Ionic Equilibrium — Page 3 | JEE Chemistry

Q21

The solubility of AgCl will be maximum in which of the following?

A 0.01 M KCl

B 0.01 M HCl

C 0.01 M AgNO3

D Deionised water

Correct Answer

Option D

Solution

The solubility of AgCl will be maximum in deionized water.

Silver chloride (AgCl) is a sparingly soluble salt.

Its solubility in water can be represented by the equilibrium:

\text{AgCl (s)} \rightleftharpoons \text{Ag}^+ (aq) + \text{Cl}^- (aq)

In the presence of a common ion, the solubility of AgCl decreases due to the common ion effect, which is explained by Le Chatelier's Principle.

Here's an analysis of each option: Option A: 0.01 M KCl The presence of additional chloride ions ( $\text{Cl}^-$ ) from KCl will shift the equilibrium to the left, reducing the solubility of AgCl.

Option B: 0.01 M HCl Similar to KCl, HCl also provides more chloride ions, which will decrease the solubility of AgCl due to the common ion effect.

Option C: 0.01 M AgNO $_3$ AgNO $_3$ introduces additional silver ions ( $\text{Ag}^+$ ) into the solution, also shifting the equilibrium to the left, hence decreasing the solubility of AgCl.

Option D: Deionized water This option does not introduce additional ions that impact the equilibrium, so there are no additional silver or chloride ions to shift the equilibrium.

As a result, the solubility of AgCl will be highest in deionized water, as there is no common ion effect.

Therefore, the solubility of AgCl is highest in deionized water.

Q22

{K_{{a_1}}}

,

{K_{{a_2}}}

and

{K_{{a_3}}}

are the respective ionization constants for the following reactions (a), (b) and (c). (a)

{H_2}{C_2}{O_4} \mathbin{\lower.3ex\hbox{\overset{\textstyle\rightarrow}{\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}}} {H^ + } + H{C_2}O_4^ -

(b)

H{C_2}O_4^ - \mathbin{\lower.3ex\hbox{\overset{\textstyle\rightarrow}{\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}}} {H^ + } + {C_2}O_4^{2 - }

(c)

{H_2}{C_2}O_4^{} \mathbin{\lower.3ex\hbox{\overset{\textstyle\rightarrow}{\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}}} 2{H^ + } + {C_2}O_4^{2 - }

The relationship between

{K_{{a_1}}}

,

{K_{{a_2}}}

and

{K_{{a_3}}}

is given as :

A

{K_{{a_3}}}

=

{K_{{a_1}}}

+

{K_{{a_2}}}

B

{K_{{a_3}}}

=

{K_{{a_1}}}

-

{K_{{a_2}}}

C

{K_{{a_3}}}

=

{K_{{a_1}}}

/

{K_{{a_2}}}

D

{K_{{a_3}}}

=

{K_{{a_1}}}

\times

{K_{{a_2}}}

Correct Answer

Option D

Solution

\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} \rightleftharpoons 2 \mathrm{H}^{+}+\mathrm{C}_{2} \mathrm{O}_{4}^{2-} \quad \mathrm{K}_{\mathrm{a}_{3}}

\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} \rightleftharpoons \mathrm{H}^{+}+\mathrm{HC}_{2} \mathrm{O}_{4}^{-} \quad \mathrm{K}_{\mathrm{a}_{1}}

\mathrm{HC}_{2} \mathrm{O}_{4}^{-} \rightleftharpoons \mathrm{H}^{+}+\mathrm{C}_{2} \mathrm{O}_{4}^{2-} \quad \mathrm{K}_{\mathrm{a}_{2}}

\mathrm{K}_{\mathrm{a}_{3}}=\frac{\left[\mathrm{H}^{+}\right]^{2}\left[\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\right]}{\left[\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right]}

\mathrm{K}_{\mathrm{a}_{1}}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{HC}_{2} \mathrm{O}_{4}^{-}\right]}{\left[\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right]}, \mathrm{K}_{\mathrm{a}_{2}}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{C}_{2} \mathrm{O}_{4}^{-}\right]}{\left[\mathrm{HC}_{2} \mathrm{O}_{4}^{-}\right]}

\mathrm{K}_{\mathrm{a}_{3}}=\mathrm{K}_{\mathrm{a}_{1}} \times \mathrm{K}_{\mathrm{a}_{2}}

Q23

200 \mathrm{~mL}

of

0.01 \,\mathrm{M} \,\mathrm{HCl}

is mixed with

400 \mathrm{~mL}

of

0.01 \,\mathrm{M} \,\mathrm{H}_{2} \mathrm{SO}_{4}

. The

\mathrm{pH}

of the mixture is _________. Given:

\log {2}=0.30, \log 3=0.48, \log 5=0.70, \log 7=0.84, \log 11=1.04

A 1.14

B 1.78

C 2.34

D 3.02

Correct Answer

Option B

Solution

\begin{aligned} {\left[\mathrm{H}^{+}\right] } &=\frac{0.01 \times 200+2 \times 0.01 \times 400}{600} \\ &=\frac{0.01+2 \times 0.01 \times 2}{3} \\ &=\frac{0.01+0.04}{3} \\ &=\frac{5}{3} \times 10^{-2} \\ \mathrm{pH} &=-\log \left[\mathrm{H}^{+}\right] \\ &=-\log \left(\frac{5}{3} \times 10^{-2}\right) \\ &=-\left[\log \frac{5}{3}+\log 10^{-2}\right] \\ &=-[\log 5-\log 3-2] \\ &=-0.7+0.48+2 \\ &=2.48-0.7 \\ &=1.78 \end{aligned}

Q24

When the hydrogen ion concentration [H

^+

] changes by a factor of 1000, the value of pH of the solution __________

A decreases by 2 units

B increases by 2 units

C decreases by 3 units

D increases by 1000 units

Correct Answer

Option C

Solution

Let the initial concentration of $\mathrm{H}^{+}$ be 1 $\therefore \left[\mathrm{H}^{+}\right]_{\mathrm{i}}=1 \Rightarrow \mathrm{pH}=0$ It changes by 1000 units $\therefore \left[\mathrm{H}^{+}\right]_{\mathrm{f}}=10^{3} \Rightarrow \mathrm{pH}=-3$ $\therefore \mathrm{pH}$ decreases by 3 units

Q25

25 \mathrm{~mL}

of silver nitrate solution (1M) is added dropwise to

25 \mathrm{~mL}

of potassium iodide

(1.05 \mathrm{M})

solution. The ion(s) present in very small quantity in the solution is/are :

A

\mathrm{I^-}

only

B

\mathrm{K^+}

only

C

\mathrm{Ag^+}

and

\mathrm{I^-}

both

D

\mathrm{NO_3^-}

only

Correct Answer

Option C

Solution

The reaction between silver nitrate (AgNO3) and potassium iodide (KI) forms silver iodide (AgI), which is practically insoluble in water.

The reaction is as follows : AgNO3(aq) + KI(aq) → AgI(s) + KNO3(aq) Although the reaction stoichiometry indicates that iodide ions (I-) and silver ions (Ag+) are produced, silver iodide (AgI) precipitates out of the solution due to its low solubility, and very little Ag+ and I- ions remain in the solution.

Hence, their concentration in the solution will be negligible.

So, the correct answer should be : Option C : Ag+ and I- both

Q26

The molar solubility(s) of zirconium phosphate with molecular formula

\left(\mathrm{Zr}^{4+}\right)_3\left(\mathrm{PO}_4^{3-}\right)_4

is given by relation :

A

\left(\dfrac{\mathrm{K}_{\mathrm{sp}}}{5348}\right)^{\dfrac{1}{6}}

B

\left(\dfrac{\mathrm{K}_{\mathrm{sp}}}{8435}\right)^{\dfrac{1}{7}}

C

\left(\dfrac{K_{s p}}{6912}\right)^{\dfrac{1}{7}}

D

\left(\dfrac{\mathrm{K}_{\mathrm{sp}}}{9612}\right)^{\dfrac{1}{3}}

Correct Answer

Option C

Solution

$\begin{aligned} & \mathrm{K}_{\mathrm{sp}}=(3 \mathrm{~s})^3 \cdot(4 \mathrm{~s})^4 \\\\ & \mathrm{~K}_{\mathrm{sp}}=6912 \mathrm{~s}^7 \\\\ & \mathrm{~s}=\left(\dfrac{\mathrm{K}_{\mathrm{sp}}}{6912}\right)^{\dfrac{1}{7}}\end{aligned}$

Q27

Solubility of calcium phosphate (molecular mass, M) in water is

\mathrm{W_{g}}

per

100 \mathrm{~mL}

at

25^{\circ} \mathrm{C}

. Its solubility product at

25^{\circ} \mathrm{C}

will be approximately.

A

10^7\left(\dfrac{W}{M}\right)^3

B

10^3\left(\dfrac{\mathrm{W}}{\mathrm{M}}\right)^5

C

10^7\left(\dfrac{W}{M}\right)^5

D

10^5\left(\dfrac{\mathrm{W}}{\mathrm{M}}\right)^5

Correct Answer

Option C

Solution

To determine the solubility product (Ksp) of calcium phosphate, we need to consider its chemical formula and how it dissociates in water.

The formula for calcium phosphate is $\text{Ca}_3(\text{PO}_4)_2$ .

When dissolved in water, it dissociates according to the following equation:

\text{Ca}_3(\text{PO}_4)_2 (s) \rightleftharpoons 3 \text{Ca}^{2+} (aq) + 2 \text{PO}_4^{3-} (aq)

Let $S$ be the molar solubility of calcium phosphate in $\text{mol}/L$ .

According to the stoichiometry of the dissociation, if $S$ moles per liter of calcium phosphate dissolve, $3S$ moles per liter of calcium ions and $2S$ moles per liter of phosphate ions are produced.

The expression for the solubility product constant (Ksp) for calcium phosphate in water is the product of the concentrations of the ions raised to the power of their respective coefficients in the balanced equation:

K_{sp} = [\text{Ca}^{2+}]^3 [\text{PO}_4^{3-}]^2

Substituting the concentrations with $3S$ for $\text{Ca}^{2+}$ and $2S$ for $\text{PO}_4^{3-}$ , we get:

K_{sp} = (3S)^3 (2S)^2 = 27S^3 \cdot 4S^2 = 108S^5

If the solubility of calcium phosphate is $\text{W}_{g}$ per $100 \text{ mL}$ of water at $25^\circ\text{C}$ , we need to convert grams to moles to find $S$ .

Solubility in moles per liter (which is the same as moles per $1000 \text{ mL}$ ) is:

S = \frac{\text{W}}{\text{M}} \times \frac{1000 \text{ mL}}{100 \text{ mL}} = 10\frac{\text{W}}{\text{M}}

Now, substituting $S$ with $10\dfrac{\text{W}}{\text{M}}$ into the expression for $K_{sp}$ :

K_{sp} = 108 \left(10\frac{\text{W}}{\text{M}}\right)^5

This simplifies to:

K_{sp} = 108 \times 10^5 \left(\frac{\text{W}}{\text{M}}\right)^5

Since $108$ is on the order of $10^2$ , we can approximate this to:

K_{sp} \approx 10^7 \left(\frac{\text{W}}{\text{M}}\right)^5

Therefore, the correct answer based on the options provided is: Option C: $10^7 \left(\dfrac{\text{W}}{\text{M}}\right)^5$

Q28

Given below are two statements : Statement (I) : A Buffer solution is the mixture of a salt and an acid or a base mixed in any particular quantities Statement (II) : Blood is naturally occurring buffer solution whose

\mathrm{pH}

is maintained by

\mathrm{H}_2 \mathrm{CO}_3 / \mathrm{HCO}_3{ }^{\ominus}

concentrations. In the light of the above statements, choose the correct answer from the options given below :

A Both Statement I and Statement II are false

B Both Statement I and Statement II are true

C Statement I is false but Statement II is true

D Statement I is true but Statement II is false

Correct Answer

Option C

Solution

Let's analyze both statements given to determine the correct answer.

Statement (I) describes a buffer solution as a mixture of a salt and an acid or a base mixed in any particular quantities.

However, this definition is partially incorrect.

A buffer solution is more accurately defined as a mixture of a weak acid and its conjugate base or a weak base and its conjugate acid.

The purpose of a buffer solution is to maintain a stable pH when small amounts of acid (H⁺ ions) or alkali (OH⁻ ions) are added.

It is not just any mixture of a salt and an acid or base but must involve components that can react with added acid or base to minimize changes in pH.

Therefore, Statement (I) is misleading or incomplete as it omits the necessity for a weak acid and its conjugate base or a weak base and its conjugate acid to form a true buffer solution.

Statement (II) mentions that blood is a naturally occurring buffer solution whose pH is maintained by the

\mathrm{H}_2\mathrm{CO}_3/\mathrm{HCO}_3^{-}

(carbonic acid/bicarbonate) system.

This is accurate.

The carbonic acid-bicarbonate buffering system is one of the main buffering systems in human blood, helping to maintain the pH within a narrow range (typically around 7.35 to 7.45).

This buffer system works by the reversible reaction between carbon dioxide (CO₂) and water to form carbonic acid (

\mathrm{H}_2\mathrm{CO}_3

), which can then dissociate into bicarbonate ion (

\mathrm{HCO}_3^{-}

) and a hydrogen ion (H⁺).

This system effectively moderates changes in pH by either consuming or releasing H⁺ ions.

Given this analysis: Statement I is false because it inaccurately or incompletely describes a buffer solution.

Statement II is true as it correctly identifies blood as a naturally occurring buffer solution that uses the

\mathrm{H}_2\mathrm{CO}_3/\mathrm{HCO}_3^{-}

system to maintain pH. Therefore, the correct answer is: Option C: Statement I is false but Statement II is true.

Q29

Arrange the following in increasing order of solubility product :

\mathrm{Ca}(\mathrm{OH})_2, \mathrm{AgBr}, \mathrm{PbS}, \mathrm{HgS}

A

\mathrm{PbS}<\mathrm{HgS}<\mathrm{Ca}(\mathrm{OH})_2<\mathrm{AgBr}

B

\mathrm{HgS}<\mathrm{AgBr}<\mathrm{PbS}<\mathrm{Ca}(\mathrm{OH})_2

C

\mathrm{HgS}<\mathrm{PbS}<\mathrm{AgBr}<\mathrm{Ca}(\mathrm{OH})_2

D

\mathrm{Ca}(\mathrm{OH})_2<\mathrm{AgBr}<\mathrm{HgS}<\mathrm{PbS}

Correct Answer

Option C

Solution

Based on the Ksp values and salt analysis cation identification, we can say that order of Ksp value is:

\mathrm{HgS} Ksp values

\begin{aligned} & \mathrm{HgS} \rightarrow 4 \times 10^{-53} \\ & \mathrm{PbS} \rightarrow 8 \times 10^{-28} \\ & \mathrm{AgBr} \rightarrow 5 \times 10^{-13} \\ & \mathrm{Ca}(\mathrm{OH})_2 \rightarrow 5.5 \times 10^{-6} \end{aligned}$$

Q30

Which of the following happens when

\mathrm{NH}_4 \mathrm{OH}

is added gradually to the solution containing 1 M

\mathrm{A}^{2+}

and

1 \mathrm{M} \mathrm{B}^{3+}

ions? Given :

\mathrm{K}_{\text{sp }}\left[\mathrm{A}(\mathrm{OH})_2\right]=9 \times 10^{-10}

and

\mathrm{K}_{\mathrm{sp}}\left[\mathrm{B}(\mathrm{OH})_3\right]=27 \times 10^{-18}

at 298 K.

A

\mathrm{A}(\mathrm{OH})_2

will precipitate before

\mathrm{B}(\mathrm{OH})_3

B

\mathrm{A}(\mathrm{OH})_2

and

\mathrm{B}(\mathrm{OH})_3

will precipitate together

C Both

\mathrm{A}(\mathrm{OH})_2

and

\mathrm{B}(\mathrm{OH})_3

do not show precipitation with

\mathrm{NH}_4 \mathrm{OH}

D

\mathrm{B}(\mathrm{OH})_3

will precipitate before

\mathrm{A}(\mathrm{OH})_2

Correct Answer

Option D

Solution

Condition for precipitation $\mathrm{Q}_{\mathrm{ip}}>\mathrm{K}_{\text{sp }}$ For $\left[\mathrm{A}(\mathrm{OH})_2\right.$ ]

\begin{aligned} & {\left[\mathrm{A}^{2+}\right]\left[\mathrm{OH}^{-}\right]^2>9 \times 10^{-10}} \\ & {\left[\mathrm{~A}^{+2}\right]=1 \mathrm{M}} \\ & \Rightarrow\left[\mathrm{OH}^{-}\right]>3 \times 10^{-5} \mathrm{M} \end{aligned}

For $\left[\mathrm{B}(\mathrm{OH})_3\right]$

\begin{aligned} & {\left[\mathrm{B}^{3+}\right]\left[\mathrm{OH}^{-}\right]^3>27 \times 10^{-18}} \\ & {\left[\mathrm{~B}^{3+}\right]=1 \mathrm{M}} \\ & \Rightarrow\left[\mathrm{OH}^{-}\right]>3 \times 10^{-6} \mathrm{M} \end{aligned}

So, $\mathrm{B}(\mathrm{OH})_3$ will precipitate before $\mathrm{A}(\mathrm{OH})_2$