Problem 70 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

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

Step-by-Step Solution

Answer

Only statements (b) and (c) are correct.

Step 1: Analyzing statement (a)

For statement (a), calculate the pH of a \(1.0 \times 10^{-8} \mathrm{M}\) \(\mathrm{HCl}\) solution. Since the concentration is very low, we must consider the ionization of water. The ionic product of water \(\mathrm{K}_w\) is \(1.0 \times 10^{-14}\). In neutral water, \([\mathrm{H}^+] = [\mathrm{OH}^-] = 1.0 \times 10^{-7} \mathrm{M}\). Hence, the actual \([\mathrm{H}^+]\) will be \(1.0 \times 10^{-7} \mathrm{M} + 1.0 \times 10^{-8} \mathrm{M}\). Calculate the pH now: \[\mathrm{pH} = -\log_{10}(1.0 \times 10^{-7} + 1.0 \times 10^{-8})\] The pH is actually slightly less than 7, so this statement is incorrect.

Step 2: Evaluating statement (b)

The conjugate base of an acid is formed by removing a proton \((\mathrm{H}^+)\) from the acid. \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) is the given acid, so its conjugate base will be \(\mathrm{HPO}_{4}^{2-}\). Since removing a proton reduces the charge by 1, this is correct.

Step 3: Investigating statement (c)

Consider the effect of temperature on the autoprotolysis constant \(\mathrm{K}_w\) of water. The reaction \(2 \mathrm{H}_2\mathrm{O} \leftrightarrow \mathrm{H}_3\mathrm{O}^+ + \mathrm{OH}^-\) is endothermic. Raising the temperature shifts the equilibrium to produce more \(\mathrm{H}_3\mathrm{O}^+\) and \(\mathrm{OH}^-\), therefore increasing \(\mathrm{K}_w\). Thus, this statement is correct.

Step 4: Analyzing statement (d)

At the half-neutralization point in a titration of a weak monoprotic acid with a strong base, the concentration of the acid is equal to the concentration of its conjugate base. Hence, \([\mathrm{HA}] = [\mathrm{A}^-]\), leading to the equation \(\mathrm{pH} = \mathrm{pK}_a\), not \(\mathrm{pH} = (1/2)\mathrm{pK}_a\). This makes statement (d) incorrect.

Key Concepts

pH calculation

Calculating the pH of a solution involves understanding the concentration of hydrogen ions \(\mathrm{H}^+\) in the solution. For most dilute strong acids like hydrochloric acid \(\mathrm{HCl}\), the pH calculation considers the ionization of water as well. In this exercise, the concentration of \(\mathrm{HCl}\) was given as \(1.0 \times 10^{-8} \mathrm{M}\). This small concentration is actually less than that of the hydrogen ions naturally found in pure water \(\mathrm{H}_2\mathrm{O}\).
Neutral water has a concentration of hydrogen ions (\