Problem 173 - A Complete Resource Book in Chemistry for JEE Main

Question

Solid \(\mathrm{AgNO}_{3}\) is added slowly to a buffer solution of \(\mathrm{pH}\) \(=10\) to precipitate \(\mathrm{AgOH}\). The \(\left[\mathrm{Ag}^{+}\right]\)concentration in the solution is \(\left[\mathrm{K}_{s}(\mathrm{AgOH})=10^{-10}\right]\) (a) \(10^{-4} \mathrm{M}\) (b) \(10^{-5} \mathrm{M}\) (c) \(10^{-6} \mathrm{M}\) (d) \(10^{-7} \mathrm{M}\)

Step-by-Step Solution

Answer

The correct concentration is \(10^{-6} \text{ M}\). Answer: (c)

Step 1: Understand the Reaction

The problem involves the precipitation of silver hydroxide, \( \text{AgOH} \), from the dissolution of solid silver nitrate \( \text{AgNO}_3 \). In a basic solution (buffer with \( \text{pH} = 10 \)), silver ions \( \text{Ag}^+ \) react with hydroxide ions \( \text{OH}^- \) to form \( \text{AgOH} \).

Step 2: Determine the \([\text{OH}^-]\) Concentration

The \(\text{pH} = 10\) gives us the \(\text{pOH} = 14 - \text{pH} = 4\). The concentration of hydroxide ions is then \([\text{OH}^-] = 10^{-\text{pOH}} = 10^{-4}\, M\).

Step 3: Apply the Solubility Product Constant, \(K_s\)

For \(\text{AgOH}\), the solubility product constant is \(K_s = [\text{Ag}^+][\text{OH}^-] = 10^{-10}\).

Step 4: Solve for \([\text{Ag}^+]\) Concentration

Using the \(K_s\) expression, substitute \([\text{OH}^-] = 10^{-4}\, M\) into \(K_s = [\text{Ag}^+][\text{OH}^-]\):\[10^{-10} = [\text{Ag}^+](10^{-4})\]Solve for \([\text{Ag}^+]\):\[[\text{Ag}^+] = \frac{10^{-10}}{10^{-4}} = 10^{-6}\, M\]

Step 5: Select the Correct Answer

The calculated silver ion concentration, \( [\text{Ag}^+] = 10^{-6} \text{ M} \), matches option (c).

Key Concepts

Buffer solution

A buffer solution is integral in this exercise. It is a special type of solution that resists changes in its pH when small amounts of acid or base are added to it.
Buffers usually consist of a weak acid and its conjugate base or a weak base and its conjugate acid.
In this exercise, the buffer has a pH of 10, which means it is a slightly basic solution. Buffers achieve their pH-stabilizing ability through equilibria. When an acid is added to the buffer, the weak base component neutralizes it. Conversely, if a base is added, the weak acid component takes over to neutralize it. This efficient neutralization mechanism is key, especially in reactions involving pH-sensitive processes like precipitation. For practical understanding:

• Buffers are used widely in biological and chemical applications where a stable pH is critical.
• In this problem, the buffer's role is to maintain a consistent pH of 10, allowing the calculation of hydroxide ion concentration ([OH^-]) reliably.

pH calculation

Knowing how to calculate pH and related parameters is crucial for understanding chemical equilibria.
At pH 10, you can calculate the pOH using the formula: \[\text{pOH} = 14 - \text{pH}\]In this exercise, the pH is given as 10, hence:\[\text{pOH} = 14 - 10 = 4\]Next, we find the hydroxide ion concentration \(\,[\text{OH}^-]\) using the formula:\[[\text{OH}^-] = 10^{-\text{pOH}}\]Substituting the given pOH:\[[\text{OH}^-] = 10^{-4}\,\]This calculation provides the necessary information to proceed with the solubility calculations. The interconnection between pH, pOH, and ion concentrations is fundamental in chemistry and helps to predict how alterations in pH influence reactions.

Silver hydroxide precipitation

The core reaction in this exercise involves the precipitation of silver hydroxide, \(\text{AgOH}\).
When solutions of silver ions \(\,(\text{Ag}^+)\) and hydroxide ions \(\,(\text{OH}^-)\) combine, they form a precipitate of silver hydroxide when their concentrations exceed the solubility product constant, \(K_s\).In this problem, the solubility product \(K_s\) is given as \(10^{-10}\).This constant is used to predict whether a precipitate will form:\[K_s = [\text{Ag}^+][\text{OH}^-]\]By rearranging this equation, you can solve for the concentration of silver ions needed to reach the solubility limit:\[[\text{Ag}^+] = \frac{K_s}{[\text{OH}^-]} = \frac{10^{-10}}{10^{-4}} = 10^{-6}\,M\]When the concentration of \,[\text{Ag}^+]\ is lowered below \(10^{-6}\,M\), \(\text{AgOH}\) will not precipitate. Above this concentration, a solid starts forming. Understanding precipitation and solubility concepts provides insight into predicting which salts will form solids in solution under specific conditions.