Problem 101 - A Complete Resource Book in Chemistry for JEE Main

Question

\(\mathrm{M}(\mathrm{OH}) \mathrm{x}\) has \(\mathrm{Ksp}=4 \times 10^{-12}\) and solubility \(10^{-4} \mathrm{M}, \mathrm{x}\) is (a) 1 (b) 2 (c) 3 (d) 4

Step-by-Step Solution

Answer

x is 2.

Step 1: Understanding Solubility Product

The solubility product constant, \( K_{sp} \), helps determine the solubility of sparingly soluble salts. For the given compound \( \mathrm{M(OH)}_x \), the dissolution can be written as: \[ \mathrm{M(OH)}_x \rightleftharpoons \mathrm{M}^{x+} + x \mathrm{OH}^- \]

Step 2: Expressing Solubility in Terms of Dissolved Ions

If the solubility is \( 10^{-4} \ \mathrm{M} \), then the concentration of \( \mathrm{M}^{x+} \) will be \( 10^{-4} \). For \( \mathrm{OH}^- \), it will be \( x \times 10^{-4} \) because \( x \) hydroxide ions are produced per formula unit of \( \mathrm{M(OH)}_x \).

Step 3: Writing the Expression for Ksp

The \( K_{sp} \) expression is: \[ K_{sp} = [\mathrm{M}^{x+}][\mathrm{OH}^-]^x \] Plug in the concentrations in terms of solubility: \[ K_{sp} = (10^{-4})(x \times 10^{-4})^x \]

Step 4: Solving for x

We know \( K_{sp} = 4 \times 10^{-12} \): \[ 4 \times 10^{-12} = (10^{-4})(x \times 10^{-4})^x \] Solving, we have: \[ 4 \times 10^{-12} = 10^{-4} \times x^x \times 10^{-4x} \] This simplifies to: \[ 4 \times 10^{-12} = x^x \times 10^{-4(x+1)} \] Equating powers of 10: \[ -12 = -4(x+1) \] Solving: \[ x+1 = 3 \] Thus, \( x = 2 \).

Step 5: Verify the Solution

Verify by substituting \( x = 2 \) back: The equation becomes \( 4 \times 10^{-12} = (10^{-4})((2 \times 10^{-4})^2) \). Calculating gives: \( (2 \times 10^{-4})^2 = 4 \times 10^{-8} \) so the whole expression is \( 10^{-4} \times 4 \times 10^{-8} = 4 \times 10^{-12} \) which checks out with the original \( K_{sp} \).

Key Concepts

Sparingly Soluble Salts

Sparingly soluble salts are compounds that dissolve in water to a very limited extent. They don't fully dissociate in solution, unlike highly soluble salts. This means they only produce a small amount of ions in water. These ions usually hold a balance between the undissolved and dissolved states.

• These salts have a specific characteristic of low solubility.
• Their limited solubility makes them important for understanding chemical equilibrium.
• One crucial aspect to remember is that the solubility can depend on factors like temperature and the presence of other ions.

For example, if we have a salt like \[ \mathrm{M(OH)}_x \rightleftharpoons \mathrm{M}^{x+} + x \mathrm{OH}^- \]we see that when it dissolves, it doesn't provide a lot of ions in the solution.

Chemical Equilibrium

In chemistry, equilibrium represents a state in which the rates of the forward and reverse reactions are equal. In the context of sparingly soluble salts, this implies that the rate at which the salt dissolves into ions is equivalent to the rate at which ions recombine to form the solid salt.

• This balance means the concentrations of ions in solution remain constant over time.
• Equilibrium is dynamic, not static, as reactions continue to happen but at equal rates.
• It’s fundamental for calculating solubility products, especially in systems with limited solubility.

Relating to our salt example, \[ \mathrm{M(OH)}_x \rightleftharpoons \mathrm{M}^{x+} + x \mathrm{OH}^- \]the equilibrium means that any dissolution of the salt is matched by a simultaneous recombination.

Ksp Calculations

Ksp, or the solubility product constant, provides a numerical value that helps predict how much of a solute will dissolve in water. It’s specifically used for sparingly soluble salts and represents the maximum potential for a salt to dissolve at equilibrium.

• Ksp is a product of the concentrations of the ions, each raised to the power of its coefficient in the balanced equation.
• It can guide us in calculating the solubility of a salt under given conditions.
• This value is very specific for each compound and varies based on temperature.

For the given compound \( \mathrm{M(OH)}_x \), the expression will be \[ K_{sp} = [\mathrm{M}^{x+}][\mathrm{OH}^-]^x \]Using this, we found that \( x = 2 \), by setting up the calculation with the provided solubility and Ksp values, solving for the variable \( x \). It involves equating the calculated product of ion concentrations with the given \( K_{sp} \), and adjusting the balance through known mathematical steps.