Question

A \(1 L\) solution contains \(0.2 \mathrm{M} \mathrm{NH}_{4} \mathrm{OH}\) and \(0.2 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\). If \(1.0 \mathrm{~mL}\) of \(0.001 \mathrm{M} \mathrm{HCl}\) is added to it. What will be the \(\left[\mathrm{OH}^{-}\right]\) of the resulting solution \(\left(K_{b}=2 \times 10^{-5}\right)\) (a) \(2 \times 10^{-5}\) (b) \(5 \times 10^{-10}\) (c) \(2 \times 10^{-3}\) (d) None of these

Short answer

The concentration of hydroxide ions \[OH^-\rbrack in the resulting solution will be \(2 \times 10^{-3} M\), which corresponds to option (c).

Step-by-step solution

Calculate moles of HCl added

First, calculate the moles of HCl that are added to the solution. This can be done by multiplying the concentration of HCl by its volume in liters. For a 0.001 M HCl solution, this is 0.001 M * 0.001 L = 0.000001 moles (1.0e-6 moles).

Determine the acid-base reaction

The HCl will react with NH4OH to form NH4+ and Cl- ions and water. This reaction can be represented by the equation: NH4OH + HCl -> NH4Cl + H2O. Since both HCl and NH4OH are present in equal molar amounts, NH4OH will get neutralized by HCl added.

Calculate the remaining concentration of NH4OH

Since 1.0e-6 moles of HCl neutralizes the same amount of NH4OH, subtract this from the original NH4OH moles. However, since the original concentration of NH4OH is 0.2 M, even after the neutralization by 1.0e-6 moles of HCl, the change in concentration is negligible. The concentration of NH4OH essentially remains at 0.2 M.

Use the Kb expression to find [OH-]

The Kb expression for NH4OH is: Kb = [NH4+][OH-]/[NH4OH]. Since the concentration of NH4OH remains constant at 0.2 M, and because the system maintains equilibrium, we can use the Kb value to find [OH-]. Kb = 2e-5 = (x)(x)/(0.2 - x), where x = [OH-]. Assuming x << 0.2, the equation simplifies to 2e-5 = x^2 / 0.2. Solving for x gives [OH-] = sqrt(2e-5 * 0.2).

Calculate the hydroxide ion concentration

Finally, calculate the concentration of [OH-] by taking the square root of the product of Kb and the NH4OH concentration: [OH-] = sqrt(2e-5 * 0.2) = sqrt(4e-6). This gives [OH-] = 2e-3 M.

Key concepts

Acid-Base Equilibria

Understanding acid-base equilibria is essential to comprehend how acids and bases interact in solution. In the context of the given exercise, when an acid like HCl is added to a basic solution containing NH4OH, a neutralization reaction occurs. The relevant equilibrium expression that governs this process is known as the acid dissociation constant (Ka) for acids or the base dissociation constant (Kb) for bases.

These constants essentially tell us how much an acid or a base will dissociate in water to form ions. The reaction proceeds until the system reaches a state of equilibrium, where the rates of the forward reaction (acid reacting with base) and the reverse reaction (formation of acid and base from ions) are equal. This concept is a cornerstone of physical chemistry and is particularly important for problems on the JEE pertaining to acid-base reactions.

Ionic Equilibrium

Ionic equilibrium refers to the balance between the ions in a solution. When a salt like NH4Cl dissolves in water, it dissociates into NH4+ and Cl- ions. If the salt is formed from a weak acid or base, as the NH4OH in our exercise, we also have to consider the presence of weak ions that can further react with water, either accepting or donating a proton in a process called hydrolysis (which we'll explore in the next section).

The equilibrium constant for this hydrolysis (either Ka or Kb) will signify the extent to which this reaction takes place. Ionic equilibrium is a central concept in understanding how salts and their corresponding ions behave in aqueous solutions and plays a significant role in calculating pH, solubility, and ionic strength in solutions—key parameters in many physical chemistry problems, including those presented in JEE examinations.

Hydrolysis of Salts

Hydrolysis of salts is the process by which an anion or a cation of a salt reacts with water to produce either hydroxide (OH-) or hydronium (H3O+) ions, affecting the pH of the solution. The salt NH4Cl, which is formed in the neutralization reaction of HCl with NH4OH, can undergo hydrolysis in water to produce NH4+ and Cl- ions. NH4+ is the conjugate acid of the weak base NH4OH, and it will react with water in a hydrolysis reaction.

Given that NH4OH is a weak base, it has a small Kb, which implies limited dissociation in water. The exercise illuminates this concept clearly: even with the addition of a small amount of HCl, the concentration of NH4OH remains nearly unchanged. This minimal change is due to NH4OH’s weak nature and the limited scope of hydrolysis under the given conditions, providing a perfect example of hydrolysis of salts for JEE-level physical chemistry problems.