Problem 179 - A Complete Resource Book in Chemistry for JEE Main

Question

\(\mathrm{AgOH}\) is added to \(\mathrm{NaCl}\) solution to form \(\mathrm{AgCl}\) precipitate. After the precipitation, the \(\mathrm{pH}\) of the solution is 8 . The \([\mathrm{Cl}]\) is \(\left(\mathrm{K}_{\mathrm{sp}}\right.\) of \(\mathrm{AgCl}=10^{-12}, \mathrm{~K}_{\$}\) of \(\left.\mathrm{AgOH}=10^{-10}\right)\) (a) \(10^{-6} \mathrm{M}\) (b) \(10^{-4} \mathrm{M}\) (c) \(10^{-8} \mathrm{M}\) (d) \(10^{-10} \mathrm{M}\)

Step-by-Step Solution

Answer

(c) \(10^{-8} \mathrm{M}\)

Step 1: Understand the Reaction

The given scenario involves the addition of \( \mathrm{AgOH} \) to a \( \mathrm{NaCl} \) solution, resulting in the formation of \( \mathrm{AgCl} \) precipitate. \( \mathrm{AgOH} \) partially dissolves in water, while \( \mathrm{NaCl} \) dissociates completely.

Step 2: Determine Hydroxide Ion Concentration

Given that the final pH of the solution is 8, we can find the \( [\mathrm{OH}^-] \) using: \( \mathrm{pH} + \mathrm{pOH} = 14 \). Thus, \( \mathrm{pOH} = 14 - 8 = 6 \), so \( [\mathrm{OH}^-] = 10^{-6} \mathrm{M} \).

Step 3: Use Solubility Product of AgOH

Using the solubility product constant \( \mathrm{K}_{\mathrm{sp}} = 10^{-10} \) for \( \mathrm{AgOH} \), we can express \( \mathrm{K}_{\mathrm{sp}} \) as \([\mathrm{Ag}^+][\mathrm{OH}^-] = 10^{-10} \). Since \( [\mathrm{OH}^-] = 10^{-6} \), \( [\mathrm{Ag}^+] = \frac{10^{-10}}{10^{-6}} = 10^{-4} \mathrm{M} \).

Step 4: Calculate Chloride Ion Concentration after Precipitation

From the solubility product constant \( \mathrm{K}_{\mathrm{sp}} = 10^{-12} \) for \( \mathrm{AgCl} \), where \( [\mathrm{Ag}^+][\mathrm{Cl}^-] = 10^{-12} \), solve for \( [\mathrm{Cl}^-] \) using \( [\mathrm{Ag}^+] = 10^{-4} \mathrm{M} \). Thus, \( [\mathrm{Cl}^-] = \frac{10^{-12}}{10^{-4}} = 10^{-8} \mathrm{M} \).

Key Concepts

Solubility Product Constant

The solubility product constant, often abbreviated as \( K_{sp} \), is a measure of the solubility of a compound in water. It tells us how much of the compound will dissolve in water before it starts to precipitate. For example, in the case of \( \mathrm{AgOH} \), the solubility product constant \( K_{sp} \) is given as \( 10^{-10} \). This means that at equilibrium, the product of the concentrations of the dissolved ions \([\mathrm{Ag}^+] \) and \([\mathrm{OH}^-] \) equals \( 10^{-10} \).
\[ K_{sp} = [\mathrm{Ag}^+][\mathrm{OH}^-] \]
Understanding \( K_{sp} \) is critical when predicting whether a precipitate will form in a given solution. A low \( K_{sp} \) value indicates that only a small amount of the compound will dissolve.

• Higher \( K_{sp} \) = higher solubility
• Lower \( K_{sp} \) = lower solubility

In our scenario, we calculate \([\mathrm{Ag}^+] \) when \( [\mathrm{OH}^-] = 10^{-6} \) M, resulting in \([\mathrm{Ag}^+] = 10^{-4} \) M, using the expression \( \frac{K_{sp}}{[\mathrm{OH}^-]} = 10^{-4} \) M.

Precipitation Reactions

Precipitation reactions occur when ions in solution combine to form a solid. These reactions often lead to vibrant changes as the solid precipitate separates from the aqueous solution. A common visual cue is that the solution turns cloudy or forms clumps.
For example, adding \( \mathrm{AgOH} \) to a \( \mathrm{NaCl} \) solution creates \( \mathrm{AgCl} \) as a precipitate:
\[ \mathrm{Ag}^+ + \mathrm{Cl}^- \rightarrow \mathrm{AgCl(s)} \]
This is a classic precipitation reaction where two soluble salts combine and an insoluble salt, in this case, \( \mathrm{AgCl} \) forms. Precipitation reactions are dictated by the solubility rules and the \( K_{sp} \) of the substances involved. Here, \( K_{sp} = 10^{-12} \) for \( \mathrm{AgCl} \), driving the precipitation process.
Predicting such reactions involves calculating whether the ionic product exceeds the \( K_{sp} \). If it does, a precipitate will form. In this case, \([\mathrm{Cl}^-] \) is \( 10^{-8} \) M after the reaction.

pH and pOH Calculations

The \( \text{pH} \) and \( \text{pOH} \) are important measures that describe the acidity and basicity of a solution. They are connected by a simple equation:
\( \text{pH} + \text{pOH} = 14 \).
In essence, if you know one, you can easily find the other. The pH measures the concentration of hydronium ions \( [\mathrm{H}^+] \), while the pOH measures the concentration of hydroxide ions \( [\mathrm{OH}^-] \).
For a solution with \( \text{pH} = 8 \):
\[ \text{pOH} = 14 - 8 = 6 \]
This tells us that \( [\mathrm{OH}^-] = 10^{-6} \) M, indicating a basic solution. Accurate pH and pOH calculations are essential when predicting the type of precipitates and concentration of ions in solution.

• Higher pH = more basic
• Lower pH = more acidic

These calculations guide us in determining the behavior and outcome of reactions in the field of chemistry.