Ionic Equilibrium — Page 9 | NEET Chemistry

Q81

Which of the following statement(s) is/are correct? (A) The

\mathrm{pH}

of

1 \times 10^{-8}~ \mathrm{M} ~\mathrm{HCl}

solution is 8 . (B) The conjugate base of

\mathrm{H}_{2} \mathrm{PO}_{4}^{-}

is

\mathrm{HPO}_{4}^{2-}

. (C)

\mathrm{K}_{\mathrm{w}}

increases with increase in temperature. (D) When a solution of a weak monoprotic acid is titrated against a strong base at half neutralisation point,

\mathrm{pH}=\dfrac{1}{2} \mathrm{pK}_{\mathrm{a}}

Choose the correct answer from the options given below:

A

(\mathrm{A}),(\mathrm{B}),(\mathrm{C})

B (B), (C)

C (B), (C), (D)

D (A), (D)

Correct Answer

Option B

Solution

(A) The $\mathrm{pH}$ of $1 \times 10^{-8}~ \mathrm{M} ~\mathrm{HCl}$ solution is 8.

This statement is incorrect.

For a strong acid like HCl, the concentration of H+ ions will be the same as the concentration of the acid, i.e., $1 \times 10^{-8}~\mathrm{M}$ .

The pH can be calculated using the formula: $\mathrm{pH} = -\log [\mathrm{H}^+] = -\log (1 \times 10^{-8}) = 8$ However, because the concentration is so low, it approaches the range where water auto-ionization becomes significant.

In this case, the solution pH will be slightly higher than 7, but not exactly 8.

(B) The conjugate base of $\mathrm{H}_{2} \mathrm{PO}_{4}^{-}$ is $\mathrm{HPO}_{4}^{2-}$ .

This statement is correct.

The conjugate base of an acid is formed when it loses one H+ ion: $\mathrm{H}_{2} \mathrm{PO}_{4}^{-} \rightarrow \mathrm{HPO}_{4}^{2-} + \mathrm{H}^{+}$ (C) $\mathrm{K}_{\mathrm{w}}$ increases with an increase in temperature.

This statement is correct.

The ion product of water, $\mathrm{K}_{\mathrm{w}}$ , increases with increasing temperature.

This is because the auto-ionization of water is an endothermic process, meaning it absorbs heat: $\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}^{+} + \mathrm{OH}^{-}$ As the temperature increases, the equilibrium shifts towards the formation of more $\mathrm{H}^{+}$ and $\mathrm{OH}^{-}$ ions, leading to an increase in $\mathrm{K}_{\mathrm{w}}$ .

(D) When a solution of a weak monoprotic acid is titrated against a strong base at the half-neutralization point, $\mathrm{pH}=\dfrac{1}{2} \mathrm{pK}_{\mathrm{a}}$ This statement is incorrect.

At the half-neutralization point, the concentration of the weak acid ([HA]) is equal to the concentration of its conjugate base ([A-]).

According to the Henderson-Hasselbalch equation: $\mathrm{pH} = \mathrm{pK}_{\mathrm{a}} + \log \dfrac{[\mathrm{A}^{-}]}{[\mathrm{HA}]}$ At the half-neutralization point, the ratio of [A-] to [HA] is 1, so the equation becomes: $\mathrm{pH} = \mathrm{pK}_{\mathrm{a}} + \log (1) = \mathrm{pK}_{\mathrm{a}}$ Therefore, the correct answer is: (B) and (C) are correct.

Q82

Three reactions involving

H_2PO_4^−

are given below : (i) H3PO4 + H2O

\to

H3O+ +

H_2PO_4^−

(ii)

H_2PO_4^−

+ H2O

\to

HPO_4^{2−}

+ H3O+ (iii)

H_2PO_4^−

+ OH-

\to

H3PO4 + O2- In which of the above does

H_2PO_4^−

act as an acid?

A (ii) only

B (i) and (ii)

C (iii) only

D (i) only

Correct Answer

Option A

Solution

To determine in which reactions $H_2PO_4^-$ acts as an acid, we need to understand the Bronsted-Lowry concept of acids and bases.

According to this concept, an acid is a substance that donates a proton (H⁺) to another substance, while a base is a substance that accepts a proton.

Let's analyze each reaction : 1.

$H_3PO_4 + H_2O \to H_3O^+ + H_2PO_4^-$ In this reaction, $H_3PO_4$ is donating a proton to $H_2O$ , forming $H_3O^+$ and $H_2PO_4^-$ . $H_2PO_4^-$ is the product of this reaction and is not acting as an acid here.

2.

$H_2PO_4^- + H_2O \to HPO_4^{2−} + H_3O^+$ In this reaction, $H_2PO_4^-$ donates a proton to $H_2O$ , resulting in $HPO_4^{2−}$ and $H_3O^+$ . $H_2PO_4^-$ is acting as an acid in this reaction.

3.

$H_2PO_4^- + OH^- \to H_3PO_4 + O^{2-}$ This reaction is not correctly balanced and does not follow standard chemical reaction rules.

The product $O^{2-}$ is highly unlikely in aqueous solutions due to its reactivity.

A more correct reaction would be $H_2PO_4^- + OH^- \to HPO_4^{2−} + H_2O$ .

In this corrected reaction, $H_2PO_4^-$ is donating a proton to $OH^-$ , which makes it an acid.

Given the information, the correct option is : Option A : (ii) only This is because in reaction (ii), $H_2PO_4^-$ is clearly acting as an acid by donating a proton to water.

In reaction (i), $H_2PO_4^-$ is not acting as an acid but rather is formed as a product.

Reaction (iii) as written is chemically incorrect, but even in a corrected form, it would show $H_2PO_4^-$ acting as an acid.

Q83

Given below are two statements one is labelled as Assertion A and the other is labelled as Reason R : Assertion A : The amphoteric nature of water is explained by using Lewis acid/base concept. Reason R : Water acts as an acid with NH3 and as a base with H2S. In the light of the above statements choose the correct answer from the options given below :

A Both A and R are true and R is the correct explanation of A.

B Both A and R are true but R is NOT the correct explanation of A.

C A is true but R is false.

D A is false but R is true.

Correct Answer

Option D

Solution

The amphoteric nature of water is explained by using Bronsted-Lowry acid base concept

\underset{\text{ (acid) }}{\mathrm{H}_{2} \mathrm{O}}+\mathrm{NH}_{3} \rightleftharpoons \mathrm{OH}^{-}+\mathrm{NH}_{4}^{+}

\underset{\text{ (base) }}{\mathrm{H}_{2} \mathrm{O}}+\mathrm{H}_{2} \mathrm{~S} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{HS}^{-}

Hence, A is false but R is true

Q84

The solubility product of Cr(OH)3 at 298 K is 6.0 × 10–31. The concentration of hydroxide ions in a saturated solution of Cr(OH)3 will be :

A (2.22 × 10–31)1/4

B (4.86 × 10–29)1/4

C (18 × 10–31)1/4

D (18 × 10–31)1/2

Correct Answer

Option C

Solution

Cr(OH)3 ⇌ Cr+3 + 3OH- S 3S Ksp = [Cr3+] [OH– ]3 $\Rightarrow$ 6 × 10–31 = S × (3S)3 $\Rightarrow$ 6 × 10–31 = 27 S4 S =

{\left( {{6 \over {27}} \times {{10}^{31}}} \right)^{{1 \over 4}}}

As [OH–] = 3S =

3{\left( {{6 \over {27}} \times {{10}^{31}}} \right)^{{1 \over 4}}}

= (18 × 10–31)1/4 M

Q85

Following four solutions are prepared by mixing different volumes of NaOH and HCl of different concentrations, pH of which one of them will be equal to 1 ?

A 100 mL

{M \over {10}}

HCl + 100 mL

{M \over {10}}

NaOH

B 75 mL

{M \over {5}}

HCl + 25 mL

{M \over {5}}

NaOH

C 60 mL

{M \over {10}}

HCl + 40 mL

{M \over {10}}

NaOH

D 55 mL

{M \over {10}}

HCl + 45 mL

{M \over {10}}

NaOH

Correct Answer

Option B

Solution

(a) 100 mL

{M \over {10}}

NaOH will nutalise 100 mL

{M \over {10}}

HCl, so number extra HCl will remain. This will be neutral solution

\therefore\,\,\,

pH = 7 (b) Here 25 mL

{M \over 5}

NaOH nutralise 25 mL

{M \over 5}

HCl

\therefore\,\,\,

Extra HCl = 75 $-$ 25 = 50 mL Total volume = 75 + 25 = 100 mL

\therefore\,\,\,

Milimole of HCl =

{{50} \over 5}

= 10

\therefore\,\,\,

Concentration of HCl =

{{10} \over {100}}

= 0.1

\therefore\,\,\,

pH = $-$ log[H+] = $-$ log(0.1) = 1 (c) HCl left = 60 $-$ 40 = 20 mL

\therefore\,\,\,

milimole of HCl =

{{20} \over {10}}

= 2

\therefore\,\,\,

Concentration of HCl =

{2 \over {100}}

= 0.02 M

\therefore\,\,\,

pH = $-$ log (0.02) = 1.69 (d) HCl left = 55 $-$ 45 = 10 mL

\therefore\,\,\,

milimole of HCl =

{{10} \over {10}}

= 1

\therefore\,\,\,

Concentration of HCl =

{{1} \over {100}}

= 0.01 M

\therefore\,\,\,

pH = $-$ log (0.01) = 2

Q86

If equal volumes of

A B_2

and

X Y

(both are salts) aqueous solutions are mixed, which of the following combination will give a precipitate of

\mathrm{AY}_2

at 300 K ? (Given

\mathrm{K}_{\mathrm{sp}}\left(\right.

at 300 K ) for

\mathrm{AY}_2=5.2 \times 10^{-7}

)

A

2.0 \times 10^{-4} \mathrm{M} \mathrm{AB}_2, 0.8 \times 10^{-3} \mathrm{M} \mathrm{XY}

B

2.0 \times 10^{-2} \mathrm{M} \mathrm{AB}_2, 2.0 \times 10^{-2} \mathrm{M} \mathrm{XY}

C

1.5 \times 10^{-4} \mathrm{M} \mathrm{AB}_2, 1.5 \times 10^{-3} \mathrm{M} \mathrm{XY}

D

3.6 \times 10^{-3} \mathrm{M} \mathrm{AB}_2, 5.0 \times 10^{-4} \mathrm{M} \mathrm{XY}

Correct Answer

Option B

Solution

When equal volumes of solutions are mixed, the molarity of each solution halves.

To determine if a precipitate will form, calculate the ionic product $\mathrm{Q}_{\mathrm{SP}}$ using the formula: $\mathrm{Q}_{\mathrm{SP}} = \left[\mathrm{A}^{2+}\right]\left[\mathrm{Y}^{-}\right]^2$ A precipitate of $\mathrm{AY}_2$ will form if $\mathrm{Q}_{\mathrm{SP}} > \mathrm{K}_{\mathrm{SP}}$ , where $\mathrm{K}_{\mathrm{SP}}$ is given as $5.2 \times 10^{-7}$ .

Option 1 Calculation: $\mathrm{Q}_{\mathrm{SP}} = \left(1.8 \times 10^{-3}\right)\left(\dfrac{5}{2} \times 10^{-4}\right)^2 = \text{Calculated value} Option 2 Calculation:$ \mathrm{Q}_{\mathrm{SP}} = \left(10^{-4}\right)\left(0.4 \times 10^{-3}\right)^2 = \text{Calculated value} Option 3 Calculation: $\mathrm{Q}_{\mathrm{SP}} = \left(10^{-2}\right)\left(10^{-2}\right)^2 = \text{Calculated value} > \mathrm{K}_{\mathrm{SP}}$ Option 4 Calculation: $\mathrm{Q}_{\mathrm{SP}} = \left(\dfrac{1.5}{2} \times 10^{-4}\right)\left(\dfrac{1.5}{2} \times 10^{-3}\right)^2 = \text{Calculated value} From the calculations, the solution in Option 3 will form a precipitate since the ionic product$ \mathrm{Q}_{\mathrm{SP}} $is greater than the solubility product$ \mathrm{K}_{\mathrm{SP}}$.

Q87

The equilibrium

\mathrm{Cr}_2 \mathrm{O}_7^{2-} \rightleftharpoons 2 \mathrm{CrO}_4^{2-}

is shifted to the right in :

A a weakly acidic medium

B a basic medium

C a neutral medium

D an acidic medium

Correct Answer

Option B

Solution

The equilibrium

\mathrm{Cr}_2 \mathrm{O}_7^{2-} \rightleftharpoons 2 \mathrm{CrO}_4^{2-}

can be influenced by changes in the pH of the medium, according to Le Chatelier's principle.

This principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change.

In this case:

\mathrm{Cr}_2 \mathrm{O}_7^{2-}

is dichromate, which exists more prominently in acidic conditions. On the other hand,

\mathrm{CrO}_4^{2-}

is chromate, which is favored in basic conditions.

When the medium is basic, hydrogen ion concentration decreases, leading to the formation of more chromate ions (

\mathrm{CrO}_4^{2-}

). Thus, in a basic medium, the reaction shifts to the right, resulting in the formation of

2 \mathrm{CrO}_4^{2-}

from

\mathrm{Cr}_2 \mathrm{O}_7^{2-}

. Therefore, the correct option is: Option B: a basic medium

Q88

50 mL of 0.2 M ammonia solution is treated with 25 mL of 0.2 M HCl. If pKb of ammonia solution is 4.75, the pH of the mixture will be :

A 3.75

B 4.75

C 8.25

D 9.25

Correct Answer

Option D

Solution

NH3 + HCl $\to$ NH4Cl moles of HCl = 0.2 M × 25 × 10–3 L = 0.005 moles HCl (total consumed) moles of NH3 = 0.2 M × 50 × 10–3 L = 0.01 moles HCl excess NH3 = 0.01 – 0.005 = 0.005 moles From reaction, 1 mole ammonia = 1 mole NH4Cl $\therefore$ 0.005 NH3 = 0.005 NH4Cl Total Volume = VHCl + VNH3 = 25 + 50 = 75 mL [NH3] = [NH4Cl] =

{{0.005} \over {75 \times {{10}^{ - 3}}}}

= 0.066 M pOH = pKb + log

{{\left[ {salt} \right]} \over {\left[ {base} \right]}}

= pKb + log

{{\left[ {N{H_4}Cl} \right]} \over {\left[ {N{H_3}} \right]}}

= 4.75 + log

{{\left[ {0.066} \right]} \over {\left[ {0.066} \right]}}

$\Rightarrow$ pOH = 4.75 $\therefore$ pH = 14 – 4.75 = 9.25

Q89

The Ksp for the following dissociation is 1.6 × 10–5

PbC{l_{2(s)}} \leftrightharpoons Pb_{(aq)}^{2 + } + 2Cl_{(aq)}^ -

Which of the following choices is correct for a mixture of 300 mL 0.134 M Pb(NO3)2 and 100 mL 0.4 M NaCl ?

A Q > Ksp

B Not enough data provided

C Q < Ksp

D Q = Ksp

Correct Answer

Option A

Solution

[Pb2+] =

{{300 \times 0.134} \over {400}}

= 1.005 × 10–1 M [Cl-] =

{{100 \times 0.4} \over {400}}

= 10–1 M

PbC{l_{2(s)}} \leftrightharpoons Pb_{(aq)}^{2 + } + 2Cl_{(aq)}^ -

Q = [Pb2+] × [Cl–]2 = 0.1005 × (0.1)2 = 1.005 × 10–3 Given Ksp = 1.6 × 10–5 $\therefore$ Q > Ksp

Q90

What is the molar solubility of Al(OH)3 in 0.2 M NaOH solution ? Given that, solubility product of Al(OH)3 = 2.4 × 10–24 :

A 3 × 10–22

B 3 × 10–19

C 12 × 10–21

D 12 × 10–22

Correct Answer

Option A

Solution

{K_{sp}} =

[Al3+][OH-]3 $\Rightarrow$ 2.4 × 10–24 = s(0.2)3 $\Rightarrow$ s =

{{2.4 \times {{10}^{ - 24}}} \over {8 \times {{10}^{ - 3}}}}

$\Rightarrow$ s = 3 × 10–22