Problem 62 - A Complete Resource Book in Chemistry for JEE Main

Question

Which of the following statement is correct? 1\. The \(\mathrm{pH}\) of \(1.0 \times 10^{-8} \mathrm{M}\) solution of \(\mathrm{HCl}\) is 8 2\. The conjugate base of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\)is \(\mathrm{HPO}_{4}^{2-}\) 3\. Autoprotolysis constant of water increases with temperature 4\. When a solution of a weak monoprotic acid is titrated against a strong base at half neutralization point, \(\mathrm{pH}=(1 / 2) \mathrm{pKa}\). (a) 2,3 (b) \(1,2,3\) (c) 3,4 (d) \(2,3,4\)

Step-by-Step Solution

Answer

The correct option is (a) 2,3.

Step 1: Evaluate Statement 1

The concentration of a strong acid like HCl at such a low concentration (\(1.0 \times 10^{-8} \mathrm{M}\)) must be considered with the contribution of ionization from water itself. For pure water, \([H^+]\) is \(1.0 \times 10^{-7} \mathrm{M}\) and adding \(1.0 \times 10^{-8} \mathrm{M}\) of HCl will slightly increase the concentration but not enough to make the \(\mathrm{pH}\) 8. Therefore, statement 1 is incorrect.

Step 2: Evaluate Statement 2

The conjugate base of an acid is formed by removing one proton \((H^+)\) from the species. For \(\mathrm{H}_2 \mathrm{PO}_4^-\), removing one proton yields \(\mathrm{HPO}_4^{2-}\), making statement 2 correct.

Step 3: Evaluate Statement 3

The autoprotolysis constant of water \(K_w\) increases with temperature because the equilibrium constant of an endothermic reaction increases with temperature. Thus, statement 3 is correct.

Step 4: Evaluate Statement 4

At half-neutralization, \(\text{pH} = \text{pKa}\) for a weak acid being titrated by a strong base. Therefore, the relationship \(\text{pH} = (1/2)\text{pKa}\) is incorrect, hence statement 4 is incorrect.

Step 5: Choose the Correct Option

Based on the evaluation of the statements: Statement 2 is correct, Statement 3 is correct, while Statements 1 and 4 are incorrect. Therefore, the correct option is (a) 2,3.

Key Concepts

Autoprotolysis constant

The autoprotolysis constant, symbolized as \(K_w\), is a crucial concept in understanding water's behavior as both an acid and a base. In the context of water, autoprotolysis refers to the self-ionization process where water molecules donate and accept protons, forming hydronium \(H_3O^+\) and hydroxide \(OH^-\) ions. This process can be represented by the chemical equilibrium:

• \[ 2H_2O (l) \rightleftharpoons H_3O^+ (aq) + OH^- (aq) \]

The equilibrium constant for this reaction is the autoprotolysis constant, \(K_w\), which at 25°C has a value of \(1.0 \times 10^{-14}\). Since this is an endothermic reaction, \(K_w\) increases with temperature. This means that as temperature increases, more water molecules autoionize, increasing the concentrations of hydronium and hydroxide ions. Understanding \(K_w\) helps in calculations involving pH and the strength of acids and bases.

Conjugate acid-base pair

In the realm of acid-base chemistry, a conjugate acid-base pair consists of two entities that differ by one proton \((H^+)\). This concept is fundamental for identifying how substances behave in different chemical reactions. When an acid donates a proton, it forms its conjugate base, and when a base accepts a proton, it forms its conjugate acid.

For example, consider

• \(\mathrm{H}_2\mathrm{PO}_4^-\) and \(\mathrm{HPO}_4^{2-}\)

Here, \(\mathrm{H}_2\mathrm{PO}_4^-\) acts as an acid by donating a proton to become \(\mathrm{HPO}_4^{2-}\), its conjugate base. Conversely, \(\mathrm{HPO}_4^{2-}\) can accept a proton to revert to \(\mathrm{H}_2\mathrm{PO}_4^-\). Understanding conjugate acid-base pairs is vital for predicting the outcome of chemical reactions and their equilibrium positions.

pH calculation

Calculating pH is an essential skill in chemistry, reflecting the acidity or basicity of a solution. The pH is calculated using the formula:

• \[ \text{pH} = -\log [H^+] \]

This formula uses the concentration of hydrogen ions \([H^+]\) to find the pH. A lower pH indicates a more acidic solution, whereas a higher pH shows a more basic environment.

In many cases, especially for very dilute solutions, understanding water's contribution to \([H^+]\) can be pivotal. For a neutral solution at 25°C, \([H^+]\) is \(1.0 \times 10^{-7} \mathrm{M}\). However, the presence of additional acids or bases can alter this concentration. Special attention needs to be given for solutions with very low concentrations of strong acids, as the self-ionization of water may significantly contribute to the overall pH. Mastery of pH calculation is crucial for effective laboratory work and understanding chemical processes.