Question

\(\mathrm{M}(\mathrm{OH})_{x}\) has \(\mathrm{K}_{\mathrm{sp}}=4 \times 10^{-12}\) and solubility \(10^{-4} \mathrm{M}\). Hence \(\mathrm{x}\) is : (1) 1 (2) 2 (3) 3 (4) 4

Short answer

The value of \( x \) is 2.

Step-by-step solution

Write the Dissociation Equation

Consider the dissociation of \( \mathrm{M(OH)}_x \). The dissociation equation is: \[ \mathrm{M(OH)}_x \leftrightarrow \mathrm{M}^{x+} + x \mathrm{OH}^- \]

Establish Relationships for Concentrations

The solubility \( s \ (10^{-4} \mathrm{M}) \) corresponds to the concentration of \( \mathrm{M}^{x+} \) and the concentration of \( \mathrm{OH}^- \) will be \[ x \cdot s \]

Write the Expression for \( \mathrm{K_{sp}} \)

The solubility product constant expression is: \[ \mathrm{K_{sp}} = [\mathrm{M}^{x+}][\mathrm{OH}^-]^x \]

Substitute Solubility Values

Using the given values in the \( \mathrm{K_{sp}} \) expression: \[ \frac{4 \times 10^{-12}}{(10^{-4})^x} = (10^{-4}) \times (x \times 10^{-4})^x \]

Simplify the Equation and Solve for x

We equate the adjusted equation and simplify: \[ 4 \times 10^{-12} = 10^{-4(x+1)} \iff 4 \times 10^{-12} = 10^{-4x-4} \]

Match Powers of 10

Since the exponents must match, solve \( 4x + 4 = 12 \): \[ 4x = 12 - 4 \] \[ x = \frac{8}{4} \] \[ x = 2 \]

Key concepts

solubility product constant

The solubility product constant, represented as Ksp, is a key concept when dealing with solubility problems in chemistry. It indicates how much of a compound can dissolve in a solution before it starts precipitating. When a salt dissolves in water, it dissociates into its ions. The Ksp expression is the product of the concentration of these ions, each raised to the power of their coefficients in the balanced equation.
For example, for a generic salt \(\text{M(OH)}_a\) that dissociates into \(\text{M}^{a+}\) and \(\text{OH}^{-}\), the Ksp expression is written as: \[ \text{K}_{\text{sp}} = [\text{M}^{a+}][\text{OH}^{-}]^a \]
Understanding the solubility product constant is crucial for predicting whether a precipitate will form in a solution.

molar solubility

Molar solubility refers to the number of moles of a solute that can dissolve in a liter of solution before the solution becomes saturated. For a salt like \(\text{M(OH)}_x\), this is essentially its solubility in mol/L (M).
Given a Ksp value, we can determine the molar solubility by setting up an equilibrium expression and solving for the solubility.
For instance, if the molar solubility of \(\text{M(OH)}_2\) is given as \(s\), the concentrations of \(\text{M}^{2+}\) and \(\text{OH}^-\) at equilibrium would be \(s\) and \(2s\) respectively. Thus, the Ksp can be expressed as: \[ \text{K}_{\text{sp}} = [\text{M}^{2+}][\text{OH}^{-}]^2 = s \times (2s)^2 = 4s^3 \]
Knowing this helps us to find how much of a solute will dissolve under certain conditions.

chemical equilibrium

Chemical equilibrium is a state in which the rate of the forward reaction equals the rate of the reverse reaction. For solubility, it means that the rate at which the solid salt dissolves in solution is equal to the rate at which its ions precipitate back into solid salt.
In the context of \(\text{M(OH)}_x\), its equilibrium can be described by the equation: \[ \text{M(OH)}_x \rightleftharpoons \text{M}^{x+} + x \text{OH}^- \] When this equilibrium is established, the concentrations of \(\text{M}^{x+}\) and \(\text{OH}^-\) ions will remain constant. These concentrations can be used in the Ksp expression to find the solubility of the compound.
The concept of chemical equilibrium is fundamental in predicting the behavior of solutions and solubility.