</th> <th class=tg-c3ow>CO<sub>3</sub><sup>2-</sup></th> </tr> <tr> <td class=tg-9wq8></td> <td class=tg-9wq8></td> <td class=tg-c3ow>x</td> <td class=tg-c3ow></td> <td class=tg-c3ow>x</td> </tr> </tbody></table> <br> <table class=tg> <tbody><tr> <th class=tg-9wq8>CaC<sub>2</sub>O<sub>4</sub></th> <th class=tg-9wq8>→\to</th> <th class=tg-c3ow>Ca<sup>2+</sup></th> <th class=tg-c3ow>+</th> <th class=tg-c3ow>C<sub>2</sub>O<sub>4</sub><sup>2-</sup></th> </tr> <tr> <td class=tg-9wq8></td> <td class=tg-9wq8></td> <td class=tg-c3ow>y</td> <td class=tg-c3ow></td> <td class=tg-c3ow>y</td> </tr> </tbody></table> <br><br>[Ca<sup>2+</sup>] = x + y <br><br>Now, K<sub>sp</sub> (CaCO<sub>3</sub>) = [Ca<sup>2+</sup>] [CO<sub>3</sub> <sup>2-</sup>] <br><br>or 4.7 × 10<sup>–9</sup> = (x + y)x .......(1) <br><br>similarly, K<sub>sp</sub> (CaC<sub>2</sub>O<sub>4</sub>) = [Ca<sup>2+</sup>] [C<sub>2</sub>O<sub>4</sub><sup>2–</sup>] <br><br>or 1.3 × 10<sup>–9</sup> = (x + y)y .......(2) <br><br>Dividing equation (1) and (2), we get <br><br>

xy=3.6{x \over y} = 3.6

<br><br>∴\therefore x = 3.6y <br><br>Putting in equation (2) we get <br><br>y(3.6y + y) = 1.3 × 10<sup>–9</sup> <br><br>⇒\Rightarrow y = 1.68 ×\times 10<sup>-5</sup> <br><br>and x = 3.6 ×\times 1.68 ×\times 10<sup>-5</sup> = 6.048 ×\times 10<sup>-5</sup> <br><br>∴\therefore [Ca<sup>2+</sup>] = (x + y) = (6.048 ×\times 10<sup>-5</sup>) + (1.68 ×\times 10<sup>-5</sup>) <br><br>∴\therefore [Ca<sup>2+</sup>] = 7.746 × 10<sup>–5</sup> M

Q29
Accumulation of lactic acid (HC 3 H 5 O 3 ), a monobasic acid in tissues leads to pain and a feeling of fatigue. In a 0.10 M aqueous solution, lactic acid is 3.7% dissociates. The value of dissociation constant, K a , for this acid will be
A 1.4 × \times 10 −-5
B 1.4 × \times 10 −-4
C 3.7 × \times 10 −-4
D 2.8 × \times 10 −-4
Correct Answer
Option B
Solution
α=Kac\alpha = \sqrt {{{{K_a}} \over c}}

⇒\Rightarrow 0.037 =

Ka0.10\sqrt {{{{K_a}} \over {0.10}}}

⇒\Rightarrow K a = (0.037) 2 ×\times 0.10 = 1.37 ×10 –4

≃\simeq

1.4 ×\times 10 −-4

Q30
KMnO 4 can be prepared from K 2 MnO 4 as per the reaction, 3MnO 4 2−- + 2H 2 O ⇌\rightleftharpoons 2MnO 4 −- + MnO 2 + 4OH −- The reaction can go to completion by removing OH −- ions by adding
A CO 2
B SO 2
C HCl
D KOH
Correct Answer
Option A
Solution

KOH is the base thus, it gives OH – ions thus it cannot remove OH – ions from reaction mixture but it adds on the concentration of OH – ions.

So, an acid must be added but if a strong acid is added to the reaction mixture then in acidic condition the MnO 4 – formed reduces to give Mn 2+ thus, HCl which is a strong acid and SO 2 which on treating with water forms a strong H 2 SO 4 cannot be used for this purpose.

Thus, CO 2 which forms H 2 CO 3 a weak acid reacts to remove OH – but not that much acidic that MnO 4 – undergo reduction.

Thus, CO 2 is used for this reaction for completion.

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