Practice Start Test

Ionic Equilibrium

NEET Chemistry · 99 questions · Page 2 of 10 · Click an option or "Show Solution" to reveal answer

Q11
Find out the solubility of Ni(OH) 2 in 0.1M NaOH. Given that the ionic product of Ni(OH) 2 is 2 × \times 10 -15 .
A 2 × \times 10 -8 M
B 1 × \times 10 -13 M
C 1 × \times 10 8 M
D 2 × \times 10 -13 M
Correct Answer
Option D
Solution

Total [OH - ] = 2s + 0.1

≃\simeq

0.1 Ionic product = [Ni 2+ ][OH - ] 2 ⇒\Rightarrow 2 ×\times 10 -15 = s(0.1) 2 ⇒\Rightarrow s = 2 ×\times 10 -13

Q12
pH of a saturated solution of Ca(OH) 2 is 9. The solubility product (K sp ) of Ca(OH) 2 is :
A 0.125 × 10 –15
B 0.5 × 10 –10
C 0.5 × 10 –15
D 0.25 × 10 –10
Correct Answer
Option C
Solution

<table class=tg> <thead> <tr> <th class=tg-bzci>Ca(OH)<sub>2</sub></th> <th class=tg-bzci>⇌</th> <th class=tg-bzci>Ca<sup>+2</sup>(aq)</th> <th class=tg-bzci>+</th> <th class=tg-bzci>2OH<sup>-</sup>(aq)</th> </tr> </thead> <tbody> <tr> <td class=tg-bzci></td> <td class=tg-bzci></td> <td class=tg-bzci>S</td> <td class=tg-bzci></td> <td class=tg-bzci>2S</td> </tr> </tbody> </table> pH = 9 ; pOH = 5 ; [OH<sup>–</sup>] = 10<sup>–5</sup> = 2S <br><br>∴\therefore S =

10−52{{{{10}^{ - 5}}} \over 2}

<br><br>K<sub>sp</sub> = [Ca<sup>+2</sup>] [OH<sup>–</sup>]<sup>2</sup> <br><br>⇒\Rightarrow K<sub>sp</sub> = S × (2S)<sup>2</sup> <br><br>⇒\Rightarrow K<sub>sp</sub> = 4S<sup>3</sup> <br><br>⇒\Rightarrow K<sub>sp</sub> =

4×(10−52)34 \times {\left( {{{{{10}^{ - 5}}} \over 2}} \right)^3}

= 0.5 × 10<sup>–15</sup>

Q13
Which will make basic buffer?
A 100 mL of 0.1 M HCl + 200 mL of 0.1 M NH 4 OH
B 100 mL of 0.1 M HCl + 100 mL of 0.1 M NHOH
C 50 mL of 0.1 M NaOH + 25 mL of 0.1 M CH 3 COOH
D 100 mL of 0.1 M CH 3 COOH + 100 mL of 0.1 M NaOH
Correct Answer
Option A
Solution

Basic buffer is mixture of weak base and salt of weak base with strong acid. <br><br>milli mole of HCl = 100 × 0.1 = 10 milli mole <br><br>milli mole of NH<sub>4</sub>OH = 200 × 0.1 = 20 milli mole <br> <table class=tg> <thead> <tr> <th class=tg-bzci>HCl</th> <th class=tg-bzci>+</th> <th class=tg-bzci>NH<sub>4</sub>OH</th> <th class=tg-bzci>→\to</th> <th class=tg-bzci>NH<sub>4</sub>Cl</th> <th class=tg-bzci>+</th> <th class=tg-bzci>H<sub>2</sub>O</th> </tr> </thead> <tbody> <tr> <td class=tg-bzci>10</td> <td class=tg-bzci></td> <td class=tg-bzci>20</td> <td class=tg-bzci></td> <td class=tg-bzci>0</td> <td class=tg-bzci></td> <td class=tg-bzci></td> </tr> <tr> <td class=tg-bzci>0</td> <td class=tg-bzci></td> <td class=tg-bzci>20 - 10 = 10</td> <td class=tg-bzci></td> <td class=tg-bzci>10</td> <td class=tg-bzci></td> <td class=tg-bzci></td> </tr> </tbody> </table> <br><br>Here in the final solution 10 milli mole weak base NH<sub>4</sub>OH and 10 milli mole salt of weak base and strong acid NH<sub>4</sub>Cl present.

So it is a basic buffer.

Q14
Conjugate base for Bronsted acids H 2 O and HF are :
A OH – and F – , respectively
B H 3 O + and H 2 F + , respectively
C OH – and H 2 F + , respectively
D H 3 O + and F – , respectively
Correct Answer
Option A
Solution

Conjugate base of H 2 O is OH – Conjugate base of HF is F –

Q15
The solubility of BaSO 4 in water is 2.42 × 10 –3 g L –1 at 298 K. The value of its solubility product (K sp ) will be (Given molar mass of BaSO 4 = 233 g mol –1 )
A 1.08 × 10 –10 mol 2 L –2
B 1.08 × 10 –12 mol 2 L –2
C 1.08 × 10 –14 mol 2 L –2
D 1.08 × 10 –8 mol 2 L –2
Correct Answer
Option A
Solution

Given, Solubility of BaSO<sub>4</sub> = 2.42 × 10<sup>–3</sup> g L<sup>–1</sup> <br><br>Convert solubility in mol/lit. <br><br>s =

2.42×10−3233=1.04×10−5{{2.42 \times {{10}^{ - 3}}} \over {233}} = 1.04 \times {10^{ - 5}}

mol L<sup>-1</sup> <br><br> <table class=tg> <thead> <tr> <th class=tg-p8sp>BaSO4(s)</th> <th class=tg-p8sp>⇌</th> <th class=tg-p8sp>Ba<sup>2+</sup>(aq)</th> <th class=tg-p8sp>+</th> <th class=tg-p8sp>SO<sub>4</sub><sup>2-</sup>(aq)</th> </tr> </thead> <tbody> <tr> <td class=tg-p8sp></td> <td class=tg-p8sp></td> <td class=tg-p8sp>s</td> <td class=tg-p8sp></td> <td class=tg-p8sp>s</td> </tr> </tbody> </table> <br>K<sub>sp</sub> = [Ba<sup>2+</sup>][SO<sub>4</sub><sup>2-</sup>] = s<sup>2</sup> <br><br>=

(1.04×10−5)2{\left( {1.04 \times {{10}^{ - 5}}} \right)^2}

<br><br>= 1.08 ×\times 10<sup>-10</sup> mol<sup>2</sup> L<sup>–2</sup>

Q16
Following solutions were prepared by mixing different volumes of NaOH and HCl of different concentrations : A. 60 mL M10{M \over {10}} HCl + 40 mL M10{M \over {10}} NaOH B. 55 mL M10{M \over {10}} HCl + 45 mL M10{M \over {10}} NaOH C. 75 mL M5{M \over {5}} HCl + 25 mL M5{M \over {5}} NaOH D. 100 mL M10{M \over {10}} HCl + 100 mL M10{M \over {10}} NaOH pH of which one of them will be equal to 1?
A B
B A
C D
D C
Correct Answer
Option D
Solution

(A) 60 mL

M10{M \over {10}}

HCl + 40 mL

M10{M \over {10}}

NaOH Mili. moles of HCl = 60 ×\times

110{1 \over {10}}

= 6 Mili. moles of NaOH = 40 ×\times

110{1 \over {10}}

= 4 Mili. moles of HCl remaining = 6 - 4 = 2 Total volume will be 60 + 40 = 100 mL Concentration of [H + ] =

2100{{2} \over {100}}

= 2 ×\times 10 -2 ∴\therefore pH = 2 - log2 = 1.7 (B) 55 mL

M10{M \over {10}}

HCl + 45 mL

M10{M \over {10}}

NaOH Mili. moles of HCl = 55 ×\times

110{1 \over {10}}

= 5.5 Mili. moles of NaOH = 45 ×\times

110{1 \over {10}}

= 4.5 Mili. moles of HCl remaining = 5.5 - 4.5 = 1 Concentration of [H + ] =

1100{{1} \over {100}}

= 10 -2 ∴\therefore pH = 2 (C) 75 mL

M5{M \over {5}}

HCl + 25 mL

M5{M \over {5}}

NaOH Mili. moles of HCl = 75 ×\times

15{1 \over {5}}

= 15 Mili. moles of NaOH = 25 ×\times

15{1 \over {5}}

= 5 Mili. moles of HCl remaining = 15 - 5 = 10 Total volume will be 75 + 25 = 100 mL Concentration of [H + ] =

10100{{10} \over {100}}

= 10 -1 ∴\therefore pH = 1 (D) 100 mL

M10{M \over {10}}

HCl + 100 mL

M10{M \over {10}}

NaOH Mili. moles of HCl = 100 ×\times

110{1 \over {10}}

= 10 Mili. moles of NaOH = 100 ×\times

110{1 \over {10}}

= 10 Mili. moles of HCl remaining = 10 - 10 = 0 So, it is neutral solution. ∴\therefore pH = 7

Q17
Concentration of the Ag + ions in a saturated solution of Ag 2 C 2 O 4 is 2.2 × \times 10 −-4 mol L −-1 . Solubility product of Ag 2 C 2 O 4 is
A 2.66 × \times 10 −-12
B 4.5 × \times 10 −-11
C 5.3 × \times 10 −-12
D 2.42 × \times 10 −-8
Correct Answer
Option C
Solution

Ag 2 C 2 O 4(s) →\to 2Ag+ + C 2 O 4 2– S 2S S K sp = [Ag + ] 2 [C 2 O 4 2– ] = [2S] 2 [S] [Ag+] = 2S = 2.2 × 10 –4 or S = 1.1 × 10 –4 M ∴\therefore K sp = [2.2 × 10 –4 ] 2 [1.1 × 10 –4 ] = 5.3 × 10 –12

Q18
Which of the following fluro-compounds is most likely to behave as a Lewis base ?
A BF 3
B PF 3
C CF 4
D SiF 4
Correct Answer
Option B
Solution

PF 3 is lewis base because on P atom there is lone pair.

Q19
The solubility of AgCl (s) with solubility product 1.6 × \times 10 −-10 in 0.1 M NaCl solution would be
A 1.26 × \times 10 −-5 M
B 1.6 × \times 10 −-9 M
C 1.6 × \times 10 −-11 M
D zero
Correct Answer
Option B
Solution

AgCl ⇌ Ag + + Cl – s s s + 0.1 Concentration of Cl – is (s + 0.1) mol L –1 because s mol L –1 from ionization of AgCl and 0.1 mol L –1 from ionization of 0.1 M NaCl.

Now, K sp = [Ag + ][Cl – ] ⇒\Rightarrow 1.6 × 10 –10 = s (s + 0.1) ⇒\Rightarrow 1.6 × 10 –10 = s (0.1) {∵\because s << 0.1} ⇒\Rightarrow s = 1.6 × 10 –9 M

Q20
The percentage of pyridine (C 5 H 5 N) that forms pyridinium ion (C 5 H 5 N + H) ina 0.10 M aqueous pyridine solution (K b for C 5 H 5 N = 1.7 × \times 10 −-9 ) is
A 0.0060%
B 0.013%
C 0.77%
D 1.6%
Correct Answer
Option B
Solution

C 5 H 5 N + H 2 O → C 5 H 5 N + H + OH – So, the amount of pyridine that forms pyridinium ion is α\alpha.

Now, α\alpha =

Kbconc. of pyridine\sqrt {{{{K_b}} \over {conc.\,of\,pyridine}}}

=

1.7×1090.10\sqrt {{{1.7 \times {{10}^9}} \over {0.10}}}

= 1.30 ×\times 10 -4 So, percentage of pyridine that forms pyridinium ion = 1.30 × 10 –4 × 100 = 1.30 × 10 –2 = 0.013%

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