Problem 62 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

62\. 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 concentration of silver ions \([\mathrm{Ag}^+]\) is \(4.0 \times 10^{-19}\) M.

Step 1: Write the Expression for the Equilibrium Constant

For the given reaction \(\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^{-} \rightleftharpoons \mathrm{Ag}^{+}+2 \mathrm{CN}^{-}\), the expression for the equilibrium constant \(K_{eq}\) can be written as:\[\K_{eq} = \frac{\left[\mathrm{Ag}^{+}\right]\left[\mathrm{CN}^{-}\right]^2}{\left[\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^{-}\right]}\]Given that \(K_{eq} = 4.0 \times 10^{-19}\).

Step 2: Determine Initial Concentrations and Changes

Initially, the concentrations are \([\mathrm{AgNO}_3] = 0.03\, \text{M}\) and \([\mathrm{KCN}] = 0.10\, \text{M}\). Upon dissociation, cyanide ions \(\mathrm{CN}^{-}\) are present from \(\mathrm{KCN}\), so they start at \(0.10\, \text{M}\). Let \(x\) be the change in concentration for \([\mathrm{Ag}^{+}]\), leading to equilibrium concentrations of \([\mathrm{Ag}^{+}] = x\) and \([\mathrm{CN}^{-}] = 0.10 + 2x\).

Step 3: Substitute Values into the Equilibrium Expression

Substitute the expressions into the equilibrium expression for \(K_{eq}\):\[4.0 \times 10^{-19} = \frac{x(0.10 + 2x)^2}{0.03 - x}\]Assume \(x\) is small compared to starting concentrations to simplify the expression, yielding:\[4.0 \times 10^{-19} \approx \frac{x(0.10)^2}{0.03}\]

Step 4: Solve for \(x\)

Simplify and solve the equation:\[4.0 \times 10^{-19} \times 0.03 = x(0.01)\]\[x = \frac{4.0 \times 10^{-21}}{0.01}\]\[x = 4.0 \times 10^{-19}\]This \(x\) represents the concentration of \(\mathrm{Ag}^+\) ions at equilibrium.

Key Concepts

Equilibrium Constant

When we talk about the equilibrium constant, denoted as \( K_{eq} \), we're referring to a number that helps us understand the balance of a reversible reaction, like in the given reaction \( \left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^{-} \rightleftharpoons \mathrm{Ag}^{+}+2 \mathrm{CN}^{-} \). The equilibrium constant is a ratio that compares the concentrations of products to reactants at equilibrium. It gives us a snapshot of where the balance lies between those products and reactants.
For this reaction, the expression for \( K_{eq} \) is:\[K_{eq} = \frac{\left[\mathrm{Ag}^{+}\right]\left[\mathrm{CN}^{-}\right]^2}{\left[\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^{-}\right]}\]
This expression indicates that if you know the equilibrium concentrations, you can find \( K_{eq} \), or vice versa. In our case, \( K_{eq} \) at \( 25^{\circ} \mathrm{C} \) is an extremely small number, \( 4.0 \times 10^{-19} \), which implies that the reactants are heavily favored at equilibrium.

Silver Ion Concentration

Silver ion concentration is a critical part of understanding the balance in silver-related reactions. In our exercise, this concentration is denoted by \( [\mathrm{Ag}^{+}] \), and it measures how many silver ions are freely floating in the solution once equilibrium is reached.
The concentration of silver ions initially wasn't given directly, but using an algebraic approach with our equilibrium constant, we can solve for it. Assuming a small change "\( x \)" happens as the reaction shifts towards equilibrium, the concentration \([\mathrm{Ag}^{+}]\) at equilibrium is found to be \( 4.0 \times 10^{-19} \) M.
This tiny value indicates a very low presence of free silver ions in the solution due to the low \( K_{eq} \), meaning the reactants are much less likely to dissociate into products.

Initial Concentrations

Initial concentrations set the stage for how a reaction proceeds towards equilibrium. In our problem, these are the concentrations before any reaction has taken place.
- For \( \mathrm{AgNO}_{3} \), the initial concentration is given as 0.03 M. This directly contributes to the silver ion concentration.- For \( \mathrm{KCN} \), the initial concentration is 0.10 M, giving us a source of cyanide ions.
Initial concentrations help us determine how a reaction might shift; they combine with \( K_{eq} \) to predict the direction and extent of the reaction. They provide us with a basis to calculate changes in concentrations as the system approaches equilibrium. In our case, these concentrations facilitated setting up our algebraic expression to solve for the change \( x \) in ion concentration.

Dissociation Reaction

A dissociation reaction is a type of chemical reaction where a compound breaks apart into two or more components. In the given exercise, the dissociation involves the complex ion \(\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^{-}\) splitting into \(\mathrm{Ag}^{+}\) ions and \(2 \mathrm{CN}^{-}\) ions.
This reaction type is crucial, especially in aqueous solutions, because it determines how ions behave in a solution. As ions dissociate, they contribute to the conductivity of the solution and engage in further chemical processes.
Understanding the dynamics of dissociation helps predict concentrations of ions at equilibrium. The backward and forward nature means the reaction can "swing" in either direction until it stabilizes. This helps in calculating silver ion concentrations and understanding how different initial conditions can affect equilibrium outcomes.