Practice Start Test

Ionic Equilibrium

NEET Chemistry · 99 questions · Page 5 of 10 · Click an option or "Show Solution" to reveal answer

Q41
Which of the following molecules acts as a Lewis acid?
A (CH 3 ) 2 O
B (CH 3 ) 3 P
C (CH 3 ) 3 N
D (CH 3 ) 3 B
Correct Answer
Option D
Solution

CH 3 ) 3 B – is an electron deficient compound due to inclete octate of B, thus behave as a lewis acid.

Q42
The ionization constant of ammonium hydroxide is 1.77 × \times 10 −-5 at 298 K. Hydrolysis constant of ammonium chloride is
A 6.50 × \times 10 −-12
B 5.65 × \times 10 −-13
C 5.65 × \times 10 −-12
D 5.65 × \times 10 −-10
Correct Answer
Option D
Solution

Ammonium chloride is a salt of weak base and strong acid.

In this case hydrolysis constant K h can be calculated as K h =

KwKb{{{K_w}} \over {{K_b}}}

=

10−141.77×10−5{{{{10}^{ - 14}}} \over {1.77 \times {{10}^{ - 5}}}}

= 5.65 ×\times 10 −-10

Q43
What is the [OH −- ] in the final solution prepared by mixing 20.0 mL of 0.050 M HCl with 30.0 mL of 0.10 M Ba(OH) 2 ?
A 0.40 M
B 0.0050 M
C 0.12 M
D 0.10 M
Correct Answer
Option D
Solution

Number of equivalents of H + = 20.0 × 0.050 milliequivalents = 1.0 milliequivalents Number of equivalents of OH – = 2 × 30.0 × 0.10 = 6.0 milliequivalents ∴\therefore Equivalents of OH - left after neutralization = 6 – 1 = 5 milliequivalents Total volume after neutralization = 20.0 + 30.0 mL = 50 mL [OH - ] =

550{5 \over {50}}

= 0.1 M

Q44
Equal volumes of three acid solutions of pH 3, 4 and 5 are mixed in a vessel. What will be the H + ion concentration in the mixture?
A 3.7 × \times 10 −-3 M
B 1.11 × \times 10 −-3 M
C 1.11 × \times 10 −-4 M
D 3.7 × \times 10 −-4 M
Correct Answer
Option D
Solution

We know, pH = – log[H + ] For pH = 3, 3 = – log [H + ] ⇒\Rightarrow [H + ] = 10 –3 M For pH = 4 , 4 = – log [H + ] ⇒\Rightarrow [H + ] = 10 –4 M For pH = 5, 5 = – log [H + ] ⇒\Rightarrow [H + ] = 10 –5 M Total concentration of [H + ] M =

10−3(1+0.1+0.01)3{{{{10}^{ - 3}}\left( {1 + 0.1 + 0.01} \right)} \over 3}

M(V 1 +V 2 + V 3 ) = M 1 V 1 + M 2 V 2 + M 3 V 3 As V 1 = V 2 = V 3 = V ⇒\RightarrowM(3V) = (M 1 + M 2 + M 3 )V ⇒\Rightarrow 3M = (10 –3 + 10 –4 + 10 –5 ) =

1.11×10−33{{1.11 \times {{10}^{ - 3}}} \over 3}

= 3.7 ×\times 10 −-4 M

Q45
A weak acid, HA, has a K a of 1.00 × \times 10 −-5 . If 0.100 mol of this acid is dissolved in one litre of water, the percentage of acid dissociated at equilibrium is closest to
A 1.00%
B 99.9%
C 0.100%
D 99.0%
Correct Answer
Option A
Solution

For weak acid degree of dissociation, α\alpha =

KaC\sqrt {{{{K_a}} \over C}}

=

1×10−50.1=10−2\sqrt {{{1 \times {{10}^{ - 5}}} \over {0.1}}} = {10^{ - 2}}

= 1.00 %

Q46
Calculate the pOH of a solution at 25 o C that contains 1 × \times 10 −-10 M of hydronium ions, i.e. H 3 O + .
A 4.000
B 9.000
C 1.000
D 7.000
Correct Answer
Option A
Solution

Given, [H 3 O + .] = 1 ×\times 10 −-10 ⇒\Rightarrow pH = 10 Also we know, pH + pOH = 14 ⇒\Rightarrow pOH = 14 - pH = 14 - 10 = 4

Q47
The hydrogen ion concentration of a 10 −-8 M, HCl aqueous solution at 298 K (K w = 10 −-14 ) is
A 1.0 × \times 10 −-8 M
B 1.0 × \times 10 −-6 M
C 1.0525 × \times 10 −-7 M
D 9.525 × \times 10 −-8 M
Correct Answer
Option C
Solution

In HCl solution [H + ] = 10 -8 M Also from water, [H + ] = 10 -7 M ∴\therefore Total [H + ] = 10 -8 M + 10 -7 M = 1.1 ×10 –7 M

≃\simeq

1.0525 ×\times 10 −-7 M

Q48
Which of the following pairs constitutes a buffer?
A HCl and KCl
B HNO 2 and NaNO 2
C NaOH and NaCl
D HNO 3 and NH 4 NO 3
Correct Answer
Option B
Solution

HNO 2 is a weak acid and NaNO 2 is its salt thus these two constitute to form an acidic buffer.

Q49
H 2 S gas when passed through a solution of cations containing HCl precipitates the cations of second group of qualitative analysis but not those belonging to the fourth group. It is because
A presence of HCl decreases the sulphide ion concentration
B solubility product of group II sulphides is more than that of group IV sulphates
C presence of HCl increases the sulphide ion concentration
D sulphides of group IV cations are unstable in HCl.
Correct Answer
Option A
Solution

H 2 S ⇌ H + + HS – HCl ⇌ H + + Cl – In presence of HCl this ionization of H 2 S is suppressed due to the presence of extra H + ions from HCl and produces less amount of sulphide ions due to common ion effect, thus HCl decreases the solubility of H 2 S which is sufficient to precipitate II nd group radicals.

Q50
At 25 o C, the dissociation constant of a base, BOH, is 1.0 × \times 10 −-12 . The concentration of hydroxyl ions in 0.01 M aqueous solution of the base would be
A 1.0 × \times 10 −-5 mol L −-1
B 1.0 × \times 10 −-6 mol L −-1
C 2.0 × \times 10 −-6 mol L −-1
D 1.0 × \times 10 −-7 mol L −-1
Correct Answer
Option D
Solution

<table class=tg> <tbody><tr> <th class=tg-baqh></th> <th class=tg-9wq8>BOH</th> <th class=tg-9wq8>⇌</th> <th class=tg-c3ow>B<sup>+</sup></th> <th class=tg-c3ow>+</th> <th class=tg-c3ow>OH<sup>-</sup></th> </tr> <tr> <td class=tg-baqh>Initially</td> <td class=tg-9wq8>C</td> <td class=tg-9wq8></td> <td class=tg-c3ow>0</td> <td class=tg-c3ow></td> <td class=tg-c3ow>0</td> </tr> <tr> <td class=tg-baqh>At equilibbrium</td> <td class=tg-baqh>C - Cα\alpha</td> <td class=tg-baqh></td> <td class=tg-baqh>Cα\alpha</td> <td class=tg-baqh></td> <td class=tg-baqh>Cα\alpha</td> </tr> </tbody></table> <br><br>[OH<sup>-</sup>] = Cα\alpha <br><br>⇒\Rightarrow [OH<sup>-</sup>] =

KbC\sqrt {{K_b}C}

<br><br>⇒\Rightarrow [OH<sup>-</sup>] =

1×10−12×10−2\sqrt {1 \times {{10}^{ - 12}} \times {{10}^{ - 2}}}

<br><br>= 1.0 ×\times 10<sup>−-7</sup> mol L<sup>−-1</sup>

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