Ionic Equilibrium — Page 7 | JEE Chemistry

Q61

Which of the following are Lewis acids?

A BCl3 and AlCl3

B PH3 and BCl3

C AlCl3 and SiCl4

D PH3 and SiCl4

Correct Answer

Option A

Solution

The compound which have the ability to accepted at least are lone pair electron.

Structure of BCl3 is Here B is electron deficient atom, so it can accepted lone pair.

So it is a lewis acid.

Structure of AlCl3 Here, Al also a electron deficient atom so it has vacant orbital and in that vacant orbital it can take lone pair.

So it is also lewis acid.

Here in PH3 there is vacant 3d orbital but it Can't take.

Lone pair in 3d orbital because P is more electro-negative than H so around P atom negative charge density is created and tendency of accepting electron decreases.

So PH3 is not lewis acid.

Here octet of Si full ut in has a tendency of accepting lone pair in vacant 3d orbital.

This can be shown by following reaction.

So, here option (A) and (C) both are correct.

But as SiCl4 is not as strong lewis acid as BCl3 and AlCl3, So we can say option (A) is more correct.

Q62

The pH of a 0.1 molar solution of the acid HQ is 3. The value of the ionization constant, Ka of this acid is :

A 3

\times

10–1

B 1

\times

10–3

C 1

\times

10–5

D 1

\times

10–7

Correct Answer

Option C

Solution

{H^ + } = C\alpha ;\alpha = {{\left[ {{H^ + }} \right]} \over C}

or

\,\,\,\alpha = {{{{10}^{ - 3}}} \over {0.1}} = {10^{ - 2}}

Ka = C{\alpha ^2} = 0.1 \times {10^{ - 2}} \times {10^{ - 2}}

= {10^{ - 5}}

Q63

1 M NaCL and 1 M HCL are present in an aqueous solution. The solution is

A not a buffer solution with pH < 7

B not a buffer solution with pH > 7

C a buffer solution with pH < 7

D a buffer solution with pH > 7

Correct Answer

Option A

Solution

NOTE : A buffer is a solution of weak acid and its salt with strong base and vice versa.

HCL

is strong acid and

NaCL

is its salt with strong base.

pH

is less than

7

due to

HCL

.

Q64

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.

Q65

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.

Q66

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

Q67

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

Q68

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

Q69

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}}$.

Q70

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