Problem 111 - A Complete Resource Book in Chemistry for JEE Main

Question

The solubility of \(\mathrm{AgCl}\) in moles per litre when its solubility product is \(1.56 \times 10^{-10}\) at \(25^{\circ} \mathrm{C}\) is (a) \(0.576 \times 10^{-8} \mathrm{~mol} /\) litre (b) \(1.056 \times 10^{-4} \mathrm{~mol} /\) litre (c) \(1.249 \times 10^{-5} \mathrm{~mol} /\) litre (d) \(1.478 \times 10^{-6} \mathrm{~mol} /\) litre

Step-by-Step Solution

Answer

The solubility of \(\text{AgCl}\) is \(1.249 \times 10^{-5}\) mol/L, answer (c).

Step 1: Understand the Dissolution Equation

AgCl dissolves in water according to the equation: \[ \text{AgCl}_{(s)} \rightleftharpoons \text{Ag}^+_{(aq)} + \text{Cl}^-_{(aq)} \]. This reaction indicates that for each mole of \( \text{AgCl} \) that dissolves, one mole of \( \text{Ag}^+ \) and one mole of \( \text{Cl}^- \) are produced.

Step 2: Write the Expression for Solubility Product

The solubility product \( K_{sp} \) is expressed as: \[ K_{sp} = [\text{Ag}^+][\text{Cl}^-] \]. Given \( K_{sp} = 1.56 \times 10^{-10} \), both \([\text{Ag}^+]\) and \([\text{Cl}^-]\) are equal to the solubility (s) of \( \text{AgCl} \) in moles per liter since they disassociate in a one-to-one ratio.

Step 3: Set Up the Equation with Solubility (s)

Since \([\text{Ag}^+] = s\) and \([\text{Cl}^-] = s\), the expression for \( K_{sp} \) becomes: \[ K_{sp} = s^2 \]. Therefore, \[ 1.56 \times 10^{-10} = s^2 \] and we need to solve for \( s \).

Step 4: Solve for Solubility (s)

To find \( s \), take the square root of both sides: \[ s = \sqrt{1.56 \times 10^{-10}} \]. Calculate this to find \( s \).

Step 5: Calculate the Solubility

\[ s = \sqrt{1.56 \times 10^{-10}} \approx 1.249 \times 10^{-5} \text{ mol/L} \].

Step 6: Choose the Correct Answer Option

Compare the calculated solubility \((1.249 \times 10^{-5} \text{ mol/L})\) with the given options. The option matching this value is (c).

Key Concepts

Dissolution Equation

When it comes to understanding how substances like silver chloride (\( \mathrm{AgCl} \)) dissolve, we need to look at the dissolution equation. This is essentially a chemical equation that shows how the solid compound breaks apart in a solution. For \( \mathrm{AgCl} \), the equation is: \[ \text{AgCl}_{(s)} \rightleftharpoons \text{Ag}^+_{(aq)} + \text{Cl}^-_{(aq)} \].

This equation tells us that when \( \mathrm{AgCl} \) dissolves in water, each molecule will separate into one silver ion (\( \text{Ag}^+ \)) and one chloride ion (\( \text{Cl}^- \)). This one-to-one ratio is crucial for simplifying the calculation of solubility since each dissolved molecule of \( \mathrm{AgCl} \) gives one unit each of \( \text{Ag}^+ \) and \( \text{Cl}^- \). Understanding this concept lays the foundation for calculating the solubility in terms of moles per litre, which we'll dive into next.

Moles per Litre

The concept of moles per litre, often referred to as molarity, is a measure of the concentration of a solute in a solution. It is defined as the number of moles of solute divided by the volume of the solution in litres.

To calculate the solubility of \( \mathrm{AgCl} \), we use the solubility product (\( K_{sp} \)) which is given as \( 1.56 \times 10^{-10} \). For \( \mathrm{AgCl} \), the solubility (\( s \)) expresses the molarity of \( \text{Ag}^+ \) and \( \text{Cl}^- \) that results from the dissolved salt:

• For \( \text{Ag}^+ \), \( [\text{Ag}^+] = s \)
• For \( \text{Cl}^- \), \( [\text{Cl}^-] = s \)

The relationship between the ions in solution reflects the solubility product equation, \( K_{sp} = s^2 \), which represents the product of the molar concentrations of the ions at equilibrium. This is why knowing the units and understanding molarity is essential for calculating solubility.

Square Root Calculation

Finding the solubility of \( \mathrm{AgCl} \) requires solving for \( s \) from the expression \( K_{sp} = s^2 \). This is where square root calculations come in. The unique thing about this calculation is it allows us to determine the concentration of ions needed to reach equilibrium in a solution.

By substituting the given \( K_{sp} \) of \( 1.56 \times 10^{-10} \) into \( s^2 \), we solve for \( s \) by calculating:\[ s = \sqrt{1.56 \times 10^{-10}} \].

This calculation gives us the solubility of \( \mathrm{AgCl} \), which is \( 1.249 \times 10^{-5} \text{ mol/L} \), meaning at saturation, there are \( 1.249 \times 10^{-5} \) moles of \( \mathrm{AgCl} \) dissolved per litre of solution. Solving square roots might need a calculator, but it's an essential step in finding how much \( \mathrm{AgCl} \) can be prepared in a \( 1 \text{ L} \) solution.