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Ionic Equilibrium

NEET Chemistry · 99 questions · Page 4 of 10 · Click an option or "Show Solution" to reveal answer

Q31
Which of these is least likely to act as a lewis base ?
A BF 3
B PF 3
C CO
D F −-
Correct Answer
Option A
Solution

BF 3 acts as Lewis acid. It is electron pair acceptor.

Q32
Buffer solutions have constant acidity and alkalinity because
A these give unionised acid or base on reaction with added acid or alkali
B acids and alkalies in these solutions are shielded from attack by other ions
C they have large excess of H + or OH −- ions
D they have fixed value of pH
Correct Answer
Option A
Solution

For this example, CH 3 COOH ⇌ CH 3 COO – + H + ; CH 3 COONa ⇌ CH 3 COO – + Na + when few drops of HCl are added to this buffer, the H + of HCl immediatly combine with CH 3 COO – ions to form undissociated acetic acid molecules.

Thus there will be no H + ions to combine with CH 3 COO – ions to form undissociated acetic acid molecules.

Thus there will be no appreciable change in its pH value.

Like wise if few drops of NaOH are added, the OH – ions will combine with H + ions to form unionised water molecule.

Thus pH of solution will remain constant.

Q33
Equimolar solutions of the following substances were prepared separately. Which one of these will record the highest pH value ?
A BaCl 2
B AlCl 3
C LiCl
D BeCl 2
Correct Answer
Option A
Solution

All of the given salts have same anion i.e., Cl which on hydrolysis gives HCl which is a strong acid.

Now, among the salts which have cation that gives a strongest base on hydrolysis of salt have the highest pH value.

As Ba form Ba(OH) 2 which is a stronger base thus, it results in the highest pH value.

Q34
pH of a saturated solution of Ba(OH) 2 is 12. The value of solubility product (K sp ) of Ba(OH) 2 is
A 3.3 × \times 10 −-7
B 5.0 × \times 10 −-7
C 4.0 × \times 10 −-6
D 5.0× \times 10 −-6
Correct Answer
Option B
Solution

<table class=tg> <tbody><tr> <th class=tg-0lax></th> <th class=tg-9wq8>Ba(OH)<sub>2</sub></th> <th class=tg-9wq8>⇌</th> <th class=tg-c3ow>Ba<sup>2+</sup></th> <th class=tg-c3ow>+</th> <th class=tg-c3ow>2OH</th> </tr> <tr> <td class=tg-0lax>At equilibrium</td> <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>2x</td> </tr> </tbody></table> <br><br>pH = – log[H<sup>+</sup>] <br><br>12 = – log [H<sup>+</sup>] <br><br>⇒\Rightarrow [H<sup>+</sup>] = 10<sup>–12</sup> <br><br>As, [H<sup>+</sup>][OH<sup>–</sup> ] = 10<sup>–14</sup> <br><br>10<sup>–12</sup> [OH<sup>–</sup> ] = 10<sup>–14</sup> <br><br>⇒\Rightarrow [OH<sup>–</sup> ] = 10<sup>–2</sup> <br><br>As [OH<sup>–</sup> ] = 2x = 10<sup>–2</sup> then x = 5.0 × 10<sup>–3</sup> <br><br>Now, K<sub>sp</sub> = [Ba<sup>2+</sup>][OH<sup>–</sup> ]<sup>2</sup> <br><br>K<sub>sp</sub> = (5 × 10<sup>–3</sup>) (10<sup>–2</sup>)<sup>2</sup> = 5.0 × 10<sup>–7</sup>

Q35
In qualitative analysis, the metals of group I can be separated from other ions by precipitating them as chloride salts. A solution initially contains Ag + and pb 2+ at a concentration of 0.10 M. Aqueous HCl is added to this solution until the Cl −- concentration is 0.10 M. What will the concentrations of Ag + and Pb 2+ be at equilibrium ? (K sp for AgCl = 1.8 × \times 10 −-10 , K sp for PbCl 2 = 1.7 × \times 10 −-5 )
A [Ag + ] = 1.8 × \times 10 −-7 M, [Pb 2+ ] = 1.7 × \times 10 −-6 M
B [Ag + ] = 1.8 × \times 10 −-11 M, [Pb 2+ ] = 8.5 × \times 10 −-5 M
C [Ag + ] = 1.8 × \times 10 −-9 M, [Pb 2+ ] = 1.7 × \times 10 −-3 M
D [Ag + ] = 1.8 × \times 10 −-11 M, [Pb 2+ ] = 1.7 × \times 10 −-4 M
Correct Answer
Option C
Solution

K sp [AgCl] = [Ag + ][Cl - ] ⇒\Rightarrow [Ag + ] =

1.8×10−1010−1{{1.8 \times {{10}^{ - 10}}} \over {{{10}^{ - 1}}}}

= 1.8 ×\times 10 −-9 M K sp [PbCl 2 ] = [Pb 2+ ][Cl - ] 2 ⇒\Rightarrow [Pb 2+ ] =

1.7×10−510−1×10−1{{1.7 \times {{10}^{ - 5}}} \over {{{10}^{ - 1}} \times {{10}^{ - 1}}}}

= 1.7 ×\times 10 −-3 M

Q36
Which of the following is least likely to behave as Lewis base?
A H 2 O
B NH 3
C BF 3
D OH −-
Correct Answer
Option C
Solution

The species which have a lone pair of electrons to donate or a negative charge on it can act as a Lewis base.

Here, N atom in NH 3 and O atom in H2O have lone pair of electrons available for donation.

In OH – the negative charge on it results in to behave it as a Lewis base.

But BF 3 is an electron deficient species thus, it is least likely to behave as Lewis base.

Q37
A buffer solution is prepared in which the concentration of NH 3 is 0.30 M and the concentration of NH + 4 is 0.20 M. If the equilibrium constant, K b for NH 3 equals 1.8 × \times 10 −-5 , what is the pH of this solution? (log 2.7 = 0.43)
A 9.08
B 9.43
C 11.72
D 8.73
Correct Answer
Option B
Solution

pOH = pK b + log

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

= – log K b + log

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

= – log 1.8× 10 –5 + log

0.200.30{{0.20} \over {0.30}}

= 5 – 0.25 + (–0.176) = 4.57 Now, pH = 14 – pOH = 14 – 4.57 = 9.43

Q38
What is [H + ] in mol/L of a solution that is 0.20 M in CH 3 COONa and 0.10 M in CH 3 COOH? K a for CH 3 COOH = 1.8 × \times 10 −-5
A 3.5 × \times 10 −-4
B 1.1 × \times 10 −-5
C 1.8 × \times 10 −-5
D 9.0 × \times 10 −-6
Correct Answer
Option D
Solution

CH 3 COOH and CH 3 COONa constitute to form an acidic buffer. ⇒\Rightarrow pH = pKa + log

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

pH = –log(1.8 × 10 –5 ) + log

(0.20)(0.10){{\left( {0.20} \right)} \over {\left( {0.10} \right)}}

= 4.74 + log 2 = 4.74 + 0.3010 = 5.041 Now, pH = – log[H + ] ⇒\Rightarrow 5.041 = – log[H + ] ⇒\Rightarrow [H + ] = 10 –5.041 = 9.0 × 10 6 mol L –1

Q39
If pH of a saturated solution of Ba(OH) 2 is 12, the value of its K sp is
A 4.00 × \times 10 −-6 M 3
B 4.00 × \times 10 −-7 M 3
C 5.00 × \times 10 −-6 M 3
D 5.00× \times 10 −-7 M 3
Correct Answer
Option D
Solution

<table class=tg> <tbody><tr> <th class=tg-0lax></th> <th class=tg-9wq8>Ba(OH)<sub>2</sub></th> <th class=tg-9wq8>⇌</th> <th class=tg-c3ow>Ba<sup>2+</sup></th> <th class=tg-c3ow>+</th> <th class=tg-c3ow>2OH</th> </tr> <tr> <td class=tg-0lax>At equilibrium</td> <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>2x</td> </tr> </tbody></table> <br><br>pH = – log[H<sup>+</sup>] <br><br>12 = – log [H<sup>+</sup>] <br><br>⇒\Rightarrow [H<sup>+</sup>] = 10<sup>–12</sup> <br><br>As, [H<sup>+</sup>][OH<sup>–</sup> ] = 10<sup>–14</sup> <br><br>10<sup>–12</sup> [OH<sup>–</sup> ] = 10<sup>–14</sup> <br><br>⇒\Rightarrow [OH<sup>–</sup> ] = 10<sup>–2</sup> <br><br>As [OH<sup>–</sup> ] = 2x = 10<sup>–2</sup> then x = 5.0 × 10<sup>–3</sup> <br><br>Now, K<sub>sp</sub> = [Ba<sup>2+</sup>][OH<sup>–</sup> ]<sup>2</sup> <br><br>K<sub>sp</sub> = (5 × 10<sup>–3</sup>) (10<sup>–2</sup>)<sup>2</sup> = 5.0 × 10<sup>–7</sup>

Q40
In a buffer solution containing equal concentration of B −- and HB, the K b for B −- is 10 −-10 . The pH of buffer solution is
A 10
B 7
C 6
D 4
Correct Answer
Option D
Solution

pOH = pK b + log

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

⇒\Rightarrow pOH = - log K b + log

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

⇒\Rightarrow pOH = –log10 –10 + log 1 [As conc. of HB and B – are same] ⇒\Rightarrow pOH = 10 ⇒\Rightarrow pH = 14 – pOH = 14 – 10 = 4

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