Problem 63 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

For the reaction $$ \left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^{-} \rightleftharpoons \mathrm{Ag}^{+}+2 \mathrm{CN}^{-} $$ the equilibrium costant, at \(25^{\circ} \mathrm{C}\), is \(4.0 \times 10^{-19} .\) Calculate the silver ion concentration in a solution which was originally \(0.10\) molar in KCN and \(0.03\) molar in \(\mathrm{AgNO}_{3}\).

Step-by-Step Solution

Answer

The silver ion concentration at equilibrium is approximately 0.03 M.

Step 1: Understand the problem

We are given the equilibrium reaction \( [\mathrm{Ag}(\mathrm{CN})_{2}]^{-} \rightleftharpoons \mathrm{Ag}^{+} + 2 \mathrm{CN}^{-} \) with an equilibrium constant \( K = 4.0 \times 10^{-19} \). Initially, the solution is \(0.10\) M in KCN and \(0.03\) M in \(\mathrm{AgNO}_{3}\). We need to find the concentration of \(\mathrm{Ag}^{+}\) ions at equilibrium.

Step 2: Write the equilibrium expression

The equilibrium constant expression for the reaction is given by \( K = \frac{[\mathrm{Ag}^{+}][\mathrm{CN}^{-}]^2}{[\mathrm{Ag}(\mathrm{CN})_{2}]^{-}} \) and is equal to \(4.0 \times 10^{-19}\).

Step 3: Set initial concentrations

Initially, concentration for \([\mathrm{Ag}^{+}]\) is \(0.03\), for \([\mathrm{CN}^{-}]\) is \(0.10\), and we assume \([\mathrm{Ag}(\mathrm{CN})_{2}]^{-} = 0\) since it will form during dissociation.

Step 4: Define changes in concentration

Let \(x\) be the change in \([\mathrm{Ag}^{+}]\) concentration at equilibrium. The changes are: \([\mathrm{Ag}(\mathrm{CN})_{2}]^{-} = 0 + x\), \([\mathrm{Ag}^{+}] = 0.03 - x\), and \([\mathrm{CN}^{-}] = 0.10 - 2x\).

Step 5: Write expression using changes

At equilibrium, substituting the changes, we have: \( K = \frac{(0.03 - x)(0.10 - 2x)^2}{x} \).

Step 6: Assume small changes for simplification

Given the very small \(K\), \(x\) is very small compared to initial concentrations. Assume \(0.03 - x \approx 0.03\) and \(0.10 - 2x \approx 0.10\).

Step 7: Solve for x

Rearrange equilibrium expression: \( 4.0 \times 10^{-19} = \frac{0.03 \times (0.10)^2}{x} \). Solving, \(x = \frac{0.03 \times 0.01}{4.0 \times 10^{-19}} = 7.5 \times 10^{-18}\).

Step 8: Determine silver ion concentration

Therefore, the concentration of \(\mathrm{Ag}^{+}\) ions at equilibrium is approximately \(0.03 - 7.5 \times 10^{-18} \approx 0.03\).

Key Concepts

Equilibrium constant

In the context of chemical reactions, the equilibrium constant, denoted as \( K \), is a crucial numerical value that helps us understand the balance in a reversible chemical reaction at a specific temperature. It reflects the ratio of the concentration of products to reactants when the reaction is at equilibrium. This constant is vital for predicting how a reaction behaves and changes under different conditions. Consider a general chemical reaction:- \( aA + bB \rightleftharpoons cC + dD \)The equilibrium constant \( K \) is expressed as:\[ K = \frac{[C]^c[D]^d}{[A]^a[B]^b} \]Here, square brackets denote the concentration of each species in the equilibrium state, and the coefficients \( a, b, c, \) and \( d \) represent the stoichiometric coefficients of the reactants and products in the balanced chemical equation.Understanding the equilibrium constant reveals:- **If \( K \) is large**: The products are favored at equilibrium.- **If \( K \) is small**: The reactants are predominately at equilibrium.- **If \( K \) equals 1**: Neither reactants nor products are favored—it is a perfect balance.The value of \( K \) is dependent on temperature but not on the concentrations of the reactants and products taken at a non-equilibrium state. This constant helps predict the position of equilibrium and the extent of reaction progress.

Silver ion concentration

Silver ion concentration in a chemical reaction provides insight into the number of \( \text{Ag}^+ \) ions present in the solution at a given time, particularly at equilibrium. This concentration is critical when studying reactions that involve silver compounds. It helps in understanding how much free silver ion is available, which can influence the reaction dynamics and the formation of complexes.In our reaction context:- Initially, the solution contains \( 0.03 \text{ M} \) of \( \text{AgNO}_3 \), meaning the initial concentration of \( \text{Ag}^+ \) is \( 0.03 \text{ M} \).- During the reaction, some \( \text{Ag}^+ \) ions may form complexes with cyanide ions (\( \text{CN}^- \)).To find the equilibrium concentration of silver ions, we use the changes in concentrations defined with a variable \( x \). Here:- The change in \( \text{Ag}^+ \) ion concentration is characterized by \( 0.03 - x \), where \( x \) represents the small amount of silver ion that reacts to form a complex.Thus, calculating this concentration at equilibrium involves:- Identifying initial concentrations.- Estimating the shifts in concentration due to complex formation.Understanding this concept allows us to properly manage reactions involving silver ions, especially in analytical chemistry and industrial processes where silver ion presence is crucial.

Le Chatelier's Principle

Le Chatelier's Principle is a fundamental concept in chemistry that explains how a dynamic equilibrium reacts to disturbances. When a system at equilibrium experiences a change—due to concentration, temperature, or pressure—it will adjust in a manner that counteracts the effect of the applied change, striving to re-establish equilibrium.Here's how Le Chatelier's Principle applies:- **Concentration changes**: If more reactants are added, the reaction will shift toward the products to reduce the added reactant concentration.- **Pressure changes**: Applicable for gaseous reactions, if the pressure is increased, the system will favor the side with fewer gas molecules.- **Temperature changes**: For exothermic reactions, increasing the temperature shifts the equilibrium to favor the reactants, while decreasing temperature favors the products.Le Chatelier's Principle helps in:- Predicting the direction of shift when the system is disturbed.- Understanding the factors affecting the balance in a chemical system.It is critical in industrial processes where optimizing yields of desired products is essential. In the context of silver ion concentration, manipulating factors such as reactant concentration can shift the equilibrium to either increase or decrease the formation of \( \text{Ag}^+ \) ions, demonstrating the principle's practical application.