Ionic Equilibrium

$ \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} $

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

$ \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} $

Colour of Phenolphthalein before the end point = colourless.

Colour of Phenolphthalein close to equivalence point = Pink.

∴ Colour change $=$ Colourless to pink.

To determine the degree of dissociation (α) of a weak acid, we use the formula:

$ \alpha = \frac{\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 = \frac{90}{400} = 0.225 $

Therefore, the degree of dissociation of the acid is 0.225.

Phosphoric acid ($\text{H}_3\text{PO}_4$) ionizes in three distinct steps, each characterized by an ionization constant: $K_{a_1}$, $K_{a_2}$, and $K_{a_3}$. Additionally, there is an overall ionization constant denoted as $K$. Here is a breakdown of the key points:

Strength of Acidity: $\text{H}_3\text{PO}_4$ is a stronger acid compared to its subsequent ionized forms, $\text{H}_2\text{PO}_4^-$ and $\text{HPO}_4^{2-}$.

Ionization Constants:

$ \begin{array}{ll} \text{H}_3\text{PO}_4(\text{aq}) \rightleftharpoons \text{H}^+(\text{aq}) + \text{H}_2\text{PO}_4^-(\text{aq}) & K_{a_1} = 7.5 \times 10^{-3} \\ \text{H}_2\text{PO}_4^-(\text{aq}) \rightleftharpoons \text{H}^+(\text{aq}) + \text{HPO}_4^{2-}(\text{aq}) & K_{a_2} = 6.2 \times 10^{-8} \\ \text{HPO}_4^{2-}(\text{aq}) \rightleftharpoons \text{H}^+(\text{aq}) + \text{PO}_4^{3-}(\text{aq}) & K_{a_3} = 1.7 \times 10^{-12} \end{array} $

Relationship between Constants:

The relationship among the ionization constants follows the order: $K_{a_1} > K_{a_2} > K_{a_3}$.

The logarithmic expression of the overall ionization constant is given as:

$ \log K = \log K_{a_1} + \log K_{a_2} + \log K_{a_3} $

This information aligns with the conclusion that statements A, B, and C are accurate regarding the behavior of phosphoric acid and its ionization constants.

The correct answer is Option C: Phenolphthalein, colourless to pink. Here's why:

Titration of Sodium Hydroxide (NaOH) against Oxalic Acid (H₂C₂O₄)

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 (Na₂C₂O₄), 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 KMnO₄ 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.

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$^{-4}$

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

For weak acid (i.e. CH₃COOH)

$[{H^ + }] = C\alpha $

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

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

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(\frac{[\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(\frac{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

The $\mathrm{pH}$ of a salt solution formed from a weak acid and a weak base can be determined using the relationship between their dissociation constants. Specifically, the formula for the $\mathrm{pH}$ of the solution is given by :

$ \mathrm{pH} = \frac{1}{2} (\mathrm{p} K_w + \mathrm{p} K_a - \mathrm{p} K_b) $

where :

$\mathrm{p} K_w$ is the ionization constant of water, which is 14 at $25^{\circ} \mathrm{C}$,

$\mathrm{p} K_a$ is the acid dissociation constant, and

$\mathrm{p} K_b$ is the base dissociation constant.

Given that :

$\mathrm{p} K_a = 4$,

$\mathrm{p} K_b = 5$,

substitute these values into the formula :

$ \mathrm{pH} = \frac{1}{2} (14 + 4 - 5) $

Simplify the expression inside the parentheses :

$ \mathrm{pH} = \frac{1}{2} (14 + 4 - 5) = \frac{1}{2} (13) = 6.5 $

Thus, the $\mathrm{pH}$ of the salt solution at $25^{\circ} \mathrm{C}$ is:

$ \boxed{6.5} $

$A X_2$ is ionised as follows:

$\underset{S \mathrm{~mol} \mathrm{~L}^{-1}}{A X_2} \rightleftharpoons \underset{S}{A^{2+}}+\underset{2 S}{2 X^{-}}$

Solubility product of $A X_2$

$K_{\text {sp }}=\left[A^{2+}\right]\left[X^{-}\right]^2=[S] \times[2 S]^2=4 S^3$

$\begin{aligned} & \because K_{\text {sp }} \text { of } A X_2=3.2 \times 10^{-11} \\ & \therefore \quad 3.2 \times 10^{-11}=4 S^3 \end{aligned}$

or, $\quad S^3=0.8 \times 10^{-11}=8 \times 10^{-12}$

or, $\quad S=\sqrt[3]{8 \times 10^{-12}}$

$\begin{aligned} & =2 \times 10^{-4} \mathrm{~mol} / \mathrm{L} \\ \text { Solubility } & =2 \times 10^{-4} \mathrm{~mol} / \mathrm{L} \end{aligned}$