Problem 264 - A Complete Resource Book in Chemistry for JEE Main

Question

Which of the following is not true? (a) The conjugate base of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\)is \(\mathrm{HPO}_{4}^{2-}\) (b) \(\mathrm{pH}+\mathrm{pOH}=14\) for all aqueous solutions (c) the \(\mathrm{pH}\) of \(1 \times 10^{-8} \mathrm{M} \mathrm{HCl}\) is 8 . (d) 96500 coulombs of electricity when passed through a \(\mathrm{CuSO}_{4}\) solution deposit 1 gram equivalent of copper at the cathode

Step-by-Step Solution

Answer

Option (c) is not true because the pH of 1x10^-8 M HCl is slightly less than 7, not 8.

Step 1: Understand the Concept of Conjugate Bases

To find the conjugate base of an acid, remove one hydrogen ion (\(\mathrm{H}^+\)) from the formula. For \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\), the conjugate base is \(\mathrm{HPO}_{4}^{2-}\). Thus, statement (a) is true.

Step 2: Verify the Relationship Between pH and pOH

In aqueous solutions at 25°C, the relationship \(\mathrm{pH} + \mathrm{pOH} = 14\) holds true. Therefore, statement (b) is also true for all neutral, acidic, and basic solutions at this temperature.

Step 3: Calculate pH of Given HCl Solution

A solution of 1 \(\times\) 10\(^{-8}\) M HCl is extremely diluted. Thus, the concentration of hydrogen ions due to water autoionization must be considered. Solving the quadratic \([\mathrm{H}^+]^2 + (1\times10^{-8})[\mathrm{H}^+] - 1.0\times10^{-14} = 0\) gives \([\mathrm{H}^+]\) slightly more than \(1\times10^{-7}\), resulting in a \(\mathrm{pH}\) less than 7. Hence, statement (c) is false.

Step 4: Analyze the Electrochemical Deposition of Copper

According to Faraday's laws of electrolysis, 96500 coulombs deposits one mole equivalent of an ionized metal. For copper (Cu), which has a molar mass of 63.5 g/mol and charge of +2, this law applies accurately, making statement (d) true.

Key Concepts

Conjugate Base

In chemistry, a conjugate base is the species that remains after an acid has donated a proton (hydrogen ion, \(\mathrm{H}^+\)). When an acid loses a \(\mathrm{H}^+\), it transforms into its conjugate base. For example, when \(\mathrm{H}_2\mathrm{PO}_4^-\) loses a hydrogen ion, it becomes \(\mathrm{HPO}_4^{2-}\). The process of removing a \(\mathrm{H}^+\) is essential to understanding acid-base reactions.
The strength of a conjugate base is inversely related to the strength of its acid. A strong acid will have a weak conjugate base because the acid completely dissociates in solution. Conversely, a weak acid will have a relatively stronger conjugate base. This interplay is crucial in predicting the behavior of acids and bases in chemical reactions.

pH and pOH Relationship

The relationship between pH and pOH is a fundamental concept in chemistry, particularly in understanding aqueous solutions. At 25°C, the sum of pH and pOH in a solution is always 14. This is because water, a neutral substance, autoionizes into equal concentrations of \(\mathrm{H}^+\) and \(\mathrm{OH}^-\) ions, maintaining a balance. The formula used to express this relationship is \(\mathrm{pH} + \mathrm{pOH} = 14\).
This relationship holds true under normal temperature and pressure conditions and is vital for calculating the hydrogen and hydroxide ion concentrations in solutions. Importantly, knowing either the pH or pOH of a solution allows you to calculate the other, which is particularly useful in lab settings where measuring equipment might be limited.

Faraday's Laws of Electrolysis

Faraday's laws of electrolysis are key to understanding chemical changes induced by electric currents. The laws describe how the amount of substance altered at an electrode during electrolysis is proportional to the electricity that flows through the circuit. Specifically, Faraday's first law states that the mass of substances deposited or dissolved at an electrode is directly proportional to the number of coulombs of electricity passed.
For example, in copper sulfate (\(\mathrm{CuSO}_4\)), passing 96500 coulombs (also known as one Faraday) results in the deposition of one mole of copper ions. Copper has a valency of +2, hence 63.5 grams (its molar mass divided by its charge) of copper will be deposited for every 96500 coulombs of charge. Faraday's principles are foundational in electroplating, electrorefining, and battery technology.

Autoionization of Water

Autoionization of water is a self-ionizing reaction where water molecules (\(\mathrm{H}_2\mathrm{O}\)) split into hydroxide ions (\(\mathrm{OH}^-\)) and hydronium ions (\(\mathrm{H}_3\mathrm{O}^+\)). Though these ions form only in very small, almost negligible amounts under neutral conditions, they are critical in maintaining the balance of pH and pOH in solutions.
This process is represented by the equilibrium equation: \( 2\mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{H}_3\mathrm{O}^+ + \mathrm{OH}^-\). Autoionization is central to understanding water’s properties as a solvent in chemistry. It also explains why even pure water carries a very low but existent level of conductivity.

• Autoionization is temperature dependent; equilibrium shifts slightly with temperature changes.
• This phenomenon plays a critical role in pH calculations, especially in extremely dilute solutions like \(1 \times 10^{-8}\,\mathrm{M}\,\mathrm{HCl}\).