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 4.

Step-by-step solution

Write the dissociation equation

The dissociation of \(\text{M(OH)}_x\) in water can be written as: \[ \text{M(OH)}_x \rightarrow \text{M}^{x+} + x \text{OH}^- \]

Write the expression for the solubility product

The solubility product (K\textsubscript{sp}) is given by: \[ \text{K}_{\text{sp}} = [\text{M}^{x+}][\text{OH}^-]^x \]

Express the concentrations in terms of solubility

Given the solubility (s) is \(10^{-4} \text{ M}\), the concentration of \( \text{M}^{x+} \) is: \[ [\text{M}^{x+}] = 10^{-4} \text{ M} \] The concentration of \[ \text{OH}^- \] ions is: \[ [\text{OH}^-] = x \times 10^{-4} \text{ M} \]

Substitute concentrations into the K\textsubscript{sp} expression

Substituting the concentrations into the K\textsubscript{sp} expression, we get: \[ 4 \times 10^{-12} = (10^{-4})(x \times 10^{-4})^x \]

Solve for \ x \

Simplify the equation to solve for \ x \: \[ 4 \times 10^{-12} = 10^{-4} \times (x \times 10^{-4})^x \] \[ 4 \times 10^{-12} = 10^{-4 + 4x} \times x^x \] \[ 4 \times 10^{-12} = 10^{-4 + 4x} \times x^x \] \[ \text{Taking logarithm on both sides, and then solving for x, we get,} \] \[ x = 4 \]

Key concepts

Solubility

Solubility is a measure of how much of a substance (solute) can dissolve in a solvent (like water) to form a homogeneous solution. It is usually expressed in terms of concentration, such as moles per liter (Molarity). For example, the solubility of \(\text{M}(\text{OH})_x\) is given as \(10^{-4} \text{ M}\), which means that 0.0001 moles of \(\text{M}(\text{OH})_{x}\) can dissolve in one liter of water. Understanding solubility is crucial for predicting how substances interact in solutions, especially in chemical reactions and processes like precipitation.

Dissociation Equation

The dissociation equation describes how a compound separates into its ions in solution. For the compound \(\text{M}(\text{OH})_x\), the dissociation can be written as: \(\text{M}(\text{OH})_x \rightarrow \text{M}^{x+} + x \text{OH}^-\). This equation shows that one formula unit of \(\text{M}(\text{OH})_x\) produces one ion of \(\text{M}^{x+}\) and x ions of hydroxide (\(\text{OH}^-\)). The dissociation equation is essential because it helps us understand the distribution of ions in solution and how they contribute to properties like electrical conductivity and reactivity.

Solubility Product Constant

The Solubility Product Constant (K\textsubscript{sp}) is a crucial value in chemistry that helps predict the solubility of a compound. It is the product of the concentrations of the ions in a saturated solution, each raised to the power of their coefficients in the dissociation equation. For \(\text{M}(\text{OH})_x\), the expression for K\textsubscript{sp} is: \(\text{K}_{\text{sp}} = [\text{M}^{x+}][\text{OH}^-]^x\). Given the values, K\textsubscript{sp} can be used to determine how much of the compound will dissolve in water. In the example given, the K\textsubscript{sp} value is \(4 \times 10^{-12}\), and using the solubility, we can solve the equations to find that x equals 4. This constant is fundamental in predicting whether a precipitate will form in a given solution and is widely used in various fields, including pharmacology, environmental science, and engineering.