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 \(\left[\mathrm{Cl}^{-}\right]\)is \(\left(\mathrm{K}_{\mathrm{sp}}\right.\) of \(\mathrm{AgCl}=10^{-12}, \mathrm{~K}_{\mathrm{sp}}\) 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

The concentration of Cl\(^-\) is \(10^{-8} \text{ M}\).

Step 1: Understand the reaction

Determine the chemical reactions involved in the process. When AgOH is added to the NaCl solution, AgCl precipitate forms:\[ \mathrm{AgOH} + \mathrm{NaCl} \rightarrow \mathrm{AgCl} (s) + \mathrm{NaOH} \]AgOH dissolves weakly and contributes to the pH of the solution.

Step 2: Calculate the hydroxide ion concentration from pH

Given that the pH of the solution is 8, calculate the concentration of hydroxide ions, [OH\(^-\)].\[ \text{pH} = 8 \Rightarrow \text{pOH} = 14 - 8 = 6 \]\[ [\mathrm{OH}^-] = 10^{-6} \text{ M} \]

Step 3: Use the solubility product of AgOH

The solubility of AgOH can be represented by the following equilibrium expression:\[ \mathrm{AgOH} \leftrightarrows \mathrm{Ag}^+ + \mathrm{OH}^- \]Using the K\(_{sp}\) expression for AgOH:\[ K_{sp} = [\mathrm{Ag}^+][\mathrm{OH}^-] = 10^{-10} \]Substitute the [OH\(^-\)] value:\[ [\mathrm{Ag}^+] \cdot 10^{-6} = 10^{-10} \]\[ [\mathrm{Ag}^+] = 10^{-10} / 10^{-6} = 10^{-4} \text{ M} \]

Step 4: Apply the solubility product of AgCl

Now, consider the equilibrium with AgCl:\[ \mathrm{Ag}^+ + \mathrm{Cl}^- \leftrightarrows \mathrm{AgCl} (s) \]For AgCl, the K\(_{sp}\) expression is:\[ K_{sp} = [\mathrm{Ag}^+][\mathrm{Cl}^-] = 10^{-12} \]Plug in the [Ag\(^+\)] value:\[ 10^{-4} \cdot [\mathrm{Cl}^-] = 10^{-12} \]\[ [\mathrm{Cl}^-] = 10^{-12} / 10^{-4} = 10^{-8} \text{ M} \]

Key Concepts

Chemical Equilibrium

Chemical equilibrium refers to the state where the concentrations of reactants and products remain constant over time because the forward and reverse reactions occur at the same rate. For a precipitation reaction involving compounds like AgCl, it’s important to recognize the significance of the solubility product constant (K\(_{sp}\)).
K\(_{sp}\) is a specific type of equilibrium constant used for sparingly soluble salts. It helps us predict the concentration of ions in a saturated solution and determine if a precipitate will form when solutions are mixed.
When AgOH is added to NaCl, the system tries to maintain equilibrium by balancing the dissolution of AgOH and the formation of solid AgCl precipitate:

• The dynamic between the dissolved ions in the solution and the precipitate ensures that the product of the molar concentrations of the ions is equal to K\(_{sp}\).
• In this reaction, when [Ag\(^+\)] and [Cl\(^-\)] exceed their solubility product, the system shifts to form more solid precipitate, achieving equilibrium.

Understanding chemical equilibrium helps you anticipate changes when equilibrium conditions are disrupted, such as when additional reactants or products are introduced to the system.

pH Calculation

Calculating the pH of a solution helps determine its acidity or basicity and involves understanding the relationship between hydronium (H\(^+\)) and hydroxide ions (OH\(^-\)). The pH scale ranges from 0 to 14, with 7 being neutral, below 7 acidic, and above 7 basic.
In the context of the exercise, given that the pH is 8, the solution is weakly basic. To find the concentration of hydroxide ions, [OH\(^-\)], you first determine the pOH:

• The formula to find pOH is: pOH = 14 - pH.
• Therefore, when the pH is 8: pOH = 14 - 8 = 6.
• Convert pOH to [OH\(^-\)] using [OH\(^-\)] = 10\(^{-pOH}\).
• Thus, [OH\(^-\)] = 10\(^{-6}\) M.

The importance of pH in chemical equilibria lies in its ability to influence the solubility of compounds and the speciation of ions in solution.

Precipitation Reaction

A precipitation reaction occurs when two soluble salts react in solution to form an insoluble solid, or precipitate. In this exercise, the reaction between AgOH and NaCl results in the formation of AgCl, a well-known precipitate:

• The precipitate forms when the ion product exceeds the solubility product (K\(_{sp}\)) of the compound, inducing a shift in equilibrium to stabilize the system.
• This shift results in some of the dissolved ions forming a solid and leaving the solution, as observed with AgCl in this reaction.

The process can be summarized by the chemical equation: Ag\(^+\) + Cl\(^-\) \( \leftrightarrows \) AgCl\((s)\).
This equilibrium involves balancing the ionic components until a state is reached where precipitation no longer occurs, and the ionic concentrations satisfy the K\(_{sp}\) value.In practical applications, understanding how precipitation reactions work allows one to predict if a certain metal salt will precipitate under specific conditions, which is crucial in fields like chemical manufacturing and wastewater treatment.