Problem 275 - A Complete Resource Book in Chemistry for JEE Main

Question

Solubility product of silver bromide is \(5.0 \times 10^{-13}\). The quantity of potassium bromide (molar mass taken as \(120 \mathrm{~g} \mathrm{~mol}^{-1}\) ) to be added to 1 litre of \(0.05 \mathrm{M}\) solution nitrate to start the precipitation of AgBr is (a) \(1.2 \times 10^{-10} \mathrm{~g}\) (b) \(1.2 \times 10^{-9} \mathrm{~g}\) (c) \(6.2 \times 10^{-5} \mathrm{~g}\) (d) \(5.0 \times 10^{-8} \mathrm{~g}\)

Step-by-Step Solution

Answer

Add 1.2 x 10^-9 g of KBr to start precipitation.

Step 1: Understand the Precipitation Condition

For a precipitate to form, the ionic product of the silver ion \([Ag^+])\) and bromide ion \([Br^-])\) must be greater than the solubility product \(K_{sp}\) of silver bromide, given as\[K_{sp} = 5.0 \times 10^{-13} \].

Step 2: Write the Ionic Equation and Determine Ag+ Concentration

The reaction between silver nitrate \(AgNO_3\) and potassium bromide \(KBr\) leads to the reaction:\(Ag^+ + Br^- \rightleftharpoons AgBr(s)\).Initially, from the 0.05 M silver nitrate solution, \([Ag^+] = 0.05 \, \text{M}\).

Step 3: Determine Required [Br-]

To find when the precipitate will start forming, set up the equation:\[ [Ag^+][Br^-] = K_{sp} \].Plug in the values:\[ 0.05 \, \text{M} \times [Br^-] = 5.0 \times 10^{-13} \].Solving for \([Br^-]\),\[ [Br^-] = \frac{5.0 \times 10^{-13}}{0.05} = 1.0 \times 10^{-11} \, \text{M} \].

Step 4: Calculate Mass of KBr

Once you have \([Br^-] = 1.0 \times 10^{-11} \, \text{M} \), use the formula \( \text{mass} = \text{moles} \times \text{molar mass} \),where moles of \(KBr = [Br^-] \times \text{volume}\), here volume = 1 liter, and molar mass of KBr = 120 g/mol.\[ \text{mass of } KBr = 1.0 \times 10^{-11} \times 120 = 1.2 \times 10^{-9} \, \text{g} \].

Key Concepts

Ionic Equilibrium

Ionic equilibrium is a fascinating concept that explains how ions in solution balance each other to achieve a state of balance or equilibrium. In aqueous solutions, ions dissociate from their compounds and interact with other ions in the environment. This interaction continues until a stable equilibrium is achieved. It essentially means that the rate at which ions combine to form compound phases, such as solids or gases, equals the rate at which those compound phases dissociate back into ions.

In the context of the exercise, the solubility product constant, or \( K_{sp} \), plays a critical role. The solution achieves ionic equilibrium when the product of the concentrations of the ions equals the \( K_{sp} \). If the concentration surpasses this value, more of the ions recombine into a solid, thus forming a precipitate. This concept is crucial in understanding the conditions required for precipitation reactions, as it helps predict when a solid will form from a saturated solution.

• Ions exist in solutions and interact until equilibrium is reached.
• Equilibrium occurs when the rate of ion association equals dissociation.
• The \( K_{sp} \) governs when precipitation occurs.

Precipitation Reaction

A precipitation reaction occurs when two soluble salts in an aqueous solution react to form an insoluble solid called a precipitate. This reaction is central to many laboratory techniques and industrial processes. Precipitation reactions are crucial because they enable the recovery of metals, removal of toxins, and even purification of chemicals.

In the exercise provided, the silver bromide (AgBr) forms a precipitate when \([Ag^+]\) and \([Br^-]\) ions in solution meet the right conditions. The key factor is the concentration of these ions, which, when multiplied together, must exceed the solubility product, \( K_{sp} \). When this threshold is passed, the excess ions combine, leaving the solution to form a solid precipitate. This example demonstrates the importance of ionic equilibrium in determining when a precipitation reaction will occur.

• Precipitation occurs when ionic concentrations exceed the solubility product.
• Insoluble solids form in solution when specific conditions are met.
• These reactions are used for material recovery and purification.

Chemical Calculations

Chemical calculations are fundamental in understanding and solving numerous problems in chemistry, particularly in determining the amounts and concentrations of substances in reactions. In the exercise, the use of chemical calculations is crucial for determining how much potassium bromide (KBr) is necessary to trigger precipitation of silver bromide.

The process involves calculating the required bromide ion concentration \([Br^-]\), given the \([Ag^+]\) and the solubility product \( K_{sp} \). Once determined, this concentration allows us to calculate the moles of KBr needed using its molar weight. This calculation ensures you know exactly how much of the substance is necessary to reach the desired reaction condition. Understanding how to set up these calculations accurately is essential for accurately predicting and manipulating chemical reactions.

• Ensure precise calculation of ion concentrations to predict reactions.
• Utilize molar mass to determine exact substance quantities required.
• Chemical calculations help in predicting and controlling reaction outcomes.