Problem 122 - A Complete Resource Book in Chemistry for JEE Main

Question

The acid ionization (hydrolysis) constant of \(\mathrm{Zn}^{2+}\) is \(1.0\) \(\times 10^{-9} .\) Which of the following statements are correct? (i) the basic dissociation constant of \(\mathrm{Zn}(\mathrm{OH})^{+}\)is \(1.0 \times 10^{5}\) (ii) the \(\mathrm{pH}\) of \(0.001 \mathrm{M} \mathrm{ZnCl}_{2}\) solution is 6 (iii) the basic dissociation constant of \(\mathrm{Zn}(\mathrm{OH})^{+}\)is \(1.0 \times 10^{-5}\) (iv) the \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\)ion concentration in \(0.001 \mathrm{M} \mathrm{ZnCl}_{2}\) solution is \(1.0 \times 10^{-4} .\) (a) 1,2 (b) 2,3 (c) 3,4 (d) \(1,2,4\)

Step-by-Step Solution

Answer

Correct statement: (iii). Option (c) 3 is correct.

Step 1: Understand the Relation Between Acid and Basic Dissociation Constants

The relationship between the acid ionization constant \( K_a \) and the conjugate base dissociation constant \( K_b \) for a given species is given by the equation \( K_a \times K_b = K_w \), where \( K_w = 1.0 \times 10^{-14} \) at 25°C. For \( \text{Zn}^{2+} \), the given \( K_a = 1.0 \times 10^{-9} \). This formula will be used to find \( K_b \) for \( \text{Zn(OH)}^+ \).

Step 2: Calculate the Basic Dissociation Constant \( K_b \) for Zn(OH)^+

Using the formula from Step 1, we calculate \( K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{1.0 \times 10^{-9}} = 1.0 \times 10^{-5} \). This matches statement (iii).

Step 3: Assess Statement (i) for Correctness

Statement (i) posits that the basic dissociation constant of \( \mathrm{Zn(OH)}^+ \) is \( 1.0 \times 10^5 \). We have calculated it to be \( 1.0 \times 10^{-5} \). Hence, statement (i) is incorrect.

Step 4: Assess Statement (ii) Regarding pH

To check the pH of a \( 0.001 \text{ M } \mathrm{ZnCl}_2 \) solution, we calculate the concentration of \( \text{H}_3\text{O}^+ \) from \( \mathrm{Zn}^{2+} \) hydrolysis. Generally, \( \text{pH} = \log\left(\sqrt{K_a \times c}\right) \), with \( c = 0.001 \). Calculate \( \text{pH} = 7 + \frac{1}{2} \log\left(\frac{K_a}{c}\right) = 7 + \frac{1}{2} \log\left(\frac{1.0 \times 10^{-9}}{0.001}\right) = 5 \). Statement (ii) is incorrect.

Step 5: Assess Statement (iii) Again

We previously concluded that the basic dissociation constant of \( \text{Zn(OH)}^+ \) is \( 1.0 \times 10^{-5} \). Statement (iii) is correct.

Step 6: Assess Statement (iv) Concerning \( [\text{H}_3\text{O}^+] \) ion Concentration

If \( \text{pH} \) is 5 (from Step 4), then \( [\text{H}_3\text{O}^+] = 10^{-5} \), not \( 1.0 \times 10^{-4} \) as stated in (iv). Statement (iv) is incorrect.

Step 7: Final Decision Based on Correct Statements

Given the above analysis, statement (iii) is correct. Thus, the only correct option is (c) 3.

Key Concepts

Acid Ionization Constant

The **acid ionization constant** (\(K_a\)) is critical in understanding how an acid dissociates in water. It measures the strength of the acid's ionization into its ions. Higher \(K_a\) values imply stronger acids that dissociate more completely. For weak acids, the \(K_a\) value is typically a small number, indicating partial dissociation. In the exercise, the given \(K_a\) for \(\text{Zn}^{2+}\) is \(1.0 \times 10^{-9}\). This low number suggests that \(\text{Zn}^{2+}\) is a weak acid, dissociating only slightly in solution. Understanding \(K_a\) helps predict reaction outcomes and the strength of the acidity. The relationship between \(K_a\) and the basic dissociation constant \(K_b\) is fundamental in these calculations.

Basic Dissociation Constant

The **basic dissociation constant** (\(K_b\)) represents a base's ability to dissociate in water, forming hydroxide ions (\(\text{OH}^-\)). It quantifies the strength of a base, similar to \(K_a\) for acids. A higher \(K_b\) indicates a stronger base that promotes greater ionization.

• Link with \(K_a\): Given by the relationship \(K_a \times K_b = K_w = 1.0 \times 10^{-14}\)
• In the context of \(\text{Zn(OH)}^+\), we calculate \(K_b = \frac{1.0 \times 10^{-14}}{1.0 \times 10^{-9}} = 1.0 \times 10^{-5}\)

This calculation shows that \(\text{Zn(OH)}^+\) is a relatively weak base, emphasizing the need to correctly determine this constant when predicting chemical behavior.

pH Calculation

**pH Calculation** allows us to understand the acidity or alkalinity of a solution. pH is a logarithmic scale used to specify the acidity or basicity of an aqueous solution. It can be calculated with the formula:
\[pH = -\log[H_3O^+]\]For solutions of salts like \(\text{ZnCl}_2\), hydrolysis affects the \([H_3O^+]\) concentration. The formula for calculating pH from hydrolysis of cations is:
\[pH = 7 + \frac{1}{2}\log\left(\frac{K_a}{c}\right)\]Where:

• \(c = 0.001\text{ M}\) — the concentration of \(\text{ZnCl}_2\)
• \(K_a = 1.0 \times 10^{-9}\)

The pH calculation yields a value of 5, suggesting a slightly acidic solution. Understanding this helps in predicting reactions in different pH environments.

Conjugate Base

When an acid loses a proton, it forms its **conjugate base**. This base can potentially reabsorb a proton to reform the original acid. The strength of a conjugate base is inversely related to the acid strength:

• A strong acid forms a weak conjugate base.
• A weak acid forms a relatively stronger conjugate base.

In the exercise, \(\text{Zn}^{2+}\) acts as an acid, and its conjugate base is \(\text{Zn(OH)}^+\). Recognizing conjugate pairs and their properties aids in equation balancing and predicting reaction shifts.

Hydrolysis of Cations

The **hydrolysis of cations** refers to the reaction of cations with water, often leading to a change in pH. Cations like \(\text{Zn}^{2+}\) can undergo hydrolysis, affecting the acidity of a solution. This occurs when the water molecule donates a proton to form hydronium ions:

• Cations from weak bases or amphoteric elements undergo hydrolysis.
• This results in an increased \([H_3O^+]\) concentration, lowering the pH.

For \(0.001 \text{ M ZnCl}_2\), the hydrolysis led to a calculated pH of 5, emphasizing the significance of this process in influencing solution properties.