Ionic Equilibrium Questions — NEET Chemistry

Q1

In a qualitative analysis,

\mathrm{Bi}^{3+}

is detected by appearance of precipitate of

\mathrm{BiO}(\mathrm{OH})(\mathrm{s})

. Calculate pH when the following equilibrium exists at 298 K .

\mathrm{BiO}(\mathrm{OH})(\mathrm{s}) \rightleftharpoons \mathrm{BiO}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}), \mathrm{K}=4 \times 10^{-10}

(Given :

\log 2=0.3010

)

A 8.714

B 4.699

C 5.286

D 9.301

Correct Answer

Option D

Solution

\begin{aligned} & \mathrm{BiO}(\mathrm{OH})(\mathrm{s}) \rightleftharpoons \mathrm{BiO}_{\mathrm{s}}^{+}(\mathrm{aq})+\mathrm{OH}_{\mathrm{s}}^{-}(\mathrm{aq}) \\ & \mathrm{K}=\frac{\left[\mathrm{BiO}^{+}\right]\left[\mathrm{OH}^{-}\right]}{[\mathrm{BiO}(\mathrm{OH})(\mathrm{s})]} \\ & \mathrm{K}=\frac{\left[\mathrm{BiO}^{+}\right]\left[\mathrm{OH}^{-}\right]}{1} \\ & \mathrm{~K}=\mathrm{s} \times \mathrm{s}=\mathrm{s}^2 \\ & \mathrm{~s}=\sqrt{\mathrm{K}}=\sqrt{4 \times 10^{-10}}=2 \times 10^{-5} \mathrm{M} \\ & {\left[\mathrm{H}^{+}\right]=\frac{\mathrm{K}_{\mathrm{w}}}{\left[\mathrm{OH}^{-}\right]}=\frac{1}{2} \times 10^{-9}, \mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]=9+\log 2=9.301} \end{aligned}

Q2

At 298 K , a certain buffer solution contains equal concentrations of

\mathrm{X}^{-}

and

\mathrm{HX}, \mathrm{K}_{\mathrm{b}}

for

\mathrm{X}^{-}

is

10^{-10}

. What is the pH of this buffer solution?

A 2

B 4

C 6

D 10

Correct Answer

Option B

Solution

$\mathrm{HX} \rightleftharpoons \mathrm{H}^{+}+\mathrm{X}^{-}$

\begin{aligned} & \mathrm{K}_{\mathrm{a}} \times \mathrm{K}_{\mathrm{b}}=\mathrm{K}_{\mathrm{w}} \\ & \mathrm{~K}_{\mathrm{a}}=\frac{\mathrm{K}_{\mathrm{w}}}{\mathrm{~K}_{\mathrm{b}}}=\frac{10^{-14}}{10^{-10}}=10^{-4} \\ & \mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \frac{\left[\mathrm{X}^{-}\right]}{[\mathrm{HX}]} \end{aligned}

Given that : $\left[\mathrm{X}^{-}\right]=[\mathrm{HX}]$

\begin{aligned} & =4+\log (1) \\ & =4+0 \\ & =4 \end{aligned}

Q3

Phenolphthalein is used as an indicator for the titration of sodium hydroxide solution against a standard solution of oxalic acid. The colour change that is observed at an alkaline pH close to the equivalence point during this titration is:

A pinkish red to yellow

B yellow to pinkish red

C pink to colourless

D colourless to pink

Correct Answer

Option D

Solution

Colour of Phenolphthalein before the end point = colourless.

Colour of Phenolphthalein close to equivalence point = Pink.

∴ Colour change $=$ Colourless to pink.

Q4

If the molar conductivity

\left(\Lambda_{\mathrm{m}}\right)

of a

0.050 \mathrm{~mol} \mathrm{~L}^{-1}

solution of a monobasic weak acid is

90 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}

, its extent (degree) of dissociation will be [Assume

\Lambda_{+}^{\circ}=349.6 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}

and

\Lambda_{-}^{\circ}=50.4 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}

.]

A 0.225

B 0.215

C 0.115

D 0.125

Correct Answer

Option A

Solution

To determine the degree of dissociation (α) of a weak acid, we use the formula: $\alpha = \dfrac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}}$ where: $\Lambda_{\mathrm{m}}$ is the molar conductivity of the solution, provided as $90 \, \mathrm{S} \, \mathrm{cm}^2 \, \mathrm{mol}^{-1}$ . $\Lambda_{\mathrm{m}}^{\circ}$ is the limiting molar conductivity of the acid, calculated as the sum of the limiting molar conductivities of its ions.

Given: $\Lambda_{+}^{\circ} = 349.6 \, \mathrm{S} \, \mathrm{cm}^2 \, \mathrm{mol}^{-1}$ (cation) $\Lambda_{-}^{\circ} = 50.4 \, \mathrm{S} \, \mathrm{cm}^2 \, \mathrm{mol}^{-1}$ (anion) First, calculate $\Lambda_{\mathrm{m}}^{\circ}$ : $\Lambda_{\mathrm{m}}^{\circ} = \Lambda_{+}^{\circ} + \Lambda_{-}^{\circ} = 349.6 + 50.4 = 400 \, \mathrm{S} \, \mathrm{cm}^2 \, \mathrm{mol}^{-1}$ Next, use the given values to find the degree of dissociation: $\alpha = \dfrac{90}{400} = 0.225$ Therefore, the degree of dissociation of the acid is 0.225.

Q5

Which indicator is used in the titration of sodium hydroxide against oxalic acid and what is the colour change at the end point?

A Phenolphthalein, pink to yellow

B Alkaline KMnO

_4

, colourless to pink

C Phenolphthalein, colourless to pink

D Methyl orange, yellow to pinkish red colour

Correct Answer

Option C

Solution

The correct answer is Option C: Phenolphthalein, colourless to pink .

Here's why: Titration of Sodium Hydroxide (NaOH) against Oxalic Acid (H 2 C 2 O 4 ) Sodium hydroxide is a strong base, and oxalic acid is a weak acid.

The reaction between them is a neutralization reaction:

2NaOH + H_2C_2O_4 \longrightarrow Na_2C_2O_4 + 2H_2O

Choosing the Right Indicator An indicator is a substance that changes color at a specific pH range, signaling the endpoint of the titration.

The ideal indicator for a titration should have a color change near the equivalence point of the reaction.

The equivalence point is the point where the moles of acid and base are stoichiometrically equal.

In this case: At the equivalence point, the solution will be slightly basic due to the formation of the sodium oxalate salt (Na 2 C 2 O 4 ), which is the conjugate base of a weak acid.

Phenolphthalein changes color in the slightly basic pH range (around 8.2 to 10.0), making it an appropriate indicator for this titration.

Color Change Phenolphthalein is colorless in acidic solutions and turns pink in basic solutions.

So, the color change at the endpoint of this titration will be: Colorless (before equivalence point) → Pink (at equivalence point) Why Other Options Are Incorrect Option A: Phenolphthalein does not change from pink to yellow, it changes from colorless to pink.

Option B: Alkaline KMnO 4 is not a suitable indicator for this titration because it is not sensitive enough to detect the slight pH change at the equivalence point.

Option D: Methyl orange changes color in the acidic pH range (around 3.1 to 4.4) and is therefore not suitable for this titration.

Q6

The ratio of solubility of AgCl in 0.1 M KCl solution to the solubility of AgCl in water is: (Given : Solubility product of AgCl = 10

^{–10}

)

A 10

^{-4}

B 10

^{-6}

C 10

^{-9}

D 10

^{-5}

Correct Answer

Option A

Solution

To find the ratio of solubility of AgCl in 0.1 M KCl solution to the solubility of AgCl in water, we first need to understand how common ion effect influences solubility.

Let's denote the solubility product constant of AgCl as

K_{sp}

. Given

K_{sp}

of AgCl =

10^{-10}

. First, we calculate the solubility of AgCl in pure water: In water, the dissociation of AgCl can be represented as:

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

If

s

is the solubility of AgCl in water, then

[Ag^{+}] = s

[Cl^{-}] = s

Hence, the solubility product

K_{sp}

can be written as:

K_{sp} = [Ag^{+}] [Cl^{-}]

K_{sp} = s \cdot s = s^{2}

So,

s^{2} = 10^{-10}

s = 10^{-5}

Now, let's consider the solubility of AgCl in 0.1 M KCl solution. Because of the common ion effect, the presence of

Cl^{-}

ions from KCl will suppress the solubility of AgCl. Here,

[Cl^{-}]

from KCl is 0.1 M. Let the new solubility of AgCl in this solution be

s'

. Then,

[Ag^{+}] = s'

[Cl^{-}] = 0.1 + s' \approx 0.1

Since

s'

is much smaller than 0.1 M, we can approximate:

K_{sp} = [Ag^{+}] [Cl^{-}]

10^{-10} = s' \times 0.1

s' = \frac{10^{-10}}{0.1}

s' = 10^{-9}

Finally, we find the ratio of solubility in 0.1 M KCl to that in pure water:

\text{Ratio} = \frac{s'}{s} = \frac{10^{-9}}{10^{-5}} = 10^{-4}

Therefore, the correct answer is: Option A 10

^{-4}

Q7

An acidic buffer is prepared by mixing :

A weak acid and it's salt with strong base

B equal volumes of equimolar solutions of weak acid and weak base

C strong acid and it's salt with strong base

D strong acid and it's salt with weak base (The pK

_a

of acid = pK

_b

of the base)

Correct Answer

Option A

Solution

Acidic buffer is prepared by mixing weak acid and its salt with strong base.

Q8

0.01 M acetic acid solution is 1% ionised, then pH of this acetic acid solution is :

A 1

B 3

C 2

D 4

Correct Answer

Option D

Solution

For weak acid (i.e. CH 3 COOH)

[{H^ + }] = C\alpha

= 0.01 \times {1 \over {100}} = {10^{ - 4}}

M

pH = - \log {H^ + } = - \log {10^{ - 4}} = 4

Q9

The pH of the solution containing 50 mL each of 0.10 M sodium acetate and 0.01 M acetic acid is [Given pK a of CH 3 COOH = 4.57]

A 5.57

B 3.57

C 4.57

D 2.57

Correct Answer

Option A

Solution

To find the pH of the solution containing a mixture of sodium acetate and acetic acid, we can use the Henderson-Hasselbalch equation.

This equation is given by : $\text{pH} = \text{p}K_a + \log\left(\dfrac{[\text{Conjugate Base}]}{[\text{Acid}]}\right)$ Here, sodium acetate acts as the conjugate base (acetate ion, $CH_3COO^-$ ) and acetic acid (CH₃COOH) is the acid.

Given the $pK_a$ of acetic acid is 4.57, we can plug in the values.

The molarity of sodium acetate and acetic acid are given as 0.10 M and 0.01 M, respectively.

Since the volumes of the solutions are equal, the molarities can be directly used in the equation : $\text{pH} = 4.57 + \log\left(\dfrac{0.10}{0.01}\right)$ Let's calculate the pH : The pH of the solution containing 50 mL each of 0.10 M sodium acetate and 0.01 M acetic acid is approximately 5.57.

Therefore, the correct option is : Option A : 5.57

Q10

The pK b of dimethyl amine and pK a of acetic acid are 3.27 and 4.77 respectively at T (K). The correct option for the pH of dimethyl ammonium acetate solution is :

A 6.25

B 8.50

C 5.50

D 7.75

Correct Answer

Option D

Solution

Dimethylammonium acetate is a salt of weak acid and weak base whose pH can be calculated as pH = 7 +

{1 \over 2}

(pK a $-$ pK b ) pK a of acetic acid = 4.77 pK b of dimethyl amine = 3.27 pH = 7 +

{1 \over 2}

(4.77 $-$ 3.27) = 7.75